Let $T : \ell^2 \to \ell^2$ (involving complex numbers) be defined by $$ Tx := (x_1, x_1+x_2, x_3, x_3+x_4, x_5, x_5+x_6, \ldots). $$ What is $\|T\|$?
Essentially I've tried :
- To find $M \geq 0$ s.t. $\|Tx\| \leq M \|x\|$ and $\|T\| \geq M$. Cauchy-Schwarz led me nowhere. The triangle inequality gives $\|Tx\| \leq 2 \|x\|$, but I've not been able to show that $\|T\| \geq 2$ (in fact I don't think $2$ is small enough).
- To maximise $$ \frac{\|Tx\|}{\|x\|} $$ for $x \neq 0$, or equivalently $$ \|Tx\| $$ for $\|x\|=1$. I didn't succeed.
- I've tried to write $|x_k| = x_k e^{- i\theta_k}$ and see if I could simplify some things or recognize known expressions. I failed with this idea.
Apparently there is a way to use matrices, but I don't see how.
You probably know that the norm of a diagonal matrix is the supremum of absolute values of its diagonal entries. Here's a generalization of this:
Claim
Suppose $T:\ell^2\to\ell^2$ has invariant subspaces $\{V_k:k=1,2\dots\}$ which are mutually orthogonal and satisfy $\ell^2=\bigoplus_k V_k$. Let $T_k:V_k\to V_k$ be the induced operator on $V_k$, essentially the restriction of $T$ to this subspace. Then $$\|T\|=\sup_k \|T_k\|$$
Proof
It's easy to see that $\|T\|\ge \|T_k\|$ for all $k$. Conversely, given any vector $x\in \ell_2$ we can write it as $x=\sum_k x_k$ with $x_k\in V_k$. Then $$ \|Tx\|^2 = \sum_k \|T_k x_k\|^2 \le \sup_k \|T_k\|^2 \sum_k \| x_k\|^2 = \sup_k \|T_k\|^2 \|x\|^2 $$ which proves $\|T\|\le \sup_k \|T_k\|$. $\quad\Box$
Armed with the above (which is useful not only on this occasion), you can find $\|T\|$ just by examining the operator norm of $2\times 2$ matrix $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$$