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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?

Ruben
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  • You probably misremembered what the Hellinger-Toeplitz theorem says. – Daniel Fischer Dec 08 '16 at 19:17
  • The Hellinger-Toeplitz theorem deals with symmetric operators, does it not? I am not assuming that the operator is symmetric. – Ruben Dec 08 '16 at 19:19
  • In my original question: in the context in which I have seen the claim, the operator is not symmetric. – Ruben Dec 08 '16 at 19:20
  • But without the symmetry condition, the claim is wrong if we have choice. In the absence of choice, it's possible that every linear map on a Banach space is continuous. In the absence of choice, really weird things can happen. – Daniel Fischer Dec 08 '16 at 19:22
  • Assume AC: can you provide a reference for this? – Ruben Dec 08 '16 at 19:24
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    Consider $X = \ell^1(\mathbb{N})$. Let $E = { e_k : k \in \mathbb{N}}$ be the usual Schauder basis of $X$. Extend it to a Hamel basis $B$. Define $T\colon X \to X$ by setting $T(e_k) = 2^k\cdot e_k$, and $T(b) = 0$ for $b \in B\setminus E$, and linearly extending to $X$. Then $T$ is an everywhere defined unbounded linear operator on $X$. – Daniel Fischer Dec 08 '16 at 19:30
  • Thanks Daniel! That's exactly what I wanted. Cheers! – Ruben Dec 08 '16 at 19:35
  • Daniel Fischer: In your example, T is not defined for $x=(1,\frac14,\frac19,\ldots,\frac1{n^2},...)$. – pabodu Jan 02 '20 at 05:36
  • @pabodu It is defined: it is 0. – Ruben Jan 07 '20 at 01:30
  • @Ruben: Do I understand right that all infinite sequences are mapped to zero but all finite sequences are mapped to nonzero finite sequence with $k$-th coordinate increased by $2^k$ times? – pabodu Jan 09 '20 at 06:01
  • @Ruben: If yes, then from one side, $T(1,\frac12,\frac14,...) = 0$, but from the other side, $T(1,\frac12,\frac14,...)=T(e_1)+T(0,\frac12,\frac14,...)=2e_1\neq0$. Here $e_1=(1,0,0,...)$. We arrive at the contradiction and, hence $T$ is nonlinear. – pabodu Jan 09 '20 at 06:14
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    @pabodu: My mistake (I was too quick to say that $T$ is $0$ on the vector $x$ you defined). The issue is more complicated than that. $T$ is zero on $B \setminus E$, so we would need first to decide what is the Hamel basis. Obviously, it cannot be that both $(1, 1/2, 1/4, \dots)$ and $(0,1/2, 1/4, \dots)$ are in the Hamel basis $B$, hence they cannot both map to zero. – Ruben Jan 10 '20 at 16:31
  • Hamel basis is a very vague notion. I do not feel that the problem is solved unless I cannot find $T(x)$ for such typical $x$. – pabodu Jan 10 '20 at 21:23

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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.

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 $n$.

martini
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  • I've seen that answer before, but why? is there a reference for this? Does this mean that if I can explicitely construct an everywhere defined operator, it must be bounded? – Ruben Dec 08 '16 at 19:10
  • It is bounded iff it is closed. – martini Dec 08 '16 at 19:11
  • I understant that. I'm probably not explaining myself correctly: is there an example of an unbounded everywhere defined (obviously not closed) operator on a Banach space (even if it uses AC)? Is there a reference for this? – Ruben Dec 08 '16 at 19:24
  • @Ruben Added something – martini Dec 08 '16 at 19:30
  • Got it! That's exactly what I wanted. Thanks a lot. – Ruben Dec 08 '16 at 19:34
  • @martini Counterexample: Let $X=l^1$, $Y=\mathbb{R}$, $y=1$, $B'=B$ and $b_n=e_n=(0,...,0,1,0,...)$. Let $x={\frac1{n^2}}$. Then $T(x)=T(\sum \frac1{n^2} e_n)=\sum \frac1{n^2} T(e_n)=\sum \frac1n=\infty$. So, functional $T$ is not defined on the entire $l^1$. – pabodu Jan 10 '20 at 11:22
  • The comment by @pabodu contains two errors. First, the answer said that $B$ should be a basis for $X$, not merely a Schauder basis. This means that every element of $X$ should be a finite linear combination of elements of $B$. Second, in the last equation of the comment, the step where $T$ is moved inside the summation presupposes that $T$ is continuous. – Andreas Blass Aug 02 '20 at 20:08
  • In the book "A Basis Theory Primper" by Heil it states the following: "In fact, it is known that the statement “Every vector space has a Hamel basis” is one of many equivalent formulations of the Axiom of Choice." Just in case anyone is wondering what the role of the Axiom of Choice is in this instance. See also: https://math.stackexchange.com/questions/194189/a-hamel-basis-for-ellp – user510186 Jun 16 '22 at 16:08