As for the first term on the last line of (117), we notice that

$$
\begin{align}
\mathbb{E}^{\mathbb{Q}^{\theta}} \left[ \int_{0}^{T} \left| \psi_{t}^{k} - \bar{\psi}_{t} \right|^{2} dt \right] &\leq \exp(\alpha T) \mathbb{E} \left[ q_{T}^{\theta} \int_{0}^{T} \left| \psi_{t}^{k} - \bar{\psi}_{t} \right|^{2} dt \right] \nonumber \\
&= (1 - \theta) \exp(\alpha T) \mathbb{E} \left[ q_{T}^{0} \int_{0}^{T} \left| \psi_{t}^{k} - \bar{\psi}_{t} \right|^{2} dt \right] \tag{119} \\
&\quad + \theta \mathbb{E} \left[ q_{T} \int_{0}^{T} \left| \psi_{t}^{k} - \bar{\psi}_{t} \right|^{2} dt \right]. \nonumber
\end{align}
$$

By Step 1, we know that the first term on the right-hand side is less than $C(1 - \theta)$.
Using the fact that $(\psi^k)_{k \in \mathbb{N}}$ strongly converges to $\bar{\psi}$ in $L^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q})$, the second one tends to 0 as $k$ tends to $+\infty$ (uniformly in $\theta \in [0, 1)$). By (117), (118) and (119), we deduce that $\lim_{\theta \to 1} \lim_{k \to +\infty} a_2^{k, \theta} = 0$. By (116), we obtain $\lim_{k \to +\infty} \mathbb{E}[|A^k|] = 0$ when $r = 1$.

Sub-step 3c: analysis of $(B^k)_{k \in \mathbb{N}}$. Back to (113), we now study the sequence $(B^k)_{k \in \mathbb{N}}$.
Using the fact that the gradient of $\ell$ in the second variable is at most of linear growth,
see A4, and once again the fact that $(\psi^k)_{k \in \mathbb{N}}$ strongly converges to $\psi$ in $L^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q})$,
we directly obtain

$$
\lim_{k \to +\infty} B^k = \lim_{k \to +\infty} \left( \mathbb{E} \left[ \int_0^T q_s \ell(s, \psi_s^k) ds \right] - \mathbb{E} \left[ \int_0^T q_s \ell(s, \bar{\psi}_s) ds \right] \right) = 0.
$$

Sub-step 3d: conclusion. From the last three sub-steps, we deduce that $\mathbb{E}[A^k + B^k]$ in the right-hand side of (113) tends to 0. We deduce that $\liminf_{k \to +\infty} \mathcal{J}(\psi^k) \ge \mathcal{J}(\bar{\psi})$, which concludes the step and the proof. $\square$

**Lemma 23.** There exists a solution to problem $(\tilde{P}')$, i.e. there exists $(\bar{\psi}, \bar{q}) \in \mathcal{A}_{c_2} \times \tilde{\mathcal{Q}}_{c_1}$ such that

$$
\min_{\psi \in \mathcal{A}_{c_2}} \max_{q \in \tilde{\mathcal{Q}}_{c_1}} \tilde{\mathcal{J}}(q, \psi) = \max_{q \in \tilde{\mathcal{Q}}_{c_1}} \min_{\psi \in \mathcal{A}_{c_2}} \tilde{\mathcal{J}}(q, \psi) = \tilde{\mathcal{J}}(\bar{q}, \bar{\psi}). \quad (120)
$$

Proof. For each $k \in \mathbb{N}$, we call $\mathcal{B}_k^\infty$ the collection of $\mathbb{F}$-progressively measurable $\mathbb{R}^n$-valued processes that are bounded by $k$ on a subset of full measure under $\mathbb{P} \otimes \text{Leb}_{[0,T]}$. We then view $\mathcal{A}_{c_2} \cap \mathcal{B}_k^\infty$ as a subset of $L^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q}^0)$, with $\mathbb{Q}^0$ being defined as in the statement of Lemma 22. By Lemma 22 again, $\mathcal{A}_{c_2} \cap \mathcal{B}_k^\infty$ is a convex and weakly compact subset of $L^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q}^0)$. By Theorem 1, which we can apply thanks to Proposition 20 and Lemma 22, we can find, for each $k \in \mathbb{N}$, a saddle point $(q^k, \psi^k) \in \tilde{\mathcal{Q}}_{c_1} \times (\mathcal{A}_{c_2} \cap \mathcal{B}_k^\infty)$ to the min-max problem

$$
\min_{\psi \in \mathcal{A}_{c_2} \cap \mathcal{B}_k^\infty} \max_{q \in \tilde{\mathcal{Q}}_{c_1}} \tilde{\mathcal{J}}(q, \psi) = \max_{q \in \tilde{\mathcal{Q}}_{c_1}} \min_{\psi \in \mathcal{A}_{c_2} \cap \mathcal{B}_k^\infty} \tilde{\mathcal{J}}(q, \psi) = \tilde{\mathcal{J}}(q^k, \psi^k). \quad (121)
$$

By compactness of $\tilde{\mathcal{Q}}_{c_1}$ and $\mathcal{A}_{c_2}$, for the weak topologies $\sigma(L^1(\mathbb{F}, \mathbb{R}, \mathbb{P})$, $L^\infty(\mathbb{F}, \mathbb{R}, \mathbb{P})$) and $\sigma(L^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q}^0), L^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q}^0))$, the sequence $(q^k, \psi^k)_{k \in \mathbb{N}}$ converges, up to a sub-sequence, for the product topology. The limit is denoted $(\bar{q}, \bar{\psi})$. The objective of the proof is thus to show that

$$
\tilde{\mathcal{J}}(q, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \psi), \quad (122)
$$

for all $\psi \in \mathcal{A}_{c_2}$ and $q \in \tilde{\mathcal{Q}}_{c_1}$, which is known to be equivalent to the equality (120).
Step 1: $\tilde{\mathcal{J}}(\bar{q}, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \psi)$ for any $\psi \in \mathcal{A}_{c_2}$. For $0 \le k_0 \le k$ and $\psi \in \mathcal{A}_{c_2}$, let $(\psi_t^{k_0} := \psi_t 1_{\{|\psi_t| \le k_0\}})_{t \in [0,T]}$. By (121), we have that

$$ \tilde{\mathcal{J}}(\bar{q}, \psi^k) \le \tilde{\mathcal{J}}(q^k, \psi^k) \le \tilde{\mathcal{J}}(q^k, \psi^{k_0}). $$

Taking infimum and supremum limits in the last inequality, we have

$$ \liminf_{k \to +\infty} \tilde{\mathcal{J}}(\bar{q}, \psi^k) \le \limsup_{k \to +\infty} \tilde{\mathcal{J}}(\bar{q}, \psi^k) \le \limsup_{k \to +\infty} \tilde{\mathcal{J}}(q^k, \psi^{k_0}). \quad (123) $$

We first handle the term on the right-hand side. By concavity of $\mathcal{G}$ in the first variable, we have

$$ \tilde{\mathcal{J}}(q^k, \psi^{k_0}) \le \tilde{\mathcal{J}}(\bar{q}, \psi^{k_0}) + a^k + b^k, \quad (124) $$

where

$$ a^k := \mathbb{E} \left[ \delta_q \mathcal{G} \left( \bar{q}_T, X_T^{\psi^{k_0}} \right) (q_T^k - \bar{q}_T) + \int_0^T (q_t^k - \bar{q}_t) \ell(t, \psi_t^{k_0}) dt \right], \quad b^k := \tilde{\mathcal{S}}(\bar{q}) - \tilde{\mathcal{S}}(q^k). $$

By boundedness of $\psi^{k_0}$, the integrand $\ell(t, \psi_t^{k_0})$ is bounded, uniformly in $t$ and $\omega$. Moreover, $|X_T^{\psi^{k_0}}|^{2-r}$ has exponential moments of any order, which implies, from Lemma 40 and Remark 21, that $\mathbb{E}[q_T|X_T^{\psi^{k_0}}|^{2-r}] < +\infty$. And then, by the growth Assumption A6 on $\delta_q \mathcal{G}$, we deduce that that $\delta_q \mathcal{G}(q_T, X_T^{\psi^{k_0}})$ has exponential moments of any order. Recalling from (102) that

$$ \sup_{t \in [0,T]} \mathbb{E}[h(\bar{q}_t)] < +\infty, \quad \sup_{k \in \mathbb{N}} \sup_{t \in [0,T]} \mathbb{E}[h(q_t^k)] < +\infty, $$

Lemma 43 yields

$$ \lim_{k \to +\infty} a^k = 0. $$

By weak lower semi-continuity of $\tilde{\mathcal{S}}$ on $\tilde{\mathcal{Q}}_{c_1}$ (see Step 3 in the proof of Proposition 20), we also have

$$ \limsup_{k \to +\infty} b^k \le 0. $$

Therefore, inserting the last two displays in (124) and then returning back to (123), we obtain

$$ \liminf_{k \to +\infty} \tilde{\mathcal{J}}(\bar{q}, \psi^k) \le \tilde{\mathcal{J}}(\bar{q}, \psi^{k_0}). $$

By weak lower semi-continuity of $\mathcal{A}_{c_2} \ni \psi \mapsto \tilde{\mathcal{J}}(\bar{q}, \psi)$ (see Lemma 22, using the fact that $\bar{q} \in \mathcal{Q}_{c_1}$), the last inequality yields

$$ \tilde{\mathcal{J}}(\bar{q}, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \psi^{k_0}). \quad (125) $$

It remains to pass to the limit in the right-hand side. By regularity of $\mathcal{G}$, we can write

$$ \tilde{\mathcal{J}}(\bar{q}, \psi^{k_0}) = \tilde{\mathcal{J}}(\bar{q}, \psi) + \mathbb{E} \left[ A^{k_0} + B^{k_0} \right], \quad (126) $$

where

$$ \begin{align} A^{k_0} &:= \int_0^1 \delta_X \mathcal{G}(q_T, X_T^{\lambda, \psi^{k_0}}) \cdot (X_T^{\psi^{k_0}} - X_T^{\psi}) \mathrm{d}\lambda, \tag{127} \\ B^{k_0} &:= \int_0^T \bar{q}_s (\ell(s, \psi_s^{k_0}) - \ell(s, \psi_s)) \mathrm{d}s, \nonumber \end{align} $$
and $X_T^{\lambda, \psi^{k_0}} := \lambda X_T^{\psi^{k_0}} + (1-\lambda)X_T^{\psi}$. At this point, we are in a situation very similar to (113), except for the fact that $\delta_X \mathcal{G}$ in the definition of $A^{k_0}$ is computed at $X^{\lambda, \psi^{k_0}}$. Apart from this, the context is the same. In particular, with the same abuse of notation as in the third step of the proof of Lemma 22, $\psi^{k_0}$ converges to $\psi$ (as $k_0 \to +\infty$) in $L^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}})$, where $\bar{\mathbb{Q}}$ is defined by $\bar{\mathbb{Q}}(E) = \mathbb{E} \int_0^T 1_E(t) \bar{q}_t dt$. Indeed, the extended version of (14) (for elements $q$ in $\tilde{\mathbb{Q}}_{c_1}$, see again Remark 21) yields

$$ \mathbb{E} \left[ \bar{q}_T \int_0^T |\psi_t|^2 dt \right] < +\infty. $$

Recalling that $(\psi_t^{k_0} := \psi 1_{\{|\psi_t| \le k_0\}})_{t \in [0,T]}$, we deduce from dominated convergence theorem that $\psi^{k_0}$ indeed converges to $\psi$ (as $k_0 \to +\infty$) in $L^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}})$. Following the exact same reasoning as in Step 3 of Lemma 22, we then obtain that

$$ \lim_{k_0 \to +\infty} \mathbb{E} [A^{k_0} + B^{k_0}] = 0. \qquad (128) $$

Combining the last result with (125) and (126), we deduce that $\tilde{\mathcal{J}}(q, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \bar{\psi})$, which completes the first step.

Step 2: $\tilde{\mathcal{J}}(q, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \bar{\psi})$ for any $q \in \tilde{\mathbb{Q}}_{c_1}$. Because the proof follows arguments that are similar to those in the first step, we just give a sketch of it. For integers $0 \le k_0 \le k$, we deduce, again from (121), that

$$ \tilde{\mathcal{J}}(q, \psi^k) \le \tilde{\mathcal{J}}(q^k, \psi^k) \le \tilde{\mathcal{J}}(q^k, \bar{\psi}^{k_0}), $$

for any $q \in \tilde{\mathbb{Q}}_{c_1}$ and where $\bar{\psi}^{k_0} := \bar{\psi} 1_{\{|\bar{\psi}| \le k_0\}}$. By definition of the infimum and supremum limits, we have

$$ \liminf_{k \to +\infty} \tilde{\mathcal{J}}(q, \psi^k) \le \limsup_{k \to +\infty} \tilde{\mathcal{J}}(q^k, \bar{\psi}^{k_0}), $$

Following the same arguments as in Step 1, we have that

$$ \limsup_{k \to +\infty} \tilde{\mathcal{J}}(q^k, \bar{\psi}^{k_0}) = \tilde{\mathcal{J}}(\bar{q}, \bar{\psi}^{k_0}), \quad \liminf_{k \to +\infty} \tilde{\mathcal{J}}(q, \psi^k) \ge \tilde{\mathcal{J}}(q, \bar{\psi}). $$

Combining the last inequalities yields

$$ \tilde{\mathcal{J}}(q, \bar{\psi}) \le \tilde{\mathcal{J}}(\bar{q}, \bar{\psi}^{k_0}). $$

Finally, using (126) in the right-hand side and then taking the limit $k_0 \to +\infty$, the conclusion of the step and the proof follows by (128). $\square$

## 5.2 Nature's control problem

Given two constants $c_1, c_2 > 0$, we address the restricted Nature control problem

$$ \sup_{q' \in \mathbb{Q}_{c_1}} \mathcal{J}(\bar{\psi}, q'), \qquad (\mathbb{P}_{\mathrm{N},c_1}) $$

under the assumption that, for some $\bar{q}$, the pair $(\bar{q}, \bar{\psi}) \in \mathbb{Q}_{c_1} \times \mathcal{A}_{c_2}$ is a saddle point of the problem (P'). In particular, the supremum in $(\mathbb{P}_{\mathrm{N},c_1})$ is equal to $\mathcal{J}(\bar{q}, \bar{\psi})$ and our goal becomes to characterize $\bar{q}$ when $\bar{\psi}$ is given.
As a corollary of our analysis, we show that, when $c_1$ is sufficiently large, $\bar{q}$ is in fact the unique minimizer of the unrestricted Nature control problem

$$ \sup_{q' \in \mathcal{Q}} \mathcal{J}(q', \bar{\psi}), \qquad (P_N) $$

which, in contrast with $(P_{N,c_1})$, is set over the entire set $\mathcal{Q}$.

A key step in relaxing the constraint imposed on $q'$ in $(P_{N,c_1})$, and thereby passing from $\mathcal{Q}_{c_1}$ to $\mathcal{Q}$, is to show that the component $\bar{q}$ of any saddle point $(\bar{q}, \bar{\psi})$ of $(P')$ actually lies in the interior of $\mathcal{Q}_{c_1}$, provided that $c_1$ is sufficiently large. This is the content of the following result, proved in Subsection 5.2.1.

**Proposition 24.** There exists a constant $c'_1 > 0$, only depending on the data, such that, for any $c_1 > c'_1$ and any saddle point $(q, \psi)$ to $(P')$ over $\mathcal{Q}_{c_1} \times \mathcal{A}_{c_2}$, the component $q$ of the saddle point necessarily belongs to $\mathcal{Q}_{c'_1}$.

This a priori bound then allows us to apply perturbative arguments to characterize the solutions of $(P_{N,c_1})$, and subsequently of $(P_N)$, by means of a stochastic maximum principle. The results are summarized in the main statement below:

**Theorem 25.** Let $\bar{\psi} \in \mathcal{A}_{c_2}$ and assume $c_1 > c'_1$ with $c'_1$ as in the statement of Proposition 24.

1. If, for some $c_0 \in (0, c_1)$, there exists $q \in \mathcal{Q}_{c_0}$ that solves $(P_{N,c_1})$ (i.e., that maximizes $q' \mapsto \mathcal{J}(q', \bar{\psi})$ over $\mathcal{Q}_{c_1}$), then there exists a pair $(Y, Z)$ such that $(q, Y, Z)$ belongs to the set $\mathcal{Q}$ defined in (26), and solves the FBSDE $(\text{Opt}_N)$ (with $\psi = \bar{\psi}$ therein).

2. Conversely, assume that there exists a solution $(\bar{q}, \bar{Y}, \bar{Z}) \in \mathcal{Q}$ to the FBSDE $(\text{Opt}_N)$, then $(\bar{q}, \bar{\psi})$ is the unique optimizer of Nature's control problem $(P_N)$.

3. In particular, if, for some $\bar{q} \in \mathcal{Q}_{c_1}$, $(\bar{q}, \bar{\psi})$ is a saddle point of the problem $(P')$, then $\bar{q}$ is the unique solution to $(P_N)$ (i.e., is the unique maximizer of $q' \mapsto \mathcal{J}(q', \bar{\psi})$ over the entire $\mathcal{Q}$).

The proof is based on a series of lemmas, which are proved in the next paragraph.

*Sketch of the proof of Theorem 25.* Taking for granted the statement of Proposition 24 and the results proven in the forthcoming Subsections 5.2.3 and 5.2.2, Theorem 25 can be established as follows.

The necessary condition is addressed in Subsection 5.2.2. We prove in Lemma 28 that, for any optimizer $q$ of $(P_{N,c_1})$ that belongs to $\mathcal{Q}_{c_0}$ for some $c_0 \in (0, c_1)$, the BSDE (24) has a solution $(Y, Z) \in D(\mathbb{F}, \mathcal{Q}) \times (\cap_{\beta(0,1)} M^\beta(\mathbb{F}, \mathbb{R}^d, \mathcal{Q}))$ that satisfies the optimality condition in the last line in $(\text{Opt}_N)$. This shows that $(q, Y, Z)$ belongs to $\mathcal{Q}$ and solves $(\text{Opt}_N)$, and this proves the first assertion in the statement of Theorem 25.

The second assertion (i.e., the converse) is a direct consequence of Lemma 30.

It remains to establish the third assertion. Given $\bar{q} \in \mathcal{Q}_{c_1}$ such that $(\bar{q}, \bar{\psi})$ is a saddle point of $(P')$, we know from Proposition 24 that $\bar{q} \in \mathcal{Q}_{c'_1}$ for some $c'_1 \in (0, c_1)$. By the first assertion in the statement of Theorem 25, we deduce that, there exists a pair $(Y, Z)$ such that $(\bar{q}, Y, Z)$ solves $(\text{Opt}_N)$. By the second assertion, we deduce that $\bar{q}$ is the unique maximizer of $q' \mapsto \mathcal{J}(q', \bar{\psi})$ over the entire set $\mathcal{Q}$. $\square$

Throughout the subsection, the parameters $c_1$ and $c_2$ appearing in $(P_{N,c_1})$ are fixed.
### 5.2.1 A priori estimate

The purpose of this subsection is to establish the following a priori estimate, from which Proposition 24 follows as a direct consequence:

**Lemma 26.** There exist two constants $c'_1 > 0$ and $C \ge 0$, only depending on the data and independent of $c_1, c_2$, such that, for any $(q, \psi) \in Q_{c_1} \times A_{c_2}$, with $c_1 > c'_1$, satisfying

$$ \mathcal{J}(q, 0) \ge \mathcal{J}(q, \psi) \ge \mathcal{J}(q^0, \psi), \quad (129) $$

where $q^0$ solves (84), it holds $q \in Q_{c'_1}$ and $\sup_{t \in [0, T]} \mathbb{E}[h(q_t)] \le C$.

Assuming that $c_1$ satisfies (83), $q^0$ in the statement belongs to $Q_{c_1}$. Notice also that the condition (129) is stated for an arbitrary pair $(q, \psi) \in Q_{c_1} \times A_{c_2}$, but is automatically satisfied by the saddle point $(\bar{q}, \bar{\psi})$ introduced in the beginning of Subsection 5.3, see $(\mathbb{P}_{N, c_1})$.

*Proof. Step 1.* We first establish a bound for $\mathcal{S}(q)$ in terms of $\mathbb{E}[h(q_T)]$. Recalling that $\mathcal{J}(q, \psi) = \mathcal{R}(q, \psi) - \mathcal{S}(q)$, by inequality (129) we have

$$ \begin{aligned} \mathcal{S}(q) = -\mathcal{J}(q, 0) + \mathcal{R}(q, 0) &\le -\mathcal{J}(q^0, \psi) + \mathcal{R}(q, 0) \\ &= -\mathcal{J}(q^0, \psi) + \mathcal{G}(X_T^0, q_T) + \mathbb{E}\left[\int_0^T q_s \ell(s, 0) ds\right]. \end{aligned} \quad (130) $$

We first provide a lower bound for $\mathcal{J}(q^0, \psi)$. Recalling (84), we have

$$ \mathcal{J}(q^0, \psi) = \mathcal{G}(q_T^0, X_T^\psi) + \mathbb{E}\left[\int_0^T q_t^0 \ell(s, \psi_s) ds\right] + \mathbb{E}\left[\int_0^T q_s^0 f(s, 0, 0) ds\right]. $$

Using the convexity of $\mathcal{G}$ in the variable $X$ (see (A6)) and the $L^{-1}$-strong convexity of $\ell$ in the variable $\psi$ (see A4), we deduce that there exists a constant $C$, independent of $\psi$, such that

$$ \begin{aligned} \mathcal{J}(q^0, \psi) &\ge \mathcal{G}(q^0, 0) - C \left(1 + \mathbb{E}\left[q_T^0 |X_T^\psi|\right]\right) + \frac{1}{2L} \mathbb{E}\left[\int_0^T q_s^0 |\psi_s|^2 ds\right] + \mathbb{E}\left[\int_0^T q_s^0 f(s, 0, 0) ds\right]. \end{aligned} $$

Using the assumptions A1 and A2, and rewriting the dynamics of $X^\psi$ under the equivalent probability measure $\mathcal{E}_T(\int_0^T \partial_z f(t, 0, 0) \cdot dW_t)\mathbb{P}$, we have, for a new value of $C$,

$$ \mathbb{E}[q_T^0 | X_T^\psi] \le C \left( 1 + \mathbb{E} \left[ \int_0^T q_s^0 |\psi_s|^2 ds \right]^{1/2} \right). $$

Then, by combining the last two displays, there exists (a new) constant $C > 0$, only depending on the data and independent of $c_1, c_2$ and $q$, such that

$$ \mathcal{J}(q^0, \psi) \ge -C. \quad (131) $$

Back to (130), we now make use of the duality inequality (13). By the latter, together with the growth Assumption A6 and the bound (131), we get

$$ \begin{aligned} \mathcal{S}(q) &\le -\mathcal{J}(q^0, \psi) + L \left(1 + \mathbb{E}\left[(1+q_T)|X_T^0|^{2-r}\right]\right) + \mathbb{E}\left[\int_0^T q_s \ell(s, 0) ds\right] \\ &\le C + \frac{1}{\vartheta} \mathbb{E}[h(q_T)] + \mathbb{E}\left[\exp\left(\vartheta\left(\xi + \int_0^T \ell_s^0 ds\right)\right)\right], \end{aligned} \quad (132) $$
where $\xi := L|X_T^0|^2$, $\ell_s^0 := \ell(s, 0)$ and $\vartheta > \upsilon := \beta e^{\alpha T}$. We recall that $\alpha$ and $\beta$ are given by (A3). Moreover, using the lower bound (16) for the dual driver, we have (similar to (103))

$$S(q) \geq -\mathbb{E}\left[\int_{0}^{T} q_s |f_s^0| ds\right] + \frac{1}{2\beta}\mathbb{E}\left[\int_{0}^{T} q_s |Z_s^*|^2 ds\right]. \quad (133)$$

Combining the last two inequalities (132) and (133), we obtain

$$\frac{1}{2\beta}\mathbb{E}\left[\int_{0}^{T} q_s |Z_s^*|^2 ds\right] \leq C_1 + \frac{1}{\vartheta}\mathbb{E}[h(q_T)] + \mathbb{E}\left[\exp\left(\vartheta\left(\xi + \int_{0}^{T} \ell_s^0 ds\right)\right)\right], \quad (134)$$

where $C_1 > 0$ is a finite constant, defined by $C_1 := C + T\|f^0\|_{L^\infty(\mathbb{F})}$.

Step 2. We now provide another bound for the entropy which will lead us to the expected result when combined with the conclusion of the first step. To do so, we let $(\tilde{q}_t := q_t \exp(-\alpha t))_{t \in [0,T]}$. Similar to (104), we have, by Itô's formula and for any stopping time $\tau$ such that $\int_0^\tau q_s |Z_s^*|^2 ds$ and $\sup_{t \in [0, \tau]} |q_t|$ belong to $L^\infty(\mathcal{F}_T)$,

$$h(\tilde{q}_\tau) = -1 + \int_0^\tau \tilde{q}_s \ln(\tilde{q}_s)(Y_s^* - \alpha)ds + \frac{1}{2} \int_0^\tau \tilde{q}_s |Z_s^*|^2 ds + H_T^\tau, \quad (135)$$

where $(H_T^\tau := \int_0^{t\wedge\tau} \tilde{q}_s \ln(\tilde{q}_s) Z_s^* \cdot dW_s)_{t \in [0,T]}$. Since $(\sqrt{\tilde{q}_t} \ln(\tilde{q}_t))_{0 \le t \le \tau}$ belongs to $L^\infty(\mathbb{F})$, $H^\tau$ is a square integrable martingale. By boundedness of the partial derivative of the driver with respect to its first variable, i.e. $|Y_s^*| \le \alpha$, we have that

$$\int_{0}^{T} \tilde{q}_s \ln(\tilde{q}_s)(Y_s^* - \alpha)ds \leq \int_{0}^{T} \tilde{q}_s \ln(\tilde{q}_s)(Y_s^* - \alpha)1_{\{\tilde{q}_s < 1\}}ds \leq 2\alpha e^{-1}T.$$

Returning to (135), inserting the above inequality and taking expectation, we get

$$\frac{1}{\beta}\mathbb{E}[h(\tilde{q}_{\tau})] \leq C_2 + \frac{1}{2\beta}\mathbb{E}\left[\int_{0}^{\tau} \tilde{q}_s |Z_s^*|^2 ds\right], \quad (136)$$

where $C_2 = 2\alpha\beta^{-1}e^{-1}T - 1$. By a standard localization argument, we choose $\tau$ along a non-decreasing sequence of stopping times $(\tau_k)_{k \in \mathbb{N}}$ converging to $T$, such that $\int_0^{\tau_k} q_s |Z_s^*|^2 ds$ and $\sup_{t \in [0, \tau_k]} |q_t|$ belong to $L^\infty(\mathcal{F}_T)$ for each $k \in \mathbb{N}$. This is possible to construct such a sequence because, by finiteness of $\mathbb{E}[h(q_T)]$, we have

$$\mathbb{E}\left[\int_{0}^{T} q_s |Z_s^*|^2 ds\right] < +\infty, \quad \text{and} \quad \mathbb{E}[q_T^*] < +\infty,$$

with the second inequality following from $L \log L$-Doob's inequality. Observing that $h(\tilde{q}_\tau)$ is lower bounded by $-1/e - e$, we deduce from (136) and Fatou's lemma that

$$\frac{1}{\beta}\mathbb{E}[h(\tilde{q}_T)] \leq C_2 + \frac{1}{2\beta}\mathbb{E}\left[\int_{0}^{T} \tilde{q}_s |Z_s^*|^2 ds\right]. \quad (137)$$

Moreover, because $\tilde{q}_T = \exp(-\alpha T)q_T$,

$$h(\tilde{q}_T) = \exp(-\alpha T)h(q_T) - \alpha T \exp(-\alpha T)q_T. \quad (138)$$

By the duality inequality (13), we have for any $\theta > 1$,

$$\alpha T \exp(-\alpha T)q_T \leq \frac{1}{\theta} \exp(-\alpha T)h(q_T) + \exp(\theta\alpha T). \quad (139)$$
Then, combining (138) and (139) yields

$$ \left(1 - \frac{1}{\theta}\right) \exp(-\alpha T) h(q_T) \leq h(\tilde{q}_T) + \exp(\theta \alpha T). $$

Combining the last inequality with (137), we obtain

$$ \left(1 - \frac{1}{\theta}\right) \frac{1}{v} \mathbb{E}[h(q_T)] \leq C_2 + \frac{1}{2\beta} \mathbb{E} \left[ \int_0^T \tilde{q}_s |Z_s^*|^2 ds \right] + \exp(\theta \alpha T) \quad (140) $$

$$ \leq C_2 + \frac{1}{2\beta} \mathbb{E} \left[ \int_0^T q_s |Z_s^*|^2 ds \right] + \exp(\theta \alpha T), \quad (141) $$

where we recall that $v = \beta \exp(\alpha T)$, see (132).

