Sending $n \to \infty$, by Lemma D.9 we obtain

$$ \langle \mu, \tau(x) \rangle = -\langle u, D\tau(x)v \rangle. $$

Since $\mu \in DN_D^b(x, u)v$ was arbitrary, this completes the proof. □

Now, we will compute the contingent curvature for closed convex cones. For this purpose, we prepare two auxiliary results.

A.46. **Lemma.** [6, Lemma 4.2.5] Suppose that $\mathcal{D}$ is a closed convex cone. Then we have

$$ T_D^b(x) = \mathcal{D} + \operatorname{lin}\{x\} \quad \text{and} \quad \mathcal{N}_D^b(x) = \mathcal{D}^\circ \cap \{x\}^\perp \quad \forall x \in \mathcal{D}. $$

For the next auxiliary result recall the inclusions (A.9).

A.47. **Lemma.** Suppose that $\mathcal{D}$ is a closed convex cone. Let $x \in \mathcal{D}$ and $v \in \mathcal{N}_D^c(x)^\perp$ be arbitrary. Then there exists $\epsilon > 0$ such that $x+tv \in \mathcal{D}$ and $\mathcal{N}_D^b(x) \subset \mathcal{N}_D^b(x+tv)$ for all $t \in [0, \epsilon]$.

*Proof.* By Lemma A.46, for $x \in \mathcal{D}$ and $v \in \mathcal{N}_D^c(x)^\perp$ we have $v = \lambda x + y$ for some $\lambda \in \mathbb{R}$ and $y \in \mathcal{D}$. If $\lambda = 0$, then we have $x+tv = x+ty \in \mathcal{D}$ for all $t \ge 0$. Therefore, we may assume $\lambda \ne 0$. Setting $\epsilon := \frac{1}{2|\lambda|}$, we obtain $|t\lambda| \le \frac{1}{2}$, and hence

$$ x + tv = (1 + t\lambda)x + ty \in \mathcal{D} \quad \forall t \in [0, \epsilon]. $$

Now, let $u \in \mathcal{N}_D^b(x)$ be arbitrary. By Lemma A.17 we have $v \in \mathcal{N}_D^b(x)^\perp$, and hence $\langle u, v \rangle = 0$. Moreover, by Lemma A.46 we have $u \in \mathcal{D}^\circ$ and $\langle u, x \rangle = 0$, and it follows that $\langle u, x + tv \rangle = 0$. Thus, using Lemma A.46 again we arrive at $u \in \mathcal{N}_D^b(x+tv)$, completing the proof. □

A.48. **Proposition.** Suppose that $\mathcal{D}$ is a closed convex cone. Let $\tau: H \to H$ be a mapping of class $C^1$ such that (A.10) is fulfilled. Then for all $(x, u) \in \text{Graph}(\mathcal{N}_D^b)$ and $v \in \mathcal{N}_D^c(x)^\perp$ we have

$$ \text{Curv}_{\mathcal{D}}(x, u)(v, \tau(x)) = 0. $$

*Proof.* By Lemma A.47 we have $0 \in DN_D^b(x, u)v$. Thus, by Proposition A.45 the assertion follows. □

Now, we will provide the required results about $\varphi$-convex sets.

A.49. **Definition.** The closed set $\mathcal{D}$ is called $\varphi$-convex if there exists a continuous function $\varphi: \mathcal{D} \times \mathcal{D} \to \mathbb{R}_+$ such that for all $x, y \in \mathcal{D}$ and $u \in \mathcal{N}_D^\sigma(x)$ we have

$$ \langle u, y - x \rangle \le \varphi(x, y) \|u\| \|y - x\|^2. $$

The following characterization of $\varphi$-convexity will be useful.

A.50. **Lemma.** [17, Prop. 6.2] The closed set $\mathcal{D}$ is $\varphi$-convex if and only if there exists a continuous function $\psi: \mathcal{D} \to \mathbb{R}_+$ such that for all $x, y \in \mathcal{D}$ and $u \in \mathcal{N}_D^p(x)$ we have

$$ \langle u, y - x \rangle \le \psi(x) \|u\| \|y - x\|^2. $$

A.51. **Definition.** The set $\mathcal{D}$ is called locally $\varphi$-convex if for each $x \in \mathcal{D}$ there exists a closed neighborhood $C \subset H$ of $x$ such that $\mathcal{D} \cap C$ is $\varphi$-convex.

Let us recall the following well-known result about convex sets.

A.52. **Lemma.** Suppose that the closed subset $\mathcal{D}$ is convex, and let $x_0 \in H$ be arbitrary. Then for all $x \in \mathcal{D}$ the following statements are equivalent:

(i) $\|x_0 - x\| = d_{\mathcal{D}}(x_0)$.

(ii) $\langle x_0 - x, y - x \rangle \le 0$ for all $y \in \mathcal{D}$.
The next result shows that convex sets are indeed examples of $\varphi$-convex sets.

A.53. **Lemma.** Suppose that the closed subset $\mathcal{D}$ is convex. Then for all $x, y \in \mathcal{D}$ and $u \in \mathcal{N}_{\mathcal{D}}^p(x)$ we have $\langle u, y - x \rangle \le 0$.

*Proof.* Let $x \in \mathcal{D}$ and $u \in \mathcal{N}_{\mathcal{D}}^p(x)$ be arbitrary. By Lemma A.27 we have $t\|u\| = d_{\mathcal{D}}(x + tu)$ for some $t > 0$. Hence, setting $x_0 := x + tu$ we have $\|x_0 - x\| = d_{\mathcal{D}}(x_0)$. Thus, by Lemma A.52 we obtain

$$ \langle u, y - x \rangle = \frac{1}{t} \langle x_0 - x, y - x \rangle \le 0 \quad \text{for all } y \in \mathcal{D}, $$

completing the proof. □

A.54. **Proposition.** Consider the following four statements:

(i) $\mathcal{D}$ is convex.

(ii) For each $x \in \mathcal{D}$ there exists a closed neighborhood $C \subset H$ of $x$ such that $\mathcal{D} \cap C$ is convex.

(iii) $\mathcal{D}$ is locally $\varphi$-convex.

(iv) We have $\mathcal{N}_{\mathcal{D}}^p(x) = \mathcal{N}_{\mathcal{D}}^c(x)$ for each $x \in \mathcal{D}$.

Then we have the implications (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv).

*Proof.* (i) $\Rightarrow$ (ii): This implication is obvious.

(ii) $\Rightarrow$ (iii): In view of Lemma A.53, this follows from Lemma A.50 by choosing $\psi = 0$.

(iii) $\Rightarrow$ (iv): Taking into account Proposition A.32, this is a consequence of [17, Prop. 6.2]. □

A.55. **Proposition.** Suppose that $\mathcal{D}$ is locally $\varphi$-convex. Then for a mapping $a: H \to H$ the following statements are equivalent:

(i) We have $\langle u, a(x) \rangle \le 0$ for all $x \in \mathcal{D}$ and $u \in \mathcal{N}_{\mathcal{D}}^{1, \text{prox}}(x)$.

(ii) We have $\langle u, a(x) \rangle \le 0$ for all $x \in \mathcal{D}$ and $u \in \mathcal{N}_{\mathcal{D}}^c(x)$.

(iii) We have $a(x) \in T_{\mathcal{D}}^c(x)$ for all $x \in \mathcal{D}$.

*Proof.* (i) $\Rightarrow$ (ii): Let $x \in \mathcal{D}$ and $u \in \mathcal{N}_{\mathcal{D}}^c(x)$ be arbitrary. By Proposition A.54 we have $u \in \mathcal{N}_{\mathcal{D}}^p(x)$. Hence, there are $v \in \mathcal{N}_{\mathcal{D}}^{1, \text{prox}}(x)$ and $\lambda \ge 0$ such that $u = \lambda v$, and it follows that $\langle u, a(x) \rangle = \lambda \langle v, a(x) \rangle \le 0$.

(ii) $\Rightarrow$ (i): This implication is obvious.

(ii) $\Leftrightarrow$ (iii): This equivalence is a consequence of Lemma A.17. □

## APPENDIX B. SUBMANIFOLDS WITH BOUNDARY

In this appendix we provide the required results about submanifolds with boundary. For further details, we refer to [25, Sec. 3]. Let $H$ be a Hilbert space and let $m \in \mathbb{N}$ be a positive integer. Throughout this section, we will use the notation

$$ \mathbb{R}_+^m = \mathbb{R}_+ \times \mathbb{R}^{m-1} = \{y \in \mathbb{R}^m : y_1 \ge 0\}. $$

We consider the relative topology on $\mathbb{R}_+^m$. Let $V$ be an open subset in $\mathbb{R}_+^m$, i.e., there exists an open set $\tilde{V} \subset \mathbb{R}^m$ such that $\tilde{V} \cap \mathbb{R}_+^m = V$. We denote by

$$ \partial V = \{y \in V : y_1 = 0\} $$

the set of all boundary points of $V$. Let $k \in \mathbb{N} \cup \{\infty\}$ be arbitrary.

B.1. **Definition.** A map $\phi: V \subset \mathbb{R}_+^m \to H$ is called a $C^k$-map, if there is an open set $\tilde{V} \subset \mathbb{R}^m$ together with a $C^k$-map $\tilde{\phi}: \tilde{V} \to H$ such that $\tilde{V} \cap \mathbb{R}_+^m = V$ and $\tilde{\phi}|_V = \phi$.

For a $C^k$-map $\phi: V \subset \mathbb{R}_+^m \to H$ and $y \in V$ we define the derivative $D\phi(y) := D\tilde{\phi}(y)$. Note that this definition does not depend on the choice of $\tilde{\phi}$.
### B.2. Definition.
A nonempty subset $\mathcal{M} \subset H$ is called an $m$-dimensional $C^k$-submanifold with boundary of $H$ if for every $x \in \mathcal{M}$ there exist an open neighborhood $U \subset H$ of $x$, a subset $V \subset \mathbb{R}_+^m$ which is open with respect to the relative topology, and a mapping $\phi \in C^k(V, H)$ such that:

(1) The mapping $\phi: V \to U \cap \mathcal{M}$ is a homeomorphism.

(2) $D\phi(y) \in L(\mathbb{R}^m, H)$ is one-to-one for all $y \in V$.

