TITLE: Defining bounded operator on orthonormal basis.
QUESTION [1 upvotes]: Let $\{e_i : i \in I\}$ be an orthonormal basis for a Hilbert space $H$. Suppose we are given a collection of vectors $\{\xi_i\}_{i \in I}$ with $\sum_{i \in I}\|\xi_i\| < \infty$. I want to prove that there exists a unique bounded linear operator $T: H \to H$ such that $Te_i = \xi_i$.
This is a generalisation of an exercise in Conway's "A course in functional analysis".
Attempt: Uniqueness is clear because $T$ is uniquely determined on the linear span of $\{e_i: i \in I\}$, which is dense in $H$.
Let us focus on existence. Write $x= \sum_{i \in I} \langle x, e_i\rangle e_i$ where the series converges in norm. Then we want to define
$$Tx:= \sum_{i \in I} \langle x,e_i\rangle \xi_i$$
but it is not clear that this series converges. However,
$$\sum_{i \in I} \|\langle x, e_i\rangle \xi_i\| = \sum_i |\langle x, e_i\rangle|\|\xi_i\| \le \left(\sum_i |\langle x, e_i\rangle|^2\right)^{1/2}\left(\sum_i \|\xi_i\|^2 \right)^{1/2}< \infty \quad (*)$$
by Bessel's inequality and since $\sum_i\|\xi_i\| < \infty \implies \sum_i\|\xi_i\|^2 < \infty$.
Hence, the above series converges absolutely and thus $Tx$ makes sense. Since the inner product is linear in the first component, it follows that $T$ is linear. It remains to check that $T$ is bounded.
The equation $(*)$ also shows that $\|Tx \| \leq \|x\| \left(\sum_i \|\xi_i\|^2 \right)^{1/2}$ and thus $T$ is bounded. This establishes the result.
Questions: Is my proof correct? Am I correct that I can further weaken the hypothesis in the question by only requiring that $\sum_i\|\xi_i\|^2 < \infty$?

REPLY [0 votes]: Your proof is fine and yes, you can change the hypothesis to the weaker condition $\sum_i \|\xi_i\|^{2} <\infty$.
{"set_name": "stack_exchange", "score": 1, "question_id": 4038004}