**Step 3: Conclusion.** Combining (134) and (140), we get in the end

$$ \left( \left(1 - \frac{1}{\theta}\right) \frac{1}{v} - \frac{1}{\vartheta} \right) h(q_T) \leq C + \exp(\theta \alpha T) + \mathbb{E} \left[ \exp \left( \vartheta \left( \xi + \int_0^T \ell_s^0 ds \right) \right) \right], $$

where $C = C_1 + C_2$. Choosing $\theta > (1 - v/\vartheta)^{-1}$, recalling that $\vartheta > v$ by definition (see again (132)), we deduce that there exists a finite constant $c'_1 > 0$ independent of $c_1$ such that $\sup_{t \in [0,T]} \mathbb{E}[h(q_t)] \leq c'_1$ provided that $\xi + \int_0^T \ell_s^0 ds \in L_{\exp(\mathcal{F}_T)}^{1,\vartheta}$. The latter can be verified by combining Lemma 42 with assumptions A4 and A5. The argument was already outlined in Remark 7. On the one hand, A4 says that $\ell^0$ is bounded by $L$. In particular, it suffices to show that $\xi$, which is here equal to $L|X_T^0|^2^{-r}$, belongs to $L_{\exp(\mathcal{F}_T)}^{1,\vartheta}$, for some $\vartheta > v$. By Lemma 42, this is always true when $r = 1$. When $r = 0$, we need $4\vartheta L \|\Gamma\|_{L^{\infty}(\mathbb{F}, \mathbb{R}^{n \times n})}^2 \|\Gamma^{-1}\|_{L^{\infty}(\mathbb{F}, \mathbb{R}^{n \times n})}^2 \|\nu\|_{L^{\infty}(\mathbb{F}, \mathbb{R}^{n \times d})}^2 T < 1$. By A5, this is indeed possible to choose $\vartheta$ satisfying the latter while ensuring the condition $\vartheta > v$. This completes the proof.

Finally, using (132) one last time, we conclude that $S(q) \leq c'_1$, for a possibly new (but still independent of $c_1$) value of $c'_1$. $\square$

### 5.2.2 Necessary condition

We show the necessary condition, i.e. the first assertion, in the statement of Theorem 25. Because $\bar{\psi} \in \mathcal{A}_{c_2}$ is fixed throughout the subsection, we omit it in most of the notations and merely write

$$ \mathcal{J}(q) := \mathcal{J}(q, \bar{\psi}), \quad \mathcal{R}(q) := \mathcal{R}(q, \bar{\psi}), \quad \ell_s := \ell(s, \bar{\psi}_s), \quad (142) $$

for $q \in Q_{c_1}$. We denote by $(Y^*, Z^*)$ the representatives of $q$, as defined in (4), i.e. $(Y^*, Z^*)$ is an $\mathbb{R} \times \mathbb{R}^d$-valued $\mathbb{F}$-progressively measurable pair satisfying $S(q) < c_1$. With $q$, we also associate the equivalent probability measure $\mathbb{Q}$ given by $d\mathbb{Q} = \mathcal{E}_T(\int_0^T Z_s^* \cdot dW_s)d\mathbb{P}$. We also consider the adjoint BSDE, with unknown $(Y, Z)$,

$$ \left\{ \begin{array}{l} -dY_t = (Y_t^* Y_t + Z_t^* \cdot Z_t - f^*(t, Y_t^*, Z_t^*) + \ell_t) dt - Z_t \cdot dW_t, \quad t \in [0, T], \\ Y_T = \delta_q \mathcal{G}(q_T, X_T^\psi). \end{array} \right. \quad (143) $$

Notice that this BSDE is not the one appearing in the first-order system (Opt_N), since at this stage of the proof, the relationship (23) between $(Y, Z)$ and $(Y^*, Z^*)$ is not yet known. The additional property (23) forms part of the necessary condition and is shown to hold under the assumption that $q \in Q_{c_0}$ for some $c_0 \in (0, c_1)$, with $q$ being a maximizer of $q' \mapsto \mathcal{J}(q')$ over $Q_{c_1}$; see Lemma 28. For the time being, we establish the following well-posedness result:
**Lemma 27.** There exists a unique solution $(Y, Z) \in D(\mathbb{F}, \mathbb{Q}) \times (\cap_{\beta \in (0,1)} M^{\beta}(\mathbb{Q}, \mathbb{F}, \mathbb{R}^d))$ to (143). It is given by the formula (with the shorthand notation $\delta_q\mathcal{G}(q_T)$ in place of $\delta_q\mathcal{G}(q_T, X_T^\psi)$):

$$Y_t = q_t^{-1} \mathbb{E} \left[ q_T \delta_q \mathcal{G}(q_T) + \int_t^T q_s (\ell_s - f^*(s, Y_s^*, Z_s^*)) \mathrm{d}s \mid \mathcal{F}_t \right], \quad t \in [0, T]. \quad (144)$$

*Proof.* The proof is mostly taken from [22]. We give it for the sake of completeness. We recall that $q_t = \Lambda_t \mathcal{E}_t$, for any $t \in [0, T]$, where $(\Lambda_t := \exp(\int_0^t Y_s^* \mathrm{d}s))_{t \in [0, T]}$ and $(\mathcal{E}_t := \mathcal{E}_t(\int_0^t Z_s^* \cdot \mathrm{d}W_s))_{t \in [0, T]}$.

**Step 1:** We claim that there exists a pair $(\tilde{Y}, \tilde{Z}) \in D(\mathbb{F}, \mathbb{Q}) \times (\cap_{\beta \in (0,1)} M^{\beta}(\mathbb{Q}, \mathbb{F}, \mathbb{R}^d))$ such that, for every $t \in [0, T]$,

$$\tilde{Y}_t = \Lambda_T \delta_q \mathcal{G}(q_T) + \int_t^T \Lambda_s (\ell_s - f^*(s, Y_s^*, Z_s^*)) \mathrm{d}s - \int_t^T \tilde{Z}_s \cdot \mathrm{d}\tilde{W}_s, \quad (145)$$

where $(\tilde{W}_t := W_t - \int_0^t Z_s^* \mathrm{d}s)_{t \in [0, T]}$ is a Brownian motion under the equivalent probability measure $\mathbb{Q} = \mathcal{E}_T \mathbb{P}$. Existence of the pair $(\tilde{Y}, \tilde{Z})$ is proven in two steps.

Throughout, the letter $C$ denotes a generic constant that only depends on the assumptions listed in Subsection 3.1 and that is, in particular, independent of $q$ and $\bar{\psi}$. The first observation is that

$$\mathbb{E}^{\mathbb{Q}} \left[ \Lambda_T |\delta_q \mathcal{G}(q_T)| + \int_0^T \Lambda_s |\ell_s - f^*(s, Y_s^*, Z_s^*)| \mathrm{d}s \right] < +\infty, \quad (146)$$

which is a consequence of the following three bounds. First, by the growth Assumption A6, we have

$$\mathbb{E}^{\mathbb{Q}} [\Lambda_T |\delta_q \mathcal{G}(q_T)|] = \mathbb{E} [q_T |\delta_q \mathcal{G}(q_T)|] \le L \exp(\alpha T) (1 + 2\mathbb{E} [q_T |X_T^{\bar{\psi}}|^{2-r}]).$$

By Lemma 40, the last term satisfies the inequality

$$\mathbb{E} \left[ q_T \left| X_T^{\bar{\psi}} \right|^{2-r} \right] \le C (1 + \mathcal{S}(q) + \mathcal{S}^*(\bar{\psi})).$$

Recalling that $q \in \mathcal{Q}_{c_1}$ and $\bar{\psi} \in \mathcal{A}_{c_2}$, we obtain

$$\mathbb{E}^{\mathbb{Q}} [\Lambda_T |\delta_q \mathcal{G}(q_T)|] < +\infty.$$

By A4 and a direct application of the duality inequality (14), we also have

$$\mathbb{E}^{\mathbb{Q}} \left[ \int_0^T \Lambda_s |\ell_s| \mathrm{d}s \right] \le C + C \mathbb{E} \left[ \int_0^T q_s |\bar{\psi}_s|^2 \mathrm{d}s \right] \le C + C (\mathcal{S}(q) + \mathcal{S}^*(\bar{\psi})) < +\infty.$$

It remains to observe from 2 that $f^*(t, y^*, z^*) \ge -f_t^0 := -f(t, 0, 0)$, which implies $|f^*(t, y^*, z^*)| \le f^*(t, y^*, z^*) + |f_t^0| + f_t^0$, and then

$$\begin{aligned} \mathbb{E}^{\mathbb{Q}} \left[ \int_0^T \Lambda_s |f^*(s, Y_s^*, Z_s^*)| \mathrm{d}s \right] &\le \mathbb{E} \left[ \int_0^T q_s (f^*(s, Y_s^*, Z_s^*) + f_s^0 + |f_s^0|) \mathrm{d}s \right] \\ &\le C (1 + \mathcal{S}(q)) < +\infty. \end{aligned} \quad (147)$$
The last three displays imply (146).

As announced, we now follow [22, Section 6]. To do so, we consider $(\xi^k, \ell^k)_{k \in \mathbb{N}}$ such that, for each $k \in \mathbb{N}$, $\xi^k$ is a bounded $\mathcal{F}_T$-measurable random variable and $\ell^k = (\ell_t^k)_{t \in [0, T]}$ is a bounded $\mathbb{F}$-progressively measurable with the property that

$$ \lim_{k \to +\infty} \mathbb{E}^{\mathbb{Q}} \left[ \Lambda_T |\xi^k - \delta_q \mathcal{G}(q_T)| + \int_0^T \Lambda_s |\ell_s^k - (\ell_s - f^*(s, Y_s^*, Z_s^*))|ds \right] = 0. $$

Then, for each $k \in \mathbb{N}$, we can define $(\tilde{Y}^k, \tilde{Z}^k)$ such that (we recall from [2, Theorem 2.4] that the martingale representation theorem holds under $\mathbb{Q}$, with respect to $\tilde{W}$)

$$ \tilde{Y}_t^k = \Lambda_T \xi^k + \int_t^T \Lambda_s \ell_s^k ds - \int_t^T \tilde{Z}_s^k \cdot d\tilde{W}_s, \quad t \in [0, T]. $$

There is no difficulty to see that

$$ \tilde{Y}_t^k = \mathbb{E}^{\mathbb{Q}} \left[ \Lambda_T \xi^k + \int_t^T \Lambda_s \ell_s^k ds \middle| \mathcal{F}_t \right], \quad t \in [0, T]. $$

Since the term inside the conditional expectation appearing in the right-hand side is bounded, for each $k \in \mathbb{N}$, uniformly in $t \in [0, T]$, we easily deduce that the process $(\tilde{Y}^k, \tilde{Z}^k)$ satisfies the conclusion of the statement. In fact, item 2 is even satisfied in a stronger sense, as $\beta$ can be taken in $(0, +\infty)$.

The key step is to prove that the sequences $(\tilde{Y}^k)_{k \in \mathbb{N}}$ and $(\tilde{Z}^k)_{k \in \mathbb{N}}$ are Cauchy sequences in well-chosen spaces. As for $(\tilde{Y}^k)_{k \in \mathbb{N}}$, we notice that, for any $\mathbb{F}$-stopping time $\tau$ with values in $[0, T]$, for any $m, k \in \mathbb{N}$,

$$ \mathbb{E}^{\mathbb{Q}} \left[ |\tilde{Y}_{\tau}^k - \tilde{Y}_{\tau}^m| \right] \le \mathbb{E}^{\mathbb{Q}} \left[ \Lambda_T |\xi^k - \xi^m| + \int_0^T \Lambda_s |\ell_s^k - \ell_s^m| ds \right], $$

and the right-hand side tends to 0, as $m$ and $k$ tend to $+\infty$. This shows that

$$ \lim_{N \to \infty} \sup_{m,k \ge N} \sup_{\tau} \mathbb{E}^{\mathbb{Q}} \left[ |\tilde{Y}_{\tau}^k - \tilde{Y}_{\tau}^m| \right] = 0, $$

with $\tau$ in the left-hand side being implicitly understood as a generic stopping time with values in $[0, T]$. Then, the analysis carried out in [22] (together with the references cited therein) permits us to show that there exists a process $\tilde{Y}$ satisfying item 1 in the statement such that

$$ \lim_{k \to \infty} \sup_{\tau} \mathbb{E}^{\mathbb{Q}} \left[ |\tilde{Y}_{\tau}^k - \tilde{Y}_{\tau}| \right] = 0. $$

It then remains to handle the martingale integrand $(\tilde{Z}^k)_{k \in \mathbb{N}}$. Writing

$$ \int_0^t (\tilde{Z}_s^k - \tilde{Z}_s^m) \cdot d\tilde{W}_s = (\tilde{Y}_t^k - \tilde{Y}_t^m) - (\tilde{Y}_0^k - \tilde{Y}_0^m) - \int_0^t \Lambda_s (\ell_s^k - \ell_s^m) ds, $$

for $t \in [0, T]$, we deduce from [22, Lemma 6.1] that, for any $\beta \in (0, 1)$,

$$ \begin{aligned} & \mathbb{E}^{\mathbb{Q}} \left[ \sup_{t \in [0,T]} \left| \int_0^t (\tilde{Z}_s^k - \tilde{Z}_s^m) \cdot d\tilde{W}_s \right|^\beta \right] \\ & \le C_\beta \mathbb{E}^{\mathbb{Q}} \left[ \left| \int_0^T (\tilde{Z}_s^k - \tilde{Z}_s^m) \cdot d\tilde{W}_s \right|^\beta \right] \\ & \le C_\beta \left( \mathbb{E}^{\mathbb{Q}} \left[ \Lambda_T |\xi^k - \xi^m| \right] + \mathbb{E}^{\mathbb{Q}} \left[ |Y_0^k - Y_0^m| \right] + \mathbb{E} \left[ \int_0^T \Lambda_s |\ell_s^k - \ell_s^m| ds \right] \right)^\beta, \end{aligned} $$
for a constant $C_β > 0$ only depending on $β$. As a consequence, we obtain

$$ \lim_{N \to \infty} \sup_{m,k \ge N} \mathbb{E}^{\mathbb{Q}} \left[ \sup_{t \in [0,T]} \left| \int_0^t (\tilde{Z}_s^k - \tilde{Z}_s^m) \cdot d\tilde{W}_s \right|^{\beta} \right] = 0. $$

And then, by Bürkholder-Davis-Gundy inequality, it holds

$$ \lim_{N \to \infty} \sup_{m,k \ge N} \mathbb{E}^{\mathbb{Q}} \left[ \left( \int_0^T |\tilde{Z}_s^k - \tilde{Z}_s^m|^2 ds \right)^{\beta/2} \right] = 0, $$

and the existence of $\tilde{Z}$ as in the statement follows from a new application of Cauchy's convergence criterion in complete spaces.

**Step 2:** We now establish uniqueness of the pair $(Y, Z)$. Multiplying any solution by $(\Lambda_t)_{t \in [0,T]}$, uniqueness of the pair $(Y, Z)$ is in fact equivalent to uniqueness of the pair $(\tilde{Y}, \tilde{Z})$ in the expansion (145) (within the same space as in the statement).

By uniform integrability of the collection $(\tilde{Y}_\tau)_\tau$, when $\tau$ runs over the set of F-stopping times with values in $[0, T]$ and by a standard localization argument, we deduce that, necessarily,

$$ \tilde{Y}_t = \mathbb{E}^{\mathbb{Q}} \left[ \Lambda_T \delta_q \mathcal{G}(q_T) + \int_t^T \Lambda_s (\ell_s - f^*(s, Y_s^*, Z_s^*)) ds \middle| \mathcal{F}_t \right], \quad t \in [0, T]. \quad (148) $$

This establishes the uniqueness of $\tilde{Y}$. We then rewrite the equation for $(\tilde{Y}, \tilde{Z})$ in the form

$$ \tilde{Y}_t - \int_0^t \Lambda_s (\ell_s - f^*(s, Y_s^*, Z_s^*)) ds = \int_0^t \tilde{Z}_s \cdot d\tilde{W}_s, \quad t \in [0, T]. $$

The right-hand side is a local martingale (by assumption). Since the left-hand side is given, we deduce that $\tilde{Z}$ is unique. Recalling the two formulas $\mathbb{Q} = \mathcal{E}_T(\int_0^t Z_s^* dW_s)\mathbb{P}$ and $(q_t = \Lambda_t \mathcal{E}_t(\int_0^t Z_s^* \cdot dW_s))_{t \in [0,T]}$, we easily derive (144) from (148). $\square$

At this stage, the notion of a solution to equation (143) is well defined and understood in the sense of Lemma 27. We are now in a position to establish the first-order condition for Nature.

**Lemma 28.** For a given $c_0 \in (0, c_1)$, assume that there exists a maximizer $q \in \mathcal{Q}_{c_0}$ to the problem $(\mathbb{P}_{N,c_1})$ (the latter being set over $\mathcal{Q}_{c_1}$). Then, denoting by $(Y, Z)$ the solution of (143), the triple $(q, Y, Z)$ satisfies the first-order condition $(\text{Opt}_N)$.

**Proof.** Step 1: localization procedure. Generally speaking, our main objective is to prove that, for prescribed directions $(y^*, z^*) \in L^\infty(\mathbb{F}) \times L^\infty(\mathbb{F}, \mathbb{R}^d)$, $\mathbb{P}$-almost surely, for almost every $t \in [0, T]$,

$$ f^*(t, Y_t^* + y_t^*, Z_t^* + z_t^*) - f^*(t, Y_t^*, Z_t^*) - (Y_t y_t^* + Z_t \cdot z_t^*) \ge 0. \quad (149) $$

From this, we will eventually derive (23) and then (OptN).

Although the proof of (149) follows seemingly standard arguments, it requires some non-trivial adjustments. We proceed by contradiction assuming that the left-hand side on (149) is negative on an event $E$ of positive measure under $\text{Leb}_{[0,T]} \otimes \mathbb{P}$, namely

$$ \exists \varrho > 0, \quad (\text{Leb}_{[0,T]} \otimes \mathbb{P})(E) > 0, \\ \text{with } E := \{f^*(t, Y_t^* + y_t^*, Z_t^* + z_t^*) - f^*(t, Y_t^*, Z_t^*) - (Y_t y_t^* + Z_t \cdot z_t^*) \le -\varrho\}. \quad (150) $$
For a given $A > 0$, we also consider the stopping time

$$ \tau_A := \inf \left\{ t \in [0, T], \frac{1}{q_t} + \left| \int_0^t z_s^* \cdot Z_s^* ds \right| + \left| \int_0^t z_s^* \cdot Z_s ds \right| + \left| \int_0^t z_s^* \cdot dW_s \right| \ge A \right\}, \quad (151) $$

with the usual convention that the stopping time is equal to $+\infty$ if the set inside the infimum is empty. It is easy to prove that

$$ \lim_{A \to +\infty} \mathbb{P}(\{\tau_A \le T\}) = 0, $$

from which we deduce that we can choose $A$ large enough such that

$$ (\mathrm{Leb}_{[0,T]} \otimes \mathbb{P}) (\{(t, \omega) \in E, t \le \tau_A(\omega)\}) > 0. \quad (152) $$

We then define the new ‘localized’ directions

$$ y_s^{\star,E}(\omega) := y_s^*(\omega) \mathbb{1}_E(s, \omega), \quad z_s^{\star,A,E}(\omega) := z_s^*(\omega) \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s), \quad (153) $$

for $s \in [0, T]$ and $\omega \in \Omega$. For an intensity $\varepsilon \in (0, 1]$, we consider the solution $(q_t^\varepsilon := \Lambda_t^\varepsilon \mathcal{E}_t^\varepsilon)_{t \in [0,T]}$ of the equation (4) driven by the pair

$$ (Y^{\star, \varepsilon}, Z^{\star, \varepsilon}) := (Y^{\star}, Z^{\star}) + \varepsilon (y^{\star, E}, z^{\star, A, E}) $$

and where

$$ \Lambda_t^\varepsilon = \exp \left( \int_0^t Y_s^{\star, \varepsilon} ds \right), \quad \mathcal{E}_t^\varepsilon = \mathcal{E}_t \left( \int_0^t Z_s^{\star, \varepsilon} \cdot dW_s \right), \quad t \in [0, T]. $$

(For simplicity, we omit to precise the dependence on $A$ and $E$.)

We also introduce the process $q'$ which will be proved to be the variational process of $q$ in the direction $(y^{\star, E}, z^{\star, A, E})$. It is defined as

$$ q_t' := q_t \left( \int_0^t (y_s^{\star, E} - Z_s^* \cdot z_s^{\star, A, E}) ds + \int_0^t z_s^{\star, A, E} \cdot dW_s \right), \quad t \in [0, T]. \quad (154) $$

Using the definition of $z^{\star, A, E}$, we can check that, for all $t \in [0, T]$, $|q_t'| \le Cq_t$, for a constant $C$ independent of $\varepsilon$ and $t$. Moreover, $q'$ solves the equation

$$ q_t' = \int_0^t (q_s' Y_s^* + q_s y_s^{\star, E}) ds + \int_0^t (q_s' Z_s^* + q_s z_s^{\star, A, E}) \cdot dW_s, \quad t \in [0, T]. \quad (155) $$

We then let

$$ \Delta q_s^\varepsilon := \varepsilon^{-1} \delta q_s^\varepsilon - q_s', \quad \text{with} \quad \delta q_s^\varepsilon := q_s^\varepsilon - q_s, $$

$$ \delta f_s^{\star, \varepsilon} := f_s^{\star, \varepsilon} - f_s^*, \quad \text{with} \quad \left\{ \begin{array}{l} f_s^* := f^*(s, Y_s^*, Z_s^*), \\ f_s^{\star, \varepsilon} := f^*(s, Y_s^{\star, \varepsilon}, Z_s^{\star, \varepsilon}) \end{array} \right. $$

for any $s \in [0, T]$. We observe that, at this stage, $f^{\star, \varepsilon}$ may take the value $+\infty$.
Step 2: $q^\epsilon \in Q_{c_1}$ for $\epsilon$ small enough. We show that $q^\epsilon$ is an admissible controlled process for $(P_{N,c_1})$. The proof relies on the explicit formula for $q^\epsilon$. For each time $t \in [0, T]$, we have

$$ q_t^\epsilon = q_t \exp \left( \epsilon \left( \int_0^t y_s^{\star,E} ds + \int_0^t z_s^{\star,A,E} \cdot dW_s - \int_0^t (Z_s^\star \cdot z_s^{\star,A,E} + \frac{1}{2}\epsilon|z_s^{\star,A,E}|^2) ds \right) \right) \\ =: q_t \exp (\epsilon \varphi_t^\epsilon). $$

We can find a constant $C$, independent of $\epsilon$, such that, with probability 1, $\sup_{t \in [0,T]} |\varphi_t^\epsilon| \le C$. This follows from the fact that $y^\star$ and $z^\star$ are bounded and from the definitions of $\tau_A$, $y^{\star,E}$ and $z^{\star,A,E}$ (see (151) and (153)). And then, for any $\epsilon \in (0, 1]$ and any $t \in [0, T]$,

$$ |q_t^\epsilon - q_t| \le C\epsilon \exp(C)q_t \le C\epsilon \exp(C) \sup_{s \in [0,T]} q_s. \quad (156) $$

To establish the desired result, we expand the term $S(q^\epsilon)$ as

$$ S(q^\epsilon) = \mathbb{E} \left[ \int_0^T q_s f_s^{\star,\epsilon} ds \right] + \mathbb{E} \left[ \int_0^T \delta q_s^\epsilon f_s^{\star,\epsilon} ds \right]. \quad (157) $$

We study the two terms on the right-hand side separately. We start with the first one. By convexity of $f^\star$ in the variables $y^\star$ and $z^\star$ and by definition of the set $E$, see (150) and (152), we have (because $\epsilon \in (0, 1]$)

$$ \begin{aligned} & \frac{1}{\epsilon} \delta f_s^{\star, \epsilon}(\omega) \\ & \le \frac{1}{\epsilon} \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s) (f^\star(s, Y_s^\star + \epsilon y_s^\star, Z_s^\star + \epsilon z_s^\star) - f^\star(s, Y_s^\star, Z_s^\star)) \quad (158) \\ & \le \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s) (y_s^\star Y_s^\star + z_s^\star \cdot Z_s^\star - \varrho) \\ & = y_s^{\star, E} Y_s^\star + z_s^{\star, A, E} \cdot Z_s^\star - \varrho \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s), \end{aligned} $$

for the same real $\varrho \ge 0$ as in (150). Multiplying both sides by $q$ and $\epsilon$, adding $q_s f_s^\star$ on both sides, integrating from 0 to $T$ and using the fact that $\sup_{t \in [0, \tau_A]} |\int_0^t z_s^\star \cdot Z_s ds| \le A$, we deduce that (allowing the value of $C$ to vary from line to line as long as it remains independent of $\epsilon$)

$$ \mathbb{E} \left[ \int_0^T q_s f_s^{\star, \epsilon} ds \right] \le C\epsilon + S(q). $$

We now turn to the second term on the right-hand side of (157). From the inequality (156), we directly deduce that

$$ \mathbb{E} \left[ \int_0^T \delta q_s^\epsilon f_s^{\star, \epsilon} ds \right] \le C\epsilon \mathbb{E} \left[ \int_0^T q_s |f_s^{\star, \epsilon}| ds \right] \le C\epsilon + C\epsilon \mathbb{E} \left[ \int_0^T q_s f_s^{\star, \epsilon} ds \right] \le C\epsilon, $$

where we used the bound $|f_t^{\star, \epsilon}| \le f_t^{\star, \epsilon} + C$, see (147), the value of the constant $C$ being allowed to change from one term to another.

