[proofplan] We reduce the dynamic Schrödinger bridge to its endpoint entropic transport problem, because the Brownian reference separates into an endpoint Brownian kernel and a conditional Brownian bridge law. After multiplying the endpoint entropy by $\varepsilon$, the coupling-dependent part is the quadratic transport cost plus an $\varepsilon$-weighted entropy term, and the standard small-noise $\Gamma$-convergence theorem for entropic optimal transport identifies the limit problem as quadratic optimal transport. Compact support gives tightness and makes the quadratic cost continuous along weakly convergent endpoint couplings. Minimality of $\pi^\varepsilon$ then forces every weak subsequential limit to attain the quadratic optimal transport value. [/proofplan]

[step:Reduce the bridge minimization to an endpoint entropic transport problem]For $\varepsilon>0$, define the Brownian transition density $q_\varepsilon:\mathbb R^d\times\mathbb R^d\to(0,\infty)$ by \begin{align*} q_\varepsilon(x,y)=(2\pi\varepsilon)^{-d/2}\exp\left(-\frac{|x-y|^2}{2\varepsilon}\right). \end{align*} Define the endpoint reference measure $Q^\varepsilon\in\mathcal P(\mathbb R^d\times\mathbb R^d)$ by \begin{align*} dQ^\varepsilon(x,y)=q_\varepsilon(x,y)\,d\mu_0(x)\,d\mathcal L^d(y). \end{align*} Let $\Pi(\mu_0,\mu_1)$ denote the set of Borel probability measures on $\mathbb R^d\times\mathbb R^d$ with first marginal $\mu_0$ and second marginal $\mu_1$. We use the static-dynamic equivalence for Schrödinger bridges. The space $C([0,1];\mathbb R^d)$ is Polish, so regular conditional laws for the Borel map $(X_0,X_1):C([0,1];\mathbb R^d)\to\mathbb R^d\times\mathbb R^d$ exist. The Brownian reference $R_{\mu_0}^{\varepsilon}$ admits a Borel family of conditional endpoint bridge laws $R_{\mu_0}^{\varepsilon,x,y}$ over $Q^\varepsilon$ because it is a Brownian Markov measure with strictly positive transition density $q_\varepsilon$. Therefore, for every $P\in\mathcal P(C([0,1];\mathbb R^d))$ with $(X_0)_\#P=\mu_0$ and $(X_1)_\#P=\mu_1$, if $\gamma=(X_0,X_1)_\#P$, then \begin{align*} H(P\mid R_{\mu_0}^{\varepsilon}) \ge H(\gamma\mid Q^\varepsilon), \end{align*} with equality when $P$ is obtained by mixing the conditional Brownian bridge laws $R_{\mu_0}^{\varepsilon,x,y}$ according to $\gamma$. Hence the endpoint coupling $\pi^\varepsilon=(X_0,X_1)_\#P^\varepsilon$ minimizes \begin{align*} \gamma\mapsto H(\gamma\mid Q^\varepsilon) \end{align*} over $\Pi(\mu_0,\mu_1)$.[/step]

