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Let X and Y be Banach Spaces and $T \in B(X,Y)$ (the space of bounded linear functions between $X$ and $Y$). I have to show that if $T(X)$ is not closed in $Y$ then $T(X)$ is of the first category, i.e. It is the countable union of nowhere dense sunsets of $Y$.

This was a question on an exam I had today and I was completely stumped. It was preceded by asking us to state the Open Mapping Theorem but I'm not sure this applies. Literally my only insight is that if $T(X)$ is not closed then $T^{*}(Y)$ is not closed -but I don't think this is enlightening either. Thanks in advance.

SEWillB
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  • I suspect that, more than the open mapping theorem itself, you should apply its method of proof. See if you can find some ideas here: https://terrytao.wordpress.com/2009/02/01/245b-notes-9-the-baire-category-theorem-and-its-banach-space-consequences/#more-1603 – Giuseppe Negro May 31 '17 at 17:06

2 Answers2

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I am confused because what is required here is almost the version of the open mapping theorem that I was taught as an undergraduate:

Suppose that $X$ and $Y$ are Banach spaces and $T\colon X\to Y$ is a bounded linear operator. If the range of $T$ is not meagre in $Y$, then $T$ is surjective.

See here for the proof. In your case, you simply apply contraposition of this statement as certainly in your case $T$ is not surjective, having non-closed range.

Tomasz Kania
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  • Interesting, we learnt the following: Suppose that $T$ is a surjective bounded linear map between Banach spaces $X$ and $Y$, then $T$ is open. – SEWillB Jun 02 '17 at 19:04
  • @SEWillB, at least in Poland, this version with category is very popular as this is exactly Rudin's approach (see his Functional analysis book). Please let me know if you want to know more or if this link is sufficient. – Tomasz Kania Jun 02 '17 at 19:08
  • No this is sufficient thank you – SEWillB Jun 02 '17 at 19:14
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I like Tomek's answer a lot. But here is another answer that is essentially the same but is phrased slightly differently.

First observe that the proof of the usual open mapping theorem still works if instead of $Y$ Banach and $T$ surjective we assume that $Y$ is a normed space and $T(X)$ is not meager in $Y$. Then $T$ is an open mapping from $X$ to $Y$. In particular, $T(X)$ is open in $Y$. Now use the following assertion (whose proof is an easy exercise): If $V$ is a subspace of a normed space $W$, and $V$ contains an open subset of $W$, then $V=W$.

MichaelGaudreau
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