Finite Gaussian increment comparison

Take independent copies of $X,Y$, with covariance entries $(C_X{)}_{ij}=\mathbb{E}(X_i{X}_j)$ and $(C_Y{)}_{ij}=\mathbb{E}(Y_i{Y}_j)$, and represent them as independent linear images of standard Gaussian coordinates using [A1]. For $0<t<1$ put $Z(t)=\sqrt{t}Y+\sqrt{1-t}X$ and $F_{\beta }(z)={\beta }^{-1}\log {\sum }_{i=1}^N{e}^{\beta z_i}$ with $w_i=e^{\beta z_i}\mathrm{/}{\sum }_j{e}^{\beta z_j}$. The soft maximum has at most linear growth, so its Gaussian slice is integrable. On each compact subinterval of $(0,1)$, $\mathrm{\mid }{\mathrm{\partial }}_t{F}_{\beta }(Z(t))\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\nabla }{F}_{\beta }{\mathrm{\parallel }}_2(\mathrm{\parallel }Y{\mathrm{\parallel }}_2\mathrm{/}(2\sqrt{t})+\mathrm{\parallel }X{\mathrm{\parallel }}_2\mathrm{/}(2\sqrt{1-t}))$ has an integrable uniform bound, so [F3] passes the derivative under the finite Gaussian integral. For the coordinate integration by parts, fix all coordinates but $r=g_j$, and let $\phi (r)=(2\pi {)}^{-1\mathrm{/}2}{e}^{-r^2\mathrm{/}2}$. The functions to which it is applied are first derivatives of $F_{\beta }$ composed with linear maps; they and their $r$ derivatives are smooth and bounded for fixed $\beta$. On $[-R,R]$, [F5] with $u=f$ and $v=\phi$ gives ${\int }_{-R}^{R}rf(r)\phi (r)dr={\int }_{-R}^R{f}^{\mathrm{\prime }}(r)\phi (r)dr-[f(r)\phi (r){]}_{-R}^R$. The boundary term tends to zero since $f$ is bounded and $\phi (\pm R)\to 0$. The integrands are dominated by constant multiples of $\mathrm{\mid }r\mathrm{\mid }\phi (r)$ and $\phi (r)$, respectively, so [F4] takes $R\to \mathrm{\infty }$. Both majorants are integrable. Their bounds are uniform in the other coordinates, and [F6] integrates the resulting identity over those coordinates. Applying this coordinatewise to a linear image $W=Lg$ gives $\mathbb{E}[W_{i}f(W)]={\sum }_j\mathrm{Cov}(W_i,W_j)\mathbb{E}[{\mathrm{\partial }}_{j}f(W)]$, including singular covariance matrices. Apply it separately to $X,Y$ in the derivative to obtain $\frac{d}{dt}\mathbb{E}{F}_{\beta }(Z(t))=\frac{1}{2}{\sum }_{i,j}(C_Y-C_X{)}_{ij}\mathbb{E}[{\mathrm{\partial }}_{ij}{F}_{\beta }(Z(t))]$. Since ${\mathrm{\partial }}_{ij}{F}_{\beta }=\beta (w_i{\mathbf{1}}_{i=j}-w_i{w}_j)$, symmetry rewrites this as $\frac{\beta }{4}\mathbb{E}{\sum }_{i,j}{w}_i{w}_j(\mathbb{E}(Y_i-Y_j{)}^2-\mathbb{E}(X_i-X_j{)}^2)\ge 0$ by the assumed increment order.

Integrate the nonnegative derivative on $(0,1)$. As $t\downarrow 0$ or $t\uparrow 1$, $Z(t)$ converges pointwise to $X$ or $Y$ and $\mathrm{\mid }{F}_{\beta }(Z(t))\mathrm{\mid }$ is bounded by a constant plus $\mathrm{\parallel }X{\mathrm{\parallel }}_2+\mathrm{\parallel }Y{\mathrm{\parallel }}_2$, so [F4] gives endpoint continuity and $\mathbb{E}{F}_{\beta }(Y)\ge \mathbb{E}{F}_{\beta }(X)$. For every $z\in {\mathbb{R}}^N$, ${\max }_i{z}_i\le F_{\beta }(z)\le {\max }_i{z}_i+\log N\mathrm{/}\beta$. Letting $\beta \to \mathrm{\infty }$ proves the asserted expected-maximum comparison.