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A Proof of Lemma 1
To simplify the notation throughout this proof, for each  $j \in \{1, \dots, k\}$  denote  $\phi^j = \phi_{\psi^j}$ . We have
$$
E _ {\lambda , \tau} ^ {\boldsymbol {\nu}, w} (\boldsymbol {\psi} ^ {*}) - E _ {\lambda , \tau} ^ {\boldsymbol {\nu}, w} (\boldsymbol {\psi}) = \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu^ {j}} \left[ \left(\psi^ {*}\right) ^ {j} (Y) - \psi^ {j} (Y) \right] - \tau \log \frac {Z _ {\psi^ {*}}}{Z _ {\psi}}. \tag {14}
$$
Observe that for any  $x\in \mathcal{X}$  it holds that
$$
\frac {d \mu_ {\psi}}{d \mu_ {\psi^ {*}}} (x) = \frac {Z _ {\psi^ {*}}}{Z _ {\psi}} \exp \left(- \frac {\sum_ {j = 1} ^ {k} w _ {j} (\phi^ {j} (x) - (\phi^ {*}) ^ {j} (x))}{\tau}\right).
$$
$$
\begin{array}{l} \tau \log \frac {Z _ {\psi^ {*}}}{Z _ {\psi}} = \tau \log \mathbf {E} _ {X \sim \mu_ {\psi^ {*}}} \left[ \frac {Z _ {\psi^ {*}}}{Z _ {\psi}} \right] \\ = \tau \log \mathbf {E} _ {X \sim \mu_ {\psi^ {*}}} \left[ \frac {d \mu_ {\psi}}{d \mu_ {\psi^ {*}}} (x) \exp \left(\frac {\sum_ {j = 1} ^ {k} w _ {j} \left(\phi^ {j} (x) - \left(\phi^ {*}\right) ^ {j} (x)\right)}{\tau}\right) \right] \\ = \tau \log \mathbf {E} _ {X \sim \mu_ {\psi}} \left[ \exp \left(\frac {\sum_ {j = 1} ^ {k} w _ {j} \left(\phi^ {j} (x) - \left(\phi^ {*}\right) ^ {j} (x)\right)}{\tau}\right) \right] \\ = \sup  _ {\mu \ll \mu_ {\psi}} \left\{\mathbf {E} _ {X \sim \mu} \left[ \sum_ {j = 1} ^ {k} w _ {j} \left(\phi^ {j} (x) - \left(\phi^ {*}\right) ^ {j} (x)\right) \right] - \tau \mathrm {K L} (\mu , \mu_ {\psi}) \right\}, \tag {15} \\ \end{array}
$$
where in the final expression we have applied the Donsker-Varadhan variational principle (i.e., convex-conjugate duality between KL-divergence and cumulant generating functions); therein, the supremum runs over probability measures  $\mu$  absolutely continuous with respect to  $\mu_{\psi}$ , and it is attained by  $\mu$  defined as
$$
\begin{array}{l} \mu (d x) \propto \exp \left(\frac {1}{\tau} \sum_ {j = 1} ^ {k} w _ {j} \left(\phi^ {j} (x) - (\phi^ {*}) ^ {j} (x)\right)\right) \mu_ {\boldsymbol {\psi}} (d x) \\ \propto \exp \left(\frac {1}{\tau} \sum_ {j = 1} ^ {k} w _ {j} \left(\phi^ {j} (x) - \left(\phi^ {*}\right) ^ {j} (x)\right)\right) \exp \left(- \frac {1}{\tau} \sum_ {j = 1} ^ {k} w _ {j} \phi^ {j} (x)\right) \pi_ {\text {r e f}} (d x) \\ \propto \exp \left(- \frac {1}{\tau} \sum_ {j = 1} ^ {k} w _ {j} (\phi^ {*}) ^ {j} (x)\right) \pi_ {\mathrm {r e f}} (d x) = \pi_ {\psi^ {*}} (d x). \\ \end{array}
$$
That is, the supremum in (15) is attained by  $\mu = \mu_{\psi^*}$ . Hence, the identity (14) becomes
$$
