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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.

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  • $\begingroup$ The votes to close may have come because of ambiguity in the notation: by L($\ell^2$) do you mean all linear maps from $\ell^2\to \ell^2$, or just the bounded ones? In the latter case, your question has an immediate positive answer, so I assume you meant to merely assume D is well-defined as a linear map (in which case Danil Skurudin's argument suffices) $\endgroup$
    – Yemon Choi
    Commented Jun 6, 2022 at 3:12

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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 $.

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  • $\begingroup$ Thank you! That is a much much simpler approach than my proof. $\endgroup$
    – Matthias
    Commented Jun 3, 2022 at 8:00

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