Finally, plugging the last two estimates into (157) yields the desired result provided that $\epsilon$ is small enough.

Step 3: strong convergences of $\delta q^\epsilon$ and $\Delta q^\epsilon$. As a direct consequence of the inequality (156), we deduce that, for $A$ fixed, $\mathbb{P}$ almost surely, $\sup_{t \in [0,T]} |q_t^\epsilon - q_t| \to 0$ as $\epsilon \to 0$.
Below, we also establish the almost sure uniform convergence of $\Delta q^{\epsilon}$ to 0. To do so, one can refine the argument presented in Step 2 and provide a second-order (in $\epsilon$) expansion of $q_t^{\epsilon}$, writing, for all $t \in [0, T]$,

$$|q_t^{\epsilon} - q_t(1 + \epsilon\varphi_t^{\epsilon})| \le q_t |\exp(\epsilon\varphi_t^{\epsilon}) - (1 + \epsilon\varphi_t^{\epsilon})| \le C^2\epsilon^2 \exp(C)q_t. \quad (159)$$

Now, we use the fact that $(q_t\varphi_t^\epsilon)_{t \in [0,T]}$ and $(q'_t)_{t \in [0,T]}$ are close one from each other. Indeed,

$$d[q_t\varphi_t^\epsilon] = (q_t\varphi_t^\epsilon Y_t^* + q_t y_t^{*,E} - \frac{1}{2}\epsilon|z_t^{*,A,E}|^2) dt + (q_t\varphi_t^\epsilon Z_t^* + q_t z_t^{*,A,E}) \cdot dW_t, \quad t \in [0, T].$$

And then, thanks to (155),

$$d[q'_t - q_t\varphi_t^\epsilon] = ([q'_t - q_t\varphi_t^\epsilon] Y_t^* + \frac{1}{2}\epsilon|z_t^{*,A,E}|^2) dt + ([q'_t - q_t\varphi_t^\epsilon] Z_t^*) \cdot dW_t, \quad t \in [0, T],$$

with 0 as initial condition. It is standard to deduce that

$$q'_t - q_t\varphi_t^\epsilon = q_t\epsilon \int_0^t q_s^{-1}|z_s^{*,A,E}|^2ds, \quad t \in [0, T].$$

Returning back to (159), we deduce from the above identity (together with the fact that $q_s^{-1} \le A$ for $s \le \tau_A$) that

$$|\delta q_t^{\epsilon} - \epsilon q'_t| \le C\epsilon^2 q_t, \quad t \in [0, T],$$

for a new value of $C$ (still independent of $\epsilon$). Dividing by $\epsilon$, we get

$$|\Delta q_t^{\epsilon}| \le C\epsilon q_t, \quad t \in [0, T]. \quad (160)$$

**Step 4: derivative of the terminal and running costs.** In this step, we address the limit (as $\epsilon$ tends to 0) of (as explained above, we omit the dependence on $\bar{\psi}$ in the notations)

$$\frac{1}{\epsilon}(\mathcal{R}(q^{\epsilon}) - \mathcal{R}(q)) = \frac{1}{\epsilon}\left(\mathcal{G}(q_T^{\epsilon}) - \mathcal{G}(q_T) + \mathbb{E}\left[\int_0^T \delta q_s^{\epsilon} \ell_s ds\right]\right). \quad (161)$$

We first compute the derivative of $\mathcal{G}$ along $q_T^{\epsilon}$. We write

$$\begin{aligned}
& \frac{1}{\epsilon} [\mathcal{G}(q_T^{\epsilon}) - \mathcal{G}(q_T)] \\
&= \frac{1}{\epsilon} \int_0^1 \mathbb{E} [\delta_q \mathcal{G}(q_T + \lambda \delta q_T^{\epsilon}) \delta q_T^{\epsilon}] d\lambda \\
&= \int_0^1 \mathbb{E} [\delta_q \mathcal{G}(q_T + \lambda \delta q_T^{\epsilon}) q'_T] d\lambda + \int_0^1 \mathbb{E} [\delta_q \mathcal{G}(q_T + \lambda \delta q_T^{\epsilon}) \Delta q_T^{\epsilon}] d\lambda. \quad (162)
\end{aligned}$$

Here, we recall from A6 that

$$|\delta_q \mathcal{G}(q_T + \lambda \delta q_T^{\epsilon})| \le L (1 + \mathbb{E}[(q_T + \lambda \delta q_T^{\epsilon})|X_T|^{2-r}] + |X_T|^{2-r}).$$
Recalling that $|\delta q_t^\varepsilon| \le C\varepsilon q_t$ by (156), we deduce from Lemma 40 that the expectation on the right-hand side is bounded uniformly in $\varepsilon$ by

$$ \mathbb{E}[(q_T + \lambda \delta q_T^\varepsilon) |X_T|^{2-r}] \le C(1 + \mathbb{E}[q_T |X_T|^{2-r}]) < +\infty. $$

Also, from the first step, we know that $q_T' \le Cq_T$. Then by Lemma 40 again, we deduce that $\mathbb{E}[(1 + |X_T|^{2-r})q_T'] < +\infty$. By dominated convergence and by (156), we deduce that the first term on the right-hand side of (162) converges to

$$ \lim_{\varepsilon \to 0} \int_0^1 \mathbb{E} [\delta_q \mathcal{G}(q_T + \lambda \delta q_T^\varepsilon) q_T'] \, d\lambda = \mathbb{E} [\delta_q \mathcal{G}(q_T) q_T'] . $$

As for the second term on the right-hand side of (162), we can proceed in the same way, but using in addition (160). We obtain

$$ \lim_{\varepsilon \to 0} \int_0^1 \mathbb{E} [\delta_q \mathcal{G}(q_T + \lambda \delta q_T^\varepsilon) \Delta q_T^\varepsilon] = 0. $$

By combining the last two displays, we obtain

$$ \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} (\mathcal{G}(q_T^\varepsilon) - \mathcal{G}(q_T)) = \mathbb{E}[\delta_q \mathcal{G}(q_T) q_T'] = \mathbb{E}[Y_T q_T'], $$

where we recall that $Y_T$ is the terminal condition given by the second line on (143).

We now turn to the second term in (161). Recalling that $\mathcal{S}^*(\bar{\psi}) < +\infty$, we deduce from the bound (160) and the duality inequality (14) that

$$ \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \mathbb{E} \left[ \int_0^T \delta q_s^\varepsilon \ell_s \, ds \right] = \mathbb{E} \left[ \int_0^T q_s' \ell_s \, ds \right]. $$

Combining the last two limits, we obtain

$$ \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} (\mathcal{R}(q^\varepsilon) - \mathcal{R}(q)) = \mathbb{E} \left[ Y_T q_T' + \int_0^T q_s' \ell_s \, ds \right]. \quad (163) $$

We then expand the first term on the right-hand side by means of Itô's formula, using the BSDE (143) satisfied by $(Y_t)_{t \in [0,T]}$ and the equation (155) satisfied by $q'$. We get for all $t \in [0, T]$,

$$ \mathrm{d}[q_t' Y_t] = q_t' (-Y_t^* Y_t - Z_t^* \cdot Z_t + f^*(t, Y_t^*, Z_t^*) - \ell_t) \mathrm{d}t \\ + Y_t (q_t' Y_t^* + q_t y_t^{*,E}) \mathrm{d}t + Z_t \cdot (q_t' Z_t^* + q_t z_t^{*,A,E}) \mathrm{d}t + H_t \cdot \mathrm{d}W_t, $$

with $(H_t := q_t' Z_t + Y_t(q_t' Z_t^* + q_t z_t^{*,A,E}))_{t \in [0,T]}$. Here, we need a new localization sequence to handle the local martingale. We define, for any $c > 0$, $\zeta_c := \inf\{t \in [0, T], |Y_t| + \int_0^t |Z_s|^2 ds \ge c\}$ (with $\inf\emptyset = +\infty$). Because $Y$ has continuous trajectories and $\mathbb{P}(\{\int_0^T |Z_t|^2 dt < +\infty\}) = 1$, it holds $\zeta_c \to +\infty$ almost surely, as $c \to +\infty$. Then, recalling the identity $q_0' = 0$ and the shorthand notation $f_t^*$ for $f^*(t, Y_t^*, Z_t^*)$, we can rearrange the above expansion and obtain

$$ q'_{\zeta_c \wedge T} Y_{\zeta_c \wedge T} + \int_0^{\zeta_c \wedge T} q'_s \ell_s \, ds = \int_0^{\zeta_c \wedge T} (q'_s f_s^* + q_s Y_s y_s^{*,E} + q_s Z_s \cdot z_s^{*,A,E}) \, ds \\ + \int_0^{\zeta_c \wedge T} H_s \cdot \mathrm{d}W_s. $$
Using the three bounds $|q'_t| \le Cq_t$, for all $t \in [0, T]$, $\|z^{\star,A,E}\|_{L^{\infty}(\mathbb{F},\mathbb{R}^d)} < +\infty$ and $\mathbb{E}[q_T \int_0^T |Z_s^\star|^2 ds] < +\infty$, together with the definition of the stopping time $\zeta_c$, we can prove that $\mathbb{E}[\int_0^{\zeta_c \wedge T} H_s \cdot dW_s] = 0$, from which we deduce

$$ \mathbb{E} \left[ q'_{\zeta_c \wedge T} Y_{\zeta_c \wedge T} + \int_0^{\zeta_c \wedge T} q'_s \ell_s ds \right] = \mathbb{E} \left[ \int_0^{\zeta_c \wedge T} (q'_s f_s^* + q_s Y_s y_s^{\star, E} + q_s Z_s \cdot z_s^{\star, A, E}) ds \right]. \quad (164) $$

The point is to let $c$ tend to $+\infty$ on both sides. Thanks to the following three inequalities (the first line follows from the two bounds $|q'_t| \le Cq_t$ and $|f_t^*| \le f_t^* + C$, for $t \in [0, T]$, see (147)), the second one from the bound $\sup_{t \in [0,T]} \mathbb{E}[q_T|Y_t|] < +\infty$, see Lemma 27 and the definition of the space $D(\mathbb{F}, \mathbb{Q})$ in Section 2, and the third one from (151),

$$ \mathbb{E} \left[ \int_0^T |q'_s f_s^*| ds \right] \le C \mathbb{E} \left[ \int_0^T q_s f_s^* ds \right] + C = C S(q) + C < +\infty, $$

$$ \mathbb{E} \left[ \int_0^T q_s |Y_s y_s^*| ds \right] \le C \mathbb{E} \left[ q_T \int_0^T |Y_s| ds \right] < +\infty, $$

$$ \mathbb{E} \left[ \int_0^T q_s |Z_s \cdot z_s^{\star, A, E}| ds \right] \le \mathbb{E} \left[ q_T \int_0^T |Z_s \cdot z_s^{\star, A, E}| ds \right] \le A \mathbb{E} [q_T] < +\infty, $$

we deduce, by dominated convergence theorem, that

$$ \lim_{c \to +\infty} \mathbb{E} \left[ \int_0^{\zeta_c \wedge T} (q'_s f_s^* + q_s Y_s y_s^{\star, E} + q_s Z_s \cdot z_s^{\star, A, E}) ds \right] \\ = \mathbb{E} \left[ \int_0^T (q'_s f_s^* + q_s Y_s y_s^{\star, E} + q_s Z_s \cdot z_s^{\star, A, E}) ds \right]. $$

Similarly,

$$ \lim_{c \to +\infty} \mathbb{E} \left[ \int_0^{\zeta_c \wedge T} q'_s \ell_s ds \right] = \mathbb{E} \left[ \int_0^T q'_s \ell_s ds \right]. $$

It remains to pass to the limit in $\mathbb{E}[Y_{\zeta_c \wedge T} q'_{\zeta_c \wedge T}]$ as $c$ tends to $+\infty$. Almost surely, $Y_{\zeta_c \wedge T} q'_{\zeta_c \wedge T} \to Y_T q'_T$. Moreover, using the bound $|q'_t| \le Cq_t$, we deduce that, for any event $F \in \mathcal{F}_T$, $\mathbb{E}[1_F |Y_{\zeta_c \wedge T} q'_{\zeta_c \wedge T}|] \le C \mathbb{E}^{\mathbb{Q}}[1_F |Y_{\zeta_c \wedge T}|]$. Recalling that the family $(Y_{\zeta_c \wedge T})_{c>0}$ is uniformly integrable under $\mathbb{Q}$, see Lemma 27, we notice that the last term can be rendered as small as desired by choosing $\mathbb{Q}(F)$ small enough, and thus by choosing $\mathbb{P}(F)$ small enough. This shows that the collection $(|Y_{\zeta_c \wedge T} q'_{\zeta_c \wedge T}|)_{c>0}$ is uniformly integrable. Therefore, $\mathbb{E}[Y_{\zeta_c \wedge T} q'_{\zeta_c \wedge T}] \to \mathbb{E}[Y_T q'_T]$. Letting $c$ tend to $+\infty$ in (164), we get

$$ \mathbb{E} \left[ q'_T Y_T + \int_0^T q'_s \ell_s ds \right] = \mathbb{E} \left[ \int_0^T (q'_s f_s^* + q_s Y_s y_s^{\star, E} + q_s Z_s \cdot z_s^{\star, A, E}) ds \right]. $$

And then returning back to (163), we get

$$ \lim_{\epsilon \to 0} \frac{1}{\epsilon} (\mathcal{R}(q^{\epsilon}) - \mathcal{R}(q)) = \mathbb{E} \left[ \int_0^T (q'_s f_s^* + q_s (Y_s y_s^{\star, E} + Z_s \cdot z_s^{\star, A, E})) ds \right]. \quad (165) $$
Step 5: sub-derivative of the entropic cost. We now address the subgradient of $\epsilon \mapsto S(q^\epsilon)$. By definition, we have (using the notation introduced in the first step)

$$ 
\begin{aligned}
S(q^\epsilon) - S(q) &= \mathbb{E} \left[ \int_0^T (q_s^\epsilon f_s^{\star, \epsilon} - q_s f_s^\star) \mathrm{d}s \right] \\
&= \mathbb{E} \left[ \int_0^T q_s^\epsilon \delta f_s^{\star, \epsilon} \mathrm{d}s \right] + \mathbb{E} \left[ \int_0^T \delta q_s^\epsilon f_s^\star \mathrm{d}s \right],
\end{aligned} 
\quad (166) $$

where, by construction,

$$ 
\begin{aligned}
\delta f_s^{\star, \epsilon}(\omega) &= \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s) (f^\star(s, Y_s^\star + \epsilon y_s^{\star, E}, Z_s^\star + \epsilon z_s^{\star, A, E}) - f^\star(s, Y_s^\star, Z_s^\star)) \\
&= \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s) (f^\star(s, Y_s^\star + \epsilon y_s^\star, Z_s^\star + \epsilon z_s^\star) - f^\star(s, Y_s^\star, Z_s^\star)).
\end{aligned} 
$$

By (158),

$$ \frac{1}{\epsilon} \delta f_s^{\star, \epsilon}(\omega) \le y_s^{\star, E} Y_s^\star + z_s^{\star, A, E} \cdot Z_s^\star - \varrho \mathbb{1}_E(s, \omega) \mathbb{1}_{[0, \tau_A(\omega)]}(s), $$

for some the same real $\varrho \ge 0$ as in (150). Since the right-hand side is integrable under $\mathbb{Q}$, we deduce that (using the bound $q_s^\epsilon \le Cq_s$)

$$ 
\begin{aligned}
\limsup_{\epsilon \to 0} \frac{1}{\epsilon} \mathbb{E} \left[ \int_0^T q_s^\epsilon \delta f_s^{\star, \epsilon} \mathrm{d}s \right] \\
&\le \limsup_{\epsilon \to 0} \mathbb{E} \left[ \int_0^T q_s^\epsilon (y_s^{\star, E} Y_s^\star + z_s^{\star, A, E} \cdot Z_s^\star) \mathrm{d}s \right] - \varrho \mathbb{E} \left[ \int_0^T q_s^\epsilon \mathbb{1}_E(s, \cdot) \mathbb{1}_{[0, \tau_A]}(s) \mathrm{d}s \right]. \\
&\le \mathbb{E} \left[ \int_0^T q_s (y_s^{\star, E} Y_s^\star + z_s^{\star, A, E} \cdot Z_s^\star) \mathrm{d}s \right] - \varrho \mathbb{E} \left[ \int_0^T q_s \mathbb{1}_E(s, \cdot) \mathbb{1}_{[0, \tau_A]}(s) \mathrm{d}s \right].
\end{aligned} 
$$

As for the second term on the right-hand side of (166), we know from (160) that $|\delta q_s^\epsilon / \epsilon - q'_t| \le C\epsilon q_t$, which gives directly (using the bound $|f_s^\star| \le C + C f_s^\star$, see again (147))

$$ \lim_{\epsilon \to 0} \frac{1}{\epsilon} \mathbb{E} \left[ \int_0^T \delta q_s^\epsilon f_s^\star \mathrm{d}s \right] = \mathbb{E} \left[ \int_0^T q_s' f_s^\star \mathrm{d}s \right]. $$

Back to (166), we deduce that

$$ 
\begin{aligned}
\limsup_{\epsilon \to 0} \frac{1}{\epsilon} (S(q^\epsilon) - S(q)) \\
\le \mathbb{E} \left[ \int_0^T (q_s' f_s^\star + q_s (y_s^{\star, E} Y_s^\star + z_s^{\star, A, E} \cdot Z_s^\star) - \varrho q_s \mathbb{1}_E(s, \cdot) \mathbb{1}_{[0, \tau_A]}(s)) \mathrm{d}s \right].
\end{aligned} 
$$

Recalling from Lemma 18 that $q$ is strictly positive, we deduce from (150) and (152) that

$$ \mathbb{E} \left[ \int_0^T q_s \mathbb{1}_E(s, \cdot) \mathbb{1}_{[0, \tau_A]}(s) \mathrm{d}s \right] > 0, $$

and thus

$$ \limsup_{\epsilon \to 0} \frac{1}{\epsilon} (S(q^\epsilon) - S(q)) < \mathbb{E} \left[ \int_0^T (q_s' f_s^\star + q_s (y_s^{\star, E} Y_s^\star + z_s^{\star, A, E} \cdot Z_s^\star)) \mathrm{d}s \right]. \quad (167) $$
Step 6: conclusion. We now come back to the definition of $\mathcal{J}$, recalling that, for any $\varepsilon \in (0, 1]$,

$$ \frac{1}{\varepsilon}(\mathcal{J}(q^{\varepsilon}) - \mathcal{J}(q)) = \frac{1}{\varepsilon}(\mathcal{R}(q^{\varepsilon}) - \mathcal{R}(q)) - \frac{1}{\varepsilon}(\mathcal{S}(q^{\varepsilon}) - \mathcal{S}(q)). \quad (168) $$

Combining (165), (167) and (168), we obtain

$$ \liminf_{\varepsilon \to 0} \frac{1}{\varepsilon} (\mathcal{J}(q^{\varepsilon}) - \mathcal{J}(q)) > 0. $$

However, by optimality of $q$ and because $q^{\varepsilon}$ is admissible for $(\mathbb{P}_{N,c_1})$ for $\varepsilon$ small enough, we have

$$ \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} (\mathcal{J}(q^{\varepsilon}) - \mathcal{J}(q)) \le 0, $$

which gives a contradiction with the penultimate line. This contradicts the assumption made in (150), as a result of which we deduce that, almost surely, for almost every $t \in [0, T]$,

$$ f^*(t, Y_t^* + y_t^*, Z_t^* + z_t^*) - f^*(t, Y_t^*, Z_t^*) - (Y_t y_t^* + Z_t \cdot z_t^*) \ge 0. $$

By construction, the above holds true when the perturbation $(y^*, z^*)$ is bounded, but this assumption can be easily dropped by means of a truncation argument. In particular, we can choose $(y^*, z^*) = (\partial_y f(t, Y_t, Z_t) - Y_t^*, \partial_z f(t, Y_t, Z_t) - Z_t^*)_{t \in [0,T]}$. With this choice, we obtain (with full measure under $\text{Leb}_{[0,T]} \otimes \mathbb{P}$)

$$ \begin{aligned} & f^*(t, \partial_y f(t, Y_t, Z_t), \partial_z f(t, Y_t, Z_t)) - f^*(t, Y_t^*, Z_t^*) \\ & \ge Y_t \partial_y f(t, Y_t, Z_t) + Z_t \cdot \partial_z f(t, Y_t, Z_t) - Y_t Y_t^* - Z_t \cdot Z_t^*. \end{aligned} $$

Recalling that

$$ \begin{aligned} & f(t, Y_t, Z_t) + f^*(t, \partial_y f(t, Y_t, Z_t), \partial_z f(t, Y_t, Z_t)) \\ & = Y_t \partial_y f(t, Y_t, Z_t) + Z_t \cdot \partial_z f(t, Y_t, Z_t), \end{aligned} $$

we deduce (again, with full measure under $\text{Leb}_{[0,T]} \otimes \mathbb{P}$)

$$ Y_t Y_t^* + Z_t \cdot Z_t^* = f^*(t, Y_t^*, Z_t^*) + f(t, Y_t, Z_t), $$

which is known to imply $(Y_t^*, Z_t^*) = (\partial_y f(t, Y_t, Z_t), \partial_z f(t, Y_t, Z_t))$. $\square$

We complement the necessary condition established in Lemma 28 with a lower bound on the process $Y$. This bound plays a key role in the sufficient condition proved later in Lemma 30. In fact, this bound is similar to the one obtained in [50, Theorem 2.1]. Unfortunately, the bound established in [50] only holds for one specific solution of the quadratic equation $(\text{Opt}_N)$ (obtained by taking the limit on truncated equations); in the absence of uniqueness for the quadratic equation, it is not possible to apply [50] to our case.

**Lemma 29.** For a given $c_0 \in (0, c_1)$, let $q \in Q_{c_0}$ be a maximizer to the problem $(\mathbb{P}_{N,c_1})$ (with $\bar{\psi} \in A_{c_2}$ being fixed) and $(Y, Z)$ be the solution of (143). Then, $\mathbb{P}$-a.s., for any $t \in [0, T]$,

$$ Y_t \ge \tilde{Y}_t, $$
where $(\tilde{Y}, \tilde{Z})$ solves the BSDE

$$ egin{cases} -d\tilde{Y}_t = (\ell_t + f(t, 0, 0) + (\partial_y f(t, 0, 0)\tilde{Y}_t + \partial_z f(t, 0, 0) \cdot \tilde{Z}_t)) dt \\ - \tilde{Z}_t \cdot dW_t, & t \in [0, T], \\ \tilde{Y}_T = \delta_q \mathcal{G}(q_T). \end{cases} \tag{169} $$

In addition, for any other $\tilde{q} \in Q_{c_1}$, the random variables $(\tilde{q}_\tau Y_\tau^- := \tilde{q}_\tau \min(-Y_\tau, 0))_\tau$, with $\tau$ running over the set of $[0, T]$-valued $F$-stopping times, are uniformly integrable under $\mathbb{P}$.

**Proof. Step 1.** Recall that $q$ denotes a maximizer to the problem $(\mathbb{P}_{N,c_1})$. Then, for a given $t \in (0, T]$ and an arbitrary event $E \in \mathcal{F}_t$, we define $q^{t,E}$ by letting

$$ q_s^{t,E} := \begin{cases} q_s & \text{if } s \in [0, t], \\ q_s \mathbb{1}_{E^c} + q_t Q_{t,s} \mathbb{1}_E & \text{if } s \in (t, T], \end{cases} $$

where

$$ Q_{t,s} := \exp \left( \int_t^s \partial_y f(r, 0, 0) dr \right) \mathcal{E}_s \left( \int_t^r \partial_z f(r, 0, 0) \cdot dW_r \right), \quad s \in [t, T]. $$

Equivalently, this means that

$$ d q_s^{t,E} = q_s^{t,E} Y_s^{t,E} ds + q_s^{t,E} Z_s^{t,E} \cdot dW_s, \quad s \in [0, T], $$

where

$$ (Y_s^{t,E}, Z_s^{t,E}) := \begin{cases} (Y_s^*, Z_s^*) & \text{if } s \in [0, t], \\ (Y_s^*, Z_s^*) \mathbb{1}_{E^c} + (\partial_y f(s, 0, 0), \partial_z f(s, 0, 0)) \mathbb{1}_E & \text{if } s \in (t, T]. \end{cases} $$

Using the fact that $(\partial_y f(r, 0, 0))_{r \in [0, T]}$ and $(\partial_z f(r, 0, 0))_{r \in [0, T]}$ are bounded, we easily deduce that, for any $p \ge 1$, there exists a (deterministic) constant $C_p$, independent of $t$, such that, $\mathbb{P}$-a.s,

$$ \mathbb{E} \left[ \sup_{s \in [t, T]} Q_{t,s}^p \middle| \mathcal{F}_t \right] \le C_p, $$

from which we deduce that

$$ \sup_{s \in [t, T]} \mathbb{E} [h(q_s^{t,E})] < +\infty. \tag{170} $$

In fact, we claim that there exists $\delta_0 > 0$, possibly depending on $t$, such that, for $\mathbb{P}(E) \le \delta_0$, the process $q^{t,E}$ belongs to $Q_{c_1}$. Indeed, we have (using Fenchel-Legendre duality to get the last line)

$$ \begin{align*} S(q^{t,E}) &= \mathbb{E} \left[ \int_0^T q_s^{t,E} f^*(s, Y_s^{t,E}, Z_s^{t,E}) ds \right] \\ &= \mathbb{E} \left[ \int_0^T q_s f^*(s, Y_s^*, Z_s^*) ds \right] + \mathbb{E} \left[ \int_t^T q_s \mathbb{1}_{E^c} f^*(s, Y_s^*, Z_s^*) ds \right] \\ &\quad + \mathbb{E} \left[ \int_t^T q_t Q_{t,s} \mathbb{1}_E f^*(s, \partial_y f(s, 0, 0), \partial_z f(s, 0, 0)) ds \right] \\ &\le S(q) - \mathbb{E} \left[ \int_t^T q_t Q_{t,s} \mathbb{1}_E f_s^0 ds \right]. \end{align*} $$
And then, using again the fact that $(f_s^0 = f(s, 0, 0))_{s \in [0, T]}$ is bounded and recalling that $q \in Q_{c_1'}$ with $c_1' < c_1$, we deduce that there exists a constant $C > 0$, such that

$$ \mathcal{S}(q^{t,E}) \le \mathcal{S}(q) + C\mathbb{E}[q_t 1_E] \le c_1' + C\mathbb{E}[q_t 1_E]. $$

Then for $\mathbb{P}(E)$ small enough, $c_1' + C\mathbb{E}[q_t 1_E]$ is less than or equal to $c_1$, which implies that $q^{t,E}$ belongs to $Q_{c_1'}$.