[guided]The first point is that the entropy minimization on path space contains no extra endpoint information beyond the law of $(X_0,X_1)$. The reference law $R_{\mu_0}^{\varepsilon}$ has endpoint law $Q^\varepsilon$, where \begin{align*} dQ^\varepsilon(x,y)=q_\varepsilon(x,y)\,d\mu_0(x)\,d\mathcal L^d(y), \end{align*} and \begin{align*} q_\varepsilon(x,y)=(2\pi\varepsilon)^{-d/2}\exp\left(-\frac{|x-y|^2}{2\varepsilon}\right). \end{align*} The conditional law of the Brownian path given $(X_0,X_1)=(x,y)$ is the Brownian bridge from $x$ to $y$ with variance parameter $\varepsilon$. We now use the chain rule for relative entropy under the measurable map \begin{align*} (X_0,X_1):C([0,1];\mathbb R^d)\to\mathbb R^d\times\mathbb R^d. \end{align*} If $P$ has endpoint law $\gamma=(X_0,X_1)_\#P$, then entropy decomposes into an endpoint entropy plus an average conditional entropy: \begin{align*} H(P\mid R_{\mu_0}^{\varepsilon}) = H(\gamma\mid Q^\varepsilon) + \int_{\mathbb R^d\times\mathbb R^d} H(P^{x,y}\mid R_{\mu_0}^{\varepsilon,x,y})\,d\gamma(x,y), \end{align*} where $P^{x,y}$ and $R_{\mu_0}^{\varepsilon,x,y}$ denote regular conditional laws given $(X_0,X_1)=(x,y)$. The second term is nonnegative by the nonnegativity of relative entropy. Therefore \begin{align*} H(P\mid R_{\mu_0}^{\varepsilon})\ge H(\gamma\mid Q^\varepsilon). \end{align*} Equality is achieved by choosing the conditional law $P^{x,y}$ to be exactly the Brownian bridge conditional law $R_{\mu_0}^{\varepsilon,x,y}$ for $\gamma$-almost every $(x,y)$. Thus minimizing over path measures with prescribed endpoint marginals is equivalent to minimizing $H(\gamma\mid Q^\varepsilon)$ over $\gamma\in\Pi(\mu_0,\mu_1)$. Since $P^\varepsilon$ is a path-space minimizer, its endpoint law $\pi^\varepsilon$ is a static entropic minimizer.[/guided]

[step:Rewrite the scaled endpoint entropy as quadratic cost plus a vanishing entropy penalty] Define the product reference measure $M:=\mu_0\otimes\mathcal L^d$ on $\mathbb R^d\times\mathbb R^d$. If $\gamma\in\Pi(\mu_0,\mu_1)$ and $H(\gamma\mid Q^\varepsilon)<\infty$, then $\gamma\ll Q^\varepsilon$. Since $q_\varepsilon>0$ everywhere, this implies $\gamma\ll M$, and the Radon-Nikodym derivatives satisfy \begin{align*} \frac{d\gamma}{dQ^\varepsilon}(x,y) = \frac{d\gamma}{dM}(x,y)\frac{1}{q_\varepsilon(x,y)} \end{align*} for $\gamma$-almost every $(x,y)$. Hence \begin{align*} \varepsilon H(\gamma\mid Q^\varepsilon) = \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\gamma(x,y) + \varepsilon H(\gamma\mid M) + \frac{d\varepsilon}{2}\log(2\pi\varepsilon). \end{align*} The final term is independent of $\gamma$. Therefore $\pi^\varepsilon$ also minimizes \begin{align*} F_\varepsilon:\Pi(\mu_0,\mu_1)\to(-\infty,\infty], \qquad F_\varepsilon(\gamma) = \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\gamma(x,y) + \varepsilon H(\gamma\mid M). \end{align*} [/step]

[step:Use finite entropy of the terminal marginal to produce finite competitors] Since $\mu_1=\rho_1\mathcal L^d$ and \begin{align*} \int_{\mathbb R^d}\rho_1(y)\log\rho_1(y)\,d\mathcal L^d(y)<\infty, \end{align*} the product coupling \begin{align*} \gamma_0:=\mu_0\otimes\mu_1 \end{align*} belongs to $\Pi(\mu_0,\mu_1)$ and satisfies $\gamma_0\ll M$. Moreover, \begin{align*} H(\gamma_0\mid M) = \int_{\mathbb R^d}\rho_1(y)\log\rho_1(y)\,d\mathcal L^d(y)<\infty. \end{align*} Because $\mu_0$ and $\mu_1$ are compactly supported, the quadratic cost integral of $\gamma_0$ is finite. Hence $F_\varepsilon(\gamma_0)<\infty$ for every $\varepsilon>0$, so the static entropic problem has finite value. This also gives finite value for the corresponding path-space Schrödinger bridge problem through the Brownian-bridge reconstruction in the previous step. [/step]

