Upper bound on the maxima of ratio of expectation of quantities under Gaussian measure Let $\lambda,\eta >0$ be given, and $u:\mathbb{R}\rightarrow \mathbb{R}$ be a real valued function. Define 
$$\Delta(u)= \frac{\int u(h) \exp(-\eta u(h))\exp(-\frac{\lambda}{2}h^2)~\mathrm{d}h}{\int \exp(-\eta u(h))\exp(-\frac{\lambda}{2}h^2)~\mathrm{d}h}$$
I want to find an upper bound on $P_0$, which is defined as   $$P_0=\sup_{u:\mathbb{R}\rightarrow \mathbb{R}} \Delta(u),$$ subject to the following two constraints.


*

*$u(h^*)=0$, for some given $h^*\in \mathbb{R}$.

*$u$ is Lipschitz in the Euclidean metric with Lipschitz constant bounded by $L$.


I have been unsuccessful at solving this problem. Here are the things that I have attempted. 
Suppose we want an upper bound for the quantity.
$$P_1= \max_{x_2,\ldots x_n} \frac{\sum_{i=1}^n p_i x_i \exp(-\eta x_i)}{\sum_{i=1}^n p_i \exp(-\eta x_i)}$$ where $x_1$ is fixed at 0, and the vector $[p_1,\ldots,p_n]$ is a probability vector, i.e. $p_i\geq 0,\sum_{i=1}^n p_i=1$. The maxima for the optimization problem involved in the calculation of $P_1$, is attained when $x_2=\ldots=x_n=x$. One can then show, via straightforward calculations, that $P_1\leq \frac{1}{\eta}\log(1/p_1)$. Calculation of an upper bound on $P_0$, can be seen as something similar to calculating an upper bound on $P_1$, just that we now have a continuous problem.I am guessing that the following function $u_0(h)$  might be the optimizer of $\Delta(u)$.
$$
u_0(h)=
\begin{cases}
a_1(h-h^{*})~\text{if}~   h^{*}\leq h\leq h_{+}\\
a_1(h_{+}-h^{*})~\text{if}~ h\geq h_{+}\\
a_2(h-h^{*})~\text{if}~ h_{-}\leq h\leq h^{*}\\
a_2(h_{-}-h^{*})~\text{if}~ h\leq h_{-},
\end{cases}
$$
where $0\leq a_1\leq L,-L\leq a_2\leq 0$, are appropriate constants.
 This way I can mimic the behavior of the solution of problem $P_1$, for our continuous problem. However, I have no idea as to how to prove that $u_0(h)$ is the optimal solution for $P_0$. 
A second attempt, would be to  write the individual integrals in the numerator and denominator of $\Delta(u)$ as discrete summation, and then pass to the limit. The advantage of doing this is that we immediately convert our problem into a discrete problem, and then utilize what we know about upper bounding $P_1$, in order to get an upper bound on $P_0$. However I have not been able to implement this technique, and I would deeply appreciate, if one could comment, if such a technique is valid and sketch the main details.
P.S. I have a feeling that this problem is not hard, and perhaps has been studied before. However, I was neither able to get a reference, nor was able to solve it by myself. 
 A: I guess exercise 4.4.5 from the Estimates of Integrals chapter (Section "Positive integrals") of Stewart's "Calculus" (parallel universe edition) may be helpful.
Let $\mu$ be a probability measure on $[0,+\infty)$. Let $S$ be the supremum of
$$
\Phi(u)=\frac{\int_0^\infty ue^{-u}\,d\mu}{\int_0^\infty e^{-u}\,d\mu}
$$ 
over all non-negative functions $u$ with Lipschitz constant $1$ satisfying $u(0)=0$.
Then, up to an absolute constant factor,
$$
S\approx \max\left(M,\int_0^\infty \min(1,x)\,d\mu\right)
$$
where $M$ is determined from the equation $\int_0^M e^{-x}\,d\mu=e^{-M}\int_M^\infty\,d\mu$
The relevant place in the Hint section of the aforementioned book reads as follows:
Consider $u=\min(x,M)$ and $u=\min(x,1)$ to get an estimate from below.
If $M>1$, observe that replacing $u$ by $\min(u,M)$ increases the numerator and can increase the denominator at most twice. If $M<1$, show that the denominator is comparable to $1$ regardless of $u$.
The full solution manual cost about 10 times as much as the textbook itself when I visited the Elsewher website last time and there was no free preview, so I don't know what is written there.  
