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My instructor has given me a proof of Banach-Steinhaus theorem which is also known as Uniform Boundedness Theorem which I find hard to grasp. Here it is $:$

Uniform Boundedness Theorem $:$

Let $(X, \| \cdot \|)$ be a Banach space and $(Y,\|\cdot\|)$ be a normed linear space. Let $\mathcal A \subset B(X,Y)$ be a pointwise bounded family of bounded linear transformations from $X$ to $Y$. Then the family $\mathcal A$ is uniformly bounded.

Proof $:$ Let $T \in \mathcal A$. Then for each $n \geq 1$ the set $\{x \in X : \|Tx\| \leq n \}$ is a closed set in $X$. $\implies$ $X \setminus \{x \in X : \|Tx\| \leq n \}$ is an open set in $X$ i.e. the set $\{x \in X : \|Tx\| > n \}$ is an open set in $X$. So $\underset {T \in \mathcal A} {\bigcup} \{x \in X : \|Tx\| > n \}$ is an open set in $X$ i.e. $\{x \in X : \underset {T \in \mathcal A} {\sup} \|Tx\| > n \}$ is an open set