Im reading the proof for Banach–Steinhaus theorem. I'll explain my question:
The theorem is as following:
Let $ V $ be a Banach space and $ W $ a normed vector space. Assume that for $ \mathcal{F}\subseteq B\left(V,W\right) $, a set of bounded linear transformations $ T: V \to W $, there exists a set $ \varOmega $ of second Baire category, such that:
$ \forall v\in\varOmega:\sup_{T\in\mathcal{F}}||Tv||<\infty $
Then, $ \underset{T\in\mathcal{F}}{\sup}||T||<\infty $.
Where $ ||T|| $ denotes $ \sup_{\underset{||v||\leq1}{v\in V}}||Tv|| $.
Now in the proof we define a sequence of sets $ A_n $ as following:
$ A_{n}=\left\{ v\in V:\sup_{T\in\mathcal{F}}||Tv||\leq n\right\} $
And we notice that
$ A_{n}=\bigcap_{T\in\mathcal{F}}\left(T^{-1}\left(B_{c}\left(\underline{0},n\right)\right)\right) $
Where $ B_c $ denotes a closed ball.
Since $ T $ is bounded for any $ T\in \mathcal{F} $ we also know that $ T $ is continuous and thus $ A_n $ is an intersection of closed sets, and specifically $ A_n $ is a closed set for any $ n $.
Now, here's what I dont get: According to any source that I found, we know claim that
$ \varOmega\subseteq\bigcup_{n=1}^{\infty}A_{n} $ (and others may take $\varOmega=V $) And using Baire Category Theorem we know that there exists $n_0 $ such that $ \text{Int}\left(A_{n}\right)\neq\emptyset $.
My question is why do we need exactly Baire Category Theorem for? (Also since I think we dont need it I also dont get why would we want $ V $ to be a Banach space).
If we assume that $ \varOmega $ is a set of second Baire category, meaning it is not a set of the first Baire category, and if we agree that a set is a set of first Baire category by definition, if it is a subset of a union of closed sets with empty interior, then assuming for contradiction that $ \text{Int}\left(A_{n}\right)=\emptyset $ for any $ n\in \mathbb{N} $, gives us that, by definition, $ \varOmega$ is a set of first Baire category.
So I'll be happy if someone can clarify why those assumption are needed. Thanks in advance
Best Answer
Because you already assume you have such a $\Omega$ of second category, you don't need BCT in the proof anymore.
But in practice, to apply the theorem, you can maybe show $V=\Omega$ but then you need $V$ to be a Banach space to conclude $\Omega$ is of second category.