2
$\begingroup$

Assume that $\psi(x)$ is bounded and integrable on $x \in [0,\infty)$ with $\int_0^\infty \psi(x) dx = 0$, and suppose that $K \colon (0,\infty) \to (0,\infty)$ is some kernel function satisfying $K(x) \to 0$ as $x \to \infty$. I am wondering under what type of additional conditions on $K$ should we impose so that we can upper bound $\int_0^\infty |\psi(x)|^2 dx$ by some functionals of $\int_0^\infty \int_0^\infty K(x+y)\psi(x)\psi(y) dxdy$ ? To be more precise, will such bound be available with the choice $K(x) = \frac{1}{x}$?

Edit: here $\psi$ is some fixed function (which is actually the difference between two cumulative distribution functions $F$ and $G$, whose probability densities $f$ and $g$ have the same mean value, which leads to the constraint $\int_0^\infty \psi(x)\, dx = \int_0^\infty (F(x)-G(x))\, dx = 0$ on $\psi$.) As a side remark, for the choice $K(x) = 1/x$ we have $$\int_0^\infty \int_0^\infty K(x+y)\psi(x)\psi(y) dxdy = \int_0^\infty \left(\int_0^\infty \psi(x)\,\mathrm{e}^{-x\xi}\,dx\right)^2 d\xi \geq 0$$

$\endgroup$
10
  • $\begingroup$ We cannot have $$ \int_0^\infty \vert \psi(x) \vert^2 dx \leq C \int_0^\infty \int_0^\infty \frac{\psi(x) \psi(y)}{x+y} dx dy $$ for some $C$ independent of $\psi$ (satisfying our conditions). Let's interpret our $\psi$ as a function $\mathbb{R}\rightarrow \mathbb{R}$ with support in $[0,\infty)$. Then consider $\psi_n(x)=\psi(x-n)$. We have $\Vert \psi_n \Vert_2 = \Vert \psi \Vert_2$. However, the double integral tends to zero. In fact it is bounded by $$ \frac{1}{2n} \left( \int_0^\infty \vert \psi(x) \vert dx \right)^2. $$ What type of inequality would you hope to get? $\endgroup$
    – Severin Schraven
    Commented Dec 8, 2023 at 8:58
  • $\begingroup$ @SeverinSchraven Hello, thank you for your comments. I have edited my post accordingly and I have emphasized that $\psi$ should be a fixed function. $\endgroup$
    – Fei Cao
    Commented Dec 8, 2023 at 15:50
  • $\begingroup$ If $\psi$ is fixed, then you just make the constant $C$ above super large, no? $\endgroup$
    – peek-a-boo
    Commented Dec 8, 2023 at 15:56
  • $\begingroup$ @peek-a-boo Almost. We only would need to prove that the thing on the right is bigger than zero. $\endgroup$
    – Severin Schraven
    Commented Dec 8, 2023 at 19:01
  • $\begingroup$ @SeverinSchraven oh true initially I glossed over the restriction $\int_0^{\infty}\psi=0$, so I thought it was fine to naively assume $\psi$ maintains a fixed sign, so the thing on the right is automatically non-negative (and positive if $\psi$ is not identically $0$). $\endgroup$
    – peek-a-boo
    Commented Dec 8, 2023 at 19:10

0

Reset to default

You must log in to answer this question.