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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 $1<p<\infty$.

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? – woucrouc Jun 9 '12 at 14:27
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. – woucrouc Jun 9 '12 at 14:45
I meant "this form", typo. $T'$ has the same norm as $T$. – Davide Giraudo Jun 9 '12 at 14:51
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. – woucrouc Jun 9 '12 at 16:13
@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]. – user31373 Jun 9 '12 at 19:14

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