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Define $$g(n) = \int_{-\infty}^{\infty} dx |h(x)|^{n} e^{-f(x)}.$$

For $n\geq 2$, must $g(n)$ be sandwiched between $g(n-1)$ and $g(n+1)$?

That is, for each $n \geq 2$, must we have at least one of $g(n-1) \leq g(n) \leq g(n+1)$ or $g(n+1) \leq g(n) \leq g(n-1)$, where the choice of the two inequalities is allowed to be $n$ dependent?

This seems to be nontrivial, since there's no natural order relation between $g(n)$ and $g(n+1)$. I'm also unsure if Jensen's inequality would be useful in the cases where $e^{-f(x)}$ is integrable and hence provides a probability measure via $\frac{e^{-f(x)}}{\int_{-\infty}^{\infty} dy e^{-f(y)}}$; then one knows $\langle |h(x)|^{n+1}\rangle^{\frac{n}{n+1}} \geq \langle |h(x)|^{n}\rangle$ according to that probability measure; but the relation to $g(n)$ is obscured.

The motivation for this question is to try to answer another, Can a convergent perturbation series converge to the wrong answer for this type of integral?, as it would allow one to use the convergence of the there-defined $S(g)$ to show the convergence of $J(g)$.

user196574
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2 Answers2

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By Cauchy-Schwarz: $$\left(\int_{-\infty}^{+\infty} |h(x)|^{\frac n 2 + \frac {n+2} 2}e^{-f(x)}dx\right)^2 \leq \int_{-\infty}^{+\infty} |h(x)|^{n}e^{-f(x)}dx\cdot\int_{-\infty}^{+\infty} |h(x)|^{n+2}e^{-f(x)}dx$$ Thus $g(n+1)^2\leq g(n)g(n+2)$ which implies that $\frac{g(n+1)}{g(n)}$ is non-decreasing.

So if $g(1)\geq g(0)$, then $g(n+1)\geq g(n)$ for all $n$. So it's sandwiched.

The question remains (for me at least), when $g(1) < g(0)$. In some cases $\frac{g(n+1)}{g(n)}$ remains below $1$. Is it always the case?

Stefan Lafon
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    This is very clear and helpful and inspired me. Consider the sequence $g(n) = \int_{0}^{1.6} x^n$, which obeys $g(0) = 1.6$, $g(1) = .8$, and $g(2) = .8533333$. $g(1)<g(0)$ but $g(2)/g(1) > 1$. We can likely choose $f(x)$ appropriately to approximate these integrals arbitrarily well, so this is a counter example where $g(1)$ is not between $g(0)$ and $g(2)$. I know I stated for $n\geq2$, but that can be done with e.g. $n=1.1$ for which $g(n)$ will decrease to a minimum before picking back up. Please check what I'm saying is sane and if you add it to the answer, I'll mark it as accepted. – user196574 Oct 28 '21 at 22:30
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Consider $f(x)=0$ and $$h(x)=0.9999 \cdot 1_{[1,10^5]} + 1.001 \cdot 1_{[0, 10^{-5}]}.$$ The first couple of iterations the integral will drop, but finally it will go to infinity (just from the second indicator function). We have $$g(n)= (0.9999)^n (10^5-1) + (1.001)^n 10^{-5}.$$

So, no, it is in general not true.

Severin Schraven
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