1
$\begingroup$

Possible Duplicate:
Cesàro operator is bounded for $1\lt p\lt\infty$

Let $1

I'm not sure at all, why $Fx\in l^p$. I figured that for $p=1$ $\lVert F\lVert=\infty$ since for $x=(1,0,0,\dots)$ the norm satisfies $\lVert x\lVert=1$ and $(Fx)_n=\frac{1}{n}$ and hence $\lVert Fx\lVert=\infty$. But I don't know how to estimate the norm for $p>1$. I had the idea that $x=(1,0,0,\dots)$ also maximizes these norms, but I'm not sure about that.

  • 0
    Thank you! Didn't think of Hardy's inequality!2012-05-31

0 Answers 0