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 $f$.


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.
