How can I argue that $L\{(x_1,x_2,...,x_n,...)\} = \{x_1, x_2/2, x_3/3 ... x_n/n ...\}$, $L: \ell_2 \rightarrow \ell_2$ is bounded operator?

Question

How can I argue that $L\{(x_1,x_2,...,x_n,...)\} = \{x_1, x_2/2, x_3/3 ... x_n/n ...\}$, $L: \ell_2 \rightarrow \ell_2$ is bounded operator?

I think that it's intuitively since $\ell_2$ is sequences s.t.

$$\sum_{i=1}^{\infty} |x_i|^2 < \infty$$

Then

$$\|x\|=\sqrt{\sum_{i=1}^{\infty} |x_i|^2} < \infty$$ intuitively, but the operator boundedness requires showing $M> 0$ s.t.

$$\| Lx\|_Y \leq M \|x\|_X$$

and I don't know how such $M$ could be picked.

Particularly, how can I know that

$$\sqrt{\sum_{i=1}^{\infty} |x_i/i|^2} \leq \sqrt{\sum_{i=1}^{\infty} |x_i|^2}$$

Answer 1

You seem to have most of the answer, it's just an issue of organizing the information in a rigorous manner. We know that $L((x_1, x_2, \ldots )) = (x_1, x_2/2, \ldots, x_n/n, \ldots)$ so we can compute the operator norm as $$||L||_{op} = \sup_{||x||=1}||Lx||_{l^2} = \sup_{||x|| = 1} ||(x_1, x_2 / 2, \ldots, x_n/n, \ldots)||_{l^2} \leq \sup_{||x||=1} ||x||_{l^2} = 1$$ with the inequality coming from the fact that, as some comments above have mentioned, if $a_n \leq b_n$ for all $n \in \mathbb{N}$ then $\sum_{n=1}^\infty a_n \leq \sum_{n=1}^\infty b_n$.

Then, you've shown the operator norm of $L$ is less than or equal to 1, which is sufficient to conclude that $L \in B(l^2(\mathbb{N}))$.