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Assumption

Suppose the function $K: \mathbb R \to \mathbb R_{\geq 0}$ is symmetric about $0$ and the following holds \begin{align} \int_{-\infty}^\infty K(z)dz &= 1,\\ \int_{-\infty}^\infty K^2(z)dz &< \infty,\\ \int_{-\infty}^\infty z^2 K(z)dz &< \infty,\\ \int_{-\infty}^\infty |z|^3 K(z)dz &< \infty. \end{align}

Problem

Can we show that the following is true in this case? \begin{align} \int_{-\infty}^\infty z^2 K^2(z)dz < \infty \end{align}

Intuitively it seems correct, but I have no idea how to show it mathematically.

ytnb
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1 Answers1

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There was an error in the first version of this answer. In fact, there was also a fundamental error in the second version (it holds that $L^\infty\subseteq L^1$, but this means $\Vert f\Vert_{L^1}\leq\Vert f\Vert_{L^\infty}$ and not the other way round as the inclusion suggests. I am really sorry for the confusion. I should have paid more attention!)

Note that condition 3 implies condition 2, so we can get rid of one of them. Instead we need that $K$ is essentially bounded, I suppose, i.e., $K\leq C$ for some $C>0$ Lebesgue-almost everywhere. Then the answer I provided is correct. Otherwise there should be a counter example (I couldn't come up with one, but it should be clear that we need a $K$ with $K(z)\geq C$ for $z\in A$ for measurable $A\subset\mathbb R$. Then $$\int_{\mathbb R} K(z)^2z^2\,\mathrm dx = \int_{A}K(z)^2z^2\,\mathrm dz + \underbrace{\int_{\mathbb R\setminus A}K(z)^2z^2\,\mathrm dz}_{\leq M<\infty}.$$ So on this set $A$ we must have that the integral over $K(z)^2z^2$ diverges, while the integral over $K(z)^2$ and $K(z)z^3$ converge. A good choice fo $A$ may be the some neighborhood of $0$ with $K$ being some function of the reciprocal of $z$. However, I found it to be difficult to derive an explicit expression for $K$ as the $z$ terms drive the expression to zero while the reciprocal drives the expression to infinity. A careful choice that balances both is hence necessary).

If you are willing to assume that $\Vert K\Vert_{L^\infty}< \infty$, then the original answer applies:

$$\int z^2 K(z)^2\,\mathrm dz \leq \Vert g\Vert_{L^1}\Vert K\Vert_{L^\infty}$$ according to Hölder inequality, where

  • $g(z) = K(z)z^2$,
  • $\Vert f\Vert_{L^1} := \int \vert f\vert\,\mathrm dx$
  • $\Vert f\Vert_{L^\infty} := \inf\{C\geq 0 : \text{$\vert f\vert \leq C$ Lebesgue-almost everywhere}\}.$

The third assumption gives $\Vert g\Vert_{L^1}<\infty$ and $\Vert K\Vert_{L^\infty}<\infty$ has to be assumed.

Syd Amerikaner
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  • Can Holder's inequality be used with $p=\infty$? – ytnb Nov 24 '23 at 07:47
  • And, Why is $||K||{L^\infty}\leq ||K||{L^1}$ valid? – ytnb Nov 24 '23 at 07:49
  • It does (see https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality). The other inequality is called Lp-inequality (see https://math.stackexchange.com/questions/66029/lp-and-lq-space-inclusion). Intuitively, the integral is an infinite sum, so it should be clear that the largest value of a function ($\Vert f\Vert_{L^\infty}$) is larger than the the sum of all values ($\Vert f\Vert_{L^1}$). This should also help seeing why the Hölder inequality in the $p=\infty$ case holds. – Syd Amerikaner Nov 24 '23 at 14:03
  • I’m not sure I agree with the statement that $\Vert K\Vert_{L^\infty}\leq\Vert K\Vert_{L^1}$. How do you address $K$ being spiky near infinity with narrower and narrower triangles? Or $K$ having something like $1/\sqrt{x}$ for $0<x\leq1$? In other words, I think there are necessary conditions for the sup-norm to satisfy that inequality that $K$ doesn’t possess. – Clayton Nov 24 '23 at 16:06
  • It should hold that $L^q\subset L^p$ for $1\leq p \leq q\leq\infty$. Then for $p = 1$ and $q = \infty$, we have that if $f\in L^1$ it must that $f\in L^\infty$, no? From this included the inequality (maybe a finite constant is involved which I missed?) – Syd Amerikaner Nov 24 '23 at 16:39
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    You need the measure of the space to be finite, no? – Clayton Nov 27 '23 at 02:22
  • Indeed, you are right. This setup looked very much like a probability space which made me forget about this assumption. – Syd Amerikaner Nov 27 '23 at 04:46
  • But we should be able to recover the probability space I was thinking about – Syd Amerikaner Nov 27 '23 at 05:05