For simplicity we stick to the euclidean case! Let $\mu(dx)=\r(x)\,\l(dx)$ be a probability measure on $\R^n$ with density $\r$ with respect to Lebesgue-measure $\l$, $p< q$ and $f:\R^n\rar\R_0^+$ measurable. We are aiming at constructing another 'minimal' function $g:\R^n\rar\R_0^+$ such that: $$ \Big(\int f^q\,d\mu\Big)^{1/q}\leq\Big(\int g^p\,d\mu\Big)^{1/p} $$ or equivalently: $$ \Big(\int f^q\,d\mu\Big)^{p/q}\leq\int g^p\,d\mu $$ In order to accomplish this we utilize the Prekopa-Leindler inequality on $\R^n$: Suppose $s\in(0,1)$ and $F,G$ are none-negative measurable functions on $\R^n$ such that $$ \forall x,y\in\R^n:\quad G((1-s)x+s y)\r((1-s)x+s y)\geq\r(x)^{1-s}F(y)^s\r(y)^s $$ Then $$ \Big(\int F\,d\mu\Big)^{s} =\Big(\int F\r\,d\l\Big)^{s}\Big(\int\r\,d\l\Big)^{1-s} \leq\int G\r\,d\l =\int G\,d\mu $$ We obviously have to put $s=p/q$, $F=f^q$ and $g=G^{1/p}$. Putting $z=(1-s)x+s y$ we get: $x=(z-s y)/(1-s)$ and thus a 'minimal' function $G$ is given by: $$ G(z) =\sup\Big\{\r((z-s y)/(1-s))^{1-s}F(y)^s\r(y)^s\r(z)^{-1}: y\in\R^n\Big\} $$ Therefore: \begin{eqnarray*} g(z) &=&\sup\Big\{\r\Big(\frac{z-s y}{1-s}\Big)^{(1-s)/p}F(y)^{s/p}\r(y)^{s/p}\r(z)^{-1/p}: y\in\R^n\Big\}\\ &=&\sup\Big\{\r\Big(\frac{qz-py}{q-p}\Big)^{1/p-1/q}\frac{\r(y)^{1/q}}{\r(z)^{1/p}}\,f(y): y\in\R^n\Big\} \end{eqnarray*} Notice that normalizing constants do not matter: if $\r(x)=h(x)/Z$ for some none-negative integrable function $h$ (and $Z\colon=\int h(x)\,dx$), then the 'minimal' function $g$ is still given by $$ g(z)=\sup\Big\{h\Big(\frac{qz-py}{q-p}\Big)^{1/p-1/q}\frac{h(y)^{1/q}}{h(z)^{1/p}} \,f(y): y\in\R^n\Big\}~. $$ For example take $h(x)=e^{-\norm x^2/2}$, then \begin{eqnarray*} &=&-\norm{qz-py}^2/2pq(q-p)-\norm y^2/2q+\norm z^2/2p\\ &=&-(q/p(q-p)-1/p)\norm z^2/2-(p/q(q-p)+1/q)\norm y^2/2+\la z,y\ra/(q-p)\\ &=&-(1/2(q-p)\norm z^2+1/2(q-p)\norm y^2-2\la z,y\ra/2(q-p))\\ &=&-\norm{z-y}^2/2(q-p) \end{eqnarray*} Taking logarithms we get: $$ -\log g(z)=\inf\Big\{\frac{\norm{z-y}^2}{2(q-p)}-\log f(y): y\in\R^n\Big\}~. $$ i.e. $-\log g$ is the inf-convolution of $x\mapsto\norm x^2/2(q-p)$ and $-\log f$, which is - by chance? a solution to the Hamilton-Jacobi equation $$ \pa_tu=-\norm{\nabla u}^2/2,\quad u(0,x)=-\log f(x) $$ at $t=q-p$.