The mapping $\phi$ is called a local parametrization of $\mathcal{M}$ around $x$.

In what follows, let $\mathcal{M}$ be an $m$-dimensional $C^k$-submanifold with boundary of $H$.

### B.3. Definition.
The boundary $\partial\mathcal{M}$ is defined as the set of all points $x \in \mathcal{M}$ such that $\phi^{-1}(x) \in \partial V$ for some local parametrization $\phi: V \to U \cap \mathcal{M}$ around $x$.

### B.4. Remark.
The boundary $\partial\mathcal{M}$ is a submanifold without boundary of dimension $m-1$, and parametrizations of $\partial\mathcal{M}$ are provided by restricting parametrizations $\phi: V \to U \cap \mathcal{M}$ to the boundary $\partial V$.

By [23, Lemma 6.1.1] and [25, Lemma 3.3] the following definitions of the tangent spaces $T_x\mathcal{M}$ and $(T_x\mathcal{M})_+$ do not depend on the choice of the parametrization.

### B.5. Definition.
Let $x \in \mathcal{M}$ be arbitrary, and let $\phi: V \subset \mathbb{R}_+^m \to U \cap \mathcal{M}$ be a local parametrization around $x$.

(1) The tangent space to $\mathcal{M}$ at $x$ is the subspace

$$T_x\mathcal{M} := D\phi(y)\mathbb{R}^m, \quad y = \phi^{-1}(x) \in V.$$

(2) For $x \in \partial\mathcal{M}$ we can distinguish a half space in $T_x\mathcal{M}$, namely the closed convex cone of all inward pointing directions in $\mathcal{M}$, given by

$$(T_x\mathcal{M})_+ := D\phi(y)\mathbb{R}_+^m, \quad y = \phi^{-1}(x) \in \partial V.$$

### B.6. Remark.
Let $x \in \mathcal{M}$ be arbitrary and let $\phi: V \subset \mathbb{R}_+^m \to U \cap \mathcal{M}$ be a local parametrization around $x$. Then we have

$$T_x\partial\mathcal{M} = D\phi(y)\partial\mathbb{R}_+^m, \quad y = \phi^{-1}(x) \in \partial V.$$

In particular, we see that

$$T_x\partial\mathcal{M} = (T_x\mathcal{M})_+ \cap -(T_x\mathcal{M})_+ \subset (T_x\mathcal{M})_+.$$

From now on, we assume that the submanifold $\mathcal{M}$ is closed as a subset of $H$.

### B.7. Lemma.
For each $x \in \mathcal{M}$ the following statements are true:

(1) If $x \in \mathcal{M} \setminus \partial\mathcal{M}$, then we have $T_x\mathcal{M} \subset T_{\mathcal{M}}^c(x)$.

(2) If $x \in \partial\mathcal{M}$, then we have $(T_x\mathcal{M})_+ \subset T_{\mathcal{M}}^c(x)$.

*Proof.* Let $v \in T_x\mathcal{M}$ be arbitrary. If $x \in \partial\mathcal{M}$, then we even assume that $v \in (T_x\mathcal{M})_+$. Let $\phi: V \to U \cap \mathcal{M}$ be a local parametrization around $x$. We set $y := \phi^{-1}(x) \in V$. Then $\phi$ extends to a mapping $\tilde{\phi}: \tilde{V} \to H$ of class $C^k$ such that $\tilde{V} \cap \mathbb{R}_+^m = V$, and we may assume that $\tilde{V} \subset \mathbb{R}^m$ is an open, convex neighborhood of $y$. We set $w := D\phi(y)^{-1}v \in \mathbb{R}^m$. If $x \in \partial\mathcal{M}$, then we have $y \in \partial V$ and $w \in \mathbb{R}_+^m$.

We wish to show that $v \in T_{\mathcal{M}}^c(x)$ by using Proposition A.4. For this purpose, let $(t_n)_{n \in \mathbb{N}} \subset (0, \infty)$ with $t_n \to 0^+$ be arbitrary, and let $(x_n)_{n \in \mathbb{N}} \subset \mathcal{M}$ with $x_n \to x$ be arbitrary. Since $U$ is an open neighborhood of $x$, we may assume that $x_n \in U \cap \mathcal{M}$ for each $n \in \mathbb{N}$. We define the sequence $(y_n)_{n \in \mathbb{N}} \subset V$ as $y_n := \phi^{-1}(x_n) \in V$. Then we have $y_n \to y$, and hence $y_n + t_n w \to y$. Recall that in case $x \in \partial\mathcal{M}$ we have $w \in \mathbb{R}_+^m$. Therefore, and since $V \subset \mathbb{R}_+^m$ is an open neighborhood of $y$ with respect to the relative topology, we may assume that $y_n + t_n w \in V$ for each $n \in \mathbb{N}$.
Since $\tilde{V}$ is open and convex, we may apply Taylor's theorem (see, e.g. [3, Thm. 2.4.15]), which provides the existence of a continuous mapping $R : \tilde{V} \times \tilde{V} \to L(\mathbb{R}^m, H)$ such that $R(z, z) = 0$ for all $z \in \tilde{V}$, and for each $n \in \mathbb{N}$ we have

$$ \begin{aligned} \phi(y_n + t_n w) &= \phi(y_n) + D\phi(y_n)t_n w + R(y_n, y_n + t_n w) \cdot t_n w \\ &= x_n + t_n (D\phi(y_n)w + R(y_n, y_n + t_n w)w). \end{aligned} $$

We define the sequence $(v_n)_{n \in \mathbb{N}} \subset H$ as

$$ v_n := D\phi(y_n)w + R(y_n, y_n + t_n w)w, \quad n \in \mathbb{N}. $$

Then we have

$$ \lim_{n \to \infty} v_n = D\phi(y)w = v, $$

and for each $n \in \mathbb{N}$ we have

$$ x_n + t_n v_n = \phi(y_n + t_n w) \in \mathcal{M}. $$

Consequently, by Proposition A.4 we deduce that $v \in T_{\mathcal{M}}^c(x)$. $\square$

<figure><img src="image_16.png" /></figure>

### B.8. Lemma. For each $x \in \mathcal{M}$ the following statements are true:

(1) If $x \in \mathcal{M} \setminus \partial \mathcal{M}$, then we have $T_{\mathcal{M}}^{\sigma}(x) \subset T_x \mathcal{M}$.

(2) If $x \in \partial \mathcal{M}$, then we have $T_{\mathcal{M}}^{\sigma}(x) \subset (T_x \mathcal{M})_+$.

*Proof.* Let $v \in T_x \mathcal{M}$ be arbitrary, and let $\phi : V \to U \cap \mathcal{M}$ be a local parametrization around $x$. We set $y := \phi^{-1}(x) \in V$. Then $\phi$ extends to a mapping $\phi : \tilde{V} \to H$ of class $C^k$ such that $\tilde{V} \cap \mathbb{R}_+^m = V$, and we may assume that $\tilde{V} \subset \mathbb{R}^m$ is an open, convex neighborhood of $y$. If $x \in \partial \mathcal{M}$, then we have $y \in \partial V$.

By Definition A.2 there are sequences $(t_n)_{n \in \mathbb{N}} \subset (0, \infty)$ with $t_n \to 0^+$ and $(v_n)_{n \in \mathbb{N}} \subset H$ with $v_n \xrightarrow{\sigma} v$ such that $x_n := x + t_n v_n \in \mathcal{M}$ for each $n \in \mathbb{N}$. The sequence $(v_n)_{n \in \mathbb{N}}$ is bounded, and hence $t_n v_n \to 0$, implying that $x_n \to x$. Thus, defining the sequence $(y_n)_{n \in \mathbb{N}} \subset V$ as $y_n := \phi^{-1}(x_n)$ for each $n \in \mathbb{N}$, we obtain $y_n \to y$. Now, we define the sequence $(w_n)_{n \in \mathbb{N}} \subset \mathbb{R}^m$ as

$$ w_n := \frac{y_n - y}{t_n}, \quad n \in \mathbb{N}. $$

If $x \in \partial \mathcal{M}$, then we have $y \in \partial V$, and hence $w_n \in \mathbb{R}_+^m$ for each $n \in \mathbb{N}$. Moreover, for each $n \in \mathbb{N}$ we have

$$ x + t_n v_n = x_n = \phi(y_n) = \phi(y + t_n w_n). $$

Since $\tilde{V}$ is open and convex, we may apply Taylor's theorem (see, e.g. [3, Thm. 2.4.15]), which provides the existence of a continuous mapping $R : \tilde{V} \times \tilde{V} \to L(\mathbb{R}^m, H)$ such that $R(z, z) = 0$ for all $z \in \tilde{V}$, and for each $n \in \mathbb{N}$ we have

$$ \begin{aligned} \phi(y + t_n w_n) &= \phi(y) + D\phi(y)(t_n w_n) + R(y, y + t_n w_n) \cdot t_n w_n \\ &= x + t_n (D\phi(y)w_n + R(y, y + t_n w_n)w_n). \end{aligned} $$

Now, it follows that

$$ v_n = \frac{\phi(y + t_n w_n) - x}{t_n} = D\phi(y)w_n + R(y, y + t_n w_n)w_n, \quad n \in \mathbb{N}. $$

We define $z_n := D\phi(y)w_n$ for each $n \in \mathbb{N}$. Then we have $z_n \in T_x \mathcal{M}$ for each $n \in \mathbb{N}$, and in case $x \in \partial \mathcal{M}$ we even have $z_n \in (T_x \mathcal{M})_+$ for each $n \in \mathbb{N}$. Furthermore, we have $t_n w_n = y_n - y \to 0$, and therefore we obtain

$$ v = \sigma_- \lim_{n \to \infty} v_n = \sigma_- \lim_{n \to \infty} z_n = \lim_{n \to \infty} z_n. $$

Since $T_x \mathcal{M}$ and $(T_x \mathcal{M})_+$ are closed subsets of $H$, we deduce that $v \in T_x \mathcal{M}$, and in case $x \in \partial \mathcal{M}$ we even have $v \in (T_x \mathcal{M})_+$. $\square$

<figure><img src="image_17.png" /></figure>
The following auxiliary result is a slight reformulation of [25, Lemma 3.7]. For a subset $A \subset H$ we use the notation

$$A^{-} := \{y \in H : \langle y, x \rangle \le 0 \text{ for all } x \in A\}.$$

**B.9. Lemma.** For each $x \in \partial\mathcal{M}$ there exists a unique vector $n_x \in (T_x\partial\mathcal{M})^{\perp}$ with $-n_x \in (T_x\mathcal{M})_+$ and $\|n_x\| = 1$ such that

$$(B.1) \qquad T_x \mathcal{M} = T_x \partial \mathcal{M} \oplus \operatorname{lin}\{n_x\}.$$

Moreover, for each $x \in \partial\mathcal{M}$ we have

$$(B.2) \qquad T_x \partial \mathcal{M} = T_x \mathcal{M} \cap \{n_x\}^{\perp},$$

$$(B.3) \qquad (T_x \mathcal{M})_+ = T_x \mathcal{M} \cap \{n_x\}^{-}.$$

**B.10. Definition.** For each $x \in \partial\mathcal{M}$ we call $n_x$ the outward pointing normal vector to $\partial\mathcal{M}$ at $x$.

