Norm of a continuous linear operator

Question

Let $A$ be a separable Hilbert space with orthonormal basis $\{a_n\}_1^\infty$ and $\{x_n\}$ a sequence s.t. $\sum_\limits{n=1}^\infty||x_n||^2<\infty$.

1. How do I show that there exists a continuous map $f:A\rightarrow A$ with $fe_n=x_n$?
2. How do I show that $f$ is unique?
3. What is an upper bound for $||f||$?

What I know:
1. For continuity I think we need $||fx||\leq C||x||$ for a constant $C$ and all $x\in A$. How would I show this?
2. I think we have to suppose here that there exists a $g$ s.t. $ge_n=x_n$ and show that $g=f$. How does this go?
3. The norm of a linear operator is $\sup\{||fx||:||x||\leq1\}$. So I think we will have to express the upper bound of $||f||$ in terms of $x_n$, but I don't quite see how.

Answer 1

• If $f$ exists using linearity and continuity it must satisfy: $$f:y:=\sum_{n=0}^{\infty}y_na_n\mapsto\sum_{n=0}^{\infty}y_nx_na_n.$$

• $f$ is well-defined, since using Parseval's identity: $$\|f(y)\|^2=\sum_{n=0}^{\infty}\|y_nx_n\|^2\leqslant\sum_{n=0}^{\infty}\|x_n\|^2\times\|y\|^2<\infty.$$ By the way, one see that $f$ is continuous and one has: $$\|f\|\leqslant\sqrt{\sum_{n=0}^{\infty}\|x_n\|^2}.$$