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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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    Thank you! I thought this argument just applied for $p=2$. Does this help in any way with the computation of the norm?2012-06-09
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    What do you mean by "norm of the form"? This doesn't mean that $\lVert T' \lVert=\lVert T \lVert$ I guess. My intuition tells me that $\lim_{p\rightarrow \infty} \lVert T' \lVert=\infty$ whereas for $T$ the limit is clearly 1.2012-06-09
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    I meant "this form", typo. $T'$ has the same norm as $T$.2012-06-09
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    I don't think this is true. I did some computation in Mathematica using the truncated haromonic series $x_k=\frac{1}{k^{1/p}}$: `Table[{p, N[Sum[Sum[1/k^(1 + 1/p), {k, n, 500}]^p, {n, 1, 500}]^(1/p)/ Sum[1/k, {k, 1, 500}]^(1/p)]}, {p, 1, 5}]`. The result was: `{{1, 1.}, {2, 1.70325}, {3, 2.27878}, {4, 2.75247}, {5, 3.14595}}` which makes something like $\lVert T'\lVert=p$ more plausible.2012-06-09
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    @woucrouc when Mathematica contradicts mathematics, I side with the latter. :) Here's a proof of $\|T^*\|\le \|T\|$ for any operator $T\colon X\to Y$. Take any $\psi \in Y^*$; since $T^*\psi$ is the composition $\psi\circ T$, we have $\|T^*\psi\|\le \|\psi\|\|T\|$. [BTW I use asterisks for dual objects, they are easier on the eyes than primes].2012-06-09
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    Probably I don't understand how the argument applies to spaces without scalar product (as $l^p$ for $p\not = 2$). Can you explain?2012-06-09
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    @LeonidKovalev Sorry didn't want to post this as an answer. Your argument obviously doesn't work for $T$ on $l^{\infty}$: $\lVert T\lVert=1$ whereas $\lVert T^* \lVert =\infty$ since for $y=(1,1,\dots)$ we have $(T^* y)_1=\sum_{k=1}^{\infty} \frac{1}{k}$. What am I missing?2012-06-09
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    @woucrouc Note the spaces on which these operators act: if $T\colon X\to Y$, then $T^*\colon Y^*\to X^*$.2012-06-09
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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