Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f$.

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.