**Step 2.** Throughout, we assume that $t$ is fixed (in $(0, T]$) and $\mathbb{P}(E) \le \delta_0$, with $\delta_0$ as in the first step. Since $q^{t,E} \in Q_{c_1'}$, and by optimality of $q$ on $Q_{c_0}$, we deduce that

$$ \mathcal{G}(q_T) + \mathbb{E}\left[\int_0^T q_s \ell_s ds\right] - \mathcal{S}(q) \ge \mathcal{G}(q_T^{t,E}) + \mathbb{E}\left[\int_0^T q_s^{t,E} \ell_s ds\right] - \mathcal{S}(q^{t,E}). \quad (171) $$

It is clear that

$$ \mathbb{E}\left[\int_0^T (q_s^{t,E} - q_s) \ell_s ds\right] = \mathbb{E}\left[1_E \int_t^T (q_t Q_{t,s} - q_s) \ell_s ds\right], \quad (172) $$

and

$$ \begin{aligned} \mathcal{S}(q^{t,E}) - \mathcal{S}(q) &= \mathbb{E}\left[1_E \int_t^T \left[q_s^{t,E} f^*(s, \partial_y f(s, 0, 0), \partial_z f(s, 0, 0)) - q_s f^*(s, Y_s^*, Z_s^*)\right] ds\right] \quad (173) \\ &= -\mathbb{E}\left[1_E \int_t^T \left[q_s^{t,E} f(s, 0, 0) + q_s f^*(s, Y_s^*, Z_s^*)\right] ds\right]. \end{aligned} $$

We now turn to the difference between the two boundary conditions in (171). From the regularity property A7 (for $\mathcal{G}$), we have

$$ \mathcal{G}(q_T^{t,E}) = \mathcal{G}(q_T) + \mathbb{E}\left[(q_T^{t,E} - q_T) \delta_q \mathcal{G}(q_T)\right] + o\left(\mathbb{E}\left[(1 + |X_T^\psi|^{2-r})|q_T^{t,E} - q_T|\right]\right), $$

where $o(r)/r \to 0$ as $r$ tends to 0 (the rate being independent of $E$). Recalling that $q^{t,E}$ and $q$ respectively belong to $Q_{c_1}$ and $Q_{c_0}$, we know from the growth Assumption A6 on $\mathcal{G}$ and from Lemma 40 that the first expectation on the right-hand side is well defined. Applying once again Lemma 40, we deduce that second expectation is also well defined. By definition of $q^{t,E}$, we have

$$ \begin{aligned} \mathcal{G}(q_T^{t,E}) = \mathcal{G}(q_T) &+ \mathbb{E}[1_E (q_t Q_{t,T} - q_T) \delta_q \mathcal{G}(q_T)] \\ &+ o\left(\mathbb{E}\left[(1 + |X_T^\psi|^{2-r})|q_T^{t,E} - q_T|\right]\right). \end{aligned} \quad (174) $$

Putting together (171), (172), (173) and (174), we obtain

$$ \begin{aligned} & \mathbb{E}[1_E (q_t Q_{t,T} - q_T) \delta_q \mathcal{G}(q_T)] + \mathbb{E}\left[1_E \int_t^T (q_t Q_{t,s} - q_s) \ell_s ds\right] \\ & + \mathbb{E}\left[1_E \int_t^T (q_t Q_{t,s} f(s, 0, 0) + q_s f^*(s, Y_s^*, Z_s^*)) ds\right] \quad (175) \\ & \le o\left(\mathbb{E}\left[(1 + |X_T^\psi|^{2-r})|q_T^{t,E} - q_T|\right]\right). \end{aligned} $$

Now, we recall from (144) that

$$ q_t Y_t = \mathbb{E}\left[q_T \delta_q \mathcal{G}(q_T) + \int_t^T q_s (\ell_s - f^*(s, Y_s^*, Z_s^*)) ds \middle| \mathcal{F}_t\right]. $$
And then, we can rewrite (175) as

$$
\begin{aligned}
& \mathbb{E}[1_E q_t (Q_{t,T} \delta_q \mathcal{G}(q_T) - Y_t)] + \mathbb{E} \left[ 1_E \int_t^T q_t Q_{t,s} \ell_s ds \right] \\
& + \mathbb{E} \left[ 1_E \int_t^T Q_{t,s} f(s, 0, 0) ds \right] \le o \left( \mathbb{E} \left[ (1 + |X_T^\psi|^{2-r}) |q_T^{t,E} - q_T| \right] \right).
\end{aligned}
$$

Also, it is easy to see from (169) that $\tilde{Y}_t$ can be represented as

$$
\tilde{Y}_t = \mathbb{E} \left[ Q_{t,T} \delta_q \mathcal{G}(q_T) + \int_t^T Q_{t,s} (\ell_s + f(s, 0, 0)) \, ds \middle| \mathcal{F}_t \right], \quad (176)
$$

which gives

$$
\mathbb{E} \left[ 1_E q_t (\tilde{Y}_t - Y_t) \right] \le o \left( \mathbb{E} \left[ (1 + |X_T^\psi|^{2-r}) |q_T^{t,E} - q_T| \right] \right).
$$

We notice that the expectation in the right-hand side can be rewritten as

$$
\mathbb{E} \left[ (1 + |X_T^\psi|^{2-r}) |q_T^{t,E} - q_T| \right] = \mathbb{E} \left[ 1_E (1 + |X_T^\psi|^{2-r}) |q_T - q_T Q_{t,T}| \right].
$$

Writing $o(r) = r\eta(r)$, with $\eta \ge 0$ and $\lim_{r\to 0} \eta(r) = 0$, and letting $R_{t,T} := (1 + |X_T^\psi|^{2-r})|q_T - q_T Q_{t,T}|$, we get

$$
\mathbb{E} \left[ 1_E \left\{ q_t (\tilde{Y}_t - Y_t) - R_{t,T} \eta \left( \mathbb{E} [1_E R_{t,T}] \right) \right\} \right] \le 0. \quad (177)
$$

The above is true for a given $t \in (0, T]$, for any event $E \in \mathcal{F}_t$ satisfying $\mathbb{P}(E) \le \delta_0$.
The function $\eta$ is independent of $E$. Moreover, we notice from (170) (with $E = \Omega$
therein) and Lemma 40 that $\mathbb{E}[R_{t,T}] < +\infty$.

Step 3. We now argue by contradiction to prove that $Y_t \ge \tilde{Y}_t$. Assume indeed that, for some $\varepsilon > 0$,

$$
\pi := \mathbb{P} \left( \left\{ q_t (\tilde{Y}_t - Y_t) \ge \varepsilon \right\} \right) > 0.
$$

Then, we can find $A > 0$ such that the event

$$
E_0 := \{ q_t(\tilde{Y}_t - Y_t) \ge \varepsilon \} \cap \{ R_{t,T} \le A \}
$$

satisfies $\mathbb{P}(E_0) \ge \pi/2$. By a standard uniform integrability argument (using the fact
that $\mathbb{E}[R_{t,T}] < +\infty$), notice also that there exists $\delta > 0$ such that

$$
\mathbb{P}(E) \le \delta \Rightarrow \eta(\mathbb{E}[1_E R_{t,T}]) \le \frac{\varepsilon}{2A}.
$$

Decompose now $E_0$ as

$$
E_0 = \cup_{k \in \mathbb{N}} (E_0 \cap \{W_t \in I_k\}),
$$

where $(I_k)_{k \in \mathbb{N}}$ is partition of $\mathbb{R}^d$ into Borel subsets such that $\mathbb{P}(\{W_t \in I_k\}) < \delta \wedge \delta_0$
(with $\delta_0$ as in the first step), for each $k \in \mathbb{N}$.

Applying (177) with $E = E_{0,k} := E_0 \cap \{W_t \in I_k\}$ for a given $k \in \mathbb{N}$, we obtain

$$
\begin{aligned}
0 &\ge \mathbb{E} \left[ 1_{E_{0,k}} \left( q_t (\tilde{Y}_t - Y_t) - R_{t,T} \eta \left( \mathbb{E} [1_{E_{0,k}} | q_T - q_T Q_{t,T} |] \right) \right) \right] \\
&\ge \mathbb{E} \left[ 1_{E_{0,k}} \left( \varepsilon - A \frac{\varepsilon}{2A} \right) \right] = \frac{\varepsilon}{2} \mathbb{P}(E_{0,k}).
\end{aligned}
$$
This proves that $\mathbb{P}(E_{0,k}) = 0$, for each $k \in \mathbb{N}$, and then $\mathbb{P}(E_0) = 0$, which contradicts the fact that $\pi > 0$. We deduce that

$$ \forall \varepsilon > 0, \quad \mathbb{P}\left(\left\{q_t(\tilde{Y}_t - Y_t) \geq \varepsilon\right\}\right) = 0, $$

i.e.,

$$ \mathbb{P}\left(\left\{q_t(\tilde{Y}_t - Y_t) \leq 0\right\}\right) = 1. $$

Since $\mathbb{P}(\{q_t > 0\}) = 1$, we deduce that $\mathbb{P}(\{Y_t \geq \tilde{Y}_t\})$. This holds true for any $t \in (0, T]$. By continuity of the two processes $Y$ and $\tilde{Y}$, we deduce that $\mathbb{P}$-a.s., for any $t \in [0, T]$, $Y_t \geq \tilde{Y}_t$, which proves the first claim in the statement.

Step 4. It remains to establish that, for any other $\tilde{q} \in Q_{c_1}$ (with decomposition $(\tilde{Y}^*, \tilde{Z}^*)$ in (4)), the family $(\tilde{q}_\tau Y_\tau^-)_\tau$, with $\tau$ running over the set of $[0, T]$-valued $\mathbb{F}$-stopping, is uniformly integrable. From the comparison principle established in Step 3, we first notice that

$$ 0 \leq Y_t^- \leq \tilde{Y}_t^-, $$

for any $t \in [0, T]$. Then to show the desired property, we provide a lower bound for the process $\tilde{Y}$ appearing on the right-hand side (recall (169) for its definition, and (176) for its representation). We notice from the strong convexity of $\ell$, see A4, that $\ell+f(0,0)$ is lower bounded. There exists a constant $c_0 \in \mathbb{R}$ such that $\tilde{Y} \geq \tilde{Y}^0$, where the process $\tilde{Y}^0$ is defined as

$$ \tilde{Y}_t^0 = \frac{1}{q_t^0} \mathbb{E} [q_T^0 \delta_q \mathcal{G}(q_T) | \mathcal{F}_t] + c_0(T - t), $$

and where $q^0$ is the solution to (84). Then, the following inequality holds

$$ 0 \leq Y_t^- \leq \tilde{Y}_t^- \leq (\tilde{Y}_t^0)^-. $$

Now, let $E$ be an element of $\mathcal{F}_T$ and $\tau$ a stopping time with values in $[0, T]$. For $\tilde{q} \in Q_{c_1}$ as above, we have

$$ 0 \leq \mathbb{E}[1_E \tilde{q}_\tau Y_\tau^-] \leq \mathbb{E}[1_E \tilde{q}_\tau (\tilde{Y}_\tau^0)^-], \quad (178) $$

and we are left to establish that the right-hand side is finite for any $E \in \mathcal{F}_T$, and can be made small with $\mathbb{P}(E)$, uniformly with respect to $\tau$. We have

$$ \tilde{q}_\tau \tilde{Y}_\tau^0 = \frac{\tilde{q}_\tau}{q_\tau^0} \mathbb{E} [q_T^0 \delta_q \mathcal{G}(q_T) | \mathcal{F}_\tau] + c_0 \tilde{q}_\tau (T - \tau). $$

We notice that the process $(\tilde{q}_t^{0,\tau})_{t \in [0,T]}$ defined by $\tilde{q}_t^{0,\tau} := \tilde{q}_t$, for $t \in [0, \tau]$, and $\tilde{q}_t^{0,\tau} := \tilde{q}_\tau q_t^0 / q_\tau^0$, for $t \in [\tau, T]$, lies in $Q$. It is indeed the solution to

$$ d\tilde{q}_t^{0,\tau} = \tilde{q}_t^{0,\tau} \tilde{Y}_t^{*,0,\tau} dt + \tilde{q}_t^{0,\tau} \tilde{Z}_t^{*,0,\tau} \cdot dW_t, \quad t \in [0, T]; \quad \tilde{q}_0^{0,\tau} = 1, $$

where

$$ (\tilde{Y}_t^{*,0,\tau}, \tilde{Z}_t^{*,0,\tau}) := \begin{cases} (\tilde{Y}_t^*, \tilde{Z}_t^*) & \text{if } t \leq \tau, \\ (\partial_y f(t, 0, 0), \partial_z f(t, 0, 0)) & \text{if } t \in (\tau, T]. \end{cases} $$
Following the first step, we deduce that

$$ \sup_{\tau} \mathcal{S}(\tilde{q}^{0,\tau}) < +\infty. \qquad (179) $$

Moreover,

$$ \mathbb{E} \left[ \mathbb{1}_{E} \tilde{q}_{\tau} \left( \tilde{Y}_{\tau}^{0} \right)^{-} \right] \leq \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} (\delta_{q} \mathcal{G}(q_{T}))^{-} \right] + |c_{0}| T \mathbb{E}[\mathbb{1}_{E} \tilde{q}_{\tau}], \quad (180) $$

As for the second term on the right-hand side, we know from Lemma 39 that $\mathbb{E}[\tilde{q}_T^*] < +\infty$. Therefore, the second term on the right-hand side is finite and can be made small with $\mathbb{P}(E)$, uniformly with respect to $\tau$. We now address the first term on the right-hand side in (180). By the lower estimate A6 on $\delta_q \mathcal{G}$, we have

$$ \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} (\delta_{q} \mathcal{G}(q_{T}))^{-} \right] \leq L \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} (1 + |X_{T}| + \mathbb{E}[q_{T} | X_{T}|^{2-r}]) \right]. $$

By Lemma 40, we know that $\mathbb{E}[q_T | X_T|^{2-r}] < +\infty$. We deduce that there exists $C > 0$ such that

$$ \mathbb{E} \left[ \mathbb{1}_{E} \tilde{q}_{T}^{0, \tau} (\delta_{q} \mathcal{G}(q_{T}))^{-} \right] \leq C \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} (1 + |X_{T}|) \right]. \quad (181) $$

Following (180), we already know that $\mathbb{E}[\mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau}] = \mathbb{E}[\mathbb{1}_{E} \tilde{q}_{\tau}]$ tends to 0 as $\mathbb{P}(E)$ tends to 0. The remainder of the proof is devoted to establishing the same result for $\mathbb{E}[\mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} | X_T|]$. To this end, we distinguish between the two cases $r = 0$ and $r = 1$.

If $r = 0$, then by Cauchy-Schwarz inequality and Lemma 40, we have

$$ \begin{aligned} \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} | X_T \right] &\leq \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} \mathbb{P}(E | \mathcal{F}_{\tau}) \right]^{1/2} \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} | X_T |^2 \right]^{1/2} \\ &\leq C \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} \mathbb{P}(E | \mathcal{F}_{\tau}) \right]^{1/2} \left( 1 + S(\tilde{q}^{0, \tau}) + S^{*}(\bar{\psi}) \right)^{1/2} \\ &\leq C \mathbb{E} [\tilde{q}_{\tau} \mathbb{1}_{E}]^{1/2} \left( 1 + S(\tilde{q}^{0, \tau}) + S^{*}(\bar{\psi}) \right)^{1/2}, \end{aligned} \quad (182) $$

where we used the inequality $\mathbb{E}[\tilde{q}_T^{0,\tau} | \mathcal{F}_\tau] \leq C \tilde{q}_\tau$ to get the last line. The first term on the last line can be handled as the last term on (180). Combining (178), (179), (180) and (182), we easily complete the proof.

If $r = 1$, we return back to (181). In comparison with (182), the only difficult comes from the stochastic integral in the definition of $X$, as its integrand grows up linearly in $\psi$, see A2. By Girsanov theorem, we write

$$ \begin{aligned} \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} \left| \int_{0}^{T} \sigma(t, \psi_t) dW_t \right| \right] &\leq \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} \left| \int_{0}^{T} \sigma(t, \psi_t) d\tilde{W}_{t}^{0, \tau} \right| \right] \\ &\quad + C \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) q_{\tau} \left| \int_{0}^{\tau} \sigma(t, \psi_t) Z_{t}^{*} dt \right| \right] \quad (183) \\ &\quad + \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} \left| \int_{\tau}^{T} \sigma(t, \psi_t) \partial_z f(t, 0, 0) dt \right| \right], \end{aligned} $$

where $\tilde{W}^{0,\tau}$ is a Brownian motion under $\mathcal{E}_T(\int_0^t \tilde{Z}_s^{*0,\tau} ds)$, and where $C$ is a constant independent of $\tau$ (which arises because $\mathbb{E}[\tilde{q}_T^{0,\tau}]$ may not be equal to 1). We first provide an upper bound for the first term on the right-hand side. By Young's inequality,
observe that, for any $\varepsilon \in (0, 1]$,

$$ \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} \left| \int_{0}^{T} \sigma(t, \psi_{t}) \mathrm{d}\tilde{W}_{t}^{0, \tau} \right| \right] \leq \frac{1}{\varepsilon} \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} \right] \\ + \varepsilon \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} \left| \int_{0}^{T} \sigma(t, \psi_{t}) \mathrm{d}\tilde{W}_{t}^{0, \tau} \right|^{2} \right]. $$

Here,

$$ \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) \tilde{q}_{T}^{0, \tau} \right] \leq \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) q_{\tau} \right] = \mathbb{E} \left[ \mathbb{1}_{E} q_{\tau} \right] \leq \mathbb{E} \left[ \mathbb{1}_{E} q_{T}^{*} \right], $$

and

$$ \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} \left| \int_{0}^{T} \sigma(t, \psi_{t}) \mathrm{d}\tilde{W}_{t}^{0, \tau} \right|^{2} \right] \leq C \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} \int_{0}^{T} |\sigma(t, \psi_{t})|^{2} \mathrm{d}t \right] \\ \leq C \mathbb{E} \left[ \tilde{q}_{T}^{0, \tau} \int_{0}^{T} \left(1 + |\psi_{t}|^{2}\right) \mathrm{d}t \right]. $$

By (14) and (179), the above right-hand side is finite, uniformly with respect to $\tau$. By combining the last three displays, we easily deduce that the first-term on the right-hand side of (183) tends to 0 as $\mathbb{P}(E)$ tends to 0, uniformly with respect to $\tau$. Using similar arguments together with the fact that $(\partial_z f(t, 0, 0))_{t \in [0, T]}$ is bounded, we can reach the same conclusion for the third term on the right-hand side of (183). It remains to handle the second term on the right-hand side of (183). To do so, it suffices to notice that

$$ \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) q_{\tau} \left| \int_{0}^{T} \sigma(t, \psi_{t}) Z_{t}^{*} \mathrm{d}t \right| \right] \leq C \mathbb{E} \left[ \mathbb{P}(E | \mathcal{F}_{\tau}) q_{T} \left| \int_{0}^{T} \sigma(t, \psi_{t}) Z_{t}^{*} \mathrm{d}t \right| \right] \\ \leq C \mathbb{E} \left[ \sup_{t \in [0, T]} \mathbb{P}(E | \mathcal{F}_{t}) q_{T} \int_{0}^{T} (1 + |\psi_{t}|) |Z_{t}^{*}| \mathrm{d}t \right]. \quad (184) $$

By Doob's inequality, $\sup_{t \in [0, T]} \mathbb{P}(E | \mathcal{F}_t)$ tends to 0 in probability as $\mathbb{P}(E)$ tends to 0. Moreover,

$$ \mathbb{E} \left[ q_T \int_0^T (1 + |\psi_t|) |Z_t^*| dt \right] \\ \leq C \mathbb{E} \left[ q_T \int_0^T (1 + |Z_t^*|^2) dt \right] + C \mathbb{E} \left[ q_T \int_0^T (1 + |\psi_t|^2) dt \right]. $$

Thanks to (14), the second term on the right-hand side is finite. By (102), the first term is also finite. This proves that the left-hand side on (184) tends to 0 as $\mathbb{P}(E)$ tends to 0, uniformly in $\tau$. $\square$

### 5.2.3 Sufficient condition

We now turn to the proof of the sufficient condition, i.e. the second assertion in the statement of Theorem 25. We recall that $\bar{\psi} \in \mathcal{A}_{c_2}$ is given. Also, we use the same abbreviated notations as in (142).
**Lemma 30.** Assume that there exists a triple $(q, Y, Z) \in \mathcal{Q}$ satisfying the first order condition (Opt_N). Then, $q$ is the unique maximizer of the mapping $q' \in \mathcal{Q} \mapsto \mathcal{J}(q', \bar{\psi})$.

*Proof.* Throughout the proof, we omit the dependence on $\bar{\psi}$ in the various notations. For instance, we just write $\mathcal{J}(q')$ for $\mathcal{J}(q', \bar{\psi})$.

Moreover, in addition to $(q, Y^*, Z^*)$, we let $(\tilde{q}, \tilde{Y}^*, \tilde{Z}^*)$ be another arbitrary tuple of state and control in $\mathcal{Q}$, and then denote $\delta q := \tilde{q} - q$. By definition of $\mathcal{J}$ and concavity of $\mathcal{G}$ with respect to its second variable $q$ (assumption (A8)), we have

$$ \begin{aligned} \mathcal{J}(\tilde{q}) - \mathcal{J}(q) &= \mathcal{R}(\tilde{q}) - \mathcal{R}(q) - (\mathcal{S}(\tilde{q}) - \mathcal{S}(q)) \\ &\leq \mathbb{E} \left[ \delta q_T \delta q_{\mathcal{G}}(q_T) + \int_0^T \delta q_t \ell_t dt \right] - (\mathcal{S}(\tilde{q}) - \mathcal{S}(q)). \end{aligned} \quad (185) $$

Notice that the right-hand side is finite, which can be shown in the same way as in (146), using the duality inequality (14), Lemma 40 and the bound $\mathcal{S}(\tilde{q}) + \mathcal{S}(q) + \mathcal{S}^*(\bar{\psi}) < +\infty$.

**Step 1: localization.** In this first step, we proceed as in the analysis of the right-hand side on (163) and expand $(\delta q_t Y_t)_{t \in [0, T]}$ by means of Itô's formula. Recalling (143), we have

$$ \begin{aligned} \mathrm{d}[\delta q_t Y_t] ={}& \delta q_t (-Y_t^* Y_t - Z_t^* \cdot Z_t + f^*(t, Y_t^*, Z_t^*) - \ell_t) \mathrm{d}t \\ &+ Y_t (\tilde{q}_t \tilde{Y}_t^* - q_t Y_t^*) \mathrm{d}t + Z_t \cdot (\tilde{q}_t \tilde{Z}_t^* - q_t Z_t^*) \mathrm{d}t \\ &+ \delta q_t Z_t \cdot \mathrm{d}W_t + Y_t (\tilde{q}_t \tilde{Z}_t^* - q_t Z_t^*) \cdot \mathrm{d}W_t, \quad t \in [0, T]. \end{aligned} \quad (186) $$

For a given $A > 0$, we then introduce the following stopping time

$$ \tau_A := \inf \left\{ t \in [0, T], |Y_t| + q_t + 1/q_t + \tilde{q}_t + \int_0^t |Z_s|^2 ds \geq A \right\}, $$

with the standard convention that $\tau_A = +\infty$ if the set on the right-hand side is empty. With this definition in hand, we notice that, for $t \in [0, \tau_A]$

$$ \begin{aligned} |\delta q_t| &\leq A \leq A^2 q_t, \\ \tilde{q}_t &\leq A \leq A^2 q_t. \end{aligned} $$

This implies in particular that, for a constant $C_A$ depending on $A$,

$$ \begin{aligned} & \mathbb{E} \left[ \int_0^{\tau_A} |\delta q_t (-Y_t^* Y_t - Z_t^* \cdot Z_t + f^*(t, Y_t^*, Z_t^*) - \ell_t)| \mathrm{d}t \right] \\ & \leq C_A \mathbb{E} \left[ \int_0^{\tau_A} q_t (|Y_t^* Y_t| + |Z_t^* \cdot Z_t| + |f^*(t, Y_t^*, Z_t^*)| + |\ell_t|) \mathrm{d}t \right] \\ & \leq C_A \left( 1 + \mathbb{E} \left[ \int_0^{\tau_A} q_t (|Y_t^*| + |Z_t^*|^2 + |f^*(t, Y_t^*, Z_t^*)| + |\ell_t|) \mathrm{d}t \right] \right) \\ & \leq C_A \left( 1 + \mathbb{E} \left[ \int_0^T q_t (|Y_t^*| + |f^*(t, Y_t^*, Z_t^*)| + |\ell_t|) \mathrm{d}t \right] \right) < +\infty, \end{aligned} $$

with the third line following from the condition $|Y_t| + \int_0^t |Z_s|^2 ds \leq A$, and the last line following from (16), and from the fact that $\mathcal{S}(q)$ and $\mathcal{S}^*(\bar{\psi})$ are finite.
Similarly,

$$
\begin{align*}
\mathbb{E} \left[ \int_0^{\tau_A} |Y_t (\tilde{q}_t \tilde{Y}_t^* - q_t Y_t^*)| dt \right] &\le C_A \mathbb{E} \left[ \int_0^T q_t (|\tilde{Y}_t^*| + |Y_t^*|) dt \right] < +\infty, \\
\mathbb{E} \left[ \int_0^{\tau_A} |Z_t \cdot (\tilde{q}_t \tilde{Z}_t^* - q_t Z_t^*)| dt \right] &\le C_A \left( 1 + \mathbb{E} \left[ \int_0^T q_t (|\tilde{Z}_t^*|^2 + |Z_t^*|^2) dt \right] \right) < +\infty.
\end{align*}
$$

Back to (186), this proves that the terms on the first and second lines of the right-
hand side, when integrated between 0 and $\tau_A$, have a finite expectation.

