Unique extension to a bounded operator

Question

Suppose $\left\{ e_{1},e_{2},\ldots\right\} $ is an orthonormal basis for a Hilbert space $\mathcal{H}$ and for each $n$ there is a vector $Ae_{n}$ in $\mathcal{H}$ such that $\sum\left\Vert Ae_{n}\right\Vert <\infty$. Show that $A$ has a unique extension to a bounded operator on $\mathcal{H}$.

Answer 1

Assume $\mathcal{H}$ is a Hilbert space over the complex numbers. For every $x \in \mathcal{H}$ we have $$ x=\sum_{n=1}^\infty x_ne_n, $$ where $$ x_n=\langle x,e_n\rangle \quad \forall n \ge 1, $$ and $$ \sum_{n=1}^\infty|x_n|^2=\|x\|^2. $$ In particular $\lim_nx_n=0$. If $x \ne 0$, there is an $N \in \mathbb{N}$ such that $$ |x_n|\le \frac{\|x\|}{2} \quad \forall n > N. $$ Hence $$ \sup_{n \ge 1}|x_n|\le \max\{\frac{\|x\|}{2},\max_{n \le N}|x_n|\} \le \|x\|. $$ Define $$ \tilde{A}: \mathcal{H} \to \mathcal{H},\ \tilde{A}x=\sum_{n=1}^\infty x_nAe_n, $$ then $\tilde{A}e_n=Ae_n$ for every $n \ge 1$, and for $x \ne 0$ we have $$ \|\tilde{A}x\|\le \sum_{n=1}^\infty |x_n|\|Ae_n\|\le \sup_{n\ge 1}|x_n|\sum_{n=1}^\infty\|Ae_n\|\le (\sum_{n=1}^\infty\|Ae_n\|)\|x\|. $$ Thus $$ \sup_{\|x\|=1}\|\tilde{A}x\| \le \sum_{n=1}^\infty\|Ae_n\| <\infty, $$ i.e. $\tilde{A}$ defines a bounded operator on $\mathcal{H}$.

Added: Let $B$ be a bounded (linear) operator on $\mathcal{H}$ such that $Be_n=Ae_n$ for every $n \ge 1$. Then, for every $x \in \mathcal{H}$ we have $$ Bx=B(\sum_{n=1}^\infty x_ne_n)=\sum_{n=1}^\infty x_nBe_n=\sum_{n=1}^\infty x_nAe_n=\tilde{A}x, $$ i.e. $B=\tilde{A}$.