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Let $E \subset \mathbb{R}$ be a measurable subset. Assume that $\int_{E} |x|^{1/4} |f(x)|^2 dx < \infty$ and $\int_{E} x^4 |f(x)|^3 dx < \infty$, then I want to prove $f \in L^1 (E)$.

Morally the first inequality says that $f(x)$ behaves nicely at $0$ and the second inequality says that $f(x)$ behaves nicely at $\pm \infty$ but I'm struggling to prove the statement rigorously. A possible idea is to use Holder's inequality: we know that $f(x) x^{1/8} \in L^2 (E)$ and $f(x) x^{4/3} \in L^3 (E)$ and probably it can be used somehow but I don't know how. Anyways, any ideas are greatly appreciated!

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  • $\begingroup$ For integrability far from zero, it seems useful to write $ f(x) = f(x) x^r x^{-r} $ and then use Hölder inequality to compare with something finite. It is possible this problem expects something more advanced, like some kind of $ L^p $ interpolation result. I'm not sure. $\endgroup$
    – Jake Mirra
    Aug 8 '21 at 22:39
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1. On $\{x\in E: |x|\le 1\}$, write $|f(x)| = \left(|f(x)| \cdot|x|^{1/8}\right)\cdot |x|^{-1/8}$ and apply the Cauchy-Schwarz inequality.

2. On $\{x\in E: |x|>1\}$, write $|f(x)| = \left(|f(x)| \cdot |x|^{4/3}\right)\cdot |x|^{-4/3}$ and apply Hölder with the conjugate exponents $p=3$, $q=3/2$.

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  • $\begingroup$ How do you know that x^{-1/8} is in L^2 (E_1) where E_1 is this set of x in E such that |x| \le 1? I think it may happen that it contains a small interval around 0. $\endgroup$
    – iou
    Aug 8 '21 at 22:40
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    $\begingroup$ If you are working with the Lebesgue measure (which I think is implicit), then you can calculate the integral of $x^{-1/8}$ over $[0,1]$ by using the Riemann integral. $\endgroup$
    – en3trix
    Aug 8 '21 at 22:52
  • $\begingroup$ Yes, that's right but it may happen that this set contains a heigbourhood of infinity. Nevertheless, how to see that $x^{-1/8} is integral on this set? $\endgroup$
    – iou
    Aug 8 '21 at 23:02
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    $\begingroup$ @iou $\int_{[0,1]}\frac{1}{x^8}\,dx=\frac{x^{7/8}}{7/8}\bigg|_{x=0}^{x=1}$ which is certainly finite $\endgroup$
    – peek-a-boo
    Aug 9 '21 at 0:19

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