This is a problem from Conway's "A Course in Functional Analysis", problem II.1.3. Suppose $E = \{e_n\}$ is an orthonormal basis for a Hilbert space $H$ and $A$ is a linear map $H \to K$ (where $K$ is also a Hilbert space) that satisfies $\sum_n \| Ae_n\| < \infty $. Show that $A$ is bounded.
I believe the statement to be false in general. Let $H=l^2(\mathbb{N})$ and $K = \mathbb{R}$. We can extend $E$ to a Hamel basis $E'$ such that $\|e\| = 1$ for all $e \in E'$. Let $(f_n)$ be any enumerable subset of $E' \setminus E$. Then setting $Ae_n = 2^{-n}$, $Af_n = n$ and $Ae = 0$ for $e \in E' \setminus (E \cup (f_n))$ yields an unbounded operator on $H$. Is my reasoning correct?
Best Answer
Your example is fine. But the exercise does not ask what you say it asks. It says to show that $A$ has a unique bounded extension. Your example produces an unbounded extension; that doesn't preclude the existence of a bounded one.
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