It remains to check in a similar way that the stochastic integrals on the third line
of (186) have a zero expectation, when they are integrated between 0 and $\tau_A$. To do
so, we notice that

$$
\begin{align*}
\mathbb{E} \left[ \int_0^{\tau_A} |\delta q_t|^2 |Z_t|^2 dt \right] &\le C_A \mathbb{E} \left[ \int_0^{\tau_A} q_t |Z_t|^2 dt \right] < +\infty, \\
\mathbb{E} \left[ \int_0^{\tau_A} |Y_t|^2 |\tilde{q}_t \tilde{Z}_t^* - q_t Z_t^*|^2 dt \right] &\le C_A \mathbb{E} \left[ \int_0^T (\tilde{q}_t |\tilde{Z}_t^*|^2 + q_t |Z_t^*|^2) dt \right] < +\infty.
\end{align*}
$$

Therefore, we obtain

$$
\begin{align*}
& \mathbb{E} \left[ \delta q_{T \wedge \tau_A} Y_{T \wedge \tau_A} + \int_0^{T \wedge \tau_A} \delta q_t \ell_t dt \right] \\
&= \mathbb{E} \left[ \int_0^{T \wedge \tau_A} \delta q_t (-Y_t^* Y_t - Z_t^* \cdot Z_t + f^*(t, Y_t^*, Z_t^*)) dt \right] \\
&\quad + \mathbb{E} \left[ \int_0^{T \wedge \tau_A} \left( Y_t (\tilde{q}_t \tilde{Y}_t^* - q_t Y_t^*) + Z_t \cdot (\tilde{q}_t \tilde{Z}_t^* - q_t Z_t^*) \right) dt \right].
\end{align*}
$$

Introduce now the notations $(\tilde{f}_t^* = \tilde{f}^*(t, \tilde{Y}_t^*, \tilde{Z}_t^*))_{t \in [0,T]}$ and $(f_t^* = f^*(t, Y_t^*, Z_t^*))_{t \in [0,T]}$
and deduce that

$$
\begin{align*}
& \mathbb{E} \left[ \delta q_{T \wedge \tau_A} Y_{T \wedge \tau_A} + \int_0^{T \wedge \tau_A} \delta q_t \ell_t dt - \int_0^{T \wedge \tau_A} (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \right] \\
&= -\mathbb{E} \left[ \int_0^{T \wedge \tau_A} \left( \tilde{q}_t (\tilde{f}_t^* - f_t^* - (\tilde{Y}_t^* - Y_t^*) Y_t - (\tilde{Z}_t^* - Z_t^*) \cdot Z_t) \right) dt \right].
\end{align*}
$$

Then, by the first order condition (Opt_N) and the (strict) joint convexity of f*, we deduce that the right-hand side is non-positive, i.e., for any A > 0,

$$
\mathbb{E} \left[ \delta q_{T \wedge \tau_A} Y_{T \wedge \tau_A} + \int_0^{T \wedge \tau_A} \delta q_t \ell_t dt - \int_0^{T \wedge \tau_A} (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \right] \le 0. \quad (187)
$$

The key step in the rest of the proof is to let $A$ tend to $+\infty$ on the left-hand side of (187). This requires some extra care due to the rather weak integrability properties of $\tilde{q}$ and $Y$. In order to proceed, we write the integrand in the form

$$
\begin{align}
& \delta q_{T \wedge \tau_A} Y_{T \wedge \tau_A} + \int_0^{T \wedge \tau_A} \delta q_t \ell_t dt - \int_0^{T \wedge \tau_A} (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \nonumber \\
&= \tilde{q}_{T \wedge \tau_A} Y_{T \wedge \tau_A} - q_{T \wedge \tau_A} Y_{T \wedge \tau_A} + \int_0^{T \wedge \tau_A} \delta q_t \ell_t dt - \int_0^{T \wedge \tau_A} (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \tag{188} \\
&=: T^1(A) + T^2(A) + T^3(A) + T^4(A). \nonumber
\end{align}
$$
Step 2: limit $A \to +\infty$ in $T^2(A)$, $T^3(A)$ and $T^4(A)$. We claim that each of the three families of random variables $(T^2(A))_{A>0}$, $(T^3(A))_{A>0}$ and $(T^4(A))_{A>0}$ is uniformly integrable. For $(T^2(A))_{A>0}$, we observe that there exists a constant $C \ge 0$ (whose value may change from line to line) such that, for any event $E \in \mathcal{F}_T$ and any constant $A' > 0$,

$$ \begin{aligned} \mathbb{E} [ |T^2(A)| 1_E ] &\le A' \mathbb{E} [ q_{T \wedge \tau_A} 1_E ] + \mathbb{E} [ q_{T \wedge \tau_A} 1_{ \{ |Y_{T \wedge \tau_A}| \ge A' \} } |Y_{T \wedge \tau_A}| ] \\ &\le A' \mathbb{E} [ q_{T \wedge \tau_A} 1_E ] + C \mathbb{E}^{\mathbb{Q}} [ 1_{ \{ |Y_{T \wedge \tau_A}| \ge A' \} } |Y_{T \wedge \tau_A}| ], \end{aligned} $$

where $\mathbb{Q} := \mathcal{E}_T(\int_0^T Z_t^* \cdot dW_t)\mathbb{P}$. Since $Y \in D(\mathbb{F}, \mathbb{Q})$, the family $(Y_{T \wedge \tau_A})_{A>0}$ is uniformly integrable under $\mathbb{Q}$, which shows that the last term tends to 0 as $A'$ tends to $+\infty$, uniformly in $A > 0$. Therefore, to establish the uniformly integrability of the family $(T^2(A))_{A>0}$, it suffices to recall that $\mathbb{E}[q_T^*] < +\infty$ (see Lemma 39), which implies in particular that

$$ \lim_{\mathbb{P}(E) \to 0} \sup_{A>0} \mathbb{E}[q_{T \wedge \tau_A} 1_E] = 0. $$

We now establish the uniform integrability of $(T^3(A))_{A>0}$. By a standard domination argument, it suffices to notice that

$$ \mathbb{E} \left[ \int_{0}^{T} |\delta q_t \ell_t| dt \right] \le C \left( 1 + \mathbb{E} \left[ \int_{0}^{T} (q_t + \tilde{q}_t) |\bar{\psi}_t|^2 dt \right] \right) < +\infty, $$

with the last inequality following from the fact that $q$ and $\tilde{q}$ belong to $\mathbb{Q}$ and $\bar{\psi}$ to $\mathcal{A}_{c2}$ (together with the duality inequality (14)).

We handle $(T^4(A))_{A>0}$ in the same way. Indeed, by the same argument as in (147), we have

$$ \mathbb{E} \left[ \int_0^T \left( \tilde{q}_t |\tilde{f}_t^*| + q_t |f_t^*| \right) dt \right] < +\infty. $$

Combining the uniform integrability properties of $(T^2(A))_{A>0}$, $(T^3(A))_{A>0}$ and $(T^4(A))_{A>0}$ together with the time continuity of the processes appearing in (188), we deduce that

$$ \begin{aligned} & \lim_{A \to +\infty} \mathbb{E} [T^2(A) + T^3(A) + T^4(A)] \\ & = \mathbb{E} \left[ -q_T Y_T + \int_0^T \delta q_t \ell_t dt - \int_0^T (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \right]. \end{aligned} \quad (189) $$

Step 3: limit $A \to +\infty$ in $T^1(A)$. We now explain how to handle $(T^1(A))_{A>0}$ in (188). We first decompose $T^1(A)$ into non positive and non negative parts

$$ T^1(A) = \tilde{q}_{T \wedge \tau_A} (Y_{T \wedge \tau_A} + Y_{T \wedge \tau_A}^{-}) - \tilde{q}_{T \wedge \tau_A} Y_{T \wedge \tau_A}^{-}. \quad (190) $$

By Lemma 29, the family $(\tilde{q}_{T \wedge \tau_A} Y_{T \wedge \tau_A}^{-})_{A>0}$ is uniformly integrable. In particular,

$$ \lim_{A \to +\infty} \mathbb{E} [ \tilde{q}_{T \wedge \tau_A} Y_{T \wedge \tau_A}^{-} ] = \mathbb{E} [ \tilde{q}_T Y_T^{-} ]. \quad (191) $$

We also notice that, $\tilde{q}_{T \wedge \tau_A} (Y_{T \wedge \tau_A} + Y_{T \wedge \tau_A}^{-})$ takes values in $[0, +\infty)$. Therefore, Fatou's lemma gives

$$ \liminf_{A \to +\infty} \mathbb{E} [ \tilde{q}_{T \wedge \tau_A} (Y_{T \wedge \tau_A} + Y_{T \wedge \tau_A}^{-}) ] \ge \mathbb{E} [ \tilde{q}_T (Y_T + Y_T^{-}) ]. $$
Inserting the latter into (190), we obtain

$$ \liminf_{A \to +\infty} \mathbb{E}[T^1(A)] = \liminf_{A \to +\infty} \mathbb{E}[\tilde{q}_{T \wedge \tau_A} Y_{T \wedge \tau_A}] \geq \mathbb{E}[\tilde{q}_T (Y_T + Y_T^{-} - Y_T^{-})] = \mathbb{E}[\tilde{q}_T Y_T]. $$

And then, thanks to (189), this gives

$$ \liminf_{A \to +\infty} \mathbb{E} [T^1(A) + T^2(A) + T^3(A) + T^4(A)] \\ \geq \mathbb{E} \left[ \tilde{q}_T Y_T - q_T Y_T + \int_0^T (\tilde{q}_t - q_t) \ell_t dt - \int_0^T (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \right]. $$

By (187), the left-hand side is less than 0, from which we deduce that

$$ 0 \geq \mathbb{E} \left[ \tilde{q}_T Y_T - q_T Y_T + \int_0^T (\tilde{q}_t - q_t) \ell_t dt - \int_0^T (\tilde{q}_t \tilde{f}_t^* - q_t f_t^*) dt \right]. $$

By (185), the right-hand side is greater than $\mathcal{J}(\tilde{q}) - \mathcal{J}(q)$. This shows $\mathcal{J}(\tilde{q}) - \mathcal{J}(q) \leq 0$, which proves the optimality of $q$.

Uniqueness of the minimizer follows from the strict convexity of $\mathcal{J}$ in the variable $q$, see Proposition 20. □

## 5.3 Central planner's control problem

In this section, we study the problem of the central planner,

$$ \inf_{\psi \in \mathcal{A}_{c_2}} \mathcal{J}(\bar{q}, \psi), \qquad (\mathrm{P}_C) $$

under the assumption that there exists $\bar{\psi}$ such that the pair $(\bar{q}, \bar{\psi}) \in \mathcal{Q}_{c_1} \times \mathcal{A}_{c_2}$ is a saddle point of the problem (P'). We recall (27) and (28) for the definitions of the two sets $\mathcal{Q}_{c_1}$ and $\mathcal{A}_{c_2}$. We further recall (25) for the definition of the set $\mathcal{A}$. The purpose of this section is to establish the following characterization of the problem (P_C).

**Theorem 31.** There exists a constant $c'_2 > 0$, only depending on the data and $c_1$, such that, if $c_2 > c'_2$, the minimizer $\psi$ of (P_C) (over $\mathcal{A}_{c_2}$) belongs in fact to $\mathcal{A}_{c'_2}$. Conversely, if $c_2 > c'_2$, any $\psi \in \mathcal{A}_{c'_2}$ is a minimizer to the problem (P_C) over $\mathcal{A}_{c_2}$ if and only if there exists a tuple $(\psi, p, k, X) \in \mathcal{A}$ solving the FBSDE (Opt_C) (with $q$ replaced by $\bar{q}$).

*Sketch of Proof.* In Lemma 32 below, we establish the existence of a constant $c'_2 > 0$, only depending on the data and $c_1$, independent of $c_2$, such that any minimizer $\psi$ to the problem (P_C) over $\mathcal{A}_{c_2}$ is in fact in $\mathcal{A}_{c'_2}$. By assuming (without any loss of generality) that the constant $c_2$ is strictly larger than $c'_2$, we ensure that the minimizer $\psi$ is an interior solution in the sense that

$$ S^*(\psi) < c_2. $$

The conclusion of the statement follows from the necessary condition proved in Lemma 37 and the sufficient condition established in Lemma 38. □
### 5.3.1 A priori estimate

This subsection is devoted to proving the following a priori estimate for the component $\bar{\psi}$ of the saddle point $(\bar{q}, \bar{\psi})$.

**Lemma 32.** There exists a positive constant $c'_2 > 0$, only depending on $c_1$ and on the data (and in particular independent of the parameter $c_2$), such that the component $\bar{\psi}$ of any saddle point $(\bar{q}, \bar{\psi}) \in Q_{c_1} \times A_{c_2}$ to (P') (with $c_2 > c'_2$) belongs in fact to $A_{c'_2}$.

*Proof.* Let $q \in Q$. Using the convexity assumption on $\mathcal{G}$, we have

$$ \begin{aligned} \mathcal{G}(q_T, X_T^{\bar{\psi}}) &\ge \mathcal{G}(q_T, 0) + \mathbb{E}[\delta_X \mathcal{G}(q_T, 0) \cdot X_T^{\bar{\psi}}] \\ &\ge -L - L\mathbb{E}[q_T |X_T^{\bar{\psi}}|], \end{aligned} $$

where the second inequality follows from the growth Assumption A6. Therefore,

$$ \mathcal{J}(q, \bar{\psi}) + \mathcal{S}(q) \ge -L - L\mathbb{E}[q_T |X_T^{\bar{\psi}}|] + \mathbb{E}\left[\int_0^T q_t \ell(t, \bar{\psi}_t) dt\right]. $$

Since $\ell$ grows up at least quadratically fast by Assumption A4, we can find a constant $C > 0$ such that

$$ \mathbb{E}\left[\int_0^T q_t \ell(t, \bar{\psi}_t) dt\right] \ge -C + \frac{1}{2L} \mathbb{E}\left[\int_0^T q_t |\bar{\psi}_t|^2 dt\right]. $$

Moreover, we know from Lemma 41 that, for any $\varepsilon \in (0, 1)$, there exists $C_\varepsilon > 0$ such that

$$ \mathbb{E}[q_T |X_T^{\bar{\psi}}|] \le C_\varepsilon + c_\varepsilon \mathcal{S}(q) + \varepsilon \mathbb{E}[\int_0^T q_t |\bar{\psi}_t|^2 dt], $$

with $c_\varepsilon = 2\beta e^{\alpha T} \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} (\|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} + \frac{3}{\varepsilon} e^{\alpha T} \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d \times n})})$.

Therefore, choosing $\varepsilon = 1/[4 \max(1, L)]$ and combining the last three displays, we get

$$ \mathcal{J}(q, \bar{\psi}) \ge -C - c_\varepsilon \mathcal{S}(q) + \frac{1}{4 \max(1, L)} \mathbb{E} \left[ \int_0^T q_t |\bar{\psi}_t|^2 dt \right]. $$

And, then for the same constants $\varepsilon$, $c_\varepsilon$ and $C$ as above,

$$ \sup_{q \in \mathcal{Q}} \left\{ \mathbb{E} \left[ \int_0^T q_t |\bar{\psi}_t|^2 dt \right] - \gamma \mathcal{S}(q) \right\} \le 4 \max(1, L) \left[ C + \sup_{q \in \mathcal{Q}} \mathcal{J}(q, \bar{\psi}) \right]. \quad (192) $$

where we recall that $\gamma$, defined in Assumption A5, is given by

$$ \gamma = 4 \max(1, L) e^{\alpha T} \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \\ \times (\|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} + 12 \max(1, L) e^{\alpha T} \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d \times n})}). $$

Recall now that $(\bar{q}, \bar{\psi})$ is a saddle point of (P'). By the third assertion in the statement of Theorem 25, we deduce that $\bar{q}$ is a maximizer of $q' \in Q \mapsto \mathcal{J}(q', \bar{\psi})$. Therefore, the supremum in the above display can be bounded as follows

$$ \sup_{q \in \mathcal{Q}} \mathcal{J}(q, \bar{\psi}) = \mathcal{J}(\bar{q}, \bar{\psi}) \le \mathcal{J}(\bar{q}, 0), \quad (193) $$
with the last inequality following from the saddle point property of $(\bar{q}, \bar{\psi})$. Since $f^*$ is lower bounded, we have

$$ \mathcal{J}(\bar{q}, 0) = \mathcal{R}(\bar{q}, 0) - \mathcal{S}(\bar{q}) \leq C' \left( 1 + \mathbb{E} \left[ \bar{q}_T |X_T^0| + \int_0^T \bar{q}_t \ell(t, 0) dt \right] \right), $$

for a new constant $C'$. We then use (13) to upper bound the right-hand side. By Lemma 42, we know that there exists $\tau > 0$, only depending on the data such that $X_T^0 \in S_{\exp}(\mathcal{F}_T, \mathbb{R}^n)$. And then, (13) (together with the fact that $\bar{q} \in \mathcal{Q}_{c_1}$) yields

$$ \mathcal{J}(\bar{q}, 0) \leq C'(1 + c_1). $$

Returning to (192) and (193), we obtain

$$ \sup_{q \in \mathcal{Q}} \left\{ \mathbb{E} \left[ \int_0^T q_t |\bar{\psi}_t|^2 dt \right] - \gamma \mathcal{S}(q) \right\} \leq 4 \max(1, L) [C + C'(1 + c_1)]. $$

This completes the proof. □

### 5.3.2 Necessary conditions

This subsection is dedicated to a series of lemmas leading eventually to Lemma 37, which we invoked in the proof of Theorem 31 to establish the necessary condition. Given the constant $c'_2$ in Lemma 32, we assume that the constant $c_2$ in the definition of $A_{c_2}$ in $(P_C)$ satisfies $c_2 > c'_2 > 0$, which condition guarantees that any minimizer to $(P_C)$ -with $\bar{q} \in \mathcal{Q}_{c_1}$ such that, for some $\bar{\psi} \in A_{c_2}$, $(\bar{q}, \bar{\psi})$ is a saddle-point- lies ‘in the interior’ of the admissible set $A_{c_2}$ (in the sense that it belongs to $A_{c'_2}$). We insist on the fact that, similar to Subsection 5.3.1, $\bar{q}$ is fixed throughout the analysis. In coherence with the convention adopted earlier, this makes it possible to omit $\bar{q}$ in the various notations. For instance, we write $\delta_X \mathcal{G}(X_T)$ for $\delta_X \mathcal{G}(\bar{q}_T, X_T)$.

Recalling the definition of the pre-Hamiltonian $H$ in (20), and denoting $X^\psi$, for a given $\psi \in A_{c'_2}$, the associated solution to the state equation, we define the adjoint BSDE with unknown $(p, k)$,

$$ -dp_t = \nabla_x H(t, X_t^\psi, \psi_t, p_t, k_t, \bar{q}_t) dt - k_t dW_t, \quad p_T = \delta_X \mathcal{G}(\bar{q}_T, X_T^\psi). \quad (194) $$

By assumptions on the mappings $b$ and $\sigma$, the derivative of the pre-Hamiltonian simplifies to

$$ \nabla_x H(t, X_t^\psi, \psi_t, p_t, k_t, \bar{q}_t) = b_t^\top p_t. $$

In the analysis carried out below, we will also use the fact that

$$ \nabla_{\psi} H(t, x, \psi, p, k, q) = q \nabla_{\psi} \ell(t, \psi) + c_t^\top p + r \mathrm{Tr}(\sigma_t^\top k), \quad (195) $$

where $r$ can be either 0 or 1, and we recall the convention (22) for the trace

$$ \mathrm{Tr}(\sigma_t^\top k_t) = \left( \sum_{i=1}^n \sum_{j=1}^d (\sigma_t)_{i,j,\ell}(k_t)_{i,j} \right)_{\ell=1,\dots,d}. \quad (196) $$
To study the BSDE (194), we introduce the following intermediary BSDE with unknown $(\varrho, h)$,

$$ -d\varrho_t = (b_t^\top \varrho_t + \bar{Y}_t^\star \varrho_t) dt - h_t d\bar{W}_t, \quad \varrho_T = \bar{q}_T^{-1} \delta_X \mathcal{G}(\bar{q}_T, X_T^\psi), \quad (197) $$

where $(\bar{W}_t = W_t - \int_0^t \bar{Z}_s^\star ds)_{t \in [0,T]}$ is a Brownian motion under the equivalent probability measure $\bar{Q}$ defined by $\bar{Q} = \bar{\mathcal{E}}_T \mathbb{P}$, with $\bar{\mathcal{E}}_T := \mathcal{E}_T(\int_0^T \bar{Z}_s^\star \cdot dW_s)$. We recall that $(\bar{Y}^\star, \bar{Z}^\star)$ is associated to $\bar{q}$ through (4), and that $\bar{q}_T = \exp(\int_0^T \bar{Y}_s^\star ds) \bar{\mathcal{E}}_T$.

**Lemma 33.** There exists a unique solution $(\varrho, h) \in S^2(\mathbb{F}, \mathbb{R}^n, \bar{Q}) \times M^2(\mathbb{F}, \mathbb{R}^{n \times d}, \bar{Q})$ to (197). When $r = 1$, $\varrho$ belongs to $L^\infty(\mathbb{F}, \mathbb{R}^n)$ and $h$ to $L^2(\mathbb{F}, \mathbb{R}^{n \times d}, \bar{Q})$.

*Proof.* Let us recall that

$$ \|b\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} + \|\bar{Y}^\star\|_{L^\infty(\mathbb{F})} \le \alpha + L, \quad (198) $$

by Assumption A1 and since $|\bar{Y}_s^\star| \le \alpha$. In particular, the BSDE (197) is a linear BSDE, with a bounded linear coefficient. The existence and uniqueness of a solution are well established once the terminal condition has been shown to be sufficiently integrable.

*Step 1:* $r = 0$. By the growth Assumption A6 on $\delta_X \mathcal{G}$, we have

$$ |\varrho_T| = \bar{q}_T^{-1} \left| \delta_X \mathcal{G}(\bar{q}_T, X_T^\psi) \right| \le L \left( 1 + |X_T^\psi| + \mathbb{E} \left[ \bar{q}_T |X_T^\psi|^2 \right] \right). $$

Taking the square on both sides, we get (for a constant $C$ only depending on $L$ and whose value is allowed to vary from line to line)

$$ \begin{aligned} \mathbb{E}^{\bar{Q}} [|\varrho_T|^2] &\le C \left( 1 + \mathbb{E}^{\bar{Q}} [|X_T^\psi|^2] + \mathbb{E} [\bar{q}_T |X_T^\psi|^2]^2 \right) \\ &\le C \left( 1 + \mathbb{E} \left[ \int_0^T \bar{q}_s |\psi_s|^2 ds \right]^2 \right) \\ &\le C (1 + S(\bar{q})^2 + S^*(\psi)^2) < +\infty, \end{aligned} $$

where we used Lemma 40 in Appendix B in order to pass from the first to the second line. Together with (198), existence and uniqueness follow from the standard $L^2$-theory of BSDEs (we recall from [2, Theorem 2.4] that the martingale representation theorem holds under $\bar{Q}$, with respect to $\bar{W}$).

*Step 2:* $r = 1$. In this case, $|\delta_X \mathcal{G}|$ is bounded by $C \bar{q}_T (1 + \mathbb{E}[\bar{q}_T |X_T|])$, for a constant $C$ only depending on $L$. By Lemma 40, $\mathbb{E}[\bar{q}_T |X_T|] < +\infty$. Therefore, the terminal condition in (197) is square integrable under the probability measure $\bar{Q}$. The conclusion follows by the same arguments as in the first step. This concludes the proof. $\square$

**Lemma 34.** There exists a unique solution $(p, k)$ to (194) in the space $D(\mathbb{F}, \mathbb{P}, \mathbb{R}^n) \times (\cap_{\beta \in (0,1)} M^\beta(\mathbb{F}, \mathbb{P}, \mathbb{R}^{n \times d}))$. It is given by

$$ (p_t, k_t)_{t \in [0,T]} = (\bar{q}_t \varrho_t, \bar{q}_t (h_t + \varrho_t \otimes \bar{Z}_t^\star))_{t \in [0,T]}, $$

where $(\varrho_t, h_t)_{t \in [0,T]}$ is the solution to (197) and $(\varrho_t \otimes \bar{Z}_t^\star)_{t \in [0,T]}$ denotes the $n \times d$ matrix with elements $(\varrho_t^i \bar{Z}_t^{\star,j})_{i \in \{1, \dots, n\}, j \in \{1, \dots, d\}}$.
*Proof. Step 1: Existence.* Repeating the computations from the proof of Lemma 33, there exists a constant C, only depending on L, such that

$$ 
\begin{aligned}
\mathbb{E} \left[ |\delta_X \mathcal{G}(\bar{q}_T, X_T^\psi)| \right] &\le L \mathbb{E} \left[ \bar{q}_T \left( 1 + |X_T^\psi|^{1-r} + \mathbb{E} \left[ \bar{q}_T |X_T^\psi|^{2-r} \right] \right) \right] \\
&\le C \left( 1 + \mathbb{E} \left[ \int_0^T \bar{q}_s |\psi_s|^2 ds \right]^2 \right) \\
&\le C (1 + S(\bar{q})^2 + S^*(\psi)^2) < +\infty.
\end{aligned}
$$

The existence of a solution to (194), within the space specified in the statement, follows from [23, Proposition 6.4].