**B.11. Remark.** In [25] the inward pointing normal vector $\eta_x := -n_x$ was considered.

The following result shows that the submanifold $\mathcal{M}$ is a so-called $\sigma$-regular set, which means that $N_{\mathcal{D}}^{\sigma}(x) = N_{\mathcal{D}}^{c}(x)$ for each $x \in \mathcal{D}$; cf. [17, Def. 4.3]. Note that the decomposition (B.7) below refers to the direct sum decomposition $H = (T_x\mathcal{M})^{\perp} \oplus T_x\mathcal{M}$ of the Hilbert space.

**B.12. Proposition.** For each $x \in \mathcal{M}$ the following statements are true:

(1) If $x \in \mathcal{M} \setminus \partial\mathcal{M}$, then we have

$$(B.4) \qquad T_{\mathcal{M}}^{c}(x) = T_{\mathcal{M}}^{\sigma}(x) = T_{x}\mathcal{M},$$

$$(B.5) \qquad N_{\mathcal{M}}^{\sigma}(x) = N_{\mathcal{M}}^{c}(x) = (T_{x}\mathcal{M})^{\perp}.$$

(2) If $x \in \partial\mathcal{M}$, then we have

$$(B.6) \qquad T_{\mathcal{M}}^{c}(x) = T_{\mathcal{M}}^{\sigma}(x) = (T_{x}\mathcal{M})_{+},$$

$$(B.7) \qquad N_{\mathcal{M}}^{\sigma}(x) = N_{\mathcal{M}}^{c}(x) = (T_x \mathcal{M})^{\perp} \oplus \operatorname{lin}^{+}\{n_x\}.$$

*Proof.* The identities (B.4) and (B.6) are consequences of Lemmas B.7 and B.8. Recalling that $N_{\mathcal{M}}^{\sigma}(x) = T_{\mathcal{M}}^{\sigma}(x)^{\circ}$ and $N_{\mathcal{M}}^{c}(x) = T_{\mathcal{M}}^{c}(x)^{\circ}$, we deduce that $T_{\mathcal{M}}^{c}(x) = T_{\mathcal{M}}^{\sigma}(x)$ for each $x \in \mathcal{M}$. This shows (B.5), because $T_x\mathcal{M}$ is a subspace of $H$. In order to show (B.7), let $x \in \partial\mathcal{M}$ be arbitrary, and recall that

$$N_{\mathcal{M}}^{c}(x) = (T_{x}\mathcal{M})_{+}^{\circ}.$$

Let $y \in (T_x\mathcal{M})^{\perp}$ and $\lambda \ge 0$ be arbitrary. Then for each $z \in (T_x\mathcal{M})_+$ we have $\langle y, z \rangle = 0$, and by (B.3) we have $\langle n_x, z \rangle \le 0$. Therefore, we obtain

$$\langle y + \lambda n_x, z \rangle = \langle y, z \rangle + \lambda \langle n_x, z \rangle \le 0,$$

showing that $y + \lambda n_x \in N_{\mathcal{M}}^{c}(x)$.

Conversely, let $y \in N_{\mathcal{M}}^{c}(x) = (T_x\mathcal{M})_+^{\circ}$ be arbitrary. We define $\lambda := \langle y, n_x \rangle$. Then we have $\lambda \ge 0$, because $-n_x \in (T_x\mathcal{M})_+$. Decomposing $y = (y - \lambda n_x) + \lambda n_x$, we wish to show that $y - \lambda n_x \in (T_x\mathcal{M})^{\perp}$. For this purpose, let $z \in T_x\mathcal{M}$ be arbitrary. By (B.1) we have $z = w + \mu n_x$ with $w \in T_x\partial\mathcal{M}$ and $\mu \in \mathbb{R}$. Since $y \in (T_x\mathcal{M})_+^{\circ} \subset (T_x\partial\mathcal{M})^{\circ} = (T_x\partial\mathcal{M})^{\perp}$, we have $\langle y, w \rangle = 0$, and since $n_x \in (T_x\partial\mathcal{M})^{\perp}$, we have $\langle n_x, w \rangle = 0$. Recalling that $\|n_x\| = 1$, we obtain

$$\begin{align*} \langle y - \lambda n_x, z \rangle &= \langle y - \lambda n_x, w + \mu n_x \rangle = \langle y, w \rangle + \mu \langle y, n_x \rangle - \lambda \langle n_x, w \rangle - \lambda \mu \\ &= \mu(\langle y, n_x \rangle - \lambda) = 0, \end{align*}$$

showing that $y - \lambda n_x \in (T_x\mathcal{M})^{\perp}$. $\square$

<figure><img src="image_18.png" /></figure>
Recalling that in [25] the inward pointing normal vector $\eta_x := -n_x$ was considered, the following result is a reformulation of [25, Lemma 3.9].

B.13. **Lemma.** Let $\phi: V \subset \mathbb{R}_+^m \to U \cap \mathcal{M}$ be a local parametrization. Then, for every $x \in U \cap \partial\mathcal{M}$ there exists a unique number $\kappa(x) < 0$ such that

$$ \langle n_x, D\phi(y)v \rangle = \kappa(x)\langle e_1, v \rangle \quad \text{for all } v \in \mathbb{R}^m, $$

where $y = \phi^{-1}(x)$.

B.14. **Proposition.** Suppose that the submanifold $\mathcal{M}$ is of class $C^2$. Then $\mathcal{M}$ is locally $\phi$-convex, and for each $x \in \mathcal{M}$ the following statements are true:

(1) If $x \in \mathcal{M} \setminus \partial\mathcal{M}$, then we have

$$ \mathcal{N}_{\mathcal{M}}^{p}(x) = \mathcal{N}_{\mathcal{M}}^{c}(x) = (T_{x}\mathcal{M})^{\perp}. $$

(2) If $x \in \partial\mathcal{M}$, then we have

$$ \mathcal{N}_{\mathcal{M}}^{p}(x) = \mathcal{N}_{\mathcal{M}}^{c}(x) = (T_{x}\mathcal{M})^{\perp} \oplus \text{lin}^{+}\{n_{x}\}. $$

*Proof.* Let $x_0 \in \mathcal{M}$ be arbitrary, and let $\phi: V \to U \cap \mathcal{M}$ be a local parametrization around $x_0$. We set $y_0 := \phi^{-1}(x_0) \in V$. Then $\phi$ extends to a mapping $\phi: \tilde{V} \to H$ of class $C^2$ such that $\tilde{V} \cap \mathbb{R}_+^m = V$, and we may assume that $\tilde{V} \subset \mathbb{R}^m$ is an open, convex neighborhood of $y_0$. Moreover, as a consequence of the inverse mapping theorem (see [23, Prop. 6.1.1]) we may assume that $\Phi := \phi^{-1} : \phi(\tilde{V}) \to \tilde{V}$ is Lipschitz continuous.

By Taylor's theorem (see, e.g. [3, Thm. 2.4.15]) there is a continuous mapping $R: \tilde{V} \times \tilde{V} \to L_s^2(\mathbb{R}^m, H)$ with $R(v, v) = 0$ for all $v \in \tilde{V}$ such that for all $u, v \in \tilde{V}$ we have

$$ \phi(v) - \phi(u) = D\phi(u) \cdot (v-u) + \frac{1}{2} D^2\phi(u) \cdot (v-u, v-u) \\ + R(u, v) \cdot (v-u, v-u). $$

Hence, for all $x, y \in U \cap \mathcal{M}$ we obtain

$$ \begin{aligned} y - x &= \phi(\Phi(y)) - \phi(\Phi(x)) \\ &= D\phi(\Phi(x)) \cdot (\Phi(y) - \Phi(x)) + \frac{1}{2} D^2\phi(\Phi(x))(\Phi(y) - \Phi(x), \Phi(y) - \Phi(x)) \\ &\quad + R(\Phi(x), \Phi(y)) \cdot (\Phi(y) - \Phi(x), \Phi(y) - \Phi(x)). \end{aligned} $$

Now, let $C \subset U$ be a closed neighborhood of $x_0$. Denoting by $L$ the Lipschitz constant of $\Phi$, we define the continuous mapping $\varphi: (C \cap \mathcal{M}) \times (C \cap \mathcal{M}) \to \mathbb{R}_+$ as

$$ \varphi(x, y) := L^2 \left\| \frac{1}{2} D^2 \phi(\Phi(x)) + R(\Phi(x), \Phi(y)) \right\|. $$

Let $x, y \in C \cap \mathcal{M}$ and $u \in \mathcal{N}_{C \cap \mathcal{M}}^{\sigma}(x)$ be arbitrary. Suppose first that $x \in \mathcal{M} \setminus \partial\mathcal{M}$. Taking into account Proposition A.32, by Proposition B.12 we have $u \in (T_x \mathcal{M})^{\perp}$, and hence

$$ \langle u, y - x \rangle \le \varphi(x, y) \|u\| \|y - x\|^2. $$

Now, consider the case $x \in \partial\mathcal{M}$. By Proposition B.12 we have $u \in (T_x \mathcal{M})^{\perp} \oplus \text{lin}^{+}\{n_x\}$. Thus we have $u = v + \lambda n_x$ for some $v \in (T_x \mathcal{M})^{\perp}$ and $\lambda \ge 0$. Set $z := \Phi(x)$ and $w := \Phi(y)$. Then we have $z \in \partial\mathbb{R}_+^m$, and hence $w - z \in \mathbb{R}_+^m$. By Proposition B.13 we obtain

$$ \langle n_x, D\phi(z)(w-z) \rangle = \kappa(x) \langle e_1, w-z \rangle \le 0. $$
Furthermore, since $n_x \in T_x M$ with $\|n_x\| = 1$, by orthogonality we have $\|u\|^2 = \|v\|^2 + \lambda^2$, and hence $\|v\| \le \|u\|$. Consequently, we arrive at

$$ \begin{aligned} \langle u, y - x \rangle &= \langle v, y - x \rangle + \lambda \langle n_x, y - x \rangle \\ &\le \varphi(x, y) \|v\| \|y - x\|^2 \\ &\le \varphi(x, y) \|u\| \|y - x\|^2. \end{aligned} $$

This proves that $M$ is locally $\varphi$-convex. Now, the remaining statements are a consequence of Proposition A.54 and Proposition B.12. □

The following example shows that for a $C^1$-submanifold $M$ the inclusion

$$ \mathcal{N}_{M}^{p}(x) \subset \mathcal{N}_{M}^{\sigma}(x) $$

may be strict.

