[proofplan] Fix two base points $x_0,x_1$ in the positive support of the marginal $g$ and compare the fibres of $f$ above $x_0$, $x_1$, and their convex interpolation. Log-concavity of $f$ gives exactly the pointwise hypothesis of the [Prékopa-Leindler inequality](/theorems/4118) on $\mathbb{R}^m$. Applying that inequality to the three fibre functions yields the desired multiplicative lower bound for $g$ at the interpolated point. The same inequality also shows that the positive support of $g$ is convex. [/proofplan]

[step:Reduce log-concavity of the marginal to an inequality between three fibre functions]Let $x_0,x_1 \in S$, and let $t \in [0,1]$. Define the interpolated base point $x_t \in \mathbb{R}^n$ by \begin{align*} x_t := (1-t)x_0+t x_1. \end{align*} If $t=0$ or $t=1$, then \begin{align*} g(x_t)=g(x_0)^{1-t}g(x_1)^t \end{align*} by direct substitution, so assume $0<t<1$. Define measurable fibre functions \begin{align*} h_0:\mathbb{R}^m &\to [0,\infty), & y &\mapsto f(x_0,y),\\ h_1:\mathbb{R}^m &\to [0,\infty), & y &\mapsto f(x_1,y),\\ h_t:\mathbb{R}^m &\to [0,\infty), & y &\mapsto f(x_t,y). \end{align*} By definition of $g$, \begin{align*} \int_{\mathbb{R}^m} h_0(y)\,d\mathcal{L}^m(y) &= g(x_0),\\ \int_{\mathbb{R}^m} h_1(y)\,d\mathcal{L}^m(y) &= g(x_1),\\ \int_{\mathbb{R}^m} h_t(y)\,d\mathcal{L}^m(y) &= g(x_t). \end{align*} Because $x_0,x_1 \in S$, the first two integrals are positive and finite.[/step]

[guided]We want to prove a log-concavity inequality for $g$, but $g$ is defined by integrating $f$ along fibres over $\mathbb{R}^m$. The natural strategy is therefore to freeze the $\mathbb{R}^n$ variable and regard each fibre as a function of the remaining $\mathbb{R}^m$ variable. Let $x_0,x_1 \in S$, and let $t \in [0,1]$. Define \begin{align*} x_t := (1-t)x_0+t x_1. \end{align*} When $t=0$, the desired inequality reads $g(x_0)\geq g(x_0)$; when $t=1$, it reads $g(x_1)\geq g(x_1)$. Thus the only substantive case is $0<t<1$. We now define the three fibre functions \begin{align*} h_0:\mathbb{R}^m &\to [0,\infty), & y &\mapsto f(x_0,y),\\ h_1:\mathbb{R}^m &\to [0,\infty), & y &\mapsto f(x_1,y),\\ h_t:\mathbb{R}^m &\to [0,\infty), & y &\mapsto f(x_t,y). \end{align*} These functions are measurable because they are restrictions of the measurable function $f$ to affine fibres. Their integrals are exactly the marginal values: \begin{align*} \int_{\mathbb{R}^m} h_0(y)\,d\mathcal{L}^m(y) &= g(x_0),\\ \int_{\mathbb{R}^m} h_1(y)\,d\mathcal{L}^m(y) &= g(x_1),\\ \int_{\mathbb{R}^m} h_t(y)\,d\mathcal{L}^m(y) &= g(x_t). \end{align*} Since $x_0,x_1 \in S$, we know $g(x_0)>0$ and $g(x_1)>0$. The standing assumption also gives $g(x_0)<\infty$ and $g(x_1)<\infty$. These are precisely the finiteness and positivity conditions needed for the multiplicative right-hand side in the desired inequality.[/guided]

[step:Verify the Prékopa-Leindler pointwise hypothesis from log-concavity of $f$]For arbitrary $y_0,y_1 \in \mathbb{R}^m$, log-concavity of $f$ applied to the points $(x_0,y_0)$ and $(x_1,y_1)$ gives \begin{align*} h_t((1-t)y_0+t y_1) &= f((1-t)x_0+t x_1,(1-t)y_0+t y_1)\\ &\geq f(x_0,y_0)^{1-t}f(x_1,y_1)^t\\ &= h_0(y_0)^{1-t}h_1(y_1)^t. \end{align*} Thus the functions $h_0,h_1,h_t$ satisfy the pointwise hypothesis of the Prékopa-Leindler inequality on $\mathbb{R}^m$ for the parameter $t$.[/step]

