Let $f(x)$ be a strictly increasing function such that $\lim\limits\_{x\to\pm\infty}f(x)=\pm\infty$ and $\lim\limits\_{x\to+\infty}f'(x)e^{f(x)-x}=+\infty$. If $f(a)=0$ for some prescribed $a\in\Bbb R$, does a minimum exist for $$\int\_{\Bbb R}xe^{f(x)-x^2}\,dx$$ and if so, what would be the minimiser $f^\*$? The presence of the exponential terms comes from a log-normal distribution integral.


The given conditions mean that $f$ has asymptotic order $x^\alpha$ for some $\alpha\in[1,2)$ as $x\to+\infty$. The goal is to maximise the negative area on the negative reals (whilst minimising the positive area) so perhaps we may need to define $f\in C^1$ piecewise, as there is no constraint on the rate of increase on the negative side.



Does a minimiser exist for this Gaussian-like functional?
$\newcommand\R{\mathbb R}$For any $a\in\R$, there is no minimizer of
\begin{equation\*}
 I(f):=\int\_\R xe^{f(x)-x^2}\,dx \tag{-1}\label{-1}
\end{equation\*}
over all $f\in F\_a$, where $F\_a$ is the set of all strictly increasing functions $f\colon\R\to\R$ such that $\lim\_{x\to\pm\infty}f(x)=\pm\infty$, $\lim\_{x\to\infty}f'(x)e^{f(x)-x}=\infty$, and $f(a)=0$. Also,
\begin{equation\*}
 \inf\_{f\in F\_a}I(f)=0 \tag{0}\label{0}
\end{equation\*}
for any $a\in\R$.


Indeed, take any $a\in\R$ and any $f\in F\_a$.
Since $f$ is strictly increasing, we have
\begin{equation\*}
 xe^{f(x)-x^2}>xe^{f(0)-x^2} \tag{1}\label{1}
\end{equation\*}
for all nonzero real $x$.


Consider now the following two possible cases:


*Case 1: $a\le0$.* Then $f(x)<f(a)=0$ and $x<0$ for real $x<a$ and hence
\begin{equation\*}
\begin{aligned}
 \int\_{-\infty}^a xe^{f(x)-x^2}\,dx&>\int\_{-\infty}^a xe^{-x^2}\,dx. 
\end{aligned}
\tag{2}\label{2}
\end{equation\*}
Also, by \eqref{1},
\begin{equation\*}
\begin{aligned}
 \int\_a^\infty xe^{f(x)-x^2}\,dx&>\int\_a^\infty xe^{f(0)-x^2}\,dx \\ 
 &=e^{f(0)}\int\_a^\infty xe^{-x^2}\,dx \\ 
 &\ge e^{f(a)}\int\_a^\infty xe^{-x^2}\,dx \\ 
 &=\int\_a^\infty xe^{-x^2}\,dx, 
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
because $f$ is increasing, $a\le0$, $\int\_a^\infty xe^{-x^2}\,dx\ge0$,
and $f(a)=0$.
In view of \eqref{-1}, \eqref{2}, \eqref{3}, and the equality $\int\_{-\infty}^\infty xe^{-x^2}\,dx=0$,
\begin{equation\*}
 I(f)>0 \tag{4}\label{4}
\end{equation\*}
in Case 1.


*Case 2: $a>0$.* Then, again by \eqref{1},
\begin{equation\*}
\begin{aligned}
 \int\_{-\infty}^a xe^{f(x)-x^2}\,dx&>\int\_{-\infty}^a xe^{f(0)-x^2}\,dx \\ 
 &=e^{f(0)}\int\_{-\infty}^a xe^{-x^2}\,dx \\ 
 &>e^{f(a)}\int\_{-\infty}^a xe^{-x^2}\,dx \\ 
 &=\int\_{-\infty}^a xe^{-x^2}\,dx, 
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
because $f$ is increasing, $a>0$, $\int\_{-\infty}^a xe^{-x^2}\,dx<0$,
and $f(a)=0$.


Also, here $f(x)>f(a)=0$ and $x>0$ for real $x>a$ and hence
\begin{equation\*}
\begin{aligned}
 \int\_a^\infty xe^{f(x)-x^2}\,dx&>\int\_a^\infty xe^{-x^2}\,dx. 
\end{aligned}
\tag{6}\label{6}
\end{equation\*}


In view of \eqref{-1}, \eqref{5}, \eqref{6}, inequality \eqref{4} holds
in Case 2 as well.


Take now any $k$ and $A$ in $(0,\infty)$, and let
\begin{equation\*}
 f\_{a,k,A}(x):=k(x-a)\,1(x\le a+A)+(kA+2(x-A))\,1(x>a+A)
\end{equation\*}
for real $x$. Then $f\_{a,k,A}\in F\_a$ and, by dominated convergence,
\begin{equation\*}
 \lim\_{k\downarrow0}\lim\_{A\to\infty}I(f\_{a,k,A})
 =\lim\_{k\downarrow0}\int\_\R xe^{k(x-a)-x^2}\,dx
 =\int\_\R xe^{-x^2}\,dx=0. 
\end{equation\*}
So, $\inf\_{f\in F\_a}I(f)\le0$. Now \eqref{0} follows by \eqref{4}, which also shows that there is no minimizer of $I(f)$ over $f\in F\_a$. $\quad\Box$.