\begin{array}{l} E _ {\lambda , \tau} ^ {\nu , w} (\psi^ {*}) - E _ {\lambda , \tau} ^ {\nu , w} (\psi) \\ = \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu^ {j}} \left[ (\psi^ {*}) ^ {j} (Y) - \psi^ {j} (Y) \right] - \mathbf {E} _ {X \sim \mu_ {\psi^ {*}}} \left[ \sum_ {j = 1} ^ {k} w _ {j} \left(\phi^ {j} (X) - (\phi^ {*}) ^ {j} (X)\right) \right] \\ + \tau \mathrm {K L} \left(\mu_ {\psi^ {*}}, \mu_ {\psi}\right) \\ = \sum_ {j = 1} ^ {k} w _ {j} \left(\mathbf {E} _ {Y \sim \nu^ {j}} \left[ (\psi^ {*}) ^ {j} (Y) - \psi^ {j} (Y) \right] + \mathbf {E} _ {X \sim \mu_ {\psi^ {*}}} \left[ (\phi^ {*}) ^ {j} (X)\right) - \phi^ {j} (X) \right]) + \tau \mathrm {K L} (\mu_ {\psi^ {*}}, \mu_ {\psi}) \\ \geq \tau \mathrm {K L} (\mu_ {\psi},, \mu_ {\psi}), \\ \end{array}
$$
where the final inequality follows by noting that for each  $j$  the optimality of the pair  $((\phi^{*})^{j},(\psi^{*})^{j})$  for the entropic optimal transport dual objective  $E_{\lambda}^{\mu_{\psi^{*}},\nu^{j}}$  implies that
$$
\begin{array}{l} \mathbf {E} _ {Y \sim \nu^ {j}} \left[ (\psi^ {*}) ^ {j} (Y) - \psi^ {j} (Y) \right] + \mathbf {E} _ {X \sim \mu_ {\psi^ {*}}} \left[ (\phi^ {*}) ^ {j} (X) - \phi^ {j} (X) \right] \\ = E _ {\lambda} ^ {\mu , \nu^ {j}} \left(\left(\phi^ {*}\right) ^ {j}, \left(\psi^ {*}\right) ^ {j}\right) - E _ {\lambda} ^ {\mu , \nu^ {j}} \left(\phi^ {j}, \psi^ {j}\right) \geq 0. \\ \end{array}
$$
The proof of Lemma 1 is complete.
B Proof of Proposition 1
Recall that for any non-negative integer  $t$  we have
$$
\mu_ {t} (d x) = Z _ {t} ^ {- 1} \exp \left(- \frac {\sum_ {j = 1} ^ {k} w _ {j} \phi_ {t} ^ {j} (x)}{\tau}\right) \pi_ {\mathrm {r e f}} (d x).
$$
where  $Z_{t}$  is the normalizing constant defined by
$$
Z _ {t} = \int_ {\mathcal {X}} \exp \left(- \frac {\sum_ {j = 1} ^ {k} w _ {j} \phi_ {t} ^ {j} (x)}{\tau}\right) \pi_ {\mathrm {r e f}} (d x).
$$
With the notation introduced above, we have
$$
E (\boldsymbol {\psi} _ {t}) = \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu^ {j}} \left[ \boldsymbol {\psi} _ {t} ^ {j} (Y) \right] - \tau \log Z _ {t}.
$$
$$
\begin{array}{l} E (\boldsymbol {\psi} _ {t + 1}) - E (\boldsymbol {\psi} _ {t}) = \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu^ {j}} \left[ \psi_ {t + 1} ^ {j} (Y) - \psi_ {t} ^ {j} (Y) \right] - \tau \log \frac {Z _ {t + 1}}{Z _ {t}}. \\ = \eta \lambda \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu^ {j}} \left[ \log \frac {d \nu^ {j}}{d \nu_ {t} ^ {j}} (Y) \right] - \tau \log \frac {Z _ {t + 1}}{Z _ {t}}. \\ = \min {(\lambda , \tau)} \sum_ {j = 1} ^ {k} w _ {j} \mathrm {K L} (\nu^ {j}, \nu_ {t} ^ {j}) - \tau \log \frac {Z _ {t + 1}}{Z _ {t}}. \\ \end{array}
$$
Therefore, to prove Proposition 1 it suffices to show that the inequality
$$
\log \frac {Z _ {t + 1}}{Z _ {t}} \leq 0 \tag {16}
$$
holds for any  $t \geq 0$ . We will complete the proof of Proposition 1 using the following lemma, the proof of which is deferred to the end of this section.
