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For the classical Hardy operator $T\colon \ell^p\to \ell^p \quad (Tx)_n=\frac{1}{n}\sum_{k=1}^n x_k$ or the integral type $S\colon L^p\rightarrow L^p \quad (Sf)(x)=\frac{1}{x}\int_0^x f(t) dt \ \ $ the norm is well known to be $\frac{p}{p-1}$ for $1.

I did some research but did not find the result for the adjoint operator $T'\colon \ell^p\rightarrow \ell^p \quad (T'x)_n=\sum_{k=n}^{\infty} \frac{x_k}{k}$ or its integral version.

What does adjoint mean in the case of general Banachspaces at all? For $p=2$ it's easy to verify $\langle Hx,y\rangle =\langle x,H'y\rangle$.

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It's the duality braket: if $l$ is a linear map and $x$ a vector we denote $\langle l,x\rangle:=l(x)$.

We can identify an element $l$ of the dual of $\ell^p$ by an element of $\ell^q$ (where $q$ is the conjugate of $p$): define $l_n:=l(e_n)$ where $e_n$ is the sequence whose $n$-th term is $1$ and the others $0$ and check that the sequence $\{l_n\}$ represents $l$.

For the problem, let $x\in\ell^p$ and $y\in \ell^q$. We have \begin{align} \langle Tx,y\rangle&=\sum_{n=1}^{+\infty}\frac 1n\sum_{k=1}^nx_ky_n\\ &=\sum_{1\leq k\leq n<\infty}\frac{x_ky_n}n\\ &=\sum_{k=1}^{+\infty}x_k\sum_{n\geq k}\frac{y_n}n\\ &=\langle x,T^*y\rangle, \end{align} as wanted (the interversion of the sum is justified by the fact that the sum is absolutely convergent).

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    Oh yes. But I'm interested in the norm of $T^*$ as a linear operator $l^p \rightarrow l^p$. Is the following correct? I'm interested in the norm of $T^*: l^p\rightarrow l^p$. But this is the same as the norm of $(T^*)^*=T: (l^p)^*\rightarrow (l^p)^*$. But $(l^p)^*\tilde{=} l^{p'}$. This norm is known to be $\frac{p'}{p'-1}=p$. So the answer $\lVert T^* \lVert=p$ is correct.2012-06-09