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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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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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