B.15. **Example.** Consider the state space $H = \mathbb{R}^2$ and the submanifold

$$ \mathcal{M} = \{(x, |x|^{3/2}) : x \in \mathbb{R}\}. $$

Then $\mathcal{M}$ is a $C^1$-submanifold, which is not of class $C^2$. According to Proposition B.12 we have $\mathcal{N}_{\mathcal{M}}^{\sigma}(0) = \text{lin}\{e_2\}$. We will show that $\mathcal{N}_{\mathcal{M}}^{p}(0) = -\text{lin}^{+}\{e_2\}$. Indeed, each $y \in \mathcal{M}$ is of the form $y = (x, |x|^{3/2})$ for some $x \in \mathbb{R}$. Therefore, we have

$$ \frac{\langle e_2, y \rangle}{\|y\|^2} = \frac{|x|^{3/2}}{|x|^2 + |x|^3}. $$

Moreover, concerning the reciprocal we have

$$ \frac{|x|^2 + |x|^3}{|x|^{3/2}} = |x|^{1/2} + |x|^{3/2} \to 0 \quad \text{for } x \to 0. $$

Using Proposition A.28 we deduce that $-e_2 \in \mathcal{N}_{\mathcal{M}}^{p}(0)$, whereas $e_2 \notin \mathcal{N}_{\mathcal{M}}^{p}(0)$. This shows $\mathcal{N}_{\mathcal{M}}^{p}(0) = -\text{lin}^{+}\{e_2\}$.

## APPENDIX C. MARTINGALES IN BANACH SPACES

In this appendix we provide the required results about martingales in Banach spaces. Let $E$ be a Banach space, and let $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \in \mathbb{R}_+}, \mathbb{P})$ be a filtered probability space satisfying the usual conditions.

C.1. **Lemma.** Let $M$ be an $E$-valued adapted process such that $\mathbb{E}[\|M_t\|] < \infty$ for all $t \in \mathbb{R}_+$. Then the following statements are equivalent:

(i) $M$ is a martingale.

(ii) $x'(M)$ is a martingale for each $x' \in E'$.

*Proof.* An adapted process $N$ with values in some Banach space $F$ satisfying $\mathbb{E}[\|N_t\|] < \infty$ for all $t \in \mathbb{R}_+$ is a martingale if and only if $\mathbb{E}[N_t 1_A] = \mathbb{E}[N_s 1_A]$ for all $s \le t$ and $A \in \mathcal{F}_s$. Recalling that for all $x, y \in E$ we have $x = y$ if and only if $x'(x) = x'(y)$ for all $x' \in E'$, this provides the stated equivalence (i) $\Leftrightarrow$ (ii). □

C.2. **Lemma.** Let $M$ be an $E$-valued càdlàg, adapted process such that $\mathbb{E}[\|M_{\tau}\|] < \infty$ for each stopping time $\tau$. Then the following statements are equivalent:

(i) $M$ is a martingale.

(ii) For each stopping time $\tau$ we have $\mathbb{E}[M_{\tau}] = \mathbb{E}[M_0]$.

(iii) For each stopping time $\tau$ the stopped process $M^{\tau}$ is a martingale.

*Proof.* This is a consequence of Lemma C.1 and [29, Lemma I.1.44]. □

C.3. **Lemma.** Let $M$ be an $E$-valued continuous, adapted process such that $M_0$ is bounded. Then the following statements are equivalent:

(i) $M$ is a local martingale.

(ii) $x'(M)$ is a local martingale for each $x \in E'$.
Proof. We only need to prove (ii) ⇒ (i). We define the localizing sequence $(\tau_n)_{n \in \mathbb{N}}$ of stopping times as

$$ \tau_n := \inf\{t \in \mathbb{R}_+ : \|M_t\| \ge n\}, \quad n \in \mathbb{N}. $$

Let $n \in \mathbb{N}$ be arbitrary. Then the stopped process $M^{\tau_n}$ is bounded. Hence, for each $x' \in E'$ the process $x'(M^{\tau_n})$ is a real-valued continuous, bounded local martingale, and hence a martingale. Therefore, by Lemma C.1 we deduce that $M^{\tau_n}$ is a martingale. Consequently, the process $M$ is a local martingale. $\square$

### C.4. Lemma.
Let $M$ be an $E$-valued continuous, bounded local martingale. Then $M$ is a martingale.

Proof. For each $x' \in E'$ the process $x'(M)$ is a real-valued continuous, bounded local martingale, and hence a martingale. By Lemma C.1 we deduce that $M$ is a martingale. $\square$

### C.5. Proposition.
Let $M$ be an $E$-valued continuous local martingale such that $M_0$ is bounded and

$$ (C.1) \qquad \mathbb{E} \left[ \sup_{t \in [0,T]} \|M_t\|^2 \right] < \infty \quad \text{for each } T \in \mathbb{R}_+. $$

Then $M$ is a square-integrable martingale.

Proof. By (C.1) the process $M$ is square-integrable. Thus, it suffices to show that $(M_t)_{t \in [0,T]}$ is a martingale for each $T \in \mathbb{R}_+$. Let $T \in \mathbb{R}_+$ be arbitrary and define the localizing sequence $(\tau_n)_{n \in \mathbb{N}}$ of stopping times as

$$ \tau_n := \inf\{t \in [0,T] : \|M_t\| \ge n\}, \quad n \in \mathbb{N}. $$

Then $(M_t^{\tau_n})_{t \in [0,T]}$ is a continuous, bounded local martingale, and by Lemma C.4 we deduce that $(M_t^{\tau_n})_{t \in [0,T]}$ is a martingale. Let $\tau \le T$ be an arbitrary stopping time. Using Lemma C.2 we obtain

$$ \mathbb{E}[M_{\tau \wedge \tau_n}] = \mathbb{E}[M_{\tau}^{\tau_n}] = \mathbb{E}[M_0^{\tau_n}] = \mathbb{E}[M_0]. $$

Note that by (C.1) we have $\mathbb{E}[\|M_\tau\|] < \infty$. Furthermore, taking into account the continuity of the sample paths of $M$, by (C.1) and Lebesgue's dominated convergence theorem we have

$$ \lim_{n \to \infty} \mathbb{E}[\|M_{\tau \wedge \tau_n} - M_\tau\|] = 0. $$

Since the expectation operator $\mathbb{E}: L^1(\Omega, \mathcal{F}, \mathbb{P}) \to E$ given by the Bochner integral is a continuous linear operator, we obtain

$$ \mathbb{E}[M_\tau] = \lim_{n \to \infty} \mathbb{E}[M_{\tau \wedge \tau_n}] = \mathbb{E}[M_0]. $$

By Lemma C.2, this proves that $(M_t)_{t \in [0,T]}$ is a martingale. $\square$

## APPENDIX D. SMOOTH FUNCTIONS IN BANACH SPACES

In this appendix we provide the required results about smooth functions in Banach spaces. In what follows, let $E, F, G, H$ be Banach spaces, and let $k \in \mathbb{N}_0 \cup \{\infty\}$ be a nonnegative integer, which may be infinite. For a function $f: U \to F$ defined on a subset $U \subset E$ we agree on the notation $D^0 f := f$. Furthermore, we agree that in case $k = \infty$ the notation $j = 0, \dots, k$ means $j \in \mathbb{N}_0$.

### D.1. Definition.
Let $U \subset E$ be an open subset, and let $f: U \to F$ be a function of class $C^k$.

(1) We say that $f$ is of class $C_b^k$ if the functions $D^j f, j = 0, \dots, k$ are bounded.

(2) We say that $f$ is of class $C_{b,\text{loc}}^k$ if $f|_V: V \to F$ is of class $C_b^k$ for every open and bounded subset $V \subset U$.
D.2. **Proposition.** [3, Prop. 2.4.8] Suppose that $U \subset E$ is open and convex. Then every mapping $f: U \to F$ of class $C_b^1$ is Lipschitz continuous.

D.3. **Proposition** (Leibniz or product rule). Let $U \subset E$ be an open subset, let $f: U \to F$, $g: U \to G$ be differentiable mappings, and let $B \in L^{(2)}(F \times G, H)$ be a continuous bilinear operator. We define the new mapping

$$B(f,g) : U \to H, \quad x \mapsto B(f(x),g(x)).$$

Then the following statements are true:

(a) The mapping $B(f,g)$ is also differentiable, and we have

$$\text{(D.1)} \qquad D(B(f,g))(x)v = B(Df(x)v,g(x)) + B(f(x),Dg(x)v) \\ \text{for all } x \in U \text{ and } v \in E.$$

(b) If $f$ and $g$ are of class $C^k$, then $B(f,g)$ is also of class $C^k$.

(c) If $f$ and $g$ are of class $C_b^k$, then $B(f,g)$ is also of class $C_b^k$.

