Here is a question that I cannot figure out myself: Let $H$ and $K$ be separable Hilbert spaces (possibly infinite-dimensional), and $\lbrace e_i\rbrace_{i\in \mathbb{N}}$ be the orthonormal basis for $H$. $T: H \to K$ a $\textbf{linear operator (not necessarily bounded)}$ such that $$\sum_{i\in\mathbb{N}}\vert\vert T(e_i)\vert\vert_{K}^2< \infty$$ Where $\vert \vert \cdot\vert\vert_{K}$ is the norm of $K$ induced by inner product. $\textbf{Prove or disprove}$ that $T$ is a bounded linear operator.
The key problem here is that I cannot passage the limit, i.e., $T(\sum\limits_{i}e_i) = \sum\limits_{i}T(e_i)$ may not hold. Thanks in advance for any help!
Best Answer
It is somewhat surprising, but in fact the statement is not true.
Counterexample: We know that $(e_n)_{n\in \mathbb{N}}$ are linearly independent. Thus, we can extend this to a basis $(e_\lambda)_{\lambda \in \mathbb{N}\cup \Lambda}$ of $H$. As the Hamel basis of an infinite-dimensional Hilbert space is not countable (by Baire's theorem), we know that there exists a countably infinite subset $S\subset \Lambda$ and we can wlog write $S=\mathbb{Z}_{<0}$. Pick any $0\neq w_0\in K$ and define $A$ to be the linear extension of
$$ A(e_\lambda)= \begin{cases} T(e_\lambda),& \lambda \in \mathbb{N},\\ \vert \lambda \vert w_0,& \lambda \in \mathbb{Z}_{<0},\\ 0,& \text{ otherwise.} \end{cases} $$
This operator $A$ is by definition linear, it is unbounded (as $\vert A(e_{-n}) \vert = \vert n \vert \cdot \vert w_0 \vert \rightarrow \infty $ for $n\rightarrow \infty$) and we have $$ \sum_{n\in \mathbb{N}} \Vert A(e_n) \Vert_K^2 = \sum_{n\in \mathbb{N}} \Vert T(e_n) \Vert_K^2 < \infty. $$
A true version: The following statement would be correct: If $T:H \rightarrow K$ is such that for all $\sum_{n\in \mathbb{N}} a_n e_n \in H$ $$ T(\sum_{n\in \mathbb{N}} a_n e_n) = \sum_{n\in \mathbb{N}} a_n T(e_n) $$ and $$ \sum_{n\in \mathbb{N}} \Vert T(e_n) \Vert_K^2 <\infty,$$ then $T$ is a bounded operator (simply by Cauchy-Schwarz and the fact that $\Vert \sum_{n\in \mathbb{N}} a_n e_n \Vert_H^2 = \sum_{n\in \mathbb{N}} \vert a_n \vert^2$). In fact, in this setting, $T$ is a Hilbert-Schmidt operator and thus even compact.