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Let $f$ be a uniformly continuous and bounded function on $\mathbb{R}$ with \begin{equation} \int_\mathbb{R}|f(x)x^a|dx<\infty\tag{1} \end{equation} for some $a>0$. Here it is shown that the function $f(x)x^c$ can be unbounded for any $c>0$. However, in the given example the integral $\int_\mathbb{R}|f(x)x^c|dx$ (and, therefore, $\int_\mathbb{R}|f^2(x)x^c|dx$ as well) is finite for all $c>0$. Now the question is whether the condition (1) implies that for some $b>a$ $$\int_\mathbb{R}|f^2(x)x^b|dx<\infty?$$ It seems that Hölder's inequality alone does not lead to the proof of the above inequality.

Gerome
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1 Answers1

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Here is a counterexample. As in the answer under the link in OP, we are going to replicate certain function along the real line. Namely, define $$ f(x) = \sum_{n\ge 2} \Big(\frac1{\ln n} + \frac{2}{ \ln |x-n|}\Big)^2\mathbf{1}_{[n-n^{-2},n+n^{-2}]}(x). $$ Then, $$ \int_{\mathbb R}|f(x)\cdot x| dx \le \sum_{n\ge 2}\int_{n-n^{-2}}^{n+n^{-2}} \frac {x\, dx}{\ln^2 n}\le \sum_{n\ge 2} \frac {n+1}{\ln^2 n}\cdot \frac{2}{n^2}<\infty. $$ However, for any $\epsilon >0$ and any $b>0$, $$ \int_{\mathbb R}f(x)^b\cdot |x|^{1+\epsilon} dx \ge \sum_{n\ge 2}\int_{n-(en)^{-2}}^{n+(en)^{-2}}\Big(\frac1{\ln n} + \frac{2}{\ln |x-n|}\Big)^{2b}x^{1+\epsilon} dx\\\ge \sum_{n\ge 2}\Big(\frac1{\ln n} - \frac{1}{\ln n + 1}\Big)^{2b} \cdot (n-1)^{1+\epsilon} \cdot \frac{1}{(en)^2}\\ \ge \frac{1}{4e^2}\sum_{n\ge 2}\frac1{(n-1)^{1-\epsilon}(\ln n+1)^{4b}} =\infty. $$

zhoraster
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