*Step 2: Uniqueness.* Now let $(p_t, k_t)_{t \in [0,T]}$ be a solution to the backward equation (194) (within the space mentioned in the statement). Then, one can verify that $(\varrho_t, h_t)_{t \in [0,T]} := (\bar{q}_t^{-1} p_t, \bar{q}_t^{-1} (k_t - p_t \otimes \bar{Z}_t^*))_{t \in [0,T]}$ solves (197), with the stochastic integral therein being understood as a local martingale under $\bar{\mathbb{Q}}$. That said, at this stage, we do not know yet that the hence defined pair $(\varrho, h)$ belongs to the space $S^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}}) \times M^2(\mathbb{F}, \mathbb{R}^{n \times d}, \bar{\mathbb{Q}})$, which prevents us from identifying directly $(\varrho, h)$ with the (unique) solution constructed in Lemma 33. We thus proceed as follows. We denote by $(R_t)_{t \in [0,T]}$ the solution to the (random) ordinary differential equation

$$ \frac{d}{dt} R_t = -b_t^\top R_t - \bar{Y}_t^\star R_t, \quad R_0 = I_n, \qquad (199) $$

with $I_n$ denoting the $n \times n$ identity matrix. The process $(R_t)_{t \in [0,T]}$ takes values in the set of $n \times n$ invertible matrices, and

$$ \frac{d}{dt} R_t^{-1} = R_t^{-1} b_t^\top + \bar{Y}_t^\star R_t^{-1}, \quad R_0^{-1} = I_n. $$

Obviously, the process $(R_t^{-1})_{t \in [0,T]}$ is bounded by a deterministic constant. Also, it is straightforward to check that $(R_t^{-1} \varrho_t)_{t \in [0,T]}$ is a local martingale under $\bar{\mathbb{Q}}$. By a standard localization argument, we can find a non-decreasing sequence of stopping times $(\tau_k)_{k \ge 1}$, converging to $T$, such that, for any $t \in [0, T]$, and any integer $k \ge 1$,

$$ R_{t \wedge \tau_k}^{-1} \varrho_{t \wedge \tau_k} = \mathbb{E}^{\bar{\mathbb{Q}}} [R_{\tau_k}^{-1} \varrho_{\tau_k} | \mathcal{F}_t]. \qquad (200) $$

In order to pass to the limit (as $k$ tends to $+\infty$) in the above display, we check that the collection of random variables $(R_{\tau_k}^{-1} \varrho_{\tau_k})_{k \ge 1}$ is uniformly integrable under $\bar{\mathbb{Q}}$ (the same argument would show that the collection of random variables $(R_{t \wedge \tau_k}^{-1} \varrho_{t \wedge \tau_k})_{k \ge 1}$ is uniformly integrable). For any event $A$, we can find a constant $C$ such that, for any $k \ge 1$,

$$ \mathbb{E}^{\bar{\mathbb{Q}}} [1_A | R_{\tau_k}^{-1} \varrho_{\tau_k}|] \le C \mathbb{E} [1_A | q_{\tau_k} \varrho_{\tau_k}|] = C \mathbb{E} [1_A | p_{\tau_k}|]. $$

Since $p$ belongs to $D(\mathbb{F}, \mathbb{P})$, the right-hand side tends to 0, uniformly in $k$, as $\mathbb{P}(A)$ tends to 0. Writing $\mathbb{P}(A) = \mathbb{E}^{\bar{\mathbb{Q}}}[(\mathcal{E}_T)^{-1}1_A]$ (here $\bar{\mathcal{E}}_T > 0$ a.s., because $\bar{q}_T > 0$ a.s.), we deduce that $\mathbb{P}(A)$ tends to 0 as $\bar{\mathbb{Q}}(A)$ tends to 0. This shows that the right-hand side (in the above display) tends to 0, uniformly in $k$, as $\bar{\mathbb{Q}}(A)$ tends to 0, which provides the required uniform integrability property. Letting $k$ tend to $+\infty$ in (200), we deduce that, for any $t \in [0, T]$,

$$ R_t^{-1} \varrho_t = \mathbb{E}^{\bar{\mathbb{Q}}} \left[ R_T^{-1} \bar{q}_T^{-1} \delta_X \mathcal{G}(\bar{q}_T, X_T^\psi) \middle| \mathcal{F}_t \right]. $$
This provides an explicit formula for $\varrho$ and makes it possible to identify it (together with $h$) with the solution obtained in Lemma 33. It remains to see that the mapping

$$ \mathbb{R}^n \times \mathbb{R}^{n \times d} \ni (p, k) \mapsto (q^{-1}p, q^{-1}(k - p \otimes z^*)), $$

is one-to-one for any $q > 0$ and $z^* \in \mathbb{R}^n$. This proves that $(p_t, k_t)_{t \in [0, T]}$ is uniquely determined by the pair $(\varrho_t, h_t)_{t \in [0, T]}$ and is thus unique. $\square$

**Tangent processes.** Let $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n)$ be such that $\psi + \varphi \in \mathcal{A}_{c_2}$ (we recall that $\psi$ is an arbitrary element in $\mathcal{A}_{c_2}$). We introduce the following system of variational (or tangent) processes with unknown $(u, v, x)$, the latter taking values in $\mathbb{R} \times \mathbb{R}^d \times \mathbb{R}^n$,

$$ \left\{ \begin{array}{l} du_t = \bar{q}_t \varphi_t \cdot \nabla_{\psi} \ell(t, \psi_t) dt - v_t \cdot dW_t, \quad u_T = \delta_X \mathcal{G}(\bar{q}_T, X_T^\psi) \cdot x_T, \\ dx_t = (b_t x_t + c_t \varphi_t) dt + r D_\psi \sigma_t(\varphi_t) dW_t, \quad x_0 = 0, \end{array} \right. \quad (V) $$

where we recall that $D_\psi \sigma_t(\varphi_t) = (\sum_{\ell=1}^n (\sigma_t)_{i,j,\ell}(\varphi_t)\ell)_{i \in \{1, \dots, n\}, j \in \{1, \dots, d\}}$, see A2.

**Lemma 35.** Let $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n)$ such that $\psi + \varphi \in \mathcal{A}_{c_2}$. Then, there exists a unique solution $(x_t)_{t \in [0, T]}$ to the forward equation in (V), lying in $S^2(\mathbb{F}, \mathbb{R}^n) \cap S^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q})$ (even in $L^\infty(\mathbb{F}, \mathbb{R}^n)$ when $r = 0$). And, there exists a solution $(u, v) \in D(\mathbb{F}, \mathbb{P}) \times (\cap_{\beta \in (0,1)} M^\beta(\mathbb{F}, \mathbb{R}^d, \mathbb{P}))$ to the backward equation in (V).

*Proof. Step 1: Existence and uniqueness to the forward equation.* Denoting by $\Gamma$ the resolvent associated with the linear part of the forward equation, i.e.,

$$ \frac{d}{dt}\Gamma_t = b_t\Gamma_t, \quad \Gamma_0 = I_n, $$

with $I_n$ standing for the $n \times n$ identity matrix, the solution to the forward equation in (V) is explicitly given by

$$ x_t = \Gamma_t \left( \eta + \int_0^t \Gamma_s^{-1} (a_s + c_s \varphi_s) ds + r \int_0^t \Gamma_s^{-1} D_\psi \sigma_s(\varphi_s) dW_s \right), \quad t \in [0, T]. $$

Thus, $(x_t)_{t \in [0, T]}$ clearly belongs to $S^2(\mathbb{F}, \mathbb{R}^n, \mathbb{P})$ (because $\varphi$ is bounded). It further belongs to $S^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q})$ by a direct application of Lemma 40 using once again the fact that $\varphi$ is bounded (where we identify the process $(\Gamma_t^{-1} D_\psi \sigma_t(\varphi_t))_{t \in [0, T]}$, which is bounded, in this proof with the process $(\nu_t)_{t \in [0, T]}$ in the statement of Lemma 40).

*Step 2: Well-posedness of the backward equation.* By the growth assumptions A4 and A6, we have

$$ \mathbb{E} \left[ |\delta_X \mathcal{G}(\bar{q}_T, X_T^\psi) \cdot x_T| \right] \le C \mathbb{E} \left[ \bar{q}_T (1 + |X_T^\psi|^{1-r}) |x_T| \right]. $$

When $r = 1$, the right-hand side reduces to $C(1 + \mathbb{E}[\bar{q}_T | x_T])$, which is finite since $(x_t)_{t \in [0, T]} \in S^2(\mathbb{F}, \mathbb{R}^n, \mathbb{Q})$. When $r = 0$, $(x_t)_{t \in [0, T]} \in L^\infty(\mathbb{F}, \mathbb{R}^n, \mathbb{P})$, and by Lemma 40, we have

$$ \mathbb{E} \left[ |\delta_X \mathcal{G}(\bar{q}_T, X_T^\psi) \cdot x_T| \right] \le C \left( 1 + \mathbb{E} \left[ \bar{q}_T |X_T^\psi| \right] \right) < +\infty. $$
Since $\nabla_{\psi} \ell$ is at most of linear growth in $\psi$, we further have

$$ 
\begin{aligned}
\mathbb{E} \left[ \int_0^T \bar{q}_s |\varphi_s \cdot \nabla_{\psi} \ell(s, \psi_s)| \mathrm{d}s \right] &\le C \mathbb{E} \left[ \int_0^T \bar{q}_s (1 + |\varphi_s|^2 + |\psi_s|^2) \mathrm{d}s \right] \quad (201) \\
&\le C (1 + S(\bar{q}) + S^*(\psi)) < +\infty.
\end{aligned}
$$

The conclusion follows by [23, Proposition 6.4]. □

So far, we have considered $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n, \mathbb{P})$ such that $\psi + \varphi \in A_{c2}$. Since $A$ is convex, we then have, for any $\varepsilon \in [0, 1]$, $\psi + \varepsilon\varphi \in A_{c2}$. By optimality of the control $\psi$, we get

$$ \varepsilon^{-1} (\mathcal{J}(\psi + \varepsilon\varphi) - \mathcal{J}(\psi)) \ge 0. \quad (202) $$

We use the above inequality to prove the following statement:

**Lemma 36.** Let $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n, \mathbb{P})$ such that $\psi + \varphi \in A_{c2}$. With $(u, v)$ being as in the statement of Lemma 35, the following variational inequality holds true:

$$ \mathbb{E}[u_0] = \mathbb{E}\left[u_T + \int_0^T \bar{q}_s \varphi_s \cdot \nabla_{\psi} \ell(s, \psi_s) \mathrm{d}s\right] \ge 0. $$

*Proof.* Since $\bar{q}$ is fixed, we omit to indicate it explicitly in the various functionals that depend on it. For instance, we use the shorthand notation $\mathcal{G}(X_T^\psi)$ for $\mathcal{G}(\bar{q}_T, X_T^\psi)$.

On the one hand, we have, from A7,

$$ \mathcal{G}(X_T^{\psi+\varepsilon\varphi}) - \mathcal{G}(X_T^\psi) = \mathbb{E} \left[ \delta_X \mathcal{G}(X_T^\psi) \cdot (X_T^{\psi+\varepsilon\varphi} - X_T^\psi) \right] + O \left( \mathbb{E} \left[ \bar{q}_T |X_T^{\psi+\varepsilon\varphi} - X_T^\psi|^2 \right] \right). $$

Because $(X_t)_{t \in [0,T]}$ solves a linear SDE, we also have

$$ \varepsilon^{-1}(X_T^{\psi+\varepsilon\varphi} - X_T^\psi) = X_T^\varphi = x_T, $$

and thus

$$ \varepsilon^{-1} \left( \mathcal{G}(X_T^{\psi+\varepsilon\varphi}) - \mathcal{G}(X_T^\psi) \right) = \mathbb{E} \left[ \delta_X \mathcal{G}(X_T^\psi) \cdot x_T \right] + \varepsilon O \left( \mathbb{E} \left[ \bar{q}_T |x_T|^2 \right] \right). \quad (203) $$

In order to handle the right-hand side, we use the same estimates as in the second step of the proof of Lemma 35. In particular, we already know that $\mathbb{E}[|\delta_X \mathcal{G}(X_T^\psi) \cdot x_T|] < +\infty$. We also know that $\mathbb{E}[\bar{q}_T |x_T|^2] < +\infty$, from which we deduce that the term on the second line of (203) tends to 0 with $\varepsilon$. So, we obtain

$$ \lim_{\varepsilon \to 0} \varepsilon^{-1} \left( \mathcal{G}(X_T^{\psi+\varepsilon\varphi}) - \mathcal{G}(X_T^\psi) \right) = \mathbb{E} \left[ \delta_X \mathcal{G}(X_T^\psi) \cdot x_T \right]. \quad (204) $$

Since $\ell$ is assumed to be twice differentiable, with bounded second-order derivatives, we also have

$$ \varepsilon^{-1} \mathbb{E} \left[ \int_0^T \bar{q}_s (\ell(s, \psi_s + \varepsilon\varphi_s) - \ell(s, \psi_s)) \mathrm{d}s \right] = \mathbb{E} \left[ \int_0^T \bar{q}_s \varphi_s \cdot \nabla_{\psi} \ell(s, \psi_s) \mathrm{d}s \right] \\ + \varepsilon o \left( \mathbb{E} \left[ \int_0^T \bar{q}_s |\varphi_s|^2 \mathrm{d}s \right] \right). $$

Because $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n)$, we obviously have $\mathbb{E}[\int_0^T \bar{q}_s |\varphi_s|^2 \mathrm{d}s] < +\infty$. As for the first term on the right-hand side, we recall from (201) that it is bounded. Then, combining
the above display with (204), and recalling again the definition of the criterion $\mathcal{J}$, we get

$$ \lim_{\epsilon \to 0} \epsilon^{-1} (\mathcal{J}(\psi + \epsilon\varphi) - \mathcal{J}(\psi)) = \mathbb{E} \left[ \delta_X \mathcal{G}(X_T^\psi) \cdot x_T \right] + \mathbb{E} \left[ \int_0^T \bar{q}_s \varphi_s \cdot \nabla_\psi \ell(s, \psi_s) ds \right] \\ = \mathbb{E} \left[ u_T + \int_0^T \bar{q}_s \varphi_s \cdot \nabla_\psi \ell(s, \psi_s) ds \right]. $$

It remains to use to the backward equation in (V) in order to identify the last term with $\mathbb{E}[u_0]$. By localization, we can find a non-decreasing sequence of stopping times $(\tau_m)_{m \ge 1}$, converging to $T$, such that, for any $m \ge 1$,

$$ \mathbb{E} \left[ u_{\tau_m} + \int_0^{\tau_m} \bar{q}_s \varphi_s \cdot \nabla_\psi \ell(s, \psi_s) ds \right] = \mathbb{E}[u_0]. \quad (205) $$

Since $u$ belongs to $D(\mathbb{F}, \mathbb{P})$, $\mathbb{E}[u_{\tau_m}] \to \mathbb{E}[u_T]$ as $m$ tends to $+\infty$. And by (201), $\mathbb{E}[\int_0^{\tau_m} \bar{q}_s \varphi_s \cdot \nabla_\psi \ell(s, \psi) ds] \to \mathbb{E}[\int_0^T \bar{q}_s \varphi_s \cdot \nabla_\psi \ell(s, \psi) ds]$. This shows that the left-hand side in the above display converges to the right-hand side of (205). Recalling inequality (202), we complete the proof. $\square$

**Lemma 37.** Let $\psi \in A_{c_2}'$ be a minimizer to problem (P_C) and $(p, k)$ the associated solution to (194) provided by Lemma 34. Then, $(\psi, p, k, X)$ belongs to the set $\mathcal{A}$ defined in (25) and satisfies the first-order condition (Opt_C) (with $\bar{q}$ being substituted for $q$ therein).

*Proof.* We start with the following preliminary remark: the fact that $(\psi, p, k, X)$ belongs to $\mathcal{A}$ is a consequence of Lemmas 34 and 40.

Next, following the analysis carried out in Lemmas 35 and 36, we consider $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n, \mathbb{P})$ such that $\psi + \varphi \in A_{c_2}$. With $x$ as in Lemma 35, and by Itô's formula, we have

$$ p_T \cdot x_T = \int_0^T p_s \cdot dx_s + \int_0^T x_s \cdot dp_s + r \int_0^T \mathrm{Tr}((D_\psi \sigma_s(\varphi_s))^\top k_s) ds. $$

For a given $A > 0$, we also consider the stopping time

$$ \tau_A := \inf \left\{ t \in [0, T], \left| \int_0^t p_s^\top D_\psi \sigma_s(\varphi_s) dW_s \right| + \left| \int_0^t x_s^\top k_s dW_s \right| \ge A \right\}. $$

Then, by cancellation of the expectations of the stopped stochastic integrals in the expansion of $(p_t \cdot x_t)_{t \in [0, T]}$, we get

$$ \mathbb{E}[p_{T \wedge \tau_A} \cdot x_{T \wedge \tau_A}] = \mathbb{E} \left[ \int_0^{T \wedge \tau_A} p_s^\top c_s \varphi_s ds + r \int_0^{T \wedge \tau_A} \mathrm{Tr}((D_\psi \sigma_s(\varphi_s))^\top k_s) ds \right]. \quad (206) $$

We now aim to pass to the limit on both sides of the equality.

**Step 1: convergence of the left-hand side of (206).** To pass to the limit, we establish that the random variables $(p_{T \wedge \tau_A} \cdot x_{T \wedge \tau_A})_{A>0}$ are uniformly integrable, distinguishing between the two cases $r = 0$ and $r = 1$. Throughout, $E$ is a fixed subset of $\Omega$, belonging to $\mathcal{F}_T$.
When $r = 0$, the process $x = (x_t)_{t \in [0,T]}$ is bounded (see Lemma 35). Then, we can find a constant $C$, independent of $A$, such that

$$ \mathbb{E}[1_E|p_{T \wedge \tau_A} \cdot x_{T \wedge \tau_A}|] \le C \mathbb{E}[1_E|p_{T \wedge \tau_A}|]. $$

Since $p$ belongs to $D(\mathbb{F}, \mathbb{R}^n, \mathbb{P})$ by Lemma 34, the right-hand side tends to 0, uniformly with respect to $A$, as $\mathbb{P}(E)$ tends to 0, yielding the required uniform integrability property.

When $r = 1$, the process $x$ belongs to $S^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}})$ thanks to Lemma 35. By Lemmas 33 and 34, the process $p$ is equal to $\bar{q}\varrho$ with $\varrho$ belonging to $L^\infty(\mathbb{F}, \mathbb{R}^n)$. Then, we have

$$ \begin{aligned} \mathbb{E}[1_E|p_{T \wedge \tau_A} \cdot x_{T \wedge \tau_A}|] &= C \mathbb{E}[1_E \bar{q}_{T \wedge \tau_A} | \varrho_{T \wedge \tau_A} \cdot x_{T \wedge \tau_A}|] \\ &\le C \mathbb{E}[1_E \bar{q}_T^* | x_{T \wedge \tau_A}|] \\ &\le C \mathbb{E}[\bar{q}_T^* 1_E]^{1/2}, \end{aligned} $$

with the last line following from Cauchy-Schwarz inequality, and, once again, from the fact that $x \in S^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}})$. In particular, the constant $C$ on the last line is allowed to depend on the $S^2$-norm of $x$ (under $\bar{\mathbb{Q}}$) and is implicitly allowed to vary from line to line. To prove that the term on the last line of the above displays tends to 0 as $\mathbb{P}(E)$ tends to 0, it suffices to recall from Lemma 39 that $q_T^*$ is integrable, so that the right-hand side tends to 0 as $\mathbb{P}(E)$ tends to 0. This yields the expected uniform integrability property.

**Step 2: convergence on the right-hand side of (206).** Using the fact that the terms $c, \varphi$ and $D_\psi\sigma(\varphi_s)$ are uniformly bounded, and that $(p, k)$ belongs to $D(\mathbb{F}, \mathbb{R}^n, \mathbb{P}) \times (\cap_{\beta \in (0,1)} M^\beta(\mathbb{F}, \mathbb{R}^{n \times d}, \mathbb{P}))$, see Lemma 34, we can derive the following upper-bound

$$ \begin{aligned} & \int_0^{T \wedge \tau_A} |p_s^\top c_s \varphi_s| \, ds + r \int_0^{T \wedge \tau_A} \left| \mathrm{Tr}((D_\psi \sigma_s(\varphi_s))^\top k_s) \right| \, ds \\ & \le C \left( \int_0^T |p_s| \, ds + r \int_0^T |k_s| \, ds \right). \end{aligned} $$

When $r = 0$, we deduce that the left-hand side is integrable, uniformly with respect to $A$. When $r = 1$, the proof is more involved. We recall from Lemmas 33 and 34 that $|p_t| \le C\bar{q}_t$ and $|k_t| \le \bar{q}_t(|h_t| + |\bar{Z}_t^*|)$, for all $t \in [0, T]$, where $h \in L^2(\mathbb{F}, \mathbb{R}^{n \times d}, \bar{\mathbb{Q}})$. Since $\bar{Z}_t^*$ belongs to $L^2(\mathbb{F}, \mathbb{R}^d, \bar{\mathbb{Q}})$, we deduce from Cauchy-Schwarz inequality that $k$ belongs to $L^1(\mathbb{F}, \mathbb{R}^{n \times d}, \mathbb{P})$. Therefore, the left-hand side in the above display is also integrable, uniformly with respect to $A > 0$.

**Step 3: conclusion.** Thanks to the uniform integrability properties established in the last two steps, we can now pass to the limit in (206). We get

$$ \mathbb{E}[p_T \cdot x_T] = \mathbb{E} \left[ \int_0^T p_s^\top c_s \varphi_s ds + r \int_0^T \mathrm{Tr}((D_\psi \sigma_s(\varphi_s))^\top k_s) ds \right]. $$

Consider now $(u, v)$ as in the statement of Lemma 36. Using the fact that $u_T = p_T \cdot x_T$
and $\mathbb{E}[u_0] \ge 0$, together with Lemma 36, we obtain

$$
\begin{align*}
0 \le \mathbb{E}[u_0] \\
&= \mathbb{E} \left[ p_T \cdot x_T + \int_0^T \varphi_s \cdot \bar{q}_s \nabla_\psi \ell(s, \psi_s) \mathrm{d}s \right] \\
&= \mathbb{E} \left[ \int_0^T \left\{ \varphi_s \cdot \left( \bar{q}_s \nabla_\psi \ell(s, \psi_s) + c_s^\top p_s \right) + r \mathrm{Tr}((D_\psi \sigma_s(\varphi_s)))^\top k_s \right\} \mathrm{d}s \right] \tag{207} \\
&= \mathbb{E} \left[ \int_0^T \varphi_s \cdot \nabla_\psi H(s, X_s, \psi_s, p_s, k_s, \bar{q}_s) \mathrm{d}s \right],
\end{align*}
$$

where, to get the last line, we used (195) together with the identity

$$
\mathrm{Tr} \left( (D_{\psi} \sigma_{s})^{\top} k_{s} \right) = \sum_{i=1}^{n} \sum_{j=1}^{d} \sum_{\ell=1}^{n} (\sigma_{s})_{i,j,k} (\varphi_{s})_{k} (k_{s})_{j,i} = \varphi_{s} \cdot (\sigma_{s}^{\top} k_{s}).
$$

The sequence of inequalities (207) is true for any arbitrary perturbation $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n)$ such that $\psi + \varphi \in A_{c_2}$. In fact, using the property that $\psi \in A_{c'_2}$, where $c'_2 < c_2$, it is easy to see that, for any given $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n)$, there exists an $\varepsilon_0$ (depending on $\varphi$) such that $\psi + \varepsilon_0\varphi \in A_{c_2}$. Substituting $\varepsilon_0\varphi$ for $\varphi$ in (207), this proves that, for any $\varphi \in L^\infty(\mathbb{F}, \mathbb{R}^n)$,

$$
\mathbb{E} \left[ \int_0^T \varphi_s \cdot \nabla_\psi H(s, X_s, \psi_s, p_s, k_s, \bar{q}_s) \mathrm{d}s \right] \ge 0.
$$

And then, changing $\varphi$ into $-\varphi$, the above inequality is in fact an equality, from which
we get that

$$
\nabla_{\psi} H(t, X_t, \psi_t, p_t, k_t, \bar{q}_t) = 0, \quad d\mathbb{P} \otimes dt \text{-a.e. .}
$$

Since $H$ is strictly convex in the variable $\psi$, this shows that the third line in (OptC)
is satisfied and concludes the proof. $\square$

**5.3.3 Sufficient conditions**

This last subsection is dedicated to the proof of the following lemma, which we
invoked in the proof of Theorem 31 to establish the sufficient condition.

**Lemma 38.** Let $(\psi, p, k, X) \in \mathcal{A}$ be a solution to the first order condition (Opt$_C$).
Then, $\psi$ is the (unique) minimizer of the problem (P$_C$) (with $\bar{q}$ being substituted for
$q$ therein).

Proof. Let $\psi' \in A_{c_2}$ and $X' := X^{\psi'}$ be the associated state. By definition of $\mathcal{J}$ (which is here a shorthand notation for $\mathcal{J}(\bar{q}, \cdot)$), we have (with a similar shorthand notation for $\mathcal{G}(\bar{q}_T, \cdot)$)

$$
\mathcal{J}(\psi') - \mathcal{J}(\psi) = \mathcal{G}(X'_{T}) - \mathcal{G}(X_{T}) + \mathbb{E} \left[ \int_{0}^{T} \bar{q}_{s} (\ell(s, \psi'_{s}) - \ell(s, \psi_{s})) \mathrm{d}s \right]. \quad (208)
$$

By Assumption A8, the mapping $\mathcal{G}$ is convex with respect to its last variable. Therefore,

$$
\mathcal{G}(X'_{T}) - \mathcal{G}(X_{T}) \geq \mathbb{E} \left[ \delta_{X} \mathcal{G}(X_{T}) \cdot (X'_{T} - X_{T}) \right] = \mathbb{E} \left[ p_{T} \cdot (X'_{T} - X_{T}) \right]. \quad (209)
$$
For a given $A > 0$, consider the stopping time

$$ \tau_A := \inf \left\{ t \in [0, T], \left| \int_0^t p_s^\top (\sigma_s(\psi_s') - \sigma_s(\psi_s)) dW_s \right| + \left| \int_0^t (X_s' - X_s)^\top k_s dW_s \right| \ge A \right\}. $$

By (194) and Itô's formula, we have the following formula (which is the analogue of (206))

$$ \begin{aligned} & \mathbb{E} [p_{T \wedge \tau_A} \cdot (X'_{T \wedge \tau_A} - X_{T \wedge \tau_A})] \\ & = \mathbb{E} \left[ \int_0^{T \wedge \tau_A} \left( p_s^\top c_s (\psi_s' - \psi_s) + \text{Tr} \left( k_s (\sigma(s, \psi_s') - \sigma(s, \psi_s))^T \right) \right) \mathrm{d}s \right]. \end{aligned} \quad (210) $$

Following the proof of Lemma 37, we now aim to take the limit $A \to +\infty$ (but the proof is more difficult because the difference $\psi' - \psi$, which is the analogue of $\varphi$ in the proof of Lemma 37, is not bounded). The strategy is to prove that the random variables inside the expectations are uniformly integrable with respect to A.

**Step 1: left-hand side of (210).** In this step, we check that the left-hand side on (210) is uniformly integrable with respect to A. Throughout, we consider a fixed event $E \in \mathcal{F}_T$. By Lemma 33, the process $p$ is equal to the process $\bar{q}\varrho$. Therefore,

$$ \mathbb{E} [1_E |p_{T \wedge \tau_A} \cdot (X'_{T \wedge \tau_A} - X_{T \wedge \tau_A})|] = C \mathbb{E} [1_E \bar{q}_{T \wedge \tau_A} |\varrho_{T \wedge \tau_A} \cdot (X'_{T \wedge \tau_A} - X_{T \wedge \tau_A})|]. $$

We recall that $X, X' \in S^{2-r}(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}})$, see Lemma 40. In addition, by Lemma 33, the process $\varrho$ belongs to $S^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}})$ when $r = 0$ and $L^\infty(\mathbb{F}, \mathbb{R}^n)$ when $r = 1$.