Lemma 3. Let  $(\psi_t)_{t\geq 0}$  be any sequence of the form
$$
\psi_ {t + 1} ^ {j} = \psi_ {t} ^ {j} + \eta \lambda \log (\Delta_ {t} ^ {j}),
$$
where for  $j \in \{1, \dots, k\}$ ,  $(\Delta_t^j)_{t \geq 0}$  is an arbitrary sequence of strictly positive functions and  $\eta = \min(1, \tau / \lambda)$ . Then, for any  $t \geq 0$  it holds that
$$
\tau \log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \leq \min (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu_ {\psi_ {t}} ^ {j}} \left[ \Delta_ {t} ^ {j} (Y) \right].
$$
To complete the proof of Proposition 1, we will apply the above lemma with  $\Delta_t^j = \frac{d\nu^j}{d\nu_t^j}$ . Indeed, we have
$$
\begin{array}{l} \tau \log \frac {Z _ {t + 1}}{Z _ {t}} \leq \min (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu_ {t} ^ {j}} \left[ \frac {d \nu^ {j}}{d \nu_ {t} ^ {j}} (Y) \right] \\ = \min  (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu^ {j}} [ 1 ] \\ = 0. \\ \end{array}
$$
By (16), the proof of Proposition 1 is complete.
We will break down the proof with the help of the following lemma, the proof of which can be found in Section B.2.
Lemma 4. For any sequence  $(\psi_t)_{t\geq 0}$  and any  $t\geq 0$  it holds that
$$
\log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \leq \left\{ \begin{array}{l l} \frac {\lambda}{\tau} \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {X \sim \mu_ {t}} \left[ \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) ^ {\tau / \lambda} \right] & i f \tau \geq \lambda , \\ \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {X \sim \mu_ {t}} \left[ \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) \right] & i f \tau <   \lambda , \end{array} \right.
$$
where  $\phi_t = \phi_{\psi_t}$  and  $\mu_t(dx) = Z_{\psi_t}^{-1} \exp \left(-\sum_{j=1}^k w_j \phi_t^j(x) / \tau\right) \pi_{\mathrm{ref}}(dx)$ .
Observe that the sequence  $(\psi_t)_{t \geq 0}$  of the form stated in Lemma 3 satisfies, for any  $j \in \{1, \ldots, k\}$  and any  $t \geq 0$ ,
$$
\exp \left(\frac {- \phi_ {t + 1} ^ {j} + \phi_ {t} ^ {j}}{\tau}\right) = \exp \left(- \frac {\lambda}{\tau} \log \frac {d \mu_ {t}}{d \tilde {\mu} _ {t} ^ {j}}\right) = \left(\frac {d \tilde {\mu} _ {t} ^ {j}}{d \mu_ {t}}\right) ^ {\lambda / \tau},
$$
$$
\begin{array}{l} \frac {d \tilde {\mu} _ {t} ^ {j}}{d \mu_ {t}} (x) = \int_ {\mathcal {X}} \nu (d y) \exp \left(\frac {\psi_ {t + 1} ^ {j} (y) + \phi_ {t} ^ {j} (x) - c (x , y)}{\lambda}\right) \\ = \int_ {\mathcal {X}} \nu^ {j} (d y) \Delta_ {t} ^ {j} (y) ^ {\eta} \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (x) - c (x , y)}{\lambda}\right). \\ \end{array}
$$
Hence, by Lemma 4 we have
$$
\begin{array}{l} \log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \\ \leq \frac {1}{\tau} \min  (\lambda , \tau) \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {X \sim \mu_ {t}} \left[ \right. \\ \left. \left(\int_ {\mathcal {X}} \nu^ {j} (d y) \Delta_ {t} ^ {j} (y) ^ {\eta} \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (X) - c (X , y)}{\lambda}\right)\right) ^ {\max  \left(1, \lambda / \tau\right)} \right]. \tag {17} \\ \end{array}
$$
We split the remaining proof into two cases:  $\tau \geq \lambda$  and  $\tau < \lambda$ .
