Let $T:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$ defined by $$(Tx)_n = a_nx_{n+1} $$ for $n\in\mathbb{N}$, with $a=(a_n)_n$ a given sequence in $\ell^{\infty}(\mathbb{N})$.
Show that $T$ is a bounded linear operator and compute the operator norm $\|T\|$.
This is my humble attemp:
We know that the sequence space $\ell^{p}(\mathbb{N})$ consists of all infinite sequences $(x_n)_{n=1}^{+\infty}$ such that $\sum_{n=1}^{+\infty}|x_n|^p<+\infty$, with the $p$-norm $$\|x\|_p = \left(\sum_{n=1}^{+\infty}|x_n|^p\right)^{1/p}.$$ Then, in this case we have that $$\|x\|_2 =\left(\sum_{n=1}^{+\infty}|x_n|^2\right)^{1/2}.$$
Therefore, by the definition of the operator and the definition of a bounded linear operator we get $$\left(\|(Tx)_n\|_2\right)^2 = \sum_{n=1}^{+\infty}|a_nx_{n+1}|^2 = \sum_{n=1}^{+\infty}|a_n|^2|x_{n+1}|^2$$
But here is where I am stuck. I want to get an expression in the form $\left(\|(Tx)_n\|_2\right)^2\leq M \|x_{n+1}\|_2 ^2$, but I don't know how to deal with the $|a_n|^2$ terms.
In addition, for the computation of the operator norm, I want to use one of the equivalent definitions $\|T\| = \sup_{\|x\|=1} \|Tx\|$ to get $$\|T\| = \sup_{\|x\|=1} \|(Tx)_n\|_2 = \sup_{\|x\|=1}\sum_{n=1}^{+\infty}|a_n|^2|x_{n+1}|^2,$$ but again, I have troubles in dealing with the $|a_n|^2$ terms.
Best Answer
This question is Old but I reckon its still worth answering since I can't find any other related questions. We had this question in a Hilbert space course I am currently doing at Uni. It is in a slightly different form but it should be more or less the same for this question.
Proposition.
let $(a_n)_{n \in \mathbb{N}}\subset \mathbb{C}$ be a fixed bounded sequence. Then the operator \begin{align*} T:&\ell_2(\mathbb{N},\mathbb{C}) \rightarrow \ell_2(\mathbb{N},\mathbb{C})\\ &(x_n) \mapsto(a_nx_n). \end{align*} Is a bounded linear operator and $||T||_{\text{op}}=\sup_{n \in \mathbb{N}}|a_n|$.
Proof:
Throughout this proof we will endow $\ell_2$ with the usual 2-norm.
We start with Linearity.
let $\lambda \in \mathbb{C}$ and $(x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}}\in \ell_2(\mathbb{N},\mathbb{C})$ then we compute; \begin{align*} T(\{\lambda x_n+y_n\}_{n \in \mathbb{N}})&= \{a_n(\lambda x_n+y_n)\}_{n \in \mathbb{N}} & \text{[Definition of $T$]}\\ &=\{a_n\lambda x_n+a_ny_n\}_{n \in \mathbb{N}}\\ &=\{\lambda a_n x_n\}_{n \in \mathbb{N}}+\{a_ny_n\}_{n \in \mathbb{N}} & \text{[Point wise addition]}\\ &=\lambda \{a_n x_n\}_{n \in \mathbb{N}}+\{a_ny_n\}_{n \in \mathbb{N}}\\ & = \lambda T(\{x_n\}_{n \in \mathbb{N}})+T(\{y_n\}_{n \in \mathbb{N}}). \end{align*} Thus we can see that $T$ is linear.
Next we show that $T$ is bounded.
Let $(x_n)_{n \in \mathbb{N}},\in \ell_2(\mathbb{N},\mathbb{C})$ be a arbitrary sequence. Then we compute; \begin{align*} ||T(x_n)||_2 = ||(a_nx_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2}. \end{align*} Now The boundedness of $(a_n)$ becomes crucial. Since $(a_n)$ is bounded we can find $\sup_{n \in \mathbb{N}}|a_n|$ (if we knew more about $(a_n)$ we could find this exactly). Now let $M = \sup_{n \in \mathbb{N}}|a_n|$, then we have \begin{align*} ||T(x_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2} \leq. \sqrt{\sum_{i=1}^{\infty}|Mx_n|^2} = M\sqrt{\sum_{i=1}^{\infty}|x_n|^2} = M||(x_n)||_2. \end{align*} Since $(x_n)$ was arbitrary, this is true for all sequences in $\ell_2$. Consequently $T$ is bounded.
Now finally we find $||T||_{\text{op}}$: Recall that $||T||_op$ is the minimum of the bounds of $T$. To show that $||T||_{\text{op}}=\sup_{n \in \mathbb{N}}|a_n|$ we need to show the following;
If $M \geq 0$ is a finite constant such that \begin{align*} ||T(x_n)||_2 \leq M||(x_n)||_2\ \ \ \ \ \ \ \ (1) \end{align*} for all $(x_n)\in \ell_2$ then $\sup_{n \in \mathbb{N}}|a_n| \leq M$. So we fix $M \geq 0$ that satisfies $(1)$. Now for all $(x_n) \in \ell_2$ we have \begin{align*} ||T(x_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2} \leq M\sqrt{\sum_{i=1}^{\infty}|x_n|^2} = M||(x_n)||_2. \end{align*} Now this holds in particular when $||(x_n)||_2 =1$. For every $n \in \mathbb(N)$ we define the sequence \begin{align*} x_n &= (x_{n,k})_{k \in \mathbb{N}}\\ x_{n,k} &:= \delta_{k,n}. \end{align*} where $\delta_{k,n}$ is the Kronecker delta. Then we have \begin{align*} ||x_n||_2 = \sqrt{\sum_{i=1}^{\infty}|x_{n,i}|^2}=\sqrt{|x_{n,n}|^2} = 1 \end{align*} thus $x_n \in \ell_2$. Then we have \begin{align*} Tx_n &= (a_k x_{n,k})_{k \in \mathbb{N}}\\ Tx_{n,k} &:= a_k \delta_{k,n}. \end{align*} Now we also have \begin{align*} ||Tx_n||_2 = \sqrt{\sum_{i=1}^{\infty}|a_kx_{n,k}|^2}=\sqrt{|a_n|^2} =|a_n|. \end{align*} then by the way we chose $M$ we have \begin{align*} ||Tx_n||_2 \leq M||x_n||_2 \implies |a_n| \leq M. \end{align*} However remember that there was nothing special about $n$ so that $|a_n| \leq M$ for all $n \in \mathbb{N}$ and in particular this means $\sup_{n \in \mathbb{N}}|a_n| \leq M$, which proves the result.