Everywhere defined operators must be bounded?

Question

I have read in many places that as soon as you have an everywhere defined operator (on a Banach space), it must be automatically bounded, by the Closed Graph Theorem. However, I can't prove this using the Closed Graph Theorem (i.e., I can't prove it would be closed) and I can't find a reference for this. Is this true? Why?

Answer 1

You cannot prove that, as it is not true (with the axiom of choice). The statement, which is true from the closed graph theorem, is:

If $T \colon X \to Y$ is a closed operator defined on a Banach space $X$ into a Banach space $Y$, than $T$ is bounded.

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Addendum: Let $X$ be an infinite dimensional Banach space, $Y \ne 0$ be a Banach space. Then there is an unbounded $T \colon X \to Y$. Let (AC!) $B$ a basis of $X$ and $B' =\{b_n : n \in \mathbf N\}$ a countable subset, $y \in Y$ with $y \ne 0$. Define $T$ by linear extension of $$ T(b) = \begin{cases} n\|b_n\|y & b = b_n \\ 0 & b \in B \setminus B'\end{cases} $$ Then $T$ is linear $X \to Y$, and unbounded due to $$ \|T(b_n)\| = n\|b_n\|\|y\| $$ hence $\|T\| \ge n \|y\|$ for every $n$.