*Proof.* Parts (a) and (b) are immediate consequences of [3, Thm. 2.4.4], and part (c) is a consequence of the multidimensional Leibniz rule, which follows inductively from (D.1). □

D.4. **Proposition** (Chain rule). Let $U \subset E$ be an open subset, and let $f: U \to F$, $g: F \to G$ be differentiable mappings. Then the following statements are true:

(a) The composition $g \circ f: E \to G$ is also differentiable, and we have

$$D(g \circ f)(x) = Dg(f(x)) \circ Df(x), \quad x \in U.$$

(b) If $f$ and $g$ are of class $C^k$, then $g \circ f$ is also of class $C^k$.

(c) If $f$ and $g$ are of class $C_b^k$, then $g \circ f$ is also of class $C_b^k$.

(d) If $f$ and $g$ are of class $C_{b,loc}^k$, then $g \circ f$ is of class $C_{b,loc}^k$.

*Proof.* Parts (a) and (b) follow from [3, Thm. 2.4.3], and parts (c) and (d) are a consequence of the higher order chain rule; see [3, p. 88]. □

D.5. **Proposition.** Let $U \subset E$ be an open subset, let $f: U \to F$ be differentiable, and let $T \in L(F, G)$ be a continuous linear operator. Then the following statements are true:

(a) The composition $T \circ f: U \to G$ is also differentiable, and we have

$$D(T \circ f)(x)v = T(Df(x)v) \quad \text{for all } x \in U \text{ and } v \in E.$$

(b) If $f$ is of class $C^k$, then $T \circ f$ is also of class $C^k$, and for all $j = 0, \dots, k$ we have

$$\text{(D.2)} \qquad D^j(T \circ f) = T \circ D^j f.$$

(c) If $f$ is of class $C_b^k$, then $T \circ f$ is also of class $C_b^k$.

*Proof.* Parts (a) and (b) are immediate consequences of the chain rule (Proposition D.4), and part (c) follows from the formula (D.2). □

D.6. **Corollary.** Let $U \subset E$ be an open subset, let $f: U \to F$ be a mapping of class $C^k$, and let $T \in L(F)$ be a continuous linear operator such that $Tf(x) = f(x)$ for all $x \in U$. Then for all $j = 0, \dots, k$ we have

$$T \circ D^j f = D^j f.$$

*Proof.* This is an immediate consequence of Proposition D.5. □

The following result is also a consequence of the chain rule (Proposition D.4).

D.7. **Proposition.** Let $S \in L(E, F)$ be a linear operator, and let $f: F \to G$ be a differentiable function. Then the following statements are true:
(a) The composition $f \circ S : E \to G$ is also differentiable, and we have

$$D(f \circ S)(x) = Df(Sx) \circ S, \quad x \in E,$$

or, in other words,

$$D(f \circ S)(x)v = Df(Sx)Sv \quad \text{for all } x, v \in E.$$

(b) If $f$ is of class $C^k$, then $f \circ S$ is also of class $C^k$.

D.8. **Proposition.** Let $U \subset E$ be an open subset, and let $f : U \to F$ be a mapping of class $C^k$ for some finite number $k \in \mathbb{N}_0$. For each $x \in U$ there exists an open neighborhood $U_0 \subset U$ of $x$ such that $f|_{U_0}$ is of class $C_b^k$.

*Proof.* Recalling the notation $D^0f := f$, we define the finite number $r > 0$ as

$$r := 1 + \sum_{j=0}^{k} \|D^j f(x)\|.$$

Now, we define the subset $U_0 \subset U$ as

$$U_0 := \bigcap_{j=0}^{k} \{y \in U : \|D^j f(y)\| < r\}.$$

By the definition of $r$ and the continuity of $\|D^j f\|$ for $j = 0, \dots, k$, the subset $U_0$ is an open neighborhood of $x$, and $f|_{U_0}$ is of class $C_b^k$. $\square$

D.9. **Lemma.** Let $U \subset E$ be an open subset, and let $f : U \to F$ be a mapping of class $C^1$. Let $x \in U$ and $v \in E$ be arbitrary. Let $(t_n)_{n \in \mathbb{N}} \subset \mathbb{R} \setminus \{0\}$ and $(v_n)_{n \in \mathbb{N}} \subset E$ be sequences such that $t_n \to 0$ and $v_n \to v$. Then we have

$$Df(x)v = \lim_{n \to \infty} \frac{f(x + t_n v_n) - f(x)}{t_n}.$$

*Proof.* By Proposition D.8 there exists an open convex neighborhood $U_0 \subset U$ of $x$ such that $f|_{U_0}$ is of class $C_b^1$. By Proposition D.2 the mapping $f|_{U_0}$ is Lipschitz continuous. Since $t_n \to 0$, may assume that $x + t_n v \in U_0$ and $x + t_n v_n \in U_0$ for each $n \in \mathbb{N}$. Denoting by $L > 0$ the Lipschitz constant of $f|_{U_0}$, we have

$$\left\| \frac{f(x + t_n v) - f(x + t_n v_n)}{t_n} \right\| \le \frac{L |t_n| \|v - v_n\|}{|t_n|} = L \|v - v_n\| \to 0 \quad \text{as } n \to \infty,$$

and it follows that

$$\begin{align*}
Df(x)v &= \lim_{n \to \infty} \frac{f(x + t_n v) - f(x)}{t_n} \\
&= \lim_{n \to \infty} \frac{f(x + t_n v_n) - f(x)}{t_n} + \lim_{n \to \infty} \frac{f(x + t_n v) - f(x + t_n v_n)}{t_n} \\
&= \lim_{n \to \infty} \frac{f(x + t_n v_n) - f(x)}{t_n}.
\end{align*}$$

This completes the proof. $\square$

D.10. **Lemma.** Let $H$ be a Hilbert space, and let $0 < r < R < \infty$ be arbitrary. There exists a function $\varphi : H \to [0, 1]$ of class $C_b^\infty$ such that $\varphi(x) = 1$ for all $x \in H$ with $\|x\| \le r$ and $\varphi(x) = 0$ for all $x \in H$ with $\|x\| \ge R$.

*Proof.* There exists a function $f : \mathbb{R} \to [0, 1]$ of class $C^\infty$ such that $f(y) = 1$ for all $y \in \mathbb{R}$ with $|y|^2 \le r$ and $f(y) = 0$ for all $y \in \mathbb{R}$ with $|y|^2 \ge R$. Since $f$ has compact support, it is even of class $C_b^\infty$. Now, we define $g : H \to \mathbb{R}_+$ as $g(x) = \|x\|^2$, $x \in H$. Then $g$ is of class $C_{b,\text{loc}}^\infty$. Furthermore, we define $\varphi := f \circ g : H \to [0, 1]$. By Proposition D.4 the mapping $\varphi$ is of class $C_{b,\text{loc}}^\infty$, and we have $\varphi(x) = 1$ for all
$x \in H$ with $\|x\| \le r$ and $\varphi(x) = 0$ for all $x \in H$ with $\|x\| \ge R$. The latter property ensures that $\varphi$ is even of class $C_b^\infty$. $\square$

**D.11. Proposition.** Let $H$ be a Hilbert space, let $U \subset H$ be an open subset, and let $f: U \to F$ be a mapping of class $C^k$ for some finite number $k \in \mathbb{N}_0$. For each $x \in U$ there exist an open neighborhood $N(x) \subset U$ of $x$ and a mapping $\hat{f}: H \to F$ of class $C_b^k$ such that $\hat{f}|_{N(x)} = f|_{N(x)}$.

*Proof.* By Proposition D.8 there is an open neighborhood $U_0 \subset U$ of $x$ such that $f|_{U_0}$ is of class $C_b^k$. There exists $R > 0$ such that $\{y \in H : \|x - y\| < R\} \subset U_0$. We set $r := \frac{R}{2}$ and define the subset $N(x) \subset U_0$ as the open ball

$$N(x) := \{y \in H : \|x - y\| < r\}.$$

Then $N(x)$ is an open neighborhood of $x$. Furthermore, by Lemma D.10 there exists a function $\varphi: H \to [0, 1]$ of class $C_b^\infty$ such that $\varphi(x) = 1$ for all $x \in H$ with $\|x\| \le r$ and $\varphi(x) = 0$ for all $x \in H$ with $\|x\| \ge R$. Using Proposition D.3, the mapping $\hat{f} := \varphi \cdot f: H \to F$ is of class $C_b^k$, and moreover we have $\hat{f}|_{N(x)} = f|_{N(x)}$. $\square$

We will need the following version of the abstract implicit function theorem; see, e.g. [33, Thm. I.5.9].

**D.12. Theorem (Implicit Function Theorem).** Let $U \subset E, V \subset F$ be open and $f: U \times V \to G$ be of class $C^k$ for some $k \in \mathbb{N}$. We assume there are $x_0 \in U$ and $y_0 \in V$ such that $f(x_0, y_0) = 0$, and that the linear operator $D_2f(x_0, y_0) \in L(F, G)$ is an isomorphism. Then there are an open neighborhood $U_0$ of $x_0$ and a unique map $g: U_0 \to V$ of class $C^k$ such that $g(x_0) = y_0$ and

$$f(x, g(x)) = 0 \quad \forall x \in U_0.$$

We will also require the following Nagumo type theorem. Let $a: E \to E$ be a Lipschitz continuous mapping. Then existence and uniqueness of solutions for the $E$-valued ODE

$$(D.3) \qquad \begin{cases} y'(t) = a(y(t)) \\ y(0) = y \end{cases}$$

holds true.

**D.13. Theorem.** For a closed subset $\mathcal{D} \subset E$ the following statements are equivalent:

(i) $\mathcal{D}$ is invariant for the ODE (D.3).

(ii) We have $a(y) \in T_{\mathcal{D}}^b(y)$ for all $y \in \mathcal{D}$.

*Proof.* Since the subset $\mathcal{D}$ is closed, it is also locally closed in the terminology of [44]. Moreover, since the mapping $a$ is Lipschitz continuous, it satisfies the linear growth condition, and hence it is positively sublinear in the terminology of [44]. Consequently, the stated result is a consequence of [44, Thms. 1.1 and 1.2] (applied with $f \equiv 0$ and $h(t, y) = a(y)$). $\square$

## APPENDIX E. LINEAR OPERATORS IN HILBERT SPACES

In this appendix we provide the required results about linear operators in Hilbert spaces. In the sequel the symbols $H$ and $H_i$ for some $i \in \mathbb{N}$ denote separable Hilbert spaces. We use the notations $L(H_1, H_2)$, $L_1(H_1, H_2)$, $L_2(H_1, H_2)$ for the spaces of all bounded linear operators, nuclear operators and Hilbert-Schmidt operators from $H_1$ to $H_2$.

