# How can I give a bound on the $L^2$ norm of this function?

I came across this question in an old qualifying exam, but I am stumped on how to approach it:

For $f\in L^p((1,\infty), m)$ ($m$ is the Lebesgue measure), $2<p<4$, let $$(Vf)(x) = \frac{1}{x} \int_x^{10x} \frac{f(t)}{t^{1/4}} dt$$ Prove that $$||Vf||_{L^2} \leqslant C_p ||f||_{L^p}$$ for some finite number $C_p$, which depends on $p$ but not on $f$.

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Is it $f(x)$ or $f(t)$ in the integrand. –  Mhenni Benghorbal Nov 27 '12 at 4:28
You must go to MSU : ) –  Euler....IS_ALIVE Nov 27 '12 at 4:46
@MhenniBenghorbal Yes, it is f(t). Thanks for catching that! I made the edit –  phenomenalwoman4 Nov 27 '12 at 15:55

Using Hölder's inequality, we have for $x>1$, \begin{align} |V(f)(x)|&\leqslant \lVert f\rVert_{L^p}\left(\int_x^{10 x}t^{-\frac p{4(p-1)}}dt\right)^{\frac{p-1}p}\frac 1x\\ &=\lVert f\rVert_{L^p}A_p\left(x^{-\frac p{4(p-1)}+1}\right)^{\frac{p-1}p}\frac 1x\\ &=A_p\lVert f\rVert_{L^p}x^{\frac{p-1}p-\frac 14}\frac 1x\\ &=A_p\lVert f\rVert_{L^p}x^{-\frac 1p-\frac 14}. \end{align} To conclude, we have to check that $\int_1^{+\infty}x^{-\frac 2p-\frac 12}dx$ is convergent. As $p<4$, it's the case.