The case  $\tau \geq \lambda$ . When  $\tau \geq \lambda$ , we have  $\max(1, \lambda / \tau) = 1$  and  $\eta = \min(1, \tau / \lambda) = 1$ . Thus, (17) yields
$$
\begin{array}{l} \log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \\ \leq \frac {1}{\tau} \min  (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {X \sim \mu_ {t}} \left[ \int_ {\mathcal {X}} \nu^ {j} (d y) \Delta_ {t} ^ {j} (y) \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (X) - c (X , y)}{\lambda}\right) \right] \\ = \frac {1}{\tau} \min  (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \left[ \int_ {\mathcal {X}} \mu_ {t} (d x) \int_ {\mathcal {X}} \nu^ {j} (d y) \Delta_ {t} ^ {j} (y) \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (X) - c (X , y)}{\lambda}\right) \right] \\ = \frac {1}{\tau} \min  (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \left[ \int_ {\mathcal {X}} \Delta_ {t} ^ {j} (y) \nu^ {j} (d y) \int_ {\mathcal {X}} \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (X) - c (X , y)}{\lambda}\right) \mu_ {t} (d x) \right] \\ = \frac {1}{\tau} \min (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \left[ \int_ {\mathcal {X}} \Delta_ {t} ^ {j} (y) \nu^ {j} (d y) \frac {d \nu_ {\psi_ {t}} ^ {j}}{d \nu^ {j}} (y) \right] \\ = \frac {1}{\tau} \min  (\lambda , \tau) \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {\Upsilon \sim \nu_ {\psi_ {t}} ^ {j}} \left[ \Delta_ {t} ^ {j} (y) \right]. \\ \end{array}
$$
This completes the proof of Lemma 3 when  $\tau \geq \lambda$ .
The case  $\tau < \lambda$ . For  $j \in \{1, \dots, k\}$  and any  $x \in \mathcal{X}$  define the measure  $\rho_x$  by
$$
\rho_ {x} ^ {j} (d y) = \nu^ {j} (d y) \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (x) - c (x , y)}{\lambda}\right).
$$
By the definition of  $\phi_t^j$ , we have
$$
\begin{array}{l} \int_ {\mathcal {X}} \rho_ {x} ^ {j} (d y) \\ = \int_ {\mathcal {X}} \nu (d y) \exp \left(\frac {\psi_ {t} ^ {j} (y) - c (x , y)}{\lambda}\right) \exp \left(\frac {\phi_ {t} ^ {j} (x)}{\lambda}\right) \\ = \int_ {\mathcal {X}} \nu (d y) \exp \left(\frac {\psi_ {t} ^ {j} (y) - c (x , y)}{\lambda}\right) \exp \left(- \log \int_ {\mathcal {X}} \nu^ {j} (d y ^ {\prime}) \exp \left(\frac {\psi_ {t} ^ {j} \left(y ^ {\prime}\right) - c (x , y ^ {\prime})}{\lambda}\right)\right) \\ = 1 \\ \end{array}
$$
In particular,  $\rho_{x}$  is a probability measure. Hence, (17) can be rewritten as
$$
\log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \leq \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {X \sim \mu_ {t}} \left[ \mathbf {E} _ {Y \sim \rho_ {X} ^ {j}} \left[ \Delta_ {t} ^ {j} (Y) ^ {\eta} \Bigg | X \right] ^ {\lambda / \tau} \right]
$$
Because  $\lambda/\tau > 1$ , the function  $x \mapsto x^{\lambda/\tau}$  is convex. Applying Jensen's inequality to the conditional expectation and using the fact that  $\eta \lambda/\tau = 1$ , it follows that
$$
\begin{array}{l} \log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \leq \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {X \sim \mu_ {t}} \left[ \mathbf {E} _ {Y \sim \rho_ {X} ^ {j}} \left[ \Delta_ {t} ^ {j} (Y) \Big | X \right] \right] \\ = \log \sum_ {j = 1} ^ {k} w _ {j} \int_ {\mathcal {X}} \mu_ {t} (d x) \int_ {\mathcal {X}} \Delta_ {t} ^ {j} (y) \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (x) - c (x , y)}{\lambda}\right) \nu (d y). \tag {18} \\ \end{array}
$$
By the definition of  $\nu_{\psi_t}^j$  we have
$$
\frac {d \nu_ {\psi_ {t}} ^ {j}}{d \nu^ {j}} (y) = \int_ {\mathcal {X}} \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (x) - c (x , y)}{\lambda}\right) \mu_ {t} (d x).