**E.1. Proposition.** [34, Prop. B.0.7] The space $L_2(H_1, H_2)$ is also a separable Hilbert space.
Furthermore, the notation $L^+$ indicates the respective subsets of nonnegative definite operators, and $L^{++}$ indicates the respective subsets of positive definite operators. Let us recall the following well-known result.

**E.2. Lemma.** The following statements are true:

(1) For $T \in L(H_1, H_2)$, $S \in L_1(H_2, H_3)$ and $R \in L(H_3, H_4)$ we have $RST \in L_1(H_1, H_4)$ and

$$ \|RST\|_{L_1(H_1, H_4)} \le \|R\|_{L(H_3, H_4)} \|S\|_{L_1(H_2, H_3)} \|T\|_{L(H_1, H_2)}. $$

(2) For $T \in L(H_1, H_2)$, $S \in L_2(H_2, H_3)$ and $R \in L(H_3, H_4)$ we have $RST \in L_2(H_1, H_4)$ and

$$ \|RST\|_{L_2(H_1, H_4)} \le \|R\|_{L(H_3, H_4)} \|S\|_{L_2(H_2, H_3)} \|T\|_{L(H_1, H_2)}. $$

(3) For $T \in L_2(H_1, H_2)$ and $S \in L_2(H_2, H_3)$ we have $ST \in L_1(H_1, H_3)$ and

$$ \|ST\|_{L_1(H_1, H_3)} \le \|S\|_{L_2(H_2, H_3)} \|T\|_{L_2(H_1, H_2)}. $$

*Proof.* See [45, Satz VI.5.4] and [34, Rem. B.0.6.iii and Prop. B.0.8]. □

<figure><img src="image_19.png" /></figure>

**E.3. Lemma.** The following statements are true:

(1) The mapping

$$ (L(H_1, H_2), \| \cdot \|_{L(H_1, H_2)}) \to (L(H_2, H_1), \| \cdot \|_{L(H_2, H_1)}), \quad T \mapsto T^* $$
is a linear isometry.

(2) The mapping

$$ (L_1(H_1, H_2), \| \cdot \|_{L_1(H_1, H_2)}) \to (L_1(H_2, H_1), \| \cdot \|_{L_1(H_2, H_1)}), \quad T \mapsto T^* $$
is a linear isometry.

(3) The mapping

$$ (L_2(H_1, H_2), \| \cdot \|_{L_2(H_1, H_2)}) \to (L_2(H_2, H_1), \| \cdot \|_{L_2(H_2, H_1)}), \quad T \mapsto T^* $$
is a linear isometry.

*Proof.* The first statement is a consequence of [45, Satz V.5.2.d]. For $x \in H_1$ and $y \in H_2$ the adjoint of the linear operator $S \in L(H_1, H_2)$, $S = \langle \cdot, x \rangle y$ is given by $S^* = \langle \cdot, y \rangle x$. Thus, the second statement follows from the definition of the nuclear norm. For the third statement we refer to [34, Rem. B.0.6.i]. □

<figure><img src="image_20.png" /></figure>

The following auxiliary result is also well-known; see, e.g. [41, Thm. 12.10]. Recall that $A^\circ$ denotes the polar of a set $A$; see Definition A.11, and also Remark A.12.

**E.4. Lemma.** For every $T \in L(H_1, H_2)$ we have $\ker(T^*) = \mathrm{ran}(T)^\perp = \mathrm{ran}(T)^\circ$.

We will often consider the spaces $L(H) := L(H, H)$, $L_1(H) := L_1(H, H)$ and $L_2(H) := L_2(H, H)$. It is well-known that $L_1(H) \hookrightarrow L_2(H) \hookrightarrow L(H)$ with continuous embedding. More precisely, we have the following result; see, e.g. [45, Satz VI.6.2].

**E.5. Lemma.** The following statements are true:

(1) We have $L_1(H) \subset L_2(H)$ and $\|T\|_{L_2(H)} \le \|T\|_{L_1(H)}$ for all $T \in L_1(H)$.

(2) We have $L_2(H) \subset L(H)$ and $\|T\|_{L(H)} \le \|T\|_{L_2(H)}$ for all $T \in L_2(H)$.

**E.6. Lemma.** Let $T \in L(H_1)$ and $A \in L(H_1, H_2)$ be linear operators. We define $S \in L(H_2)$ as $S := ATA^*$. Then the following statements are true:

(1) If $T$ is self-adjoint, then $S$ is self-adjoint as well.

(2) If $T$ is nonnegative definite, then $S$ is nonnegative definite as well.
*Proof.* Sine $S^* = AT^*A$, the first statement is evident. If $T$ is nonnegative definite, then for all $x \in H_2$ we have

$$ \langle Sx, x \rangle = \langle ATA^*x, x \rangle = \langle TA^*x, A^*x \rangle \geq 0, $$

proving the second statement.

<figure><img src="image_21.png" /></figure>

For $T \in L_1(H)$ we denote by $\mathrm{Tr}(T)$ the trace of the nuclear operator $T$, which is defined as

$$ \mathrm{Tr}(T) := \sum_{j=1}^{\infty} \langle Te_j, e_j \rangle, $$

independent of the choice of the orthonormal basis $\{e_j\}_{j \in \mathbb{N}}$ of $H$.

E.7. **Lemma.** [45, Satz VI.5.8] The following statements are true:

(1) The trace $L_1(H) \to \mathbb{R}$, $T \mapsto \mathrm{Tr}(T)$ is a continuous linear functional with $\|\mathrm{Tr}\| = 1$.

(2) We have $\mathrm{Tr}(T) = \mathrm{Tr}(T^*)$ for all $T \in L_1(H)$.

In view of the upcoming result, let us recall the notation $|T| := (T^*T)^{1/2}$ for a compact operator $T \in L(H_1, H_2)$. Note that $|T| \in L(H_1)$ is always self-adjoint and nonnegative definite. Furthermore, note that for every self-adjoint and nonnegative definite compact operator $T \in L(H)$ we have $T = |T|$.

E.8. **Lemma.** The following statements are true:

(1) For every $T \in L_1(H)$ we have $|T| \in L_1^+(H)$ and $\|T\|_{L_1(H)} = \mathrm{Tr}(|T|)$.

(2) We have $|\mathrm{Tr}(T)| \le \|T\|_{L_1(H)}$ for all $T \in L_1(H)$.

(3) For every self-adjoint operator $T \in L_1^+(H)$ we have $\|T\|_{L_1(H)} = \mathrm{Tr}(T)$.

(4) The inner product on $L_2(H)$ is given by $\langle T, S \rangle_{L_2(H)} = \mathrm{Tr}(S^*T)$ for all $T, S \in L_2(H)$.

(5) In particular, we have $\|T\|_{L_2(H)}^2 = \|T^*T\|_{L_1(H)}$ for every $T \in L_2(H)$.

*Proof.* The statements follow from [45, Satz VI.5.5], [34, Rem. B.0.4] and [45, Satz VI.6.2.f]. □

<figure><img src="image_22.png" /></figure>

E.9. **Lemma.** Let $T, S \in L(H)$ be such that $T \in L_1(H)$ or $T, S \in L_2(H)$. Then we have $ST, TS \in L_1(H)$ and $\mathrm{Tr}(ST) = \mathrm{Tr}(TS)$.

*Proof.* In case $T \in L_1(H)$ the result follows from [45, Satz VI.5.8.c]. If $T, S \in L_2(H)$, then using Lemmas E.3, E.8 and E.7 we obtain

$$ \mathrm{Tr}(ST) = \langle T, S^* \rangle_{L_2(H)} = \langle T^*, S \rangle_{L_2(H)} = \mathrm{Tr}(S^*T^*) = \mathrm{Tr}((TS)^*) = \mathrm{Tr}(TS), $$
completing the proof. □

<figure><img src="image_23.png" /></figure>

The following result is a consequence of [34, Prop. B.0.10].

E.10. **Lemma.** Let $T \in L_2(H)$ and $S \in L(H)$ be arbitrary. Then we have

$$ \mathrm{STT}^*, T^*ST \in L_1(H) \quad \text{and} \quad \mathrm{Tr}(\mathrm{STT}^*) = \mathrm{Tr}(T^*ST). $$

In what follows, we denote by $S_1^+(H) \subset L_1(H)$ the subset of all nuclear operators which are self-adjoint and nonnegative definite. The following result shows that the mapping

$$ (L_1(H), \| \cdot \|_{L_1(H)}) \to (S_1^+(H), \| \cdot \|_{L_1(H)}), \quad T \mapsto |T| $$

is a nonlinear continuous isometry.

E.11. **Lemma.** The following statements are true:

(1) For every $T \in L_1(H)$ we have $|T| \in S_1^+(H)$ and $\| |T| \|_{L_1(H)} = \|T\|_{L_1(H)}$.
(2) For all $T, S \in L_1(H)$ we have

$$ (E.1) \qquad \| |T| - |S| \|_{L_1(H)} \le (2\|T + S\|_{L_1(H)} \|T - S\|_{L_1(H)})^{1/2}. $$

*Proof.* The first statement is an immediate consequence of Lemma E.8. Since $L_1(H)$ is the predual of the $W^*$-algebra $L(H)$, the second statement is a consequence of the Theorem on page 123 in [32]. □

E.12. **Remark.** In the finite dimensional situation $\dim H < \infty$, the inequality (E.1) is also a consequence of [11, Thm. X.2.1].

