# A question about integral operator

I have a question: Prove or disprove that: for every $f\in L^{1}\left(\mathbb{R}\right)$, $$\sup\left\{ { \int_{\mathbb{R}}\frac{\sqrt{n}}{\sqrt{\left|x-y\right|}\left(1+n^{2}\left(x-y\right)^{2}\right)}f\left(y\right)dy:n\in\mathbb{N}}\right\} <\infty,$$ for Lebesgue almost every $x\in\mathbb{R}$.

I failed in my attempt to disprove this statement (I think so!!!). Can everybody help me?

-

## 1 Answer

Let $$f_n(x):=\int_{\Bbb R}\frac{\sqrt n}{\sqrt{|x-y|}(1+n^2|x-y|^2)}f(y)dy,$$ assuming WLOG that $f\geqslant 0$. We use the substitution $t=n(x-y)$, hence $dt=-ndy$ to get $$f_n(x)=\int_{\Bbb R}\frac 1{\sqrt{|t|}}\frac 1{1+t^2}f\left(x-\frac tn\right)dt.$$ By Fubini's theorem for non-negative functions, and since $f$ is integrable, we get that $f_n$ is integrable, and in particular almost everywhere finite.

Approximate $f\in L^1$ by $g$, continuous with compact support, such that $\lVert f-g\rVert_{L^1}\leqslant 1$. Then, as the integral $\int_{\Bbb R}\frac 1{\sqrt{|t|}}\frac 1{1+t^2}dt<\infty$, to show the result when $f$ is continuous and bounded. With the latest formula, it's easier to see it.

-
You are almost done, note however that one asks about $\sup\limits_{n\geqslant1}f_n(x)$, not $f_n(x)$ for some $n$. – Did Aug 15 '12 at 23:15
Supremum get on $n$ fixed $x$. I think the fact $f_{n}\in L^{1}\left(\mathbb{R}\right)$ is easy. Beside that, we also have \begin{eqnarray*} T_{n}:L^{1}\left(\mathbb{R}\right) & \rightarrow & L^{1}\left(\mathbb{R}\right)\\ f & \rightarrow T_{n}\left(f\right)\left(x\right)= & \intop_{\mathbb{R}}\dfrac{\sqrt{n}}{\sqrt{\left|x-y\right|}\left(1+n^{2}\left(x-‌​y\right)^{2}\right)}f\left(y\right)dy\end{eqnarray*} is a continous linear operator and $\sup_{n\in\mathbb{N}}\left\Vert T_{n}\right\Vert <\infty.$ – user36548 Aug 16 '12 at 2:21