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I'm looking for a direct proof of the Closed Graph Theorem (or Bounded Inverse Theorem) from the Uniform Boundedness Principle. But I can't find one in the literature.

I'm hoping there's a nice proof of the Closed Graph Theorem of the following form. Let $T:X \to Y$ be a closed linear map between Banach spaces. Define a family $\{T_{\alpha}\}_{\alpha \in A}$ of bounded linear maps from $X$ to $Y$ (or from $X$ to another normed space $Z$) such that $\sup_{\alpha \in A} \| T_{\alpha}(x) \| < \infty$ for all $x \in X$ and $\| T \| \leq \sup_{\alpha \in A} \| T_{\alpha}\|$. Conclude using Uniform Boundedness Principle. Of course, coming up with the family $\{T_{\alpha}\}_{\alpha \in A}$ is the hard part.

Similarly/Alternatively, I'd be very happy to see a proof of the Bounded Inverse Theorem of the following form. Let $T:X \to Y$ be a bounded linear bijection between Banach spaces. Define a family $\{S_{\alpha}\}_{\alpha \in A}$ of bounded linear maps from $Y$ to $X$ such that $\sup_{\alpha \in A} \| S_{\alpha}(y) \| < \infty$ for all $y \in Y$ and $\| T^{-1} \| \leq \sup_{\alpha \in A} \| S_{\alpha}\|$. Conclude using Uniform Boundedness Principle.

I was inspired by this proof of the Uniform Boundedness Principle from the Closed Graph Theorem: https://math.stackexchange.com/a/1473367/570438 It looks at the map $\Phi(x) = (T_{\alpha}(x))_{\alpha \in A}$, which maps $X$ to the space of bounded maps in $Y^A$.

A similar question was asked here before, but without a satisfactory answer: https://math.stackexchange.com/questions/1703248/does-the-closed-graph-theorem-follow-from-banach-steinhaus?noredirect=1#comment6033062_1703248

Theorem 27.26-27.31 of Schechter's Handbook of Analysis and its Foundations gives a indirect argument. Relatedly, the argument is adapted to give a direct proof of the Open Mapping Theorem from the Uniform Boundedness Principle here: Is there a simple direct proof of the Open Mapping Theorem from the Uniform Boundedness Theorem?

I am aware of the standard arguments for the implications Open Mapping Theorem $\Leftrightarrow$ Bounded Inverse Theorem $\Leftrightarrow$ Closed Graph Theorem.

I originally posted at MSE with upvotes but no responses, so I'm cross-posting here. MSE Link: https://math.stackexchange.com/questions/2924391/direct-proof-of-closed-graph-theorem-or-bounded-inverse-theorem-from-uniform-b

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  • $\begingroup$ imsc.res.in/~kesh/trinity.pdf (S. Kesavan, IIT Madras) seems to provide a proof that UBP implies CGT on Hilbert spaces, and states "Ramaswamy and Ramasamy [2] show that the uniform boundedness theorem implies the closed graph theorem provided we assume that the target space W is reflexive, while it is enough to assume that V is Banach." Unfortunately the reference [2] is "personal communication". $\endgroup$
    – D.R.
    Commented Jan 25, 2022 at 3:55
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    $\begingroup$ @D.R. Thanks. However, the argument appears to use the Hahn-Banach theorem in an essential way (if the argument was generalized from Hilbert spaces to normed spaces). Therefore, I don't count it as a direct or simple proof that Banach-Steinhaus implies Closed Graph. In fact, I've already seen essentially the same argument in a paper by Ramm (sciencepubco.com/index.php/GJMA/article/view/5534), which is cited in the paper by Kesavan that you have pointed out. $\endgroup$ Commented Jan 25, 2022 at 4:34

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