[guided]The key point is that interpolation in the fibre variable must be paired with the same interpolation in the base variable. Take arbitrary $y_0,y_1 \in \mathbb{R}^m$. Then the interpolated point in the full space $\mathbb{R}^{n+m}$ is \begin{align*} ((1-t)x_0+t x_1,(1-t)y_0+t y_1). \end{align*} By the definition of $x_t$, this point is also \begin{align*} (x_t,(1-t)y_0+t y_1). \end{align*} Applying log-concavity of $f$ to $(x_0,y_0)$ and $(x_1,y_1)$ gives \begin{align*} h_t((1-t)y_0+t y_1) &= f(x_t,(1-t)y_0+t y_1)\\ &= f((1-t)x_0+t x_1,(1-t)y_0+t y_1)\\ &\geq f(x_0,y_0)^{1-t}f(x_1,y_1)^t\\ &= h_0(y_0)^{1-t}h_1(y_1)^t. \end{align*} This is exactly the pointwise comparison required by the Prékopa-Leindler inequality: the middle fibre $h_t$ dominates the geometric interpolation of the endpoint fibres $h_0$ and $h_1$ along every interpolated pair of fibre points.[/guided]

[step:Apply Prékopa-Leindler to obtain the marginal log-concavity inequality]By the Prékopa-Leindler inequality on $\mathbb{R}^m$ (citing a result not yet in the wiki: Prékopa-Leindler Inequality), applied to the non-negative [measurable functions](/page/Measurable%20Functions) $h_0,h_1,h_t$ and the parameter $t \in (0,1)$, the pointwise inequality from the previous step implies \begin{align*} \int_{\mathbb{R}^m} h_t(y)\,d\mathcal{L}^m(y) \geq \left(\int_{\mathbb{R}^m} h_0(y)\,d\mathcal{L}^m(y)\right)^{1-t} \left(\int_{\mathbb{R}^m} h_1(y)\,d\mathcal{L}^m(y)\right)^t. \end{align*} Substituting the definitions of the fibre integrals gives \begin{align*} g(x_t) \geq g(x_0)^{1-t}g(x_1)^t. \end{align*} Because $g(x_0)>0$ and $g(x_1)>0$, the right-hand side is positive, so $g(x_t)>0$. Hence $x_t \in S$. Since this holds for every $x_0,x_1 \in S$ and every $t \in [0,1]$, the set $S$ is convex and the displayed inequality proves that $g$ is log-concave on $S$.[/step]

[guided]We now use the functional form of Brunn-Minkowski, namely the Prékopa-Leindler inequality on $\mathbb{R}^m$ (citing a result not yet in the wiki: Prékopa-Leindler Inequality). Its hypotheses are satisfied here: $h_0,h_1,h_t:\mathbb{R}^m\to[0,\infty)$ are measurable, $0<t<1$, and the previous step proved that for every $y_0,y_1 \in \mathbb{R}^m$, \begin{align*} h_t((1-t)y_0+t y_1)\geq h_0(y_0)^{1-t}h_1(y_1)^t. \end{align*} Therefore Prékopa-Leindler gives \begin{align*} \int_{\mathbb{R}^m} h_t(y)\,d\mathcal{L}^m(y) \geq \left(\int_{\mathbb{R}^m} h_0(y)\,d\mathcal{L}^m(y)\right)^{1-t} \left(\int_{\mathbb{R}^m} h_1(y)\,d\mathcal{L}^m(y)\right)^t. \end{align*} Using the identities between these integrals and the marginal $g$, this becomes \begin{align*} g((1-t)x_0+t x_1) = g(x_t) \geq g(x_0)^{1-t}g(x_1)^t. \end{align*} This proves the log-concavity inequality. It also proves convexity of the positive support: since $x_0,x_1 \in S$, both $g(x_0)$ and $g(x_1)$ are positive, so \begin{align*} g(x_0)^{1-t}g(x_1)^t>0. \end{align*} The inequality therefore implies $g(x_t)>0$, hence $x_t \in S$. Including the already handled endpoint cases $t=0$ and $t=1$, the conclusion holds for every $t \in [0,1]$. Thus $S$ is convex and $g$ is log-concave on $S$.[/guided]