$$
Interchanging the order of integration in (18) and plugging in the above equation yields
$$
\begin{array}{l} \log \frac {Z _ {\psi_ {t + 1}}}{Z _ {\psi_ {t}}} \leq \log \sum_ {j = 1} ^ {k} w _ {j} \int_ {\mathcal {X}} \left[ \int_ {\mathcal {X}} \exp \left(\frac {\psi_ {t} ^ {j} (y) + \phi_ {t} ^ {j} (x) - c (x , y)}{\lambda}\right) \mu_ {t} (d x) \right] \Delta_ {t} ^ {j} (y) \nu^ {j} (d y) \\ = \log \sum_ {j = 1} ^ {k} w _ {j} \int_ {\mathcal {X}} \left[ \frac {d \nu_ {\psi_ {t}} ^ {j}}{d \nu^ {j}} (y) \right] \Delta_ {t} ^ {j} (y) \nu^ {j} (d y) \\ = \log \sum_ {j = 1} ^ {k} w _ {j} \mathbf {E} _ {Y \sim \nu_ {\phi_ {t}} ^ {j}} \left[ \Delta_ {t} ^ {j} (Y) \right]. \\ \end{array}
$$
This completes the proof of Lemma 3.
B.2 Proof of Lemma 4
To simplify the notation, denote  $Z_{t} = Z_{\psi_{t}}$ . Let  $x \in \mathcal{X}$  and  $t \geq 0$ . We have  $\mu_t \ll \mu_{t+1}$  with the Radon-Nikodym derivative  $d\mu_{t+1} / d\mu_t$  given by
$$
\begin{array}{l} \frac {d \mu_ {t + 1}}{d \mu_ {t}} (x) = \frac {Z _ {t}}{Z _ {t + 1}} \exp \left(\frac {- \sum_ {j = 1} ^ {k} w _ {k} (\phi_ {t + 1} ^ {j} (x) - \phi_ {t} ^ {j} (x))}{\tau}\right) \\ = \frac {Z _ {t}}{Z _ {t + 1}} \prod_ {j = 1} ^ {k} \exp \left(\frac {- \phi_ {t + 1} ^ {j} (x) + \phi_ {t} ^ {j} (x)}{\tau}\right) ^ {w _ {j}}. \\ \end{array}
$$
Multiplying both sides by  $Z_{t + 1} / Z_t$  and taking expectations with respect to  $\mu_t$  yields
$$
\begin{array}{l} \frac {Z _ {t + 1}}{Z _ {t}} = \mathbf {E} _ {X \sim \mu_ {t + 1}} \left[ \frac {Z _ {t + 1}}{Z _ {t}} \right] \\ = \mathbf {E} _ {X \sim \mu_ {t}} \left[ \frac {Z _ {t + 1}}{Z _ {t}} \frac {d \mu_ {t + 1}}{d \mu_ {t}} (X) \right] \\ = \mathbf {E} _ {X \sim \mu_ {t}} \left[ \prod_ {j = 1} ^ {k} \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) ^ {w _ {j}} \right]. \\ \end{array}
$$
In the case  $\tau < \lambda$ , the proof is complete by the Artihmetic-Geometric mean inequality (recall that  $w_{j} > 0$  for  $j = 1, \ldots, k$  and  $\sum_{j=1}^{k} w_{j} = 1$ ). On the other hand, if  $\tau \geq \lambda$  then  $x \mapsto x^{\lambda/\tau}$  is concave. Hence, it follows that
$$
\begin{array}{l} \log \frac {Z _ {t + 1}}{Z _ {t}} = \log {\bf E} _ {X \sim \mu_ {t}} \left[ \left(\prod_ {j = 1} ^ {k} \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) ^ {w _ {j} \tau / \lambda}\right) ^ {\lambda / \tau} \right] \\ \leq \log \mathbf {E} _ {X \sim \mu_ {t}} \left[ \left(\prod_ {j = 1} ^ {k} \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) ^ {w _ {j} \tau / \lambda}\right) \right] ^ {\lambda / \tau} \\ = \frac {\lambda}{\tau} \log \mathbf {E} _ {X \sim \mu_ {t}} \left[ \left(\prod_ {j = 1} ^ {k} \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) ^ {w _ {j} \tau / \lambda}\right) \right] \\ \leq \sum_ {j = 1} ^ {k} w _ {j} {\bf E} _ {X \sim \mu_ {t}} \left[ \exp \left(\frac {- \phi_ {t + 1} ^ {j} (X) + \phi_ {t} ^ {j} (X)}{\tau}\right) ^ {\tau / \lambda} \right], \\ \end{array}
$$
where the final step follows via the Arithmetic-Geometric mean inequality. This completes the proof of Lemma 4.
