Cesàro operator is bounded for $1The 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.
 A: 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 $x\notin \ell^1$.
A: 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
$$
\Vert T\Vert\leq\frac{p}{p-1}
$$
