Prekopa-Leindler inequality

\textbf{Theorem} (Prekopa-Leindler inequality (an integral form of the Brunn-Minkowski inequality)) Let $0< \lambda <1$ and $f$, $g$, and $h$ be nonnengative integrable function on $\mathbb{R}^n$ such that \begin{equation} \label{eq:15} h((1- \lambda)x + \lambda y) \geq f(x)^{1- \lambda} g(y)^{\lambda} \end{equation} for all $x, y \in \mathbb{R}^n$. Then, we have: \begin{equation} \int_{\mathbb{R}^n} h(x) dx \geq (\int_{\mathbb{R}^n})^{1- \lambda} ( \int_{\mathbb{R}^n}g(x) dx)^ \lambda \end{equation}

\begin{proof} We assume that $A$, $B$ and $A+B$ are measurable sets, and we shift them such that $A \cap B = \phi$, and so \begin{equation} \label{eq:17} A+B \geq A \cup B \end{equation} and so by almost disjointedness we get \begin{equation} \label{eq:18} \mu(A+B) \geq \mu(A) + \mu(B) \end{equation} Now we show the inequality by induction: \\ \begin{enumerate} \item Basic Step $n=1$: Define the set \begin{equation} L_h(t)={x : h(x) \geq t} \end{equation} and similarly, \begin{equation} L_f(t)={x : f(x) \geq t} \end{equation} \begin{equation} L_g(t)={y : g(y) \geq t} \end{equation} From \ref{eq:17}, we get \begin{equation} L_h(t) \supseteq \lambda L_f(t) + (1- \lambda ) L_g(t) \end{equation} and so from \ref{eq:18}, we get \begin{equation} \mu(L_h(t))\geq \mu(\lambda L_f(t)) + \mu ((1- \lambda ) L_g(t)) \end{equation} and since we are only considering the one dimensional (n=1) case, we get \begin{equation} \mu(L_h(t))\geq \lambda \mu( L_f(t)) + (1-\lambda) \mu ( L_g(t)) \end{equation} Without loss of generality, assume that $f\geq0$, then by Fubini's theorem we get \begin{equation} \int_{\mathbb{R}} h(x) dx = \int_{t \geq 0} \mu(L_h(t))dt \end{equation} \begin{equation} \begin{split} \int_{\mathbb{R}}h(x)dx &= \int_{t\geq0} \mu(L_h(t))dt\\ &\geq \int_{t\geq0}(\lambda \mu( L_f(t)) + (1-\lambda) \mu ( L_g(t)))dt \\ &= \lambda \int_{t\geq0} \mu( L_f(t)) dt + (1-\lambda) \int_{t\geq0} \mu ( L_g(t))dt\\ &=\lambda \int_{\mathbb{R}} f(x) dx + (1-\lambda) \int_{\mathbb{R}} g(y) dy \end{split} \end{equation} Now, from (\ref{eq:49}, below), we get that \begin{equation} \int_{\mathbb{R}}h(x)dx \geq (\int_{\mathbb{R}} f(x)dx)^{\lambda}(\int_{\mathbb{R}} g(x)dx)^{1-\lambda} \end{equation} \item Inductive Step: Suppose the inequality holds for $\mathbb{R}^n$, now we show it also holds true for $\mathbb{R}^{n+1}$:\\ Let $x, y \in \mathbb{R}^n$, and $\alpha, \beta \in \mathbb{R}$\\ Set \begin{equation} \label{eq:28} \gamma = \lambda \alpha + (1-\lambda) \beta \end{equation} Define the following function for any constant $c$: \begin{equation} \label{eq:29} h_c(x)=h(c,x) \end{equation} where $(c,x) \in \mathbb{R}^{n+1}$. Similarly,\\ \begin{equation} \label{eq:30} f_c(x)=f(c,x) \end{equation} \begin{equation} \label{eq:31} g_c(y)=g(c,y) \end{equation} Hence, from (\ref{eq:28}) and (\ref{eq:29}), we get: \begin{equation} \begin{split} h_{\gamma}(\lambda x +(1- \lambda)y) &= h(\lambda \alpha +(1- \lambda) \beta, \lambda x +(1- \lambda)y)\\ &=h( \lambda (\alpha , x) + (1- \lambda)(\beta, y))\\ \end{split} \end{equation} From (\ref{eq:15}), we get: \begin{equation} h_{\gamma}(\lambda x +(1- \lambda)y) \geq f( \alpha, x)^{\lambda} g(\beta, y)^{1- \lambda} \end{equation} From (\ref{eq:30}) and (\ref{eq:31}), we finally get: \begin{equation} h_{\gamma}(\lambda x +(1- \lambda)y) \geq f_{\alpha}(x)^{\lambda}g_{\beta}(y)^{1- \lambda} \end{equation} Recall that the induction hypothesis applied to the function $f_{\gamma}, f_{\alpha}, g_{\beta}$ is \begin{equation} \label{eq:35}\int_{\mathbb{R}^n}h(x)dx \geq (\int_{\mathbb{R}^n} f(x)dx)^{\lambda}(\int_{\mathbb{R}^n} g(x)dx)^{1-\lambda} \end{equation} For simplicity, set \begin{equation} \label{eq:36} \begin{split} H(\gamma) &= \int_{\mathbb{R}^n} h_{\gamma}(z)dz\\ F(\alpha) &= \int_{\mathbb{R}^n} f_{\alpha}(z)dz\\ G(\beta) &= \int_{\mathbb{R}^n} g_{\beta}(z)dz\\ \end{split} \end{equation} We rewrite (\ref{eq:35}) using (\ref{eq:36}) and (\ref{eq:28}) \begin{equation} H(\lambda \alpha + (1- \lambda ) \beta ) \geq {F(\alpha)}^{\alpha} {G(\beta)}^{1- \lambda} \end{equation} Notice that in the basic case (n=1), we proved the statement for any fixed $\alpha, \beta \in \mathbb{R}$, and so the functions $H, F,$ and $G$ satisfy the hypothesis for the one-dimensional case. Thus, we have \begin{equation} \int_{\mathbb{R}} H(\gamma) d{\gamma} \geq (\int_\mathbb{R} F(\alpha) d\alpha)^{\lambda}(\int_\mathbb{R} G(\beta) d\beta)^{1- \lambda} \end{equation} Finally, applying Fubini's theorem, we get: \begin{equation} \label{eq:39}\int_{\mathbb{R}^{n+1}}h(x)dx \geq (\int_{\mathbb{R}^{n+1}} f(x)dx)^{\lambda}(\int_{\mathbb{R}^{n+1}} g(x)dx)^{1-\lambda} \end{equation} Therefore, from steps 1 and 2 and via induction, we conclude that for any $n \in \mathbb{N}$, we have \begin{equation} \label{eq:39}\int_{\mathbb{R}^{n}}h(x)dx \geq (\int_{\mathbb{R}^{n}} f(x)dx)^{\lambda}(\int_{\mathbb{R}^{n}} g(x)dx)^{1-\lambda} \end{equation} \end{enumerate} \end{proof}

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