If $T: \ell^2 \to \ell^2 $ is defined as $Tx = (\frac{x_i}{i})$, find $\lVert T \rVert$.

Question

Here $(x_i) \in \ell^2$. All I have on paper right now is $$\rVert Tx\rVert = \sqrt{\sum_{i=1}^{\infty} \frac{x_i ^2}{i^2}}$$

and I'm not sure about what's next. Please help.

Answer 1

It should be obvious enough that $\|Tx\|\leq\|x\|$, which means that $\|T\|\leq1$. On the other hand, since $\|Te_1\|=\|e_1\|=1$, the norm is actually attained.

More detail: One typically shows that $\|T\|=C$ by showing the following two things:

1. For every $x$, $\|Tx\|\leq C\|x\|$
2. There is an $x$ such that $\|Tx\|=C\|x\|$.

Statement 1 says that the norm of $T$ is at most $C$, and statement 2 shows that the norm of $T$ is at least $C$. Indeed, the norm of $T$ is defined to be the least number $\|T\|$ such that $\|Tx\|\leq\|T\|\|x\|$ for all $x$. (Sometimes, if the norm is not attained, instead of showing 2 one might have to show

3. For every $\varepsilon>0$, there is an $x$ such that $\|Tx\|\geq(C-\varepsilon)x$. [This says that the constant $C$ is the least upper bound for $\|T\|$.]

This would happen with an operator like $Tx = \big(\frac{i}{i+1}x_i\big)$.)

For this problem, we can prove statement 1 with $C=1$ by observing that $$ \|Tx\| = \sqrt{\sum_{n=1}^\infty \frac{x_i^2}{i^2}} \leq \sqrt{\sum_{n=1}^\infty x_i^2} = \|x\|. $$ And statement 2 follows by picking a particular $x$ for which $\|Tx\|=\|x\|$; picking $x=e_1$, the first basis vector, does the trick.

Answer 2

Note that $$ \| Tx\|^2 = \sum_{j=1}^{\infty} \frac{|x_j|^2}{j^2} \leq \sum_{j=1}^{\infty} |x_j|^2 = \|x\|^2 $$ Hence $\|Tx\|^2 \leq 1 \cdot \|x\|^2$, which implies that $\|Tx\| \leq 1 \cdot \|x\|$ and therefore $\|T\|\leq 1$ (1).

On the other side, since $e_1=(1,0,\cdots)$ is such that $\|e_1\|=1$, then $$ \| T \| = \sup_{\|x\|=1}\{ \|Tx\|\} \geq \|Te_1\|=\|e_1\|=1 $$ Then $\|T\|\geq 1$ (2). Together (1) and (2) gives of course that $\|T\|=1$