Adjoint of T $\in B(\ell^{2})$

Question

I was wondering if you guys can help me with this one!

Let $T \in B(\ell^{2})$ a continuous operator. Also,

\begin{align*} (Ts)(n)=\frac{1}{n}\sum^{t}_{m=1} s(m),\hspace{1cm}n\in \mathbb{N}, s\in \ell^{2} \end{align*}

What is the adjoint of $T$?

PS: Thank you very much in advance!

Answer 1

Let $f,s\in\ell^2$ then $$\langle f,Ts\rangle =\sum_{n=1}^{\infty}f_n\frac{1}{n}\sum_{m=1}^{n}\overline{s_m}=\sum_{m=1}^{\infty}\left(\sum_{n=m}^{\infty}\frac{1}{n}f_n\right)\overline{s_m}.$$ So define $$T^{*}f(n)=\sum_{m=n}^{\infty}\frac{1}{m}f_m$$ which is well defined since $$\sum_{m=n}^{\infty}|\frac{1}{m}f_m|\leq |\{\frac{1}{m}\}_{m=1}^{\infty}|_{\ell²}|f|_{\ell^2}=\sqrt{\frac{\pi^2}{6}}|f|_{\ell^2}<\infty$$ by Cauchy-Schwartz. The first equation shows this is the adjoint operator. If $t$ is $n$ which seemed natural to me ( taking the average of the first $n$ numbers in the sequence)

Answer 2

Lets assume $$Ts(n) = \frac1n \sum_{m=1}^n s(m)$$ Then $$\langle Ts, r\rangle = \sum_{n=1}^\infty \frac 1n \sum_{m=1}^ns(m)r(n) = \sum_{\substack{n,m\in\mathbb Z_{\ge 1}\\ 1\le m\le n}} \frac 1n s(m)r(n) = \sum_{m=1}^\infty s(m) \sum_{n=m}^\infty\frac{r(n)}n$$ so the adjoint operator is at least formally

$$T^* r(m) := \sum_{n=m}^\infty\frac{r(n)}n$$ This is finite for each $m$ due to Fubini's theorem and the above computation. Is it a bounded operator on $\ell^2$? Yes, if $T$ is bounded, since $$ \|T^* r\|_{\ell^2} = \sup_{\|s\|_{\ell^2} = 1} |\langle s, T^*r\rangle | = \sup_{\|s\|_{\ell^2} = 1} |\langle Ts, r\rangle | \le \|r\|_{\ell^2} \|T\| $$

Why is $T$ bounded? Because of the discrete Hardy's inequality.