Universal bound conjecture for Gaussian entropic optimal transport

The sharp universal constant is $d/2$

Let $A,B\in\mathbb R^{d\times d}$ be positive definite. Let $\pi_\varepsilon$ be the Gaussian entropic optimal coupling, with precisely the regularization parameter $\varepsilon$ used in the cited paper, and let $\pi_0$ be the ordinary optimal Gaussian coupling.

Define the unique positive-definite optimal transport map

$$T=A^{-1/2}(A^{1/2}BA^{1/2})^{1/2}A^{-1/2}, \qquad TAT=B.$$

We prove the exact asymptotic identity

$$\boxed{\displaystyle \lim_{\varepsilon\downarrow0} \frac{\mathcal W_2^2(\pi_\varepsilon,\pi_0)}{\varepsilon} =\operatorname{tr}\!\left(T(I+T^2)^{-1}\right) =\sum_{j=1}^d\frac{t_j}{1+t_j^2} \leq\frac d2,}$$

where $t_1,\ldots,t_d>0$ are the eigenvalues of $T$. Equality holds if and only if $A=B$. Consequently, the optimal universal constant in Conjecture 3.1 is

$$C_d=\frac d2.$$

No commutativity assumption on $A$ and $B$ is required.

Properly aligned factors and the exact coupling formula

Choose invertible properly aligned factors $G,M$ of the two covariance matrices:

$$A=GG^*,\qquad B=MM^*,\qquad K=G^*M=M^*G\succ0.$$

For example, take $G=A^{1/2}$ and $M=TA^{1/2}$. Then

$$G^*M=A^{1/2}TA^{1/2} =(A^{1/2}BA^{1/2})^{1/2}\succ0.$$

Introduce the positive-definite matrices

$$C=G^*G,\qquad D=M^*M,\qquad S=C+D.$$

The source's Theorem 3.4 gives the entropic correlation operator

$$R_\varepsilon=f_\varepsilon(K), \qquad f_\varepsilon(x) =\frac{2x}{\sqrt{4x^2+\varepsilon^2}+\varepsilon} =\frac{\sqrt{4x^2+\varepsilon^2}-\varepsilon}{2x} \quad(x>0).$$

By Theorem 3.10 of the same source,

$$\mathcal W_2^2(\pi_\varepsilon,\pi_0) =2\operatorname{tr}\!\left(S-Q_\varepsilon^{1/2}\right),$$

where

$$Q_\varepsilon =C^2+D^2+CR_\varepsilon D+DR_\varepsilon C.$$

Although $\pi_0$ has singular covariance on $\mathbb R^{2d}$, the matrix being differentiated below is $Q_0=S^2\succ0$, so its positive square root is smooth.

The cancellation that removes all covariance dependence

Proper alignment implies

$$M=G^{-*}K, \qquad D=M^*M=KC^{-1}K.$$

In particular,

$$CK^{-1}D=K, \qquad DK^{-1}C=K.$$

Since $K\succ0$, the scalar formula for $f_\varepsilon$, applied through the spectral calculus, gives

$$R_\varepsilon =I-\frac{\varepsilon}{2}K^{-1} +O(\varepsilon^2).$$

Substitution into the exact coupling formula therefore yields

$$\begin{aligned} Q_\varepsilon &=C^2+D^2+CD+DC -\frac{\varepsilon}{2} \left(CK^{-1}D+DK^{-1}C\right) +O(\varepsilon^2)\\ &=S^2-\varepsilon K+O(\varepsilon^2). \end{aligned}$$

Write

$$Q_\varepsilon^{1/2}=S+\varepsilon L+O(\varepsilon^2).$$

Squaring this expansion, without assuming that $S$ and $L$ commute, gives the Sylvester equation

$$SL+LS=-K.$$

Multiplication by $S^{-1}$ followed by cyclicity of the trace shows that

$$2\operatorname{tr}(L) =-\operatorname{tr}(S^{-1}K).$$

Hence

$$\lim_{\varepsilon\downarrow0} \frac{\mathcal W_2^2(\pi_\varepsilon,\pi_0)}{\varepsilon} =\operatorname{tr}(S^{-1}K).$$

Sharp bound, equality, and the transport-map formula

The aligned-factor identity gives the exact positive-semidefinite factorization

$$S-2K =C+KC^{-1}K-2K =(C-K)C^{-1}(C-K)\succeq0.$$

Thus

$$0\prec S^{-1/2}KS^{-1/2} \preceq\frac12I,$$

and consequently

$$\operatorname{tr}(S^{-1}K) =\operatorname{tr}(S^{-1/2}KS^{-1/2}) \leq\frac d2.$$

Equality forces $S=2K$, hence $C=K$, and then $G^*M=G^*G$. Since $G$ is invertible, $M=G$, so $A=B$. Conversely, $A=B$ permits $M=G$ and attains $d/2$.

