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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?

2 Answers 2

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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.

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    You are almost done, note however that one asks about $\sup\limits_{n\geqslant1}f_n(x)$, not $f_n(x)$ for some $n$.2012-08-15
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    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.$2012-08-16
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    I think your argument not work because you can't approximate the integral after using g.2013-08-13
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    My argument is not quite complete, but we can go by this way: cutting the integral from $\int_0^n+\int_n^{\infty}$: for the second one, approximation gives what we want, and the first one goes to $0$.2013-08-13
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    Can you take some time to write it more explicitly? Thanks/2013-08-22
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    Actually, my argument does not work. However, using a Fourier analysis argument, we can see that the result is true if the Fourier transform of $f$ is integrable.2013-08-23
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    Can you write clearly about your Fourier argument? I have attained my solution for this problem:2013-08-24
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First, since $f\in L^{1}\left(\mathbb{R}\right)$, we have the Hardy–Littlewood maximal function $${\displaystyle \sup_{r>0}\dfrac{1}{2r}\int_{\left|y-x\right|

By changing variable, we just have to show $$ \sup\left\{ \intop_{\mathbb{R}}\dfrac{1}{\sqrt{\left|t\right|}\left(1+t^{2}\right)}f\left(-\dfrac{t}{n}\right)dt:\; n\in\mathbb{N}\right\} <\infty $$ with $${\displaystyle \sup_{r>0}\dfrac{1}{2r}\int_{\left|y\right|0}\int_{\left|u\right|