# Cesàro operator is bounded for $1<p<\infty$

The Cesàro operator $T\colon \ell_{p}\to\ell_{p}$ is defined by $(Tx)_{k}=\frac{1}{k}\sum_{j=1}^{k}x_{j},\: k\in\mathbb{N}$, where $x=(x_{k})_{k=1}^{\infty}$ Show that $T$ is bounded if $1<p<\infty$. I can do it for $p=\infty$, but not when it is between $1$ and $\infty$. Thank you.

-
In addition to the answers below, note that you can also directly apply the Marcinkiewicz interpolation theorem for $\mathbb{N}$ with the counting measure. –  Willie Wong Mar 14 '12 at 11:29
@WillieWong Could you please give some more detail on how are going to apply MI? –  AD. Mar 14 '12 at 20:28
@AD. $T$ is bounded $\ell_1 \to \ell_1^w$ and $\ell_\infty\to\ell_\infty$. So by MI it is bounded $\ell_p\to\ell_p$. –  Willie Wong Mar 15 '12 at 9:29

Using Hardy inequality one may see that $$\Vert T(x)\Vert_p= \left(\sum\limits_{k=1}^\infty \left|\frac{1}{k}\sum\limits_{j=1}^k x_j\right|^p\right)^{1/p}\leq \left(\sum\limits_{k=1}^\infty \left(\frac{1}{k}\sum\limits_{j=1}^k |x_j|\right)^p\right)^{1/p}\leq$$ $$\left(\left(\frac{p}{p-1}\right)^p\sum\limits_{k=1}^\infty |x_j|^p\right)^{1/p}= \frac{p}{p-1}\left(\sum\limits_{k=1}^\infty |x_j|^p\right)^{1/p}= \frac{p}{p-1}\Vert x\Vert_p$$ This means that $$\Vert T\Vert\leq\frac{p}{p-1}$$

-

Some hints: let $q$ the conjugate exponent of $p$: $1/p+1/q=1$. Write $$\left|\sum_{k=1}^Nx_k\right|^p=\left|\sum_{k=1}^Nx_kk^{1/(pq)}k^{-1/(pq)}\right|^p$$ and use Hölder's inequality to get that $$\left|\sum_{k=1}^Nx_k\right|^p\leq \sum_{k=1}^N\left|x_k\right|^pk^{1/q}N^{p/q-1q}$$ (we have to find a bound for $\sum_{k=1}^N k^{-1/(pq)}$ for example comparing with an integral). Then take the sum over $N$, change the order of summation and find a bound, comparing with an integral, of $\sum_{N\geq k}N^{-1-1/q}$ to get the result.

In fact, we used Hardy's inequality.

Note that for $p=1$ (even if it's not asked), $T$ is not well-defined since if we take $x:=(1,0,\ldots,0,\ldots)$ then $(Tx)_k=\frac 1k$ so $x\notin \ell^1$.

-
Why didn't you just apply Hardy's inequality? –  AD. Mar 14 '12 at 20:20
I just wanted to give the steps which lead to this inequality. I didn't read the proof of the link in details, so I don't know whether this approach is different or not. –  Davide Giraudo Mar 14 '12 at 20:26
Okey, now I see what you mean! Btw, it would be much more interesting to put on a weight in the norm. –  AD. Mar 14 '12 at 20:31

In the case p=1, $T$ is not defined in $l_1$, but it is well-defined on a subspace of $l_1$. So, the question is "Can $T$ be bounded?" The answer is NO. If you want, you can try to figure out how to prove that, that is interesting.

-
Yes, you are right. My proof is wrong –  Norbert Mar 15 '12 at 22:32
you see that your post is no longer make sense, since i have changed my answer. So you should delete it. –  Norbert Mar 16 '12 at 7:55
I am sorry, Norbert. I deleted my post. –  Tran Mar 16 '12 at 20:36