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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?

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up vote 1 down vote accepted

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 $n$. –  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
    
I think your argument not work because you can't approximate the integral after using g. –  user36548 Aug 13 '13 at 3:13
    
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 $0$. –  Davide Giraudo Aug 13 '13 at 22:11
    
Can you take some time to write it more explicitly? Thanks/ –  user36548 Aug 22 '13 at 18:57
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First, since $f\in L^{1}\left(\mathbb{R}\right)$, we have the Hardy–Littlewood maximal function $${\displaystyle \sup_{r>0}\dfrac{1}{2r}\int_{\left|y-x\right|<r}}\left|f\left(y\right)\right|dy<\infty$$ for a.e $x\in\mathbb{R}$. So fix $x$ like that, we will prove $\sup\left\{ f_{n}\left(x\right):n\in\mathbb{N}\right\} <\infty.$

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 $${\displaystyle \sup_{r>0}\dfrac{1}{2r}\int_{\left|y\right|<r}}\left|f\left(y\right)\right|dy<\infty.$$ Using the traditional way for approximation by simple functions: $g_{m}=\sum_{k=1}^{2^{2m}+1}\dfrac{k-1}{2^{m}}\chi_{g^{-1}\left(I_{m,k}\right)}$, where $I_{m,k}=\left[\dfrac{k-1}{2^{m}},\dfrac{k}{2^{m}}\right),k=1,\ldots,2^{2m}$, $I_{m,2^{2m}+1}=\left[2^{m},\infty\right)$ and $g\left(t\right)=\dfrac{1}{\sqrt{\left|t\right|}\left(1+t^{2}\right)}\in L^{1}\left(\mathbb{R}\right)$. Note that, $g$ is decreasing radial so we can rearrange $g_{m}$ to form $g_{m}=\sum c_{k,m}\chi_{B\left(0,r_{k,m}\right)}$, beside that $\displaystyle\sum_{k} c_{k,m}r_{k,m}<M$ ($M$ is constant depend only on $g$, since $g\left(t\right)\in L^{1}\left(\mathbb{R}\right)$.) Now, \begin{eqnarray*} \int_{\mathbb{R}}g_{m}\left(t\right)f\left(-\dfrac{t}{n}\right)dt & \leq & \sum_{k}c_{k,m}\intop_{B\left(0,r_{k,m}\right)}f\left(-\dfrac{t}{n}\right)dt\\ & = & \sum_{k}\left(c_{k,m}\times2r_{k,m}\right)\dfrac{1}{\frac{2r_{k,m}}{n}}\intop_{B\left(0,\frac{r_{k,m}}{n}\right)}f\left(u\right)du\\ & \leq & 2M\sup_{r>0}\int_{\left|u\right|<r}f\left(u\right)du \;(\textrm{not depend on $n$}). \end{eqnarray*} Finally, we can approximate $\int_{\mathbb{R}}g_{m}\left(t\right)f\left(-\dfrac{t}{n}\right)dt$ to $\int_{\mathbb{R}}g\left(t\right)f\left(-\dfrac{t}{n}\right)dt$ by Monotone convergence theorem.

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