Let us first study the case $r = 0$. By Young's inequality, we have, for any $\varepsilon > 0$,

$$ \begin{aligned} & \mathbb{E} [1_E \bar{q}_{T \wedge \tau_A} |\varrho_{T \wedge \tau_A} \cdot (X'_{T \wedge \tau_A} - X_{T \wedge \tau_A})|] \\ & \le \varepsilon^{-1} \mathbb{E} [1_E \bar{q}_{T \wedge \tau_A} |\varrho_{T \wedge \tau_A}|^2] + \varepsilon \mathbb{E} [\bar{q}_{T \wedge \tau_A} |X_{T \wedge \tau_A} - X'_{T \wedge \tau_A}|^2] \\ & \le C \varepsilon^{-1} \mathbb{E} [\mathbb{P}(E |\mathcal{F}_{T \wedge \tau_A}) \bar{q}_T |\varrho_T^*|^2] + C \varepsilon \mathbb{E} [\bar{q}_T |X_T^*|^2] + C \varepsilon \mathbb{E} [\bar{q}_T |(X')_T^*|^2]. \end{aligned} $$

Observing from Doob's inequality that $\sup_{t \in [0, T]} \mathbb{P}(E |\mathcal{F}_t)$ tends to 0 in $L^1(\mathbb{P})$ as $\mathbb{P}(E)$ tends to 0, we deduce that the left-hand side tends to 0 with $\mathbb{P}(E)$, uniformly in A. This proves the expected uniform integrability property when $r = 0$.

When $r = 1$, we have

$$ \begin{aligned} & \mathbb{E} [1_E \bar{q}_{T \wedge \tau_A} |\varrho_{T \wedge \tau_A} \cdot (X'_{T \wedge \tau_A} - X_{T \wedge \tau_A})|] \\ & \le C \mathbb{E} [1_E \bar{q}_{T \wedge \tau_A} |X'_{T \wedge \tau_A} - X_{T \wedge \tau_A}|] \\ & \le C \mathbb{E} [\mathbb{P}(E |\mathcal{F}_{T \wedge \tau_A}) \bar{q}_T |(X - X')_T^*|], \end{aligned} $$

and we conclude as in the case $r = 0$. We deduce that

$$ \lim_{A \to +\infty} \mathbb{E} [p_{T \wedge \tau_A} \cdot (X'_{T \wedge \tau_A} - X_{T \wedge \tau_A})] = \mathbb{E} [p_T \cdot (X'_T - X_T)]. \quad (211) $$

**Step 2: right-hand side of (210).** Consider now the term on the right-hand side of (210). Obviously,

$$ \begin{aligned} & \int_0^{T \wedge \tau_A} \left| p_s^\top c_s (\psi_s' - \psi_s) + \text{Tr} \left( k_s (\sigma(s, \psi_s') - \sigma(s, \psi_s))^T \right) \right| \mathrm{d}s \\ & \le \int_0^T \left| q_s \varrho_s^\top c_s (\psi_s' - \psi_s) + q_s \text{Tr} \left( (h_s + \varrho_s \otimes \bar{Z}_s^*) (\sigma(s, \psi_s') - \sigma(s, \psi_s))^T \right) \right| \mathrm{d}s. \end{aligned} $$
For our purpose, it suffices to prove the integrability of the right-hand side. When $r=0$, the volatility term is independent of the control and reduces to $\sigma(t, \psi_t) = \nu_t$, see Assumption A2. Moreover, $(\varrho, h) \in S^2(\mathbb{F}, \mathbb{R}^n, \bar{\mathbb{Q}}) \times M^2(\mathbb{F}, \mathbb{R}^{n \times d}, \bar{\mathbb{Q}})$. Then, by Cauchy-Schwarz and Young inequalities,

$$ \mathbb{E} \left[ \int_0^T \bar{q}_s |\varrho_s c_s(\psi_s' - \psi_s)| ds \right] \le C \mathbb{E} \left[ \int_0^T \bar{q}_s (|\varrho_s|^2 + |\psi_s'|^2 + |\psi_s|^2) ds \right]^{1/2} < +\infty, $$

and

$$ \mathbb{E} \left[ \int_0^T \bar{q}_s \left| \operatorname{Tr} \left( (h_s + \varrho_s \otimes \bar{Z}_s^*) (\sigma(s, \psi_s') - \sigma(s, \psi_s)) \right)^T \right| ds \right] = 0. $$

since $\sigma(s, \psi_s) = \sigma(s, \psi_s') = \nu_s$ almost surely, for all $s \in [0, T]$. Combining the last two displays yields

$$ \mathbb{E} \left[ \int_0^T \left| p_s^\top c_s(\psi_s' - \psi_s) + \operatorname{Tr} \left( k_s (\sigma(s, \psi_s') - \sigma(s, \psi_s)) \right)^T \right| ds \right] < +\infty. \quad (212) $$

Now, when $r=1$, the volatility term $\sigma(s, \psi_s') - \sigma(s, \psi_s)$ is no longer null and depends linearly on the difference $\psi_s' - \psi_s$. Moreover, the process $\varrho$ is bounded, see Lemma 33. Therefore,

$$ \mathbb{E} \left[ \int_0^T \bar{q}_s |\varrho_s c_s(\psi_s' - \psi_s) ds \right] \le C \mathbb{E} \left[ \int_0^T \bar{q}_s (|\psi_s'| + |\psi_s|) ds \right] < +\infty, $$

and, by Cauchy-Schwarz and Young inequalities, we further have

$$ \begin{aligned} & \mathbb{E} \left[ \int_0^T \bar{q}_s \left| \operatorname{Tr} \left( (h_s + \varrho_s \otimes \bar{Z}_s^*) (\sigma(s, \psi_s') - \sigma(s, \psi_s)) \right)^T \right| ds \right] \\ & \le C \mathbb{E} \left[ \int_0^T \bar{q}_s (|h_s|^2 + |Z_s^*|^2 + |\psi_s'|^2 + |\psi_s|^2) ds \right] < +\infty, \end{aligned} $$

so that (212) is also true when $r=1$. Thus, by dominated convergence theorem, we have, for $r \in \{0, 1\}$,

$$ \begin{aligned} & \lim_{A \to +\infty} \mathbb{E} \left[ \int_0^{T \wedge \tau_A} \left( p_s^\top c_s(\psi_s' - \psi_s) + \operatorname{Tr} \left( k_s (\sigma(s, \psi_s') - \sigma(s, \psi_s)) \right)^T \right) ds \right] \\ & = \mathbb{E} \left[ \int_0^T \left( p_s^\top c_s(\psi_s' - \psi_s) + \operatorname{Tr} \left( k_s (\sigma(s, \psi_s') - \sigma(s, \psi_s)) \right)^T \right) ds \right]. \end{aligned} \quad (213) $$

**Step 3: conclusion.** By (211) and (213), we can pass to the limit in (210), letting $A$ tend to $+\infty$ therein. Informally, this means that we can substitute $T$ for $T \wedge \tau_A$ in (210), from which we get

$$ \mathbb{E} [p_T \cdot (X'_T - X_T)] = \mathbb{E} \left[ \int_0^T \left( p_s^\top c_s(\psi_s' - \psi_s) + \operatorname{Tr} \left( k_s (\sigma(s, \psi_s') - \sigma(s, \psi_s)) \right)^T \right) ds \right]. $$

Finally, using that $\nabla_\psi H(t, X_t, \psi_t, p_t, k_t, \bar{q}_t) = 0$ (see (195)), we have $\bar{q}_t \nabla_\psi \ell(t, \psi_t) + c_t^\top p_t + r \operatorname{Tr}(\sigma_t^\top k_t) = 0$. Combining the above display with (208) and (209), we get

$$ \mathcal{J}(\psi') - \mathcal{J}(\psi) \ge \mathbb{E} \left[ \int_0^T \bar{q}_s (\ell(s, \psi_s') - \ell(s, \psi_s) - \nabla_\psi \ell(s, \psi_s) \cdot (\psi_s' - \psi_s)) ds \right] \ge 0, $$

where, by strict convexity of $\ell$ and strict positivity of $\bar{q}$, the last inequality is strict whenever $\psi \ne \psi'$.

□
A Finite entropy and positive measures

The aim of this appendix is to clarify the representation of positive measures with finite entropy, a task that is nontrivial due to their limited integrability properties.

**Lemma 39.** Let $q_T$ be an $\mathcal{F}_T$-measurable random variable with values in $\mathbb{R}_+$. Under the conditions $\mathbb{E}[h(q_T)] < +\infty$, $\mathbb{P}(\{q_T > 0\}) = 1$ and $\mathbb{E}[q_T] = 1$, there exists a unique progressively measurable process $Z^*$ (with values in $\mathbb{R}^d$) such that

$$q_T = 1 + \int_0^T q_s Z_s^* \cdot dW_s. \qquad (214)$$

It satisfies $\frac{1}{2}\mathbb{E}[\int_0^T q_s|Z_s^*|^2ds] = \mathbb{E}[h(q_T)]$. Moreover, $q_T$ can be represented as

$$q_T = \mathcal{E}_T \left( \int_0^T Z_s^* \cdot dW_s \right). \qquad (215)$$

Lastly, the process $(q_t := (\mathcal{E}_t(\int_0^t Z_s^* \cdot dW_s))_{t \in [0,T]})$ is true a martingale. It satisfies $\mathbb{E}[q_T^*] < +\infty$, and is the unique solution to the SDE

$$q_t = 1 + \int_0^t q_s Z_s^* \cdot dW_s, \quad t \in [0, T], \qquad (216)$$

within the class of continuous positive-valued and $\mathbb{F}$-adapted processes.

As a consequence of the above lemma, the class $\mathcal{Q}$ is parameterized by the sole given data $q$ and $Y^*$. Indeed, it suffices to apply the lemma above to the process $(\exp(-\int_0^t Y_s^*ds)q_t)_{t \in [0,T]}$ to obtain the representation in equation (4), namely

$$q_T = 1 + \int_0^T q_s Y_s^* ds + \int_0^T q_s Z_s^* \cdot dW_s.$$

*Proof. Step 1: representation of $q_T$ in the form (214).* Let $q_t = \mathbb{E}[q_T|\mathcal{F}_t]$, for $t \in [0, T]$. It is a strictly positive (continuous) martingale. By $L \log L$-Doob's maximal inequality, we deduce that $\mathbb{E}[q_T^*] < +\infty$.

For any integer $m \ge 1$, we let $q_T^m := q_T \wedge m$. By martingale representation theorem, we can write

$$q_T^m = \mathbb{E}[q_T^m] + \int_0^T Z_s^m \cdot dW_s.$$

Letting $q_T^m = \mathbb{E}[q_T^m|\mathcal{F}_t]$, for $t \in [0, T]$, we invoke $L \log L$-Doob's maximal inequality again to deduce that there exists a universal constant $C > 0$ such that

$$\mathbb{E} \left[ \sup_{t \in [0,T]} |q_t^m - q_t| \right] \le C \mathbb{E} \left[ \max \{0, |q_T^m - q_T| \ln (|q_T^m - q_T|) \} \right].$$

Obviously (since $q_T^m = q_T$ if $q_T \le m$),

$$|q_T^m - q_T| \ln (|q_T^m - q_T|) \le q_T |\ln(2q_T)|.$$

By dominated convergence theorem, we deduce that

$$0 \le \limsup_{m \to \infty} \mathbb{E} \left[ \sup_{t \in [0,T]} |q_t^m - q_t| \right] \le C \lim_{m \to \infty} \mathbb{E} \left[ \max \{0, |q_T^m - q_T| \ln (|q_T^m - q_T|) \} \right] = 0.$$
In particular,

$$ \lim_{m \to \infty} \sup_{k \in \mathbb{N}} \mathbb{E} \left[ \sup_{t \in [0, T]} |q_t^{m+k} - q_t^m| \right] = 0. $$

By Bürkolder-Davies-Gundy inequality, we further obtain that

$$ \lim_{m \to \infty} \sup_{k \in \mathbb{N}} \mathbb{E} \left[ \left( \int_0^T |Z_t^m - Z_t^{m+k}|^2 dt \right)^{1/2} \right] = 0. $$

By completeness of $L^1(\Omega \times [0, T], \mathcal{R}, \mathbb{P} \otimes \text{Leb}_{[0,T]})$ (where $\mathcal{R}$ is the progressive $\sigma$-field), we deduce that there exists a process $Z$ satisfying

$$ \mathbb{E} \left[ \left( \int_{0}^{T} |Z_{t}|^{2} dt \right)^{1/2} \right] < +\infty, $$

and

$$ q_T = 1 + \int_0^T Z_s \cdot dW_s. $$

Since $(q_t)_{t \in [0,T]}$ is continuous and strictly positive, we can let $Z_t^* = Z_t/q_t$. We then notice, from Itô's formula, that the process

$$ \left( h(q_t) - \frac{1}{2} \int_0^t q_s |Z_s^*|^2 ds \right)_{t \in [0,T]} $$

is a local martingale. In particular, one can find a sequence of stopping times $(\sigma_m)_{m \ge 1}$, converging to $T$ as $m$ tends to $+\infty$, such that for all $m \ge 1$,

$$ \mathbb{E}[h(q_{T \wedge \sigma_m})] = \frac{1}{2} \mathbb{E} \left[ \int_{0}^{T \wedge \sigma_m} q_s |Z_s^*|^2 ds \right]. \quad (217) $$

We now prove that

$$ \sup_{m \ge 1} \mathbb{E}[h(q_{T \wedge \sigma_m})] < +\infty. $$

Since

$$ q_{T \wedge \sigma_m} = \mathbb{E}[q_T | \mathcal{F}_{T \wedge \sigma_m}]. $$

and because $h$ is convex, we deduce that

$$ \mathbb{E}[h(q_{T \wedge \sigma_m})] \le \mathbb{E}[h(q_T)], $$

from which we get, by applying Fatou's lemma to the right-hand side on (217), that

$$ \frac{1}{2} \mathbb{E} \left[ \int_{0}^{T} q_s |Z_s^*|^2 ds \right] \le \mathbb{E}[h(q_T)]. $$

Conversely, by observing that the function $h$ is lower bounded and then by applying Fatou's lemma to the left-hand side on (217), we get

$$ \mathbb{E}[h(q_T)] \le \frac{1}{2} \mathbb{E} \left[ \int_0^T q_s |Z_s^*|^2 ds \right]. $$

By the last two identities, we deduce that the two terms on the above inequality are equal.
Step 2: representation of $q_T$ in the form (215). Recalling that $(q_t)_{t \in [0,T]}$, as defined in the first step, is continuous and strictly positive, we deduce that, a.s., it is lower bounded by a positive constant. This proves that, a.s.,

$$ \int_{0}^{T} |Z_{s}^{*}|^{2} ds < +\infty, $$

which makes it possible to define $(\mathcal{E}_t(\int_0^t Z_s^* \cdot dW_s))_{t \in [0,T]}$.

Moreover, one can compute the logarithm of the process $q$. By Itô's formula, we get

$$ \ln(q_T) = -\frac{1}{2} \int_0^T |Z_s^*|^2 ds + \int_0^T Z_s^* \cdot dW_s, $$

which says that

$$ q_T = \mathcal{E}_T \left( \int_0^t Z_s^* \cdot dW_s \right). $$

This gives the expected representation of $q_T$, and more generally of the process $(q_t = \mathbb{E}[q_T | \mathcal{F}_t])_{t \in [0,T]}$.

Uniqueness to (216) can be easily checked by computing $(q'_t/q_t)_{t \in [0,T]}$ and then by proving that this ratio is constant (equal to 1), for any other solution $q' = (q'_t)_{t \in [0,T]}$.

□

## B A priori estimates on a linear SDE

Let $q \in \mathcal{Q}$ and $\psi \in \mathcal{A}$. In this section we provide technical results for controlled linear SDEs, including $S^1(\mathbb{F}, \mathcal{Q})$ and $S^2(\mathbb{F}, \mathcal{Q})$ regularity of the solutions, for $\mathcal{Q} = q\mathbb{P}$. With the same notations as in A1 and A2, we consider the controlled equation (7), namely

$$ dX_t = (a_t + b_t X_t + c_t \psi_t) dt + (\nu_t + r \sigma_t(\psi_t)) dW_t, \quad X_0 = \eta. \quad (218) $$

Following (A5), we denote by $\Gamma = (\Gamma_t)_{t \in [0,T]}$ the resolvent associated with the linear part of the equation, namely the solution (with values in the space of $n \times n$ matrices) of

$$ \frac{d}{dt}\Gamma_t = b_t\Gamma_t, \quad t \in [0, T], \quad \Gamma_0 = I_n, $$

with $I_n$ standing for the $n \times n$ identity matrix. It is well-known that $\Gamma_t$ is invertible for any $t \in [0, T]$ and that the solution to (218) can be represented as

$$ X_t = \Gamma_t \left( \eta + \int_0^t \Gamma_s^{-1} (a_s + c_s \psi_s) ds + \int_0^t \Gamma_s^{-1} (\nu_s + r \sigma_s(\psi_s)) dW_s \right), \quad t \in [0, T]. $$

Recalling from assumptions A1 and A2 that $\|\eta\|_{L^\infty(\mathcal{F}_0, \mathbb{R}^n)} + \|a\|_{L^\infty(\mathbb{F}, \mathbb{R}^n)} + \|b\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} + \|c\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} + \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} + \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} < +\infty$, we deduce that there exists a constant $C > 0$, independent of $\psi$, such that

$$ X_T^* \le C \left( 1 + \int_0^T |\psi_s| ds \right) + M_T^*, \quad M_t := \Gamma_t \int_0^t \Gamma_s^{-1}(\nu_s + r\sigma_s(\psi_s)) dW_s. \quad (219) $$

**Lemma 40.** The unique solution $X$ to (218) belongs to $S^{2-r}(\mathbb{F}, \mathcal{Q}, \mathbb{R}^n)$, where $r \in \{0, 1\}$ is as in A2, and there exists a constant $C$, independent of $q$ and $\psi$, such that

$$ \mathbb{E}[q_T | X_T^*|^{2-r}] \le C (1 + S(q) + S^*(\psi)). $$
Proof. We distinguish between the two cases $r = 0$ and $r = 1$.

Step 1: case $r = 0$. We have that

$$|X_T^*|^2 \le 2C \left( 1 + \int_0^T |\psi_s|^2 ds \right) + 2 \sup_{t \in [0,T]} \left| \Gamma_t \int_0^t \Gamma_s^{-1} \nu_s dW_s \right|^2.$$

Using that $\|\nu\|_{L^\infty(\mathbb{F},\mathbb{R}^{n \times d})} + \|\Gamma^{-1}\|_{L^\infty(\mathbb{F},\mathbb{R}^{n \times n})} < +\infty$ and [91, Theorem IV.37.8], we know that $M_T^*$ (whose square appears on the right-hand side) has sub-Gaussian tails, i.e., $M_T^*$ belongs to $L_{\exp(\mathcal{F}_T, \mathbb{R})}^{2,\vartheta}$ for a certain $\vartheta > 0$. Therefore, by the duality inequalities (13) and (14), there exists a constant $C > 0$ which might increase from line to line such that

$$\begin{aligned} \mathbb{E} [q_T |X_T^*|^2] &\le C \left( 1 + \mathbb{E} \left[ q_T \int_0^T |\psi_s|^2 ds \right] + \mathbb{E}[h(q_T)] + \mathbb{E} \left[ \exp(\vartheta (M_T^*)^2) \right] \right) \\ &\le C (1 + \mathcal{S}(q) + \mathcal{S}^*(\psi)) < +\infty. \end{aligned}$$

Step 2: case $r = 1$. In this situation we have

$$|X_T^*| \le C \left( 1 + \int_0^T |\psi_s| ds \right) + |M_T^*|.$$

Introducing $(\tilde{W}_t = W_t - \int_0^t Z_s^* ds)_{t \in [0,T]}$ and applying Girsanov's theorem (see Lemma 39), we have

$$\begin{aligned} \mathbb{E} [q_T |X_T^*|] &\le C \mathbb{E} \left[ q_T \left( 1 + \int_0^T |\psi_s| ds + |M_T^*| \right) \right] \\ &\le C \mathbb{E} \left[ q_T \left( 1 + \int_0^T |\psi_s| ds + \left| \sup_{t \in [0,T]} \Gamma_t \int_0^t \Gamma_s^{-1} (\nu_s + \sigma_s(\psi_s)) d\tilde{W}_s \right| \right) \right] \\ &\quad + C \mathbb{E} \left[ q_T \left| \sup_{t \in [0,T]} \Gamma_t \int_0^t \Gamma_s^{-1} (\nu_s + \sigma_s(\psi_s)) Z_s^* ds \right| \right]. \end{aligned} \quad (220)$$

On the one hand, by Itô's isometry (under the measure $\tilde{\mathbb{Q}} := \mathcal{E}_T(\int_0^T Z_s^* \cdot dW_s)\mathbb{P}$) and the assumption A2 on $\sigma$, we have that

$$\mathbb{E} \left[ q_T \left| \sup_{t \in [0,T]} \Gamma_t \int_0^t \Gamma_s^{-1} \sigma_s(\psi_s) d\tilde{W}_s \right| \right] \le C \mathbb{E}^{\tilde{\mathbb{Q}}} \left[ \left( \int_0^T [1 + |\psi_s|^2] ds \right)^{1/2} \right]. \quad (221)$$

On the other hand, by Fenchel-Young inequality, we also have that

$$\mathbb{E} \left[ q_T \left| \sup_{t \in [0,T]} \Gamma_t \int_0^t \Gamma_s^{-1} \sigma_s(\psi_s) Z_s^* ds \right| \right] \le C \mathbb{E} \left[ q_T \int_0^T [1 + |\psi_s|^2 + |Z_s^*|^2] ds \right]. \quad (222)$$

Combining (220) with (221) and (222), we obtain that

$$\begin{aligned} \mathbb{E} [q_T |X_T^*|] &\le C \mathbb{E} \left[ q_T \left( 1 + \int_0^T |\psi_s|^2 ds + \int_0^T |Z_s^*|^2 ds \right) \right] \\ &\le C \left( 1 + \mathbb{E} \left[ q_T \int_0^T |\psi_s|^2 ds \right] + \mathbb{E} [h(q_T)] \right) \\ &\le C (1 + \mathcal{S}(q) + \mathcal{S}^*(\psi)) < +\infty, \end{aligned}$$

where the last two lines follow by duality inequality (14) between $\mathcal{S}$ and $\mathcal{S}^*$, concluding the proof. $\square$
In fact, the proof of Lemma 40 can be easily re-examined to get the following variant:

**Lemma 41.** Let $r \in \{0, 1\}$ be as in A2 and $X$ be the unique solution to (218). Then, for any $\varepsilon \in (0, 1)$, there exist two constants $C_\varepsilon > 0$ and $c_\varepsilon > 0$, independent of $q$ and $\psi$, such that

$$ \mathbb{E}[q_T |X_T^*|] \le C_\varepsilon + c_\varepsilon S(q) + \varepsilon \mathbb{E} \left[ q_T \int_0^T |\psi_s|^2 ds \right], $$

where the second constant is explicitly given by

$$ c_\varepsilon = 2\beta e^{\alpha T} \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \left( \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} + \frac{3}{\varepsilon} e^{\alpha T} \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d \times n})} \right). $$

*Proof.* It suffices to adapt the computations developed in the second step of the proof of Lemma 40 (whether $r$ is equal to 0 or 1). Throughout the proof, the value of $\varepsilon \in (0, 1)$ is fixed. In the following, we shall use repeatedly that, for any $s \in [0, T]$, $\mathbb{E}[q_T] \le \exp(\alpha T)\mathbb{E}[q_s]$ and

$$ \mathbb{E} \left[ q_T \int_0^T |\psi_s|^2 ds \right] = \mathbb{E} \left[ \int_0^T \mathbb{E}[q_T | \mathcal{F}_s] |\psi_s|^2 ds \right] \le e^{\alpha T} \mathbb{E} \left[ \int_0^T q_s |\psi_s|^2 ds \right]. $$

The first term on the right-hand side of (220) can be easily bounded, by means of Young's inequality. We obtain

$$ \begin{aligned} \mathbb{E} \left[ q_T \left( 1 + \int_0^T |\psi_s| ds \right) \right] &\le \mathbb{E} \left[ q_T \left( 1 + C_\varepsilon + \frac{\varepsilon}{3} \int_0^T |\psi_s|^2 ds \right) \right] \\ &\le C_\varepsilon + \frac{\varepsilon}{3} \mathbb{E} \left[ \int_0^T q_s |\psi_s|^2 ds \right]. \end{aligned} $$

Similarly, by Jensen's and Young's inequalities, (221) yields

$$ \begin{aligned} & \mathbb{E} \left[ q_T \left| \sup_{t \in [0,T]} \Gamma_t \int_0^t \Gamma_s^{-1} \sigma_s(\psi_s) d\tilde{W}_s \right| \right] \\ & \le \mathbb{E} \left[ q_T \left| \int_0^T \Gamma_t \Gamma_s^{-1} \sigma_s(\psi_s) ds \right|^2 \right]^{1/2} \\ & \le T \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} \mathbb{E}[q_T]^{1/2} \\ & \quad + r \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d \times n})} \mathbb{E} \left[ q_T \int_0^T |\psi_s|^2 ds \right]^{1/2} \\ & \le C_\varepsilon + \frac{\varepsilon}{3} \mathbb{E} \left[ \int_0^T q_s |\psi_s|^2 ds \right]. \end{aligned} $$
And, (222) can be rewritten as

$$
\begin{aligned}
& \mathbb{E} \left[ q_T \left| \sup_{t \in [0,T]} \Gamma_t \int_0^t \Gamma_s^{-1} \sigma_s(\psi_s) Z_s^* ds \right| \right] \\
& \le \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} \mathbb{E} \left[ q_T \int_0^T |Z_s^*| ds \right] \\
& \quad + \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d \times n})} \mathbb{E} \left[ q_T \int_0^T |\psi_s| |Z_s^*| ds \right] \\
& \le e^{\alpha T} \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})} \mathbb{E} \left[ \int_0^T q_s (1 + |Z_s^*|^2) ds \right] \\
& \quad + \frac{3}{\varepsilon} e^{2\alpha T} \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})} \|\sigma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d \times n})} \mathbb{E} \left[ \int_0^T q_s |Z_s^*|^2 ds \right] \\
& \quad + \frac{\varepsilon}{3} \mathbb{E} \left[ \int_0^T q_s |\psi_s|^2 ds \right] \\
& \le C + c_\varepsilon S(q) + \frac{\varepsilon}{3} \mathbb{E} \left[ \int_0^T q_s |\psi_s|^2 ds \right],
\end{aligned}
$$

for some fixed constant $C > 0$. Combining (219) and the three last displays, we obtain the desired inequality. □

**Lemma 42.** Let $\psi = 0$ and $X \in S^2(\mathbb{F}, \mathbb{R}^n)$ be the associated solution to the state equation (218). Then $X_T^* \in S_{\exp}^{2-r, L\theta}(\mathcal{F}_T, \mathbb{R}^n)$ when $r=1$ in A2. The result holds for $r=0$ under the condition $4\vartheta L \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})}^2 \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})}^2 \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})}^2 T < 1$.

