# Does the Closed Graph Theorem follow from Banach-Steinhaus?

Q: Is there a simple (but perhaps tricky or clever) proof of the Closed Graph Theorem (or the Open Mapping Theorem, or the result I call the Automatic Inverses Theorem below) from the Banach-Steinhaus Theorem, aka the Uniform Boundedness Principle?

It seems to me that there "should" be such an argument. Like something is automatically bounded when restricted to finite-dimmensional subspaces of something, or something like that, and then B-S shows that the norms of the whatevers are bounded, hence $T$ is bounded.

My work so far: I noticed that three big theorems are equivalent, which surprised me, and that they imply B-S, which surprised me. The one thing I sort of expected, that B-S implies CGT, is the one thing I haven't been able to prove. Just for the amusement of those of you who are amused by such things:

Note that $X$ and $Y$ will always denote Banach spaces.

There's a corollary of the Open Mapping Theorem that's so trivial it doesn't have a name; I'm going to call it the Automatic Inverses Theorem:

Theorem (AIT) If $T:X\to Y$ is a bounded linear bijection then $T^{-1}$ is bounded.

It's trivial that OMT implies AIT. A well known cute trick shows that AIT implies CGT. It's also trivial that CGT implies AIT. And it's easy to see that AIT implies OMT: Given a bounded linear surjection $T:X\to Y$, consider the induced map from $X/Z$ to $Y$, where $Z$ is the nullspace of $T$. (Details left to the reader.)

And it's not hard to deduce B-S from CGT. Suppose that $T_n:X\to Y$ is a bounded linear operator for $n=1,2,\dots$, and $||T_nx||_Y$ is bounded for every $x\in X$. Let $\ell^\infty_Y$ be the space of bounded sequences of elements of $Y$, with the norm $$||(y_1,y_2,\dots)||=\sup_n||y_n||.$$Define $T:X\to\ell^\infty_Y$ by $$Tx=(T_1x,T_2x,\dots).$$CGT shows easily that $T$ is bounded, and it follows that $||T_n||$ is bounded.

• A bit of Googling brought this up (I haven't checked it). – David Mitra Mar 18 '16 at 16:36
• @DavidMitra Thanks! This is just perfect - he (claims to) prove the result for Hilbert spaces and says it's easily extended to Banach spaces, so it gives me something to do. – David C. Ullrich Mar 18 '16 at 17:02
• I don't think they are equivalent in some sense. At least, consider how strong the axiom of choice is needed for these theorems. See, for example, this paper and here. – Yai0Phah Mar 18 '16 at 17:45
• @DavidC.Ullrich Did you ever find a satisfactory argument? – MichaelGaudreau Sep 17 '18 at 15:20
• @MichaelGaudreau Not really. Didn't worry about it much... – David C. Ullrich Sep 17 '18 at 17:17