Moreover, we denote by $S_2^+(H) \subset L_2(H)$ the subset of all Hilbert-Schmidt operators which are self-adjoint and nonnegative definite. The following result shows that the mapping

$$ (S_1^+(H), \| \cdot \|_{L_1(H)}) \to (S_2^+(H), \| \cdot \|_{L_2(H)}), \quad T \mapsto T^{1/2} $$

is a nonlinear continuous mapping such that

$$ \|T^{1/2}\|_{L_2(H)} = \sqrt{\|T\|_{L_1(H)}} \quad \text{for all } T \in S_1^+(H). $$

E.13. **Lemma.** The following statements are true:

(1) For every $T \in S_1^+(H)$ we have $T^{1/2} \in S_2^+(H)$ and

$$ \|T^{1/2}\|_{L_2(H)}^2 = \|T\|_{L_1(H)}. $$

(2) For all $T, S \in S_1^+(H)$ we have

$$ \|T^{1/2} - S^{1/2}\|_{L_2(H)}^2 \le \|T - S\|_{L_1(H)}. $$

*Proof.* For $T \in S_1^+(H)$ consider the spectral decomposition

$$ T = \sum_{j=1}^{\infty} \lambda_j \langle \cdot, e_j \rangle e_j $$

with eigenvalues $\lambda_j \ge 0$, $j \in \mathbb{N}$ and an orthonormal basis $\{e_j\}_{j \in \mathbb{N}}$ of $H$. Then we have

$$ \begin{aligned} \|T^{1/2}\|_{L_2(H)}^2 &= \sum_{j=1}^{\infty} \|T^{1/2} e_j\|^2 = \sum_{j=1}^{\infty} \langle T^{1/2} e_j, T^{1/2} e_j \rangle = \sum_{j=1}^{\infty} \langle Te_j, e_j \rangle \\ &= \mathrm{Tr}(T) = \|T\|_{L_1(H)}, \end{aligned} $$

proving the first statement. For the second statement we refer to [39, Lemma 4.1]. □

<figure><img src="image_24.png" /></figure>

As an immediate consequence of Lemmas E.11 and E.13 we obtain the following result, showing that the mapping

$$ (L_1(H), \| \cdot \|_{L_1(H)}) \to (S_2^+(H), \| \cdot \|_{L_2(H)}), \quad T \mapsto |T|^{1/2} $$

is a nonlinear continuous mapping such that

$$ \|T\|_{L_2(H)} = \sqrt{\|T\|_{L_1(H)}} \quad \text{for all } T \in L_1(H). $$

E.14. **Corollary.** The following statements are true:

(1) For every $T \in L_1(H)$ we have $|T|^{1/2} \in S_2^+(H)$ and

$$ \| |T|^{1/2} \|_{L_2(H)}^2 = \|T\|_{L_1(H)}. $$

(2) The mapping $L_1(H) \to S_2^+(H)$, $T \mapsto |T|^{1/2}$ is continuous.
For the following result we consider self-adjoint compact operators $T \in L(H)$ with spectral decomposition

$$T = \sum_{j=1}^{\infty} \lambda_j(T) \langle \cdot, e_j \rangle e_j.$$

Concerning the ordering of the eigenvalues $(\lambda_j(T))_{j \in \mathbb{N}}$ we assume that $|\lambda_j(T)| \ge |\lambda_{j+1}(T)|$ for all $j \in \mathbb{N}$, and that for all $j < k$ with $|\lambda_j(T)| = |\lambda_k(T)|$ and $\lambda_j(T) \ne \lambda_k(T)$ we have $\lambda_j(T) > 0$ and $\lambda_k(T) < 0$.

E.15. **Proposition.** For all self-adjoint nuclear operators $T, S \in L_1(H)$ we have

$$\sum_{j=1}^{\infty} |\lambda_j(T) - \lambda_j(S)| \le \|T - S\|_{L_1(H)}.$$

*Proof.* In the finite dimensional situation, this is a consequence of [11, Lemma IV.3.2], and in the general situation, the inequality follows from approximating the operators $T$ and $S$ according to their spectral decompositions. $\square$

E.16. **Lemma.** For every $T \in L(H_1, H_2)$ we have $\ker(T) = \ker(T^*T)$, and thus $\ker(T^*) = \ker(TT^*)$.

*Proof.* For each $x \in \ker(T)$ we have $T^*Tx = 0$, showing that $\ker(T) \subset \ker(T^*T)$. Furthermore, for each $x \in \ker(T^*T)$ we have

$$\|Tx\|^2 = \langle Tx, Tx \rangle = \langle T^*Tx, x \rangle = 0,$$

and hence $Tx = 0$, showing that $\ker(T^*T) \subset \ker(T)$. $\square$

<figure><img src="image_25.png" /></figure>

Now we will provide the required results about the Moore-Penrose pseudoinverse. This inverse will be introduced for linear operators $T \in L(H_1, H_2)$ with closed range. Before introducing the Moore-Penrose pseudoinverse, we prepare some auxiliary results about the closed range property.

E.17. **Lemma.** Let $T \in L(H_1, H_2)$ be a linear operator. Then $T$ has closed range if and only if $T^*$ has closed range.

*Proof.* This is a consequence of the closed range theorem; see, e.g. [41, Thm. 4.14]. $\square$

E.18. **Lemma.** Let $T \in L(H_1, H_2)$ be a closed range operator. Then we have $\mathrm{ran}(T) = \mathrm{ran}(TT^*)$. In particular $TT^*$ also has closed range.

*Proof.* It is clear that $\mathrm{ran}(TT^*) \subset \mathrm{ran}(T)$. Now, let $y \in \mathrm{ran}(T)$ be arbitrary. Then there exists $x \in H_1$ such that $Tx = y$. By Lemma E.17 the adjoint operator $T^*$ also has closed range, and hence we can consider the decomposition $x = x_1 + x_2$ according to $H_1 = \mathrm{ran}(T^*) \oplus \mathrm{ran}(T^*)^{\perp}$. Noting that $\mathrm{ran}(T^*)^{\perp} = \ker(T)$ due to Lemma E.4, we obtain $Tx_1 = Tx = y$, showing that $y \in \mathrm{ran}(TT^*)$. $\square$

E.19. **Lemma.** Let $T \in L(H_1, H_2)$ be such that $TT^*$ has closed range. Then we have $\mathrm{ran}(T) = \mathrm{ran}(TT^*)$. In particular $T$ also has closed range.

*Proof.* Using Lemma E.4 and Lemma E.16 we have

$$\mathrm{ran}(TT^*) \subset \mathrm{ran}(T) \subset \overline{\mathrm{ran}(T)} = \ker(T^*)^{\perp} = \ker(TT^*)^{\perp} = \overline{\mathrm{ran}(TT^*)} = \mathrm{ran}(TT^*),$$
proving that $\mathrm{ran}(T) = \mathrm{ran}(TT^*)$. $\square$

<figure><img src="image_26.png" /></figure>

As an immediate consequence of Lemmas E.17–E.19 we obtain the following result about closed range operators:

E.20. **Proposition.** For a linear operator $T \in L(H_1, H_2)$ the following statements are equivalent:
(i) $T$ has closed range.

(ii) $T^*$ has closed range.

(iii) $TT^*$ has closed range.

(iv) $T^*T$ has closed range.

If these equivalent conditions are fulfilled, then we have

$$ \mathrm{ran}(T) = \mathrm{ran}(TT^*) \quad \text{and} \quad \mathrm{ran}(T^*) = \mathrm{ran}(T^*T). $$

For compact operators, the closed range property can be characterized as follows.

**E.21. Theorem.** [41, Thm. 4.18.b] For a compact operator $T \in L(H_1, H_2)$ the following statements are equivalent:

(i) $T$ has closed range.

(ii) $\mathrm{ran}(T)$ is finite dimensional.

The proofs of the following two results (Theorems E.22 and E.24) are a consequence of the results from [28, Chap. II].

**E.22. Theorem.** For each operator $T \in L(H_1, H_2)$ with closed range there exists a unique operator $T^+ \in L(H_2, H_1)$ such that

$$ (1) \quad TT^+ = (TT^+)^*. $$

$$ (2) \quad T^+T = (T^+T)^*. $$

$$ (3) \quad TT^+T = T. $$

$$ (4) \quad T^+TT^+ = T^+. $$

**E.23. Definition.** For each $T \in L(H_1, H_2)$ with closed range we call $T^+$ the Moore-Penrose pseudoinverse of $T$.

**E.24. Theorem.** Let $T \in L(H_1, H_2)$ be an operator with closed range, and let $T^+ \in L(H_2, H_1)$ be another linear operator. Then the following statements are equivalent:

(i) $T^+$ the Moore-Penrose pseudoinverse of $T$.

(ii) We have $T^+Tx = x$ for all $x \in \ker(T)^\perp$ and $T^+y = 0$ for all $y \in \mathrm{ran}(T)^\perp$.

(iii) We have $TT^+ = P_{\mathrm{ran}(T)}$ and $T^+T = P_{\mathrm{ran}(T^+)}$.

The following result in particular shows that for a closed range operator $T$ the Moore-Penrose pseudoinverse $T^+$ has closed range as well.

**E.25. Theorem.** [28, Thms. 2.1.2 and 2.1.5] Let $T \in L(H_1, H_2)$ be a closed range operator. Then the following statements are true:

(1) We have $\mathrm{ran}(T^+) = \mathrm{ran}(T^*) = \mathrm{ran}(T^+T)$. In particular, the operators $T^+$ and $T^+T$ have closed range.

(2) We have $T^+ = (T^*T)^+T^* = T^*(TT^*)^+$.

**E.26. Lemma.** Let $T \in L(H_1, H_2)$ be a closed range operator. Then $T^*$ is also a closed range operator, and we have $(T^*)^+ = (T^+)^*$.

*Proof.* By Proposition E.20 the adjoint operator $T^*$ also has closed range. Setting $S := (T^*)^*$, we have $S^* = T^+$, and by Theorem E.22 we obtain

$$ T^*S = (T^+T)^* = T^+T = (T^*S)^*, $$

$$ ST^* = (TT^*)^* = TT^+ = (ST^*)^*, $$

$$ T^*ST^* = (TT^+T)^* = T^*, $$

$$ ST^*S = (T^+TT^+)^* = S. $$

In view of Theorem E.22, this completes the proof. □

<figure><img src="image_27.png" /></figure>

**E.27. Lemma.** Let $T \in L(H)$ be a self-adjoint operator with closed range. Then the following statements are true:
(1) $T^+ \in L(H)$ is also self-adjoint; that is $(T^+)^* = T^+$.

(2) We have $TT^+ = T^+T$.

(3) More generally, we have $T^n T^+ = T^+ T^n$ for each $n \in \mathbb{N}$.

(4) $T^2$ also has closed range, and we have $(T^2)^+ = (T^+)^2$.

(5) We have $T^2 = T^2(T^+)^2 T^2$.

