0
$\begingroup$

I am having trouble computing the norm of the following operator: $$T:\ell_2 \to \ell_2$$ given by $$T(x_1, x_2, x_3, \dots) = \left(x_1, \frac{x_2}{2}, \frac{x_3}{3},\dots\right).$$

$\endgroup$

1 Answer 1

3
$\begingroup$

We want to find

$$ \|T\|=\sup_{\|x\|_2=1}\|Tx\|_2. $$

We have, for every $x \in \ell_2$ with $\|x\|_2=1$, $$ \|Tx\|^2=\sum_{i=1}^{\infty} \left(\frac{x_i}{i}\right)^2 \leq \sum_{i=1}^{\infty} x_i^2 = \|x\|_2^2=1. $$ so $\|T\| \leq 1$.

By considering $x=(1, 0 \dots, 0, \dots)$ we see that $\|x\|=1$ and $Tx=x$, so $\|Tx\|_2=1$, hence $\|T\| \geq 1$, so we can conclude $\|T\|=1$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .