# Minimize $\int_0^\infty g'(x)f(x)\,dx$ where $f(x)$ has a log-normal density

I would like to optimize (minimize) the following expression in order to find the functional minimizer $$g$$ (which should be at least once differentiable): $$\int_0^\infty g'(x) f(x) \ dx$$ where $$f(x)$$ is the density of the log-normal distribution, and also subject to the following constraints: $$\forall x>0: g'(x)>0, g(x) > 0, g(0)=0, g(1)=1, \lim\limits_{x\to\infty} g(x) = +\infty, \lim\limits_{x\to\infty} g'(x) = +\infty$$

Any ideas for that? It seems that the standard Euler-Lagrange equation cannot be used (maybe this means there is no minimizer?).

Proposition. There is no minimiser.

Proof: An equivalent version of your problem is as follows.

Statement. We wish to minimise $$\int_{-\infty}^\infty xe^{q(x)-x^2}\,dx$$ where $$q(x)$$ is a strictly increasing function with $$q(a)=0$$ for a given $$a\in\Bbb R$$ and $$\lim_{x\to\pm\infty}q(x)=\pm\infty,\quad\lim_{x\to+\infty}q'(x)e^{q(x)-x}=+\infty.$$

The proof for the equivalence is provided at the end of this post.

Now, since $$\int_{-\infty}^\infty xe^{q(x)-x^2}\,dx$$ is positive by the original integral $$\int_0^\infty g'(x)f(x)\,dx>0$$, it suffices to choose a sequence of functions $$q_A(x)$$ such that the integral becomes arbitrarily close to zero.

This is shown by Iosif Pinelis on MathOverflow where we take $$q_A(x):=k(x-a)\,1_{x\le a+A}+(kA+2(x-A))\,1_{x>a+A}$$ for some $$k,A\in(0,\infty)$$. By the dominated convergence theorem, we have $$\lim_{k\to0^+}\lim_{A\to+\infty}\int_{-\infty}^\infty xe^{q_A(x)-x^2}\,dx=\lim_{k\to0^+}\int_{-\infty}^\infty xe^{k(x-a)-x^2}\,dx=\int_{-\infty}^\infty xe^{-x^2}=0$$ and so we have a zero infimum. $$\quad\square$$

Proof of Statement: Let $$g=e^h$$ so that $$g'=h'e^h$$. Then the constraints become $$h'(x)>0,h(1)=0,\lim_{x\to0}h(x)=-\infty,\lim_{x\to\infty}h(x)=+\infty,\lim_{x\to\infty}h'(x)e^{h(x)}=+\infty$$ and the functional is \begin{align}\int_0^\infty g'f\,dx&=[gf]_0^\infty-\int_0^\infty gf'\,dx\\&=\frac1{\sigma\sqrt{2\pi}}\left(\lim_{x\to\infty}e^{h(x)-\log x-\frac{(\log x-\mu)^2}{2\sigma^2}}-\int_0^\infty\frac{e^{h(x)-\frac{(\log x-\mu)^2}{2\sigma^2}}(\mu-\sigma^2-\log x)}{\sigma^2 x^2}\,dx\right)\end{align} where we assume the limit exists. Let $$h(x)=p(u)$$ where $$x=e^u$$. Then $$p'(u)>0,p(0)=0,\lim_{u\to\pm\infty}p(u)=\pm\infty,\lim_{u\to+\infty}p'(u)e^{p(u)-u}=+\infty$$ and \begin{align}\sigma\sqrt{2\pi}\int_0^\infty g'f\,dx&=\lim_{u\to\infty}e^{p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}}+\frac1{\sigma^2}\int_{-\infty}^\infty e^{p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}}(u+\sigma^2-\mu)\,du.\end{align} For both $$\int_0^\infty$$ and $$\int_{-\infty}^0$$ to be finite, we must have $$\lim_{u\to\pm\infty}p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}=-\infty\implies p(u)=o(u^2)$$ since $$p'(u)>0$$ for all $$u$$ means $$p$$ cannot have a quadratic term. So \begin{align}\sigma^3\sqrt{2\pi}\int_0^\infty g'f\,dx&=\int_{-\infty}^\infty e^{p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}}(u+\sigma^2-\mu)\,du\\&=\frac{e^{\sigma^2/2-\mu}}{2\pi\sigma^2}\int_{-\infty}^\infty ve^{q(v)-v^2}\,dv\end{align} where $$q(u)=p(u/\sqrt{2\sigma^2}-\sigma^2+\mu)$$ under the same constraints as $$p(u)$$ except $$q((\sigma^2-\mu)\sqrt{2\sigma^2})=0$$. In the statement, we simply define $$a:=(\sigma^2-\mu)\sqrt{2\sigma^2}$$. $$\quad\square$$