[step:Pass endpoint marginals to the weak limit] For every $n\in\mathbb N$, the measure $\pi^{\varepsilon_n}$ belongs to $\Pi(\mu_0,\mu_1)$. Let $p_0:\mathbb R^d\times\mathbb R^d\to\mathbb R^d$ be the first coordinate projection $p_0(x,y)=x$, and let $p_1:\mathbb R^d\times\mathbb R^d\to\mathbb R^d$ be the second coordinate projection $p_1(x,y)=y$. The maps $p_0$ and $p_1$ are continuous. Weak convergence of $\pi^{\varepsilon_n}$ to $\pi$ therefore implies \begin{align*} (p_0)_\#\pi=\lim_{n\to\infty}(p_0)_\#\pi^{\varepsilon_n}=\mu_0 \end{align*} and \begin{align*} (p_1)_\#\pi=\lim_{n\to\infty}(p_1)_\#\pi^{\varepsilon_n}=\mu_1. \end{align*} Thus $\pi\in\Pi(\mu_0,\mu_1)$. [/step]

[step:Apply the small-noise Gamma convergence of entropic transport] Define $F_0:\Pi(\mu_0,\mu_1)\to[0,\infty)$ by \begin{align*} F_0(\gamma)= \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\gamma(x,y). \end{align*} We apply the small-noise $\Gamma$-convergence theorem for static entropic optimal transport with reference measures $Q^\varepsilon$ and product base measure $M=\mu_0\otimes\mathcal L^d$. The theorem requires that the marginals are compactly supported Borel probability measures on $\mathbb R^d$, that the terminal marginal is absolutely continuous with finite Boltzmann entropy relative to $\mathcal L^d$, and that the endpoint kernel is the Brownian heat kernel with variance parameter $\varepsilon$. These hypotheses hold here by the assumptions on $\mu_0$ and $\mu_1$, by the finite entropy condition on $\rho_1$, and by the definition of $q_\varepsilon$. Therefore, on $\Pi(\mu_0,\mu_1)$ equipped with weak convergence, the functionals $F_\varepsilon$ defined above $\Gamma$-converge as $\varepsilon\downarrow0$ to $F_0$. Equivalently, for every sequence $\gamma_\varepsilon\in\Pi(\mu_0,\mu_1)$ converging weakly to $\gamma\in\Pi(\mu_0,\mu_1)$, \begin{align*} F_0(\gamma)\le \liminf_{\varepsilon\downarrow0}F_\varepsilon(\gamma_\varepsilon), \end{align*} and for every $\gamma\in\Pi(\mu_0,\mu_1)$ there exists a recovery sequence $\tilde\gamma_\varepsilon\in\Pi(\mu_0,\mu_1)$ converging weakly to $\gamma$ such that \begin{align*} \limsup_{\varepsilon\downarrow0}F_\varepsilon(\tilde\gamma_\varepsilon)\le F_0(\gamma). \end{align*} The recovery part allows competitors with infinite $H(\gamma\mid M)$: the theorem supplies finite-entropy approximating couplings with the same marginals, using compact support and the finite entropy of the terminal marginal. [/step]

[step:Identify the limiting minimal value]Define the quadratic optimal transport value \begin{align*} m_0:= \inf_{\gamma\in\Pi(\mu_0,\mu_1)} \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\gamma(x,y). \end{align*} Since $\mu_0$ and $\mu_1$ are compactly supported, every $\gamma\in\Pi(\mu_0,\mu_1)$ is supported in a common compact subset of $\mathbb R^d\times\mathbb R^d$. The cost function $c:\mathbb R^d\times\mathbb R^d\to[0,\infty)$ defined by $c(x,y)=|x-y|^2/2$ is continuous and bounded on that compact set. The $\Gamma$-convergence statement and compactness of $\Pi(\mu_0,\mu_1)$ under weak convergence imply convergence of the minimum values: \begin{align*} \lim_{\varepsilon\downarrow0}\inf_{\gamma\in\Pi(\mu_0,\mu_1)}F_\varepsilon(\gamma)=m_0. \end{align*} Because $\pi^\varepsilon$ minimizes $F_\varepsilon$, this gives \begin{align*} \lim_{n\to\infty}F_{\varepsilon_n}(\pi^{\varepsilon_n})=m_0. \end{align*}[/step]

