if $\{e_1, e_2, \cdots \}$ is an orthonormal basis for Hilbert space $H$ and for each $n$ there is a vector $Ae_n$ in $H$ such that
\begin{equation*} \sum ||Ae_n||<\infty . \end{equation*}
Show that $A$ has an unique extension to a bounded operator on $H.$
First extend $A$ to be a linear map $A$ on $X={\rm lin}\{e_n:n\in\mathbb N\}$ (the linear span of the orthonormal basis $(e_n)$). (I'm going to assume that $H$ is not finite dimensional.)
We now need to show that $A$ is continuous on $X$. Let $x\in X$. Then there exists $a_1,a_2,\dotsc$ such that $$ x=\sum_n a_ne_n. $$ We have $Ax=\sum_n A(a_ne_n) =\sum_n a_n(Ae_n)$. There is a little checking necessary to check that, when $\lVert x\rVert\leq 1$, we have $\sum_n \lvert a_n\rvert\lVert Ae_n\rVert<\sum_n\lVert Ae_n\rVert<\infty$ so that $$\lVert A\rVert_{\rm op}\leq \sum_n\lVert Ae_n\rVert<\infty.$$ Thus $A$ is continuous on $X$. Now we have a continuous linear map $A$ defined on a dense subset $X$ of $H$ so we may extend $A$ to a continuous linear map $\tilde A$ from $H$ to $H$. In other words, $\tilde A$ is a bounded linear operation on $H$. A little extra work shows that this extension is unique.