*Proof.* The first statement is an immediate consequence of Lemma E.26. Furthermore, by Theorem E.22 we have

$$TT^+ = (TT^+)^* = (T^+)^* T^* = T^+ T.$$

In order to prove the more general identity, we proceed by induction and suppose that $T^n T^+ = T^+ T^n$. By induction hypothesis we obtain

$$T^{n+1} T^+ = T T^n T^+ = T T^+ T^n = T^+ T T^n = T^+ T^{n+1}.$$

Moreover, by Proposition E.20 the operator $T^2$ also has closed range, and by Theorem E.25 and Theorem E.22 we have

$$(T^+)^2 = (T^2)^+ T \circ T(T^2)^+ = (T^2)^+ T^2 (T^2)^+ = (T^2)^+.$$

Now, using Theorem E.22 again it follows that

$$T^2 = T^2(T^2)^+ T^2 = T^2(T^+)^2 T^2,$$

completing the proof.

<figure><img src="image_28.png" /></figure>

For a closed subspace $U \subset H$ we denote by $P_U$ the orthogonal projection on $U$. Furthermore, for a linear operator $T \in L(H)$ we denote by $P_T$ the orthogonal projection on the closure of the range of $T$; that is $P_T := P_U$ with $U := \text{ran}(T)$.

E.28. **Lemma.** Let $T \in L(H)$ be such that $T$, or equivalently $TT^*$, has closed range. Then we have $P_{TT^*} T = T$.

*Proof.* By Proposition E.20 the operator $T$ has closed range if and only if $TT^*$ has closed range, and in this case we have $P_T = P_{TT^*}$. Since $P_T x = x$ for all $x \in \text{ran}(T)$, the claimed identity $P_{TT^*} T = T$ follows. □

E.29. **Proposition.** Let $U \subset H$ be an open subset, and let $C: U \to L_1(H)$ and $T: U \to L(H)$ be mappings such that $C$ is of class $C^1$. Let $\{e_j\}_{j \in \mathbb{N}}$ be an orthonormal basis of $H$. For each $j \in \mathbb{N}$ we denote by $C^j: H \to H$ and $T^j: H \to H$ the mappings given by

$$C^j(x) := C(x)e_j \quad \text{and} \quad T^j(x) := T(x)e_j \quad \text{for each } x \in U.$$

Then for all $x \in U$ the series

$$(E.2) \qquad \sum_{j=1}^{\infty} DC^j(x)T^j(x)$$

is weakly convergent, and for each $u \in H$ we have

$$(E.3) \qquad \sum_{j=1}^{\infty} \langle u, DC^j(x)T^j(x) \rangle = \text{Tr}(DC^*(x)T(x)u).$$

*Proof.* For any mapping $S: U \to L(H)$ and any $y \in H$ we denote by $Sy: U \to H$ the mapping $z \mapsto S(z)y$. Let $x \in U$ be arbitrary. Using Proposition D.5, for each
$u \in H$ we obtain

$$
\begin{align*}
\sum_{j=1}^{\infty} \langle u, DC^j(x)T^j(x) \rangle &= \sum_{j=1}^{\infty} \langle u, DC(x)T^j(x)e_j \rangle = \sum_{j=1}^{\infty} \langle (DC(x)T^j(x))^*u, e_j \rangle \\
&= \sum_{j=1}^{\infty} \langle DC^*(x)T^j(x)u, e_j \rangle = \sum_{j=1}^{\infty} \langle D(C^*u)(x)T^j(x), e_j \rangle \\
&= \sum_{j=1}^{\infty} \langle D(C^*u)(x)T(x)e_j, e_j \rangle = \operatorname{Tr}(D(C^*u)(x)T(x)) = \operatorname{Tr}(DC^*(x)T(x)u),
\end{align*}
$$

showing (E.3). Moreover, noting that due to Lemma E.7 the mapping

$H \to \mathbb{R}, \quad u \mapsto \operatorname{Tr}(\Phi u)$

is a continuous linear functional for every $\Phi \in L(H, L_1(H))$, by the Fréchet-Riesz theorem it follows that the series (E.2) is weakly convergent. $\square$

Now, we provide the required results about continuous linear functionals on Hilbert spaces. By the Riesz representation theorem we have the following well-known result.

E.30. **Lemma.** The mapping $H \to L(H, \mathbb{R})$ given by $x \mapsto \langle x, \cdot \rangle$ is an isometric isomorphism.

Let $S(H) \subset L(H)$ be the subspace of all self-adjoint operators $T \in L(H)$, and let $L_s^{(2)}(H, \mathbb{R}) \subset L^{(2)}(H, \mathbb{R})$ be the subspace of all symmetric bilinear operators $B \in L^{(2)}(H, \mathbb{R})$. By the Lax-Milgram theorem and the open mapping theorem we have the following result.

E.31. **Lemma.** The mapping $\Psi : L(H) \to L^{(2)}(H, \mathbb{R})$ given by

$(\Psi T)(x, y) = \langle Tx, y \rangle, \quad T \in L(H)$

is a linear isomorphism such that $\Psi(S(H)) = L_s^{(2)}(H, \mathbb{R})$.

E.32. **Remark.** Let $U \subset H$ be an open subset, and let $\phi : U \to \mathbb{R}$ be a mapping. We fix an arbitrary element $x \in U$.

(1) Suppose that $\phi$ is of class $C^1$. Then we have $D\phi(x) \in L(H, \mathbb{R})$. By Lemma E.30 we may regard the first order derivative as an element from $H$, which justifies the notation

$D\phi(x)v = \langle D\phi(x), v \rangle, \quad v \in H.$

(2) Suppose that $\phi$ is even of class $C^2$. Then we have $D^2\phi(x) \in L_s^{(2)}(H, \mathbb{R})$; see, e.g. [3, Prop. 2.4.14]. By Lemma E.31 we may regard the second order derivative as an operator from $S(H)$, which justifies the notation

$D^2\phi(x)(v, w) = \langle D^2\phi(x)v, w \rangle, \quad v, w \in H.$

Combining Lemma E.30 and Parseval's identity we obtain the following result.

E.33. **Lemma.** We have $H \cong L(H, \mathbb{R}) = L_2(H, \mathbb{R})$, and

$\|x\| = \|\langle x, \cdot \rangle\|_{L(H, \mathbb{R})} = \|\langle x, \cdot \rangle\|_{L_2(H, \mathbb{R})} \text{ for all } x \in H.$

As an immediate consequence we obtain:

E.34. **Proposition.** We have $H \cong L_2(H, \mathbb{R}) \cong \ell^2(\mathbb{N})$, and the mapping

$H \mapsto \ell^2(\mathbb{N}), \quad x \mapsto (\langle x, e_j \rangle)_{j \in \mathbb{N}}$

is an isometric isomorphism, where $\{e_j\}_{j \in \mathbb{N}}$ denotes any orthonormal basis of $H$.
Furthermore, the following result holds true.

E.35. **Proposition.** The following statements are true:

(1) We have $L_2(H) \cong L_2(H, L_2(H, \mathbb{R})) \cong \ell^2(\mathbb{N} \times \mathbb{N})$, and the mapping

$$ (E.4) \qquad L_2(H) \to \ell^2(\mathbb{N} \times \mathbb{N}), \quad T \mapsto (\langle Te_i, e_j \rangle)_{i,j \in \mathbb{N}} $$

is an isometric isomorphism, where $\{e_j\}_{j \in \mathbb{N}}$ denotes any orthonormal basis of $H$.

(2) An operator $T \in L_2(H)$ is self-adjoint if and only if the associated double sequence in (E.4) is symmetric.

*Proof.* The linear operator $\Phi : L_2(H) \to L_2(H, L_2(H, \mathbb{R}))$ given by

$$ (\Phi T)x = \langle Tx, \cdot \rangle, \quad T \in L_2(H) $$

is an isometric isomorphism. Indeed, let $T \in L_2(H)$ be arbitrary. By Lemma E.33 we have

$$ \begin{aligned} \|\Phi T\|_{L_2(H, L_2(H, \mathbb{R}))}^2 &= \sum_{i=1}^{\infty} \|(\Phi T)e_i\|_{L_2(H, \mathbb{R})}^2 = \sum_{i=1}^{\infty} \|\langle Te_i, \cdot \rangle\|_{L_2(H, \mathbb{R})}^2 \\ &= \sum_{i=1}^{\infty} \|Te_i\|^2 = \|T\|_{L_2(H)}^2. \end{aligned} $$

Moreover, we have

$$ \sum_{i=1}^{\infty} \|\langle Te_i, \cdot \rangle\|_{L_2(H, \mathbb{R})}^2 = \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |\langle Te_i, e_j \rangle|^2, $$

proving the first statement. For the second statement, it suffices to show that every linear operator $T \in L(H)$ such that $\langle Te_i, e_j \rangle = \langle Te_j, e_i \rangle$ for all $i, j \in \mathbb{N}$ is self-adjoint. Indeed, for all $x, y \in H$ we have

$$ \begin{aligned} \langle Tx, y \rangle &= \left\langle T \left( \sum_{i=1}^{\infty} \langle x, e_i \rangle e_i \right), \sum_{j=1}^{\infty} \langle y, e_j \rangle e_j \right\rangle \\ &= \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \langle x, e_i \rangle \langle y, e_j \rangle \langle Te_i, e_j \rangle = \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \langle x, e_i \rangle \langle y, e_j \rangle \langle Te_j, e_i \rangle \\ &= \left\langle T \left( \sum_{j=1}^{\infty} \langle y, e_j \rangle e_j \right), \sum_{i=1}^{\infty} \langle x, e_i \rangle e_i \right\rangle = \langle Ty, x \rangle, \end{aligned} $$

completing the proof.

<figure><img src="image_29.png" /></figure>

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ÉCOLE POLYTECHNIQUE, MATHEMATICAL FINANCE GROUP, CMAP DEPARTMENT, PARIS,
FRANCE

Email address: eduardo.abi-jaber@polytechnique.edu

ALBERT LUDWIG UNIVERSITY OF FREIBURG, DEPARTMENT OF MATHEMATICAL STOCHASTICS,
ERNST-ZERMELO-STRASSE 1, D-79104 FREIBURG, GERMANY

Email address: stefan.tappe@math.uni-freiburg.de