For every  $t \geq 0$  and  $j \in \{1, \dots, k\}$ , let  $\widetilde{\nu}_t^j$  be the distribution returned by the approximate Sinkhorn oracle that satisfies the properties listed in Definition 1. We follow along the lines of proof of Theorem 1.
First, we will establish an upper bound on the oscillation norm of the iterates  $\widetilde{\psi}_t$ . Indeed, by the property four in Definition 1 we have
$$
\| \tilde {\psi} _ {t + 1} ^ {j} \| _ {\mathrm {o s c}} \leq (1 - \eta) \| \tilde {\psi} _ {t} ^ {j} \| _ {\mathrm {o s c}} + \eta c _ {\infty} (\mathcal {X}).
$$
Since  $\widetilde{\psi}_0^j = 0$ , for any  $t \geq 0$  we have  $\| \widetilde{\psi}_t^j \|_{\mathrm{osc}} \leq c_\infty(\mathcal{X})$ .
Let  $\tilde{\delta}_t = E_{\lambda,\tau}^{\nu,w}(\psi^*) - E_{\lambda,\tau}^{\nu,w}(\widetilde{\psi}_t)$  be the suboptimality gap at time  $t$ . Using the concavity upper bound (10) and the property two in Definition 1 we have
$$
\begin{array}{l} \tilde {\delta} _ {t} \leq 2 c _ {\infty} (\mathcal {X}) \sum_ {j = 1} ^ {k} w _ {j} \| \nu^ {j} - \nu_ {t} ^ {j} \| _ {\mathrm {T V}} \\ \leq \varepsilon + 2 c _ {\infty} (\mathcal {X}) \sum_ {j = 1} ^ {k} w _ {j} \| \nu^ {j} - \widetilde {\nu} _ {t} ^ {j} \| _ {\mathrm {T V}} \\ \leq \varepsilon + \sqrt {2} c _ {\infty} (\mathcal {X}) \sum_ {j = 1} ^ {k} w _ {j} \sqrt {\mathrm {K L} \left(\nu^ {j} , \widetilde {\nu} _ {t} ^ {j}\right)} \\ \leq \varepsilon + \sqrt {2} c _ {\infty} (\mathcal {X}) \sqrt {\sum_ {j = 1} ^ {k} w _ {j} \mathrm {K L} \left(\nu^ {j} , \widetilde {\nu} _ {t} ^ {j}\right)}. \\ \end{array}
$$
Combining the property three stated in the Definition 1 with Lemma 3 we obtain
$$
\begin{array}{l} \tilde {\delta} _ {t} - \tilde {\delta} _ {t + 1} \geq \min  (\lambda , \tau) \sum_ {j = 1} ^ {k} w _ {j} \mathrm {K L} (v ^ {j}, \widetilde {v} _ {t} ^ {j}) - \min  (\lambda , \tau) \log \left(\sum_ {j = 1} ^ {k} w _ {j} \int_ {\mathcal {X}} \frac {d \nu_ {t}}{d \widetilde {\nu} _ {t}} (y) \nu^ {j} (d y)\right) \\ \geq \sum_ {j = 1} ^ {k} w _ {j} \mathrm {K L} (v ^ {j}, \widetilde {v} _ {t} ^ {j}) - \min (\lambda , \tau) \log \left(1 + \varepsilon^ {2} / (2 c _ {\infty} (\mathcal {X}) ^ {2})\right) \\ \geq \min (\lambda , \tau) \sum_ {j = 1} ^ {k} w _ {j} \mathrm {K L} (v ^ {j}, \widetilde {v} _ {t} ^ {j}) - \frac {\min (\lambda , \tau)}{2 c _ {\infty} (\mathcal {X}) ^ {2}} \varepsilon^ {2} \\ \geq \frac {\min  (\lambda , \tau)}{2 c _ {\infty} (\mathcal {X}) ^ {2}} \max  \left\{0, \tilde {\delta} _ {t} - \varepsilon \right\} ^ {2} - \frac {\min  (\lambda , \tau)}{2 c _ {\infty} (\mathcal {X}) ^ {2}} \varepsilon^ {2}. \\ \end{array}
$$
Provided that  $\widetilde{\delta}_t\geq 2\varepsilon$  it holds that
$$
\left(\tilde {\delta} _ {t} - 2 \varepsilon\right) - \left(\tilde {\delta} _ {t + 1} - 2 \varepsilon\right) \geq \frac {\min  (\lambda , \tau)}{2 c _ {\infty} (\mathcal {X})} \left(\tilde {\delta} _ {t} - 2 \varepsilon\right) ^ {2}.
