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 . I can do it for $p=\infty$, but not when it is between $1$ and $\infty$. Thank you.
Cesàro operator is bounded for $1
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0In addition to the answers below, note that you can also directly apply the [Marcinkiewicz interpolation theorem](http://en.wikipedia.org/wiki/Marcinkiewicz_interpolation_theorem) for $\mathbb{N}$ with the counting measure. – 2012-03-14
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0@WillieWong Could you please give some more detail on how are going to apply MI? – 2012-03-14
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1@AD. $T$ is bounded $\ell_1 \to \ell_1^w$ and $\ell_\infty\to\ell_\infty$. So by MI it is bounded $\ell_p\to\ell_p$. – 2012-03-15
2 Answers
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 $x\notin \ell^1$.
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0Why didn't you just apply Hardy's inequality? – 2012-03-14
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0I 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. – 2012-03-14
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0Okey, now I see what you mean! Btw, it would be much more interesting to put on a weight in the norm. – 2012-03-14