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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.

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

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It looks like this is connected to the Esscher transform (see and – Stephan Sturm Aug 26 '13 at 2:04
@Stephan: Thanks for pointing out to the Esscher transform. However, I am not sure how the Esscher transform can be used here, as we are not guaranteed that $u(h)$ should be a linear function of $h$. Do you think Esscher transform could be somehow used here? – gmravi2003 Aug 26 '13 at 15:05
Seems to me that you can define the numerator of $\Delta$ as an operator over functions u that satisfy your constraints, and denominator as the norm of this function space. Then an equivalent problem is finding the operator norm. Maybe some functional analytic tools can help in such a formulation ? – Piyush Grover Aug 28 '13 at 18:42

1 Answer 1

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.

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Is that website registered in the country of Erewhon? – Yemon Choi Sep 2 '13 at 17:39
@Yemon Choi Possibly. I know that country as "The Other Place" or "Tlon",_Uqbar,_Orbis_Tertius but I won't be surprised if it is named differently in your part of the world. Its actual name in the native language is quite impossible to pronounce or transliterate... :) – fedja Sep 3 '13 at 5:32
Can you tell me what is the parallel universe edition? I searched James Stewarts "Single variable calculus" book, but could not find a reference to the above mentioned problem. It would be great if you can paste the link to the right edition, and the link (or atleast the title) of the solution manual. – gmravi2003 Sep 10 '13 at 0:33
Sure. Unfortunately the access to the Elsewher website is not free and they accept only universal currency (you know the English proverb "time is money", don't you?), so the direct link may be useless at this point. A good general reading on this side of Reality is… ("Asymptotics" chapter) though they stress somewhat different things. Anyway, the hint is good enough to recover the full solution of the quoted problem and then to answer your original question :-). – fedja Sep 13 '13 at 11:55

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