$$
Let  $T$  be the first index such that  $\widetilde{\delta}_{T+1} < 2\varepsilon$  and set  $T = \infty$  if no such index exists. Then, the above equation is valid for any  $t \leq T$ . In particular, repeating the proof of Theorem 1, for any  $t \leq T$  we have
$$
\widetilde {\delta} _ {t} - 2 \varepsilon \leq \frac {2 c _ {\infty} (\mathcal {X}) ^ {2}}{\min (\lambda , \tau)} \frac {1}{t},
$$
which completes the proof of this theorem.
D Proof of Lemma 2
The first property - the positivity of the probability mass function of  $\widetilde{\nu}^j$  - is immediate from its definition.
To simplify the notation, denote in what follows
$$
K ^ {j} (x, y) = \exp \left(\frac {\phi_ {\psi^ {j}} (x) + \psi^ {j} (y) - c (x , y)}{\lambda}\right).
$$
With this notation, recall that
$$
\widehat {\nu} _ {\psi} ^ {j} (y _ {l} ^ {j}) = \frac {1}{n} \sum_ {i = 1} ^ {n} \nu^ {j} (y _ {l} ^ {j}) K (X _ {i}, y _ {l} ^ {j}).
$$
The above is a sum of  $n$  non-negative random variables bounded by one with expectation
$$
(\nu^ {\prime}) ^ {j} (y _ {l} ^ {j}) = \mathbf {E} _ {X \sim \mu_ {\psi} ^ {\prime}} \left[ \nu^ {j} (y _ {l} ^ {j}) \right]
$$
It follows by Hoeffding's inequality and the union bound that with probability at least  $1 - \delta$  the following holds for any  $j\in \{1,\ldots ,k\}$  and any  $l\in \{1,\dots ,m_j\}$ :
$$
\left| \widehat {\nu} _ {\psi} (y _ {l} ^ {j}) - (\nu^ {\prime}) ^ {j} (y _ {l} ^ {j}) \right| \leq \sqrt {\frac {2 \log \left(\frac {2 m}{\delta}\right)}{n}}.
$$
In particular, the above implies that
$$
\begin{array}{l} \| \tilde {\nu} _ {\psi} ^ {j} - \nu_ {\psi} ^ {j} \| _ {\mathrm {T V}} \leq 2 \zeta + (1 - \zeta) \| \tilde {\nu} _ {\psi} ^ {j} - \nu_ {\psi} ^ {j} \| _ {\mathrm {T V}} \\ \leq 2 \zeta + (1 - \zeta) \| \widetilde {\nu} _ {\psi} ^ {j} - (\nu^ {\prime}) ^ {j} \| _ {\mathrm {T V}} + (1 - \zeta) \| (\nu^ {\prime}) ^ {j} - \nu_ {\psi} ^ {j} \| _ {\mathrm {T V}} \\ \leq 2 \zeta + \| \widetilde {\nu} _ {\psi} ^ {j} - (\nu^ {\prime}) ^ {j} \| _ {\mathrm {T V}} + \| (\nu^ {\prime}) ^ {j} - \nu_ {\psi} ^ {j} \| _ {\mathrm {T V}} \\ \leq 2 \zeta + m _ {j} \varepsilon_ {\mu} + m _ {j} \sqrt {\frac {2 \log \left(\frac {2 m}{\delta}\right)}{n}}. \\ \end{array}
$$
Notice that the above bound can be made arbitrarily close to  $m_{j}\varepsilon_{\mu}$  by taking a large enough  $n$  and a small enough  $\zeta$ . This proves the second property of Definition 1.
To prove the third property, observe that