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.

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


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

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

  • $\begingroup$ Sorry for the question, but isn't that trivial? I mean $|x_i/i|^2<|x_i|^2$ therefore so is the sum. (So $M=1$ will work) $\endgroup$
    – Yanko
    Mar 22, 2019 at 20:33
  • $\begingroup$ Well, $|x_n/n| = |x_n|/n \leq |x_n|$ for all $n\geq 1$, for a start? @mavavilj $\endgroup$
    – Clement C.
    Mar 22, 2019 at 20:33
  • $\begingroup$ @ClementC. Yea but how to compare infinite sums? $\endgroup$
    – mavavilj
    Mar 22, 2019 at 20:35
  • $\begingroup$ The same way. If $0\leq a_n \leq b_n$ for all $n$ and $\sum_n b_n$ exists, then $\sum_n a_n \leq \sum_n b_n$. $\endgroup$
    – Clement C.
    Mar 22, 2019 at 20:36
  • 1
    $\begingroup$ This is the comparison test. $\endgroup$ Mar 22, 2019 at 20:42

1 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 $L \in B(l^2(\mathbb{N}))$.


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