# Prove that function is in $L^1$

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!

• 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. Aug 8 '21 at 22:39

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 $$p=3$$, $$q=3/2$$.
• 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. Aug 8 '21 at 22:52
• Yes, that's right but it may happen that this set contains a heigbourhood of infinity. Nevertheless, how to see that $x^{-1/8} is integral on this set? – iou Aug 8 '21 at 23:02 • @iou$\int_{[0,1]}\frac{1}{x^8}\,dx=\frac{x^{7/8}}{7/8}\bigg|_{x=0}^{x=1}\$ which is certainly finite Aug 9 '21 at 0:19