Adjoint operator for $l^2(\mathbb{N})$

Question

I am working with a problem from my exercise sheet (not mandatory). I hope that someone could held me to get started with the problem. I do not know how to get started.

Problem: Consider the Hilbert space $l^2(\mathbb{N})$. Let $\{e_n|n\in\mathbb{N}\}$ be the standard orthonormal basis of $l^2(\mathbb{N})$, where $e_n$ is the sequence whose $n$-term is $1$ and whose all other terms are $0$. Let $T:l^2(\mathbb{N})\rightarrow l^2(\mathbb{N})$ be a bounded linear operator on $l^2(\mathbb{N})$ such that, \begin{equation} Te_n=\frac{1}{n+1}e_{n+1}. \end{equation} (a) Let $T^\ast$ be the adjoint operator of $T$. For each $n$, find $T^\ast e_n$ and $T^\ast Te_n$.

(b) Find the operator norms $||T^\ast T||$ and $||T||$.

Here $l^2(\mathbb{N})$ denotes the $L^2-$space over the measure space of natural numbers $\mathbb{N}=\{0,1,2,...\}$.

Answer 1

\begin{align*} \langle Tx,y\rangle&=\langle x,T^*y\rangle\\ \langle Te_k,e_n\rangle&=\langle e_k,T^*e_n\rangle && (\text{ choose }x=e_{k} \,\,\& \,\, y=e_n)\\ \frac{1}{k+1}\langle e_{k+1},e_n\rangle&=\langle e_k,T^*e_n\rangle \end{align*} Using the fact that $\{e_n\}$ is the standard orthonormal basis, we get $$\langle e_k,T^*e_n\rangle= \begin{cases} 0 & \text{ if } k \neq n-1\\ \frac{1}{n} & \text{ if } k = n-1 \end{cases} $$ So $$T^*e_n=\frac{1}{n}{e_{n-1}} \qquad \forall n \geq 2.$$

Now you can get $T^*Te_n=T\left(\frac{e_{n+1}}{n+1}\right)=\frac{1}{n+1}T^*(e_{n+1})=\frac{1}{(n+1)^2}e_n.$

The norm part has been answered by @Lizard King so no point repeating it here.

Answer 2

So since the operator is bounded it is continuous by hypothesis you will have that it is defined for all $x\in l^2$ since $Tx=T(\sum_n \langle x,e_n\rangle e_n)=\sum_n\langle x,e_n\rangle Te_n$. To find the operator norm we notice that $||Tx||^2=\sum_n|\langle x,e_n\rangle|^2\frac{1}{(n+1)^2}\leq \frac{1}{2}\sum_n |\langle x,e_n\rangle|^2$, using the fact that the $e_n's $ are orthonormal. And so using Bessel's Inequality you will have that $||Tx||\leq \frac{1}{2}||x||$ for any $x$ so $||T||\leq \frac{1}{2}$ and since $T(e_1)=\frac{1}{2}e_2$ we will have that $||T||=\frac{1}{2}$. To calculate the other norm use the fact that $||T^*T||=||T||^2$. Someone has already covered the part about the adjoint so I won't do that .