Finally, proper alignment implies that

$$T=MG^{-1}=T^*\succ0,$$

and this is the unique positive-definite solution of $TAT=B$. Therefore

$$K=G^*TG, \qquad S=G^*(I+T^2)G.$$

Cyclicity of trace now gives the stronger intrinsic identity

$$\operatorname{tr}(S^{-1}K) =\operatorname{tr}\!\left((I+T^2)^{-1}T\right) =\sum_{j=1}^d\frac{t_j}{1+t_j^2}.$$

The exact deficit from the extremal constant is

$$\frac d2- \lim_{\varepsilon\downarrow0} \frac{\mathcal W_2^2(\pi_\varepsilon,\pi_0)}{\varepsilon} =\frac12\sum_{j=1}^d\frac{(t_j-1)^2}{1+t_j^2}.$$

This proves the full universal-bound conjecture, determines its sharp constant, characterizes every equality case, and identifies the exact rate for all positive-definite covariance pairs. The spectral-shrinkage and coupling identities used above are Theorems 3.4 and 3.10 of Ho Yun, Spectral Shrinkage of Gaussian Entropic Optimal Transport, arXiv:2512.19457v2, where the universal-bound problem is stated as Conjecture 3.1.

The sharp universal Gaussian EOT coupling bound

Statement

Let A,B be positive-definite covariance matrices on R^d. Let pi_epsilon be their Gaussian entropic optimal-transport coupling and let pi_0 be the unregularized Gaussian optimal coupling. We prove that the limit in MathDB #372212 exists and satisfies

$$\lim_{\varepsilon\downarrow0} \frac{\mathcal W_2^2(\pi_\varepsilon,\pi_0)}{\varepsilon} \leq \frac d2. \tag{1}$$

The constant is sharp. In fact, equality holds exactly when A=B. Consequently the optimal universal constant is

$$\boxed{C_d=d/2}.$$

Throughout, epsilon has the normalization used by the source and by the MathDB statement: the entropic objective contains 2 epsilon KL. Rescaling that objective rescales the parameter and therefore the displayed constant.

Aligned square roots

Choose the canonical properly aligned Green operators G_0,M_0 used by Theorem 3.10 of the source, and abbreviate them to G,M, so

$$GG^*=A,\qquad MM^*=B,\qquad C:=G^*M=M^*G\succ0.$$

All three matrices are invertible because A and B are positive definite. Put

$$X=G^*G,\qquad Y=M^*M,\qquad S=X+Y.$$

From G^*M=C we have M=G^{-*}C, and therefore

$$Y=CX^{-1}C. \tag{2}$$

In particular,

$$XC^{-1}Y=C,\qquad YC^{-1}X=C. \tag{3}$$

Differentiate the exact distance formula

The source defines

$$R_\varepsilon=f_\varepsilon(C),\qquad f_\varepsilon(t)= \frac{2t}{\sqrt{4t^2+\varepsilon^2}+\varepsilon} =\frac{\sqrt{4t^2+\varepsilon^2}-\varepsilon}{2t}.$$

Because the spectrum of C is bounded away from zero,

$$R_\varepsilon =I-\frac{\varepsilon}{2}C^{-1}+O(\varepsilon^2) \tag{4}$$

in operator norm. Theorem 3.10 gives

$$\mathcal W_2^2(\pi_\varepsilon,\pi_0) =2\operatorname{tr}\!\left(S-\sqrt{Q_\varepsilon}\right), \tag{5}$$

where

$$Q_\varepsilon =X^2+Y^2+XR_\varepsilon Y+YR_\varepsilon X.$$

Equations (3)--(4) now collapse the first-order perturbation:

$$Q_\varepsilon =S^2-\varepsilon C+O(\varepsilon^2). \tag{6}$$

For completeness, if Q(t)=S^2+tH+O(t^2) and sqrt(Q(t))=S+tZ+O(t^2), differentiating the square gives the Sylvester equation

$$SZ+ZS=H.$$

Multiplication by S^{-1}, followed by cyclicity of trace, yields

$$\operatorname{tr}Z =\frac12\operatorname{tr}(S^{-1}H). \tag{7}$$

Here S^2 is positive definite, so Q_epsilon remains positive definite near zero and its principal square root is Frechet differentiable there.

Applying (7) to (6), then differentiating (5), proves the exact formula

$$\boxed{ \lim_{\varepsilon\downarrow0} \frac{\mathcal W_2^2(\pi_\varepsilon,\pi_0)}{\varepsilon} =\operatorname{tr}(S^{-1}C). } \tag{8}$$

Thus the conjectured limsup is actually a limit.

The universal bound and equality case

Proper alignment supplies the decisive order relation:

$$S-2C =X+Y-G^*M-M^*G =(G-M)^*(G-M)\succeq0. \tag{9}$$

Conjugating (9) by S^{-1/2} gives

$$0\prec S^{-1/2}CS^{-1/2}\preceq\frac12I.$$

Taking traces and using cyclicity in (8),

$$\operatorname{tr}(S^{-1}C) =\operatorname{tr}(S^{-1/2}CS^{-1/2})\leq\frac d2,$$

which proves (1).

If equality holds, the positive-semidefinite matrix I/2-S^{-1/2}CS^{-1/2} has trace zero, hence is zero. Therefore S=2C, and (9) forces G=M, so A=B. Conversely, when A=B we may take G=M; then (8) gives tr((2X)^{-1}X)=d/2. This also proves sharpness.

Verification scope

The proof above is finite-dimensional matrix analysis and does not depend on computation. The companion script reconstructs aligned roots for deterministic positive-definite test matrices, checks identities (2)--(3) and (9), and confirms that the exact finite-epsilon formula converges to (8), including the equality case.