*Proof.* Inequality (219) (with $\psi = 0$) gives

$$X_T^* \le C + M_T^*, \quad \text{where} \quad M_t = \Gamma_t \int_0^t \Gamma_s^{-1} \nu_s dW_s.$$

Raising to the power $2-r$ and multiplying by $\vartheta L$, and then taking the exponential and the expectation on both sides, we obtain

$$\mathbb{E} [\exp(\vartheta L |X_T^*|^{2-r})] \le C \mathbb{E} [\exp(2\vartheta L |M_T^*|^{2-r})],$$

for a constant $C > 0$, which is allowed to increase. We recall from [91, Theorem IV.37.8] that, for

$$\kappa := \|\Gamma\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})}^2 \|\Gamma^{-1}\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times n})}^2 \|\nu\|_{L^\infty(\mathbb{F}, \mathbb{R}^{n \times d})}^2 T,$$

it holds $\mathbb{P}(\{M_T^* \ge y\}) \le \exp(-y^2/2\kappa)$ for all $y > 0$, from which we deduce that $\mathbb{E}[\exp(2\vartheta L(M_T^*)^{2-r})] < +\infty$ when $r=1$. When $r=0$,

$$\mathbb{E} [\exp (2\vartheta L |M_T^*|^{2-r})] \le 1 + 2\vartheta L \int_{\mathbb{R}} e^{y^2(2\vartheta L - \frac{1}{2\kappa})} dy,$$

which is finite whenever $\vartheta L\kappa < 1/4$, hence concluding the proof. □
C Uniform integrability and convergence results

This section focuses on weak convergence of random variables in the space L log L(F), when tested against random variables having a finite exponential moment, and vice versa.

**Lemma 43.** Let $(q^k)_{k \in \mathbb{N}}$ be a $\mathcal{Q}$-valued sequence, uniformly bounded in $L \log L(\mathbb{F})$, $(y^k)_{k \in \mathbb{N}}$ be an $L^\infty(\mathbb{F})$-valued sequence, uniformly bounded in $L^\infty(\mathbb{F})$, and $(\xi^k, \ell^k)_{k \in \mathbb{N}}$ be a collection of random variables in $L^1(\mathcal{F}_T) \times L^1(\mathbb{F})$ satisfying for all $\vartheta > 0$,

$$ \sup_{k \in \mathbb{N}} \mathbb{E} \left[ \exp \left( \vartheta |\xi_T^k| + \int_0^T \exp \left( \vartheta |\ell_s^k| \right) ds \right) \right] < +\infty. \quad (223) $$

Then, the sequences of random variables

$$ (q_T^k |y_T^k| |\xi^k|)_{k \in \mathbb{N}}, \quad (q_s^k |y_s^k| |\ell_s^k|)_{k \in \mathbb{N}}, $$

are respectively uniformly integrable on $\Omega$ and $\Omega \times [0, T]$, equipped with $\mathbb{P}$ and $\mathbb{P} \otimes \text{Leb}_{[0,T]}$, in the sense that

$$ \sup_{k \in \mathbb{N}} \mathbb{E} [q_T^k |y_T^k| |\xi^k|] < +\infty, \quad \sup_{k \in \mathbb{N}} \mathbb{E} \left[ \int_0^T q_s^k |y_s^k| |\ell_s^k| ds \right] < +\infty, \quad (224) $$

and for every $\epsilon > 0$, there exists $\delta > 0$ such that

$$ \forall A \in \mathcal{F}_T, \quad \mathbb{P}(A) \le \delta \Rightarrow \mathbb{E} \left[ q_T^k |y_T^k| |\xi^k| 1_A \right] \le \varepsilon, \quad (225) $$

$$ \forall A \in \mathcal{B}([0, T]) \otimes \mathcal{F}_T, \quad (\mathbb{P} \otimes \text{Leb}_{[0,T]})(A) \le \delta \Rightarrow \mathbb{E} \left[ \int_0^T q_s^k |y_s^k| |\ell_s^k| 1_A ds \right] \le \varepsilon. $$

*Proof.* Obviously, we can assume without any loss of generality that the processes $(y^k)_{k \in \mathbb{N}}$ are all equal to 1. Moreover, we just make the proof for the sequence $(q_s^k |y_s^k| |\ell_s^k|)_{k \in \mathbb{N}}$, as the proof for $(q_T^k |y_T^k| |\xi^k|)_{k \in \mathbb{N}}$ is analogous.

**Step 1:** display (224) holds. The proof of (224) follows from the duality inequality (13). If $\vartheta \ge 1$ in the latter display, the term $\ln(\vartheta)x$ therein is positive, which leaves us with

$$ x^*x \le \frac{1}{\vartheta}h(x) + \exp(\vartheta x^*), \quad (226) $$

for every $\vartheta \ge 1$ and for all $x, x^* > 0$. We now apply this inequality with $x = q_T^k(\omega)$ and $x^* = \int_0^T |\ell_s^k(\omega)| ds$ for any $\omega \in \Omega$. Choosing $\vartheta = 1$, we get

$$ \mathbb{E} \left[ q_T^k \int_0^T |\ell_s^k| ds \right] \le \mathbb{E} \left[ h \left( q_T^k \right) \right] + \mathbb{E} \left[ \exp \left( \int_0^T |\ell_s^k| ds \right) \right]. $$

By assumption (see (223)), the right-hand side is uniformly bounded with respect to $k \in \mathbb{N}$. In order to derive (224), it suffices to recall, from the definition of $\mathcal{Q}$, that there exists a constant $C > 0$ such that, for any $k \in \mathbb{N}$, $q_s^k \le C \mathbb{E}[q_T^k |\mathcal{F}_s]$ (with probability 1 under $\mathbb{P}$).

**Step 2:** display (225) holds. Fix $\epsilon > 0$, and then choose $\vartheta \ge 1$ large enough so that

$$ \frac{1}{\vartheta} \sup_{k \in \mathbb{N}} \mathbb{E} \left[ h \left( q_T^k \right) \right] \le \frac{\epsilon}{2CT}, $$
with $C$ as in the first step. By (226) (with $x = |q_T^k|$ and $x^* = |\ell_t^k|$), we deduce that, for $A \in \mathcal{B}([0,T]) \otimes \mathcal{F}_T$,

$$
\begin{align*}
& \mathbb{E} \left[ \int_0^T q_s^k |\ell_s^k| \mathbf{1}_A \, \mathrm{d}s \right] \\
&= \int_0^T \mathbb{E} \left[ q_s^k |\ell_s^k| \mathbf{1}_A(s, \cdot) \right] \, \mathrm{d}s \\
&\le \frac{C}{\vartheta} \int_0^T \mathbb{E} \left[ h(q_T^k) \mathbb{E} \left[ \mathbf{1}_A(s, \cdot) |\mathcal{F}_s\right] \right] \, \mathrm{d}s + \int_0^T \mathbb{E} \left[ \exp \left( \vartheta |\ell_s^k| \right) \mathbb{E} \left[ \mathbf{1}_A(s, \cdot) |\mathcal{F}_s\right] \right] \, \mathrm{d}s \\
&\le \frac{\epsilon}{2} + \int_0^T \mathbb{E} \left[ \exp \left( \vartheta |\ell_s^k| \right) \mathbb{E} \left[ \mathbf{1}_A(s, \cdot) |\mathcal{F}_s\right] \right] \, \mathrm{d}s.
\end{align*}
$$

By Cauchy-Schwarz' and then Jensen's inequalities,

$$
\begin{align*}
& \mathbb{E} \left[ \int_0^T q_s^k |\ell_s^k| \mathbf{1}_A \, \mathrm{d}s \right] \\
&\le \frac{\epsilon}{2} + \left[ (\mathrm{Leb}_{[0,T]} \otimes \mathbb{P}) (A) \right]^{1/2} \mathbb{E} \left[ \int_0^T \exp \left( 2\vartheta |\ell_s^k| \right) \, \mathrm{d}s \right]^{1/2} \\
&\le \frac{\epsilon}{2} + \sqrt{T} \left[ (\mathrm{Leb}_{[0,T]} \otimes \mathbb{P}) (A) \right]^{1/2} \mathbb{E} \left[ \exp \left( 2\vartheta \int_0^T |\ell_s^k| \, \mathrm{d}s \right) \right]^{1/2},
\end{align*}
$$

and we conclude by invoking (223). □

**Lemma 44.** Let $(q^k)_{k \in \mathbb{N}} \in \mathcal{Q}$ be a sequence, weakly converging to $q$ for the $\sigma(L^1, L^\infty)$ topology. For $(\xi, \ell) \in L^1(\mathcal{F}_T) \times L^1(\mathbb{F})$, assume further that the sequences $(q_T^k \xi)_{k \in \mathbb{N}}$ and $(q_s^k \ell_s)_{s \in \mathbb{N}}$ are uniformly integrable. Then,

$$
\lim_{k \to +\infty} \mathbb{E} \left[ q_T^k \xi + \int_0^T q_s^k \ell_s \, \mathrm{d}s \right] = \mathbb{E} \left[ q_T \xi + \int_0^T q_s \ell_s \, \mathrm{d}s \right]. \quad (227)
$$

*Proof.* We only show the convergence of $(q_s^k \ell_s)_{k \in \mathbb{N}}$, the proof for $(q_T^k \xi)_{k \in \mathbb{N}}$ being analogous. Let $(\tilde{\xi}, \tilde{\ell}) \in L^\infty(\mathcal{F}_T) \times L^\infty(\mathbb{F})$. By weak convergence, we have

$$
\lim_{k \to +\infty} \mathbb{E} \left[ \int_0^T q_s^k \tilde{\ell}_s \, \mathrm{d}s \right] = \mathbb{E} \left[ \int_0^T q_s \tilde{\ell}_s \, \mathrm{d}s \right]. \quad (228)
$$

Now, by uniform integrability (see Lemma 43), we also have

$$
\lim_{a \to +\infty} R^a = 0, \quad \text{with} \quad R^a := \sup_{k \in \mathbb{N}} \mathbb{E} \left[ \int_0^T q_s^k \ell_s \mathbf{1}_{\{|\ell_s| \ge a\}} \, \mathrm{d}s \right].
$$

With this notation, we have, for any $a > 0$,

$$
\limsup_{k \to +\infty} \mathbb{E} \left[ \int_{0}^{T} q_{s}^{k} \ell_{s} \, \mathrm{d}s \right] \le \limsup_{k \to +\infty} \mathbb{E} \left[ \int_{0}^{T} q_{s}^{k} \ell_{s} \mathbf{1}_{\{|\ell_{s}| \le a\}} \, \mathrm{d}s \right] + R^{a}. \quad (229)
$$

Here, we can use (228) in order to identify the superior limit in the right-hand side. And then, letting $a \to +\infty$, we deduce from (229):

$$
\limsup_{k \to +\infty} \mathbb{E} \left[ \int_0^T q_s^k \ell_s \, \mathrm{d}s \right] \le \mathbb{E} \left[ \int_0^T q_s \ell_s \, \mathrm{d}s \right]. \quad (230)
$$
Changing $\ell$ into $-\ell$, we also have

$$ \liminf_{k \to +\infty} \mathbb{E} \left[ \int_0^T q_s^k \ell_s ds \right] \geq \mathbb{E} \left[ \int_0^T q_s \ell_s ds \right]. \qquad (231) $$

Combining (230) and (231) yields that

$$ \lim_{k \to +\infty} \mathbb{E} \left[ \int_0^T q_s^k \ell_s ds \right] = \mathbb{E} \left[ \int_0^T q_s \ell_s ds \right], $$

which concludes the proof

□

# D Distance and differentiability on spaces of non-negative measures

## D.1 Generalized Wasserstein distance

We here establish the equivalence between the notion of continuity used in Section 4.1 and the notion of generalized p-Wasserstein distance introduced in [88] (see also [43]). We recall the following definition (using the notations introduced in Section 4.1, in particular the distances $d_p$ and $W_p$):

**Definition 45.** Let $p \ge 1$ and $\mu, \nu \in \mathcal{M}_p(\mathbb{R}^n)$. We call generalized p-Wasserstein distance between $\mu$ and $\nu$ the quantity

$$ W_{p,\text{ext}}(\mu, \nu) := \inf_{\tilde{\mu}, \tilde{\nu} \in \mathcal{M}_p(\mathbb{R}^n), \tilde{\mu}(\mathbb{R}^n) = \tilde{\nu}(\mathbb{R}^n)} \{W_p(\tilde{\mu}, \tilde{\nu}) + \|\mu - \tilde{\mu}\|_{\text{TV}} + \|\nu - \tilde{\nu}\|_{\text{TV}}\}. $$

The fact that $W_{p,\text{ext}}$ is a distance is established in [88, Proposition 1]. We state below the main result of this section. Following C, we recall that a subset $E \subset \mathcal{M}_p(\mathbb{R}^n)$ is said to be p-uniformly integrable if

$$ \sup_{\mu \in E} \int_{\mathbb{R}^n} (1 + |x|^p) d\mu(x) < +\infty, \quad \limsup_{a \to +\infty} \sup_{\mu \in E} \int_{\mathbb{R}^n} \mathbf{1}_{\{|x| \ge a\}} |x|^p d\mu(x) = 0. $$

We claim

**Proposition 46.** For a given $p \ge 1$, let $\psi : \mathcal{M}_p(\mathbb{R}^n) \to \mathbb{R}$. Then, $\psi$ is continuous with respect to $W_{p,\text{ext}}$ on any subset of p-uniform integrability, if and only the following two properties hold:

1. $\psi$ is continuous with respect to $d_p$, uniformly on subsets of p-uniform integrability;

2. on any isomass subset of $\mathcal{M}_p(\mathbb{R}^n)$, $\psi$ is continuous with respect to $W_p$.

Pay attention that continuity of $\psi$ is just restricted to subsets that are p-uniformly integrable: equivalently, we require that $\psi(\mu_m) \to \psi(\mu)$ for any sequence $(\mu_m)_{m \ge 1}$ that converges to $\mu$ in $\mathcal{M}_p(\mathbb{R}^n)$ with respect to $W_{p,\text{ext}}$ and that is p-uniformly integrable (recall that convergence in $\mathcal{M}_p(\mathbb{R}^n)$ with respect to $W_{p,\text{ext}}$ does not guarantee p-uniform integrability).
Proof. We first prove the implication (direct sense). We thus assume that $\psi$ is continuous with respect to $W_{p,ext}$ on any subset of $p$-uniform integrability. Obviously, any sequence that converges with respect to $W_p$ on an isomass subset of $M_p(\mathbb{R}^n)$ is $p$-uniformly integrable and converges with respect to $W_{p,ext}$. Therefore, $\psi$ is continuous with respect to $W_p$ on any isomass subset. This is item 2 in the statement. In order to prove item 1, consider a sequence $(\mu_m)_{m \ge 1}$ that converges to some limit $\mu$, in $M_p(\mathbb{R}^n)$ equipped with $d_p$. Clearly, it is $p$-uniformly integrable. Moreover, by [88, Theorem 3], $(\mu_m)_{m \ge 1}$ converges to $\mu$ with respect to $W_{p,ext}$. By continuity of $\psi$ with respect to $W_{p,ext}$, this shows that $\psi(\mu_m) \to \psi(\mu)$ as $m \to +\infty$. Continuity is uniform on any subset of $p$-uniform integrability. This follows from a standard compactness argument, as any subset of $p$-uniform integrability is relatively compact for $W_{p,ext}$.

We now establish the converse, assuming that $\psi$ satisfies items 1 and 2 in the statement. We thus consider a $p$-uniformly integrable sequence $(\mu_m)_{m \ge 1}$ that converges to some limit $\mu$, in $M_p(\mathbb{R}^n)$ equipped with $W_{p,ext}$.

If $\mu(\mathbb{R}^n) = 0$, then [88, Theorem 4] says that $(\mu^m)_{m \ge 1}$ converges to the null measure in $M_p(\mathbb{R}^n)$. By item 1 in the statement, we deduce that $\psi(\mu^m) \to \psi(\mu)$ as $m \to +\infty$, as expected.

We thus assume that $\mu(\mathbb{R}^n) > 0$. By definition of $W_{p,ext}$, we can find two sequences $(\tilde{\mu}^m)_{m \ge 1}$ and $(\tilde{\nu}^m)_{m \ge 1}$ such that, for each $m \ge 1$, $\tilde{\mu}^m$ is dominated by $\mu^m$, $\tilde{\nu}^m$ is dominated by $\mu$, and $\tilde{\mu}^m(\mathbb{R}^n) = \tilde{\nu}^m(\mathbb{R}^n)$, and

$$ \lim_{m \to \infty} [W_p(\tilde{\mu}^m, \tilde{\nu}^m) + \|\mu^m - \tilde{\mu}^m\|_{TV} + \|\mu - \tilde{\nu}^m\|_{TV}] = 0. $$

In particular, $\mu^m(\mathbb{R}^n) \to \mu(\mathbb{R}^n)$. Therefore, without any loss of generality, we can assume that $\mu^m(\mathbb{R}^n) > 0$ for any $m \ge 1$, which makes it possible to let

$$ \bar{\mu}^m(\cdot) = \frac{\mu(\mathbb{R}^n)}{\tilde{\mu}^m(\mathbb{R}^n)} \tilde{\mu}^m(\cdot), \quad \bar{\nu}^m(\cdot) = \frac{\mu(\mathbb{R}^n)}{\tilde{\mu}^m(\mathbb{R}^n)} \tilde{\nu}^m(\cdot). $$

It is easy to see that $W_p(\bar{\mu}^m, \bar{\nu}^m)$ tends to 0 as $m \to +\infty$. Moreover, the sequence $(\bar{\mu}^m)_{m \ge 1}$ is $p$-uniformly integrable because $(\mu^m)_{m \ge 1}$ is $p$-uniformly integrable, and each $\bar{\mu}^m$ is dominated by $C\mu^m$, for a constant $C$ independent of $m$. Obviously, $(\bar{\nu}^m)_{m \ge 1}$ is also $p$-uniformly integrable (because $\bar{\nu}^m$ is dominated by $C\mu$, for a possibly different value of $C$, but still independent of $m$). Since $\psi$ is $W_p$-continuous on the subset of measures with constant mass equal to $\mu(\mathbb{R}^n)$, it is in particular equicontinuous on any subset of $p$-uniformly integrable measures with constant mass equal to $\mu(\mathbb{R}^n)$. Therefore, by item 2,

$$ \lim_{m \to +\infty} |\psi(\bar{\mu}^m) - \psi(\bar{\nu}^m)| = 0. $$

It remains to see that $\|\mu^m - \bar{\mu}^m\|_{TV} \to 0$ as $m \to +\infty$. Since the two sequences $(\mu^m)_{m \ge 1}$ and $(\bar{\mu}^m)_{m \ge 1}$ are $p$-uniformly integrable, we deduce that $d_p(\mu^m, \bar{\mu}^m) \to 0$ as $m$ tends to $+\infty$. By item 1 (using the fact that continuous is uniform on subsets of uniform $p$-integrability), we deduce that

$$ \lim_{m \to +\infty} |\psi(\mu^m) - \psi(\bar{\mu}^m)| = 0. $$

Similarly,

$$ \lim_{m \to +\infty} |\psi(\mu) - \psi(\bar{\nu}^m)| = 0. $$

By combining the last three displays, we complete the proof. $\square$

<figure><img src="image_5.png" /></figure>
## D.2 Differentiability

The purpose of this subsection is to prove Lemma 14.

*Proof of Lemma 14.* To simplify, we prove the result assuming that A9 holds true without any restriction on the mass of $\mu$.

We first establish (55). Given $\mu \in \mathcal{M}_{2-r}(\mathbb{R}^n)$, we deduce from (42) (together with the continuity of the derivative in $d_{2-r}$) that, for any $t > 0$ and any $x \in \mathbb{R}^n$,

$$G(\mu + t\delta_x) = t \int_0^1 \frac{\delta G}{\delta \mu}(\mu + \theta t \delta_x, x) d\theta.$$

And then, for any integer $\ell \ge 1$, any $t_1, \dots, t_\ell > 0$ and any $x_1, \dots, x_\ell \in \mathbb{R}^n$,

$$ 
\begin{aligned}
& G\left(\mu + \sum_{i=1}^{\ell} t_i \delta_{x_i}\right) - G(\mu) \\
&= \sum_{i=1}^{\ell} \left[ G\left(\mu + \sum_{j=1}^{i} t_j \delta_{x_j}\right) - G\left(\mu + \sum_{j=1}^{i-1} t_j \delta_{x_j}\right) \right] \\
&= \sum_{i=1}^{\ell} t_i \int_0^1 \frac{\delta G}{\delta \mu}\left(\mu + \sum_{j=1}^{i-1} t_j \delta_{x_j} + \theta t_i \delta_{x_i}, x_i\right) d\theta \\
&= \int_0^1 \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}\left(\mu + \sum_{j=1}^{i-1} t_j \delta_{x_j} + \theta t_i \delta_{x_i}, y\right) d\left(\sum_{i=1}^{\ell} t_i \delta_{x_i}\right)(y) \right] d\theta.
\end{aligned}
$$

By continuity of $\delta G/\delta \mu$ in the measure argument, we deduce that

$$\frac{d}{d\varepsilon}|_{\varepsilon=0} G\left(\mu + \varepsilon \sum_{i=1}^{\ell} t_i \delta_{x_i}\right) = \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}(\mu, y) d\left(\sum_{i=1}^{\ell} t_i \delta_{x_i}\right)(y).$$

And then,

$$ 
\begin{aligned}
& G\left(\mu + \sum_{i=1}^{\ell} t_i \delta_{x_i}\right) - G(\mu) \\
&= \int_0^1 \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}\left(\mu + \theta \sum_{i=1}^{\ell} t_i \delta_{x_i}, y\right) d\left(\sum_{i=1}^{\ell} t_i \delta_{x_i}\right)(y) \right] d\theta.
\end{aligned}
$$

Now, we can approximate any given $\nu \in \mathcal{M}_{2-r}(\mathbb{R}^n)$ by measures $(\nu^\ell)_{\ell \ge 1}$ with the same mass, but with each being supported by a finite set; the approximation holds true with respect to $W_{2-r}$. For each $\ell \ge 1$, we have

$$G(\mu + \nu^\ell) - G(\mu) = \int_0^1 \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}(\mu + \theta \nu^\ell, y) d\nu^\ell(y) \right] d\theta. \quad (232)$$

We denote by $\pi^\ell$ an optimal coupling between $\nu$ and $\nu^\ell$. Using the third line in (44),
we have

$$
\begin{aligned}
& \left| \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu} (\mu + \theta \nu^\ell, y) \, d(\nu^\ell - \nu) (y) \right| \\
& \leq \int_{\mathbb{R}^n \times \mathbb{R}^n} \left| \frac{\delta G}{\delta \mu} (\mu + \theta \nu^\ell, y) - \frac{\delta G}{\delta \mu} (\mu + \theta \nu^\ell, z) \right| \, d\pi^\ell(y, z) \\
& \leq C \left( 1 + M_{2-r}(\mu + \nu^\ell) \right) \int_{\mathbb{R}^n \times \mathbb{R}^n} (1 + |y|^{1-r} + |z|^{1-r}) |y - z| \, d\pi^\ell(y, z).
\end{aligned}
$$

Observing that the moments $(M_{2-r}(\nu^\ell))_{\ell \ge 1}$ are uniformly bounded (because the convergence holds true with respect to $W_{2-r}$) and using Cauchy-Schwarz inequality to handle the last term in the right-hand side when $r=0$, we deduce that the left-hand side in the above display tends to 0 as $\ell$ tends to $+\infty$.

Using the continuity of $\delta G/\delta \mu$ in $\mu$ with respect to $W_{2-r}$ on isomass subsets, and the growth condition (44), we can pass to the limit in (232). We get

$$G(\mu + \nu) - G(\mu) = \int_{0}^{1} \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu} (\mu + \theta \nu, y) \, d\nu(y) \right] \, d\theta.$$

And then, for any $\varepsilon \in [0, 1]$,

$$
\begin{aligned}
G((1-\varepsilon)\mu + \varepsilon\nu) - G((1-\varepsilon)\mu) &= \varepsilon \int_0^1 \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}((1-\varepsilon)\mu + \varepsilon\theta\nu, y) \, d\nu(y) \right] \, d\theta \\
&= \varepsilon \int_0^1 \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}(\mu, y) \, d\nu(y) \right] \, d\theta + o(\varepsilon),
\end{aligned}
$$

where $o(\varepsilon)/\varepsilon \to 0$ as $\varepsilon$ tends to 0, with the last line following from (45). Performing a similar expansion for $\nu = \mu$, we obtain

$$\frac{d}{d\varepsilon}\Big|_{\varepsilon=0} G((1-\varepsilon)\mu + \varepsilon\nu) = \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}(\mu, y) \, d[\nu - \mu](y),$$

from which we deduce that

$$G(\nu) - G(\mu) = \int_{0}^{1} \left[ \int_{\mathbb{R}^n} \frac{\delta G}{\delta \mu}((1 - \theta)\mu + \theta\nu, y) \, d[\nu - \mu](y) \right] \, d\theta.$$

Choosing $\nu = (q'\mathbb{P})_X$ and $\mu = (q\mathbb{P}_X)$, this completes the proof of (55).

It remains to prove (56). Generally speaking, it is a consequence of [35, Proposition 5.44], applied on the space $(\Omega, \mathcal{F}, q\mathbb{P}/\mathbb{E}(q))$. Indeed, following Remark 13, we can apply [35, Proposition 5.44] to the function $X \in L^2(\Omega, \mathcal{F}, q\mathbb{P}/c) \mapsto G^{(c)}((q\mathbb{P}/c)_X)$, where $c := \mathbb{E}[q]$ and $G^{(c)}(\mu) = G(c\mu)$. We obtain (56) when $X$ satisfies $\mathbb{E}[q|X|^2] < +\infty$. When $X$ is just in $L^1(\Omega, \mathcal{F}, q\mathbb{P})$ (which is the case when $r=1$), we can approximate it, in $L^1(q\mathbb{P})$, by a sequence in $L^2(\Omega, \mathcal{F}, q\mathbb{P})$; we then apply (56) to the approximating subsequence and then pass to the limit using (44) and (46). $\square$

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