If $\int_2^\infty f(x)^2 dx $ is convergent, is it true that $\int_2^\infty f(x)x^{-3/4} dx $ is convergent?

Question

If $\int_2^\infty f(x)^2 dx $ is convergent, is it true that $\int_2^\infty f(x)x^{-3/4} dx $ is convergent?

Answer 1

Let $p=q=2$ and apply Hölder's inequality: $$ \int_2^\infty|f(x)|x^{-{3\over4}}dx\le\sqrt{\int_{2}^{\infty}f(x)^2dx\underbrace{\int_{2}^{\infty}x^{-1.5}dx}_{=\sqrt2}}<\infty $$ The integral is absolutely convergent and hence convergent.

Answer 2

$\int_{2}^{\infty}f(x)x^{-3/4}\,dx\leq \left ( \int_{2}^{\infty}|f(x)|^2\,dx \right )^{1/2}\,\left (\int_{2}^{\infty}x^{-3/2}\right )^{1/2}$ By Cauchy Schwarz inequality. And the result follows.

Answer 3

Edit: Assuming $f$ is integrable.

Yes. This is the inner product of $f$ with $g(x)=x^{-3/4}$, which are both members of $L_2(2,\infty)$. One way to see convergence is with the Cauchy-Schwarz inequality:

$\int_2^\infty f(x)g(x)dx \leq (\int_2^\infty f^2(x)dx\int_2^\infty g^2(x)dx)^{1/2}<\infty$

Answer 4

Assuming that $f$ is a Lebesgue-measurable function : Let $A=\{x:|f(x)| x^{3/4}|\geq 1\}.$ Then $x\in A\implies |f(x)x^{-3/4}|\leq |f(x)^2|,$ and $x\in A^c\implies |f(x)x^{-3/4}|<x^{-3/2}. $ So $|f(x)x^{-3/4}|\leq |f(x)^2|+x^{-3/2}$ for all $x.$

See Zachary Selk's comment and my reply regarding my assumption.