[guided]The role of $\Gamma$-convergence is to transfer minimizers of the perturbed entropic problems to minimizers of the limiting quadratic problem. Let \begin{align*} m_\varepsilon:=\inf_{\gamma\in\Pi(\mu_0,\mu_1)}F_\varepsilon(\gamma). \end{align*} Because $\pi^\varepsilon$ is a minimizer, $F_\varepsilon(\pi^\varepsilon)=m_\varepsilon$. We first get the upper bound on limiting minimum values. Choose any $\eta\in\Pi(\mu_0,\mu_1)$. By the recovery part of the small-noise $\Gamma$-convergence theorem, there exists $\eta_\varepsilon\in\Pi(\mu_0,\mu_1)$ with $\eta_\varepsilon$ converging weakly to $\eta$ and \begin{align*} \limsup_{\varepsilon\downarrow0}F_\varepsilon(\eta_\varepsilon)\le F_0(\eta). \end{align*} Since $m_\varepsilon\le F_\varepsilon(\eta_\varepsilon)$ for every $\varepsilon>0$, we obtain \begin{align*} \limsup_{\varepsilon\downarrow0}m_\varepsilon\le F_0(\eta). \end{align*} Taking the infimum over all $\eta\in\Pi(\mu_0,\mu_1)$ gives \begin{align*} \limsup_{\varepsilon\downarrow0}m_\varepsilon\le m_0. \end{align*} For the lower bound, take any sequence $\varepsilon_k\downarrow0$. Since $\mu_0$ and $\mu_1$ are compactly supported, the family $\Pi(\mu_0,\mu_1)$ is tight, and every sequence of couplings has a weakly convergent subsequence. Let $\gamma_k\in\Pi(\mu_0,\mu_1)$ satisfy \begin{align*} F_{\varepsilon_k}(\gamma_k)\le m_{\varepsilon_k}+\varepsilon_k. \end{align*} After passing to a subsequence, assume $\gamma_k$ converges weakly to some $\gamma\in\Pi(\mu_0,\mu_1)$. The liminf part of $\Gamma$-convergence gives \begin{align*} F_0(\gamma)\le\liminf_{k\to\infty}F_{\varepsilon_k}(\gamma_k). \end{align*} Since $m_0\le F_0(\gamma)$, we get \begin{align*} m_0\le\liminf_{k\to\infty}F_{\varepsilon_k}(\gamma_k). \end{align*} The near-minimality inequality also gives \begin{align*} F_{\varepsilon_k}(\gamma_k)\le m_{\varepsilon_k}+\varepsilon_k. \end{align*} Because $\varepsilon_k\to0$, taking liminf on the right yields \begin{align*} \liminf_{k\to\infty}F_{\varepsilon_k}(\gamma_k)\le\liminf_{k\to\infty}m_{\varepsilon_k}. \end{align*} Therefore \begin{align*} m_0\le\liminf_{k\to\infty}m_{\varepsilon_k}. \end{align*} Together with the upper bound, this proves \begin{align*} \lim_{\varepsilon\downarrow0}m_\varepsilon=m_0. \end{align*} Substituting the minimizing endpoint coupling $\pi^\varepsilon$ gives \begin{align*} \lim_{n\to\infty}F_{\varepsilon_n}(\pi^{\varepsilon_n})=m_0. \end{align*}[/guided]

[step:Conclude optimality of the weak endpoint limit] Apply the liminf inequality from the small-noise $\Gamma$-convergence theorem to the weakly convergent sequence $\pi^{\varepsilon_n}\to\pi$. Since $\pi\in\Pi(\mu_0,\mu_1)$, we have \begin{align*} \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\pi(x,y) = F_0(\pi) \le \liminf_{n\to\infty}F_{\varepsilon_n}(\pi^{\varepsilon_n}). \end{align*} The previous step identifies the right-hand side as $m_0$. Therefore \begin{align*} \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\pi(x,y)\le m_0. \end{align*} The reverse inequality follows from the definition of $m_0$, because $\pi\in\Pi(\mu_0,\mu_1)$. Hence \begin{align*} \int_{\mathbb R^d\times\mathbb R^d}\frac{|x-y|^2}{2}\,d\pi(x,y)=m_0. \end{align*} Thus $\pi$ is an optimal coupling of $\mu_0$ and $\mu_1$ for the cost $(x,y)\mapsto |x-y|^2/2$, as claimed. [/step]