Let $D$ be an operator defined by $D(c\_n)\_{n\in\mathbb{N}} = (a\_n c\_n)\_{n\in\mathbb{N}}$. It's a well known fact that $D$ is well defined as an operator from $l^2(\mathbb{N})$ to $l^2(\mathbb{N})$ if $(a\_n)\_{n\in\mathbb{N}} \in l^\infty(\mathbb{N})$.


Does the converse hold, i.e. if $D\in L(l^2(\mathbb{N}))$ then $(a\_n)\_{n\in\mathbb{N}} \in l^\infty(\mathbb{N})$?


For me this seems like an easy to answer question, but I did not find any proofs or statements in the literature. I think I found a proof by myself but I want to be sure about that and I have the feeling that sequences like $(\log n)\_{n\in\mathbb{N}}$ make problems.



Diagonaloperators on $l^2(N)$
If $ (a\_n) \notin l^{\infty} $, we can find subsequence $ (a\_{n\_i}) $ such that $ |a\_{n\_i}| > 2^i $. Then we can take $ (c\_n) \in l^2 $ like this:
$$ c\_k = \frac{1}{i}, \text{ if } k = n\_i \text{ for some } i, $$
$$ c\_k = 0 \text{ otherwise.} $$


Then I hope it's clear that $ (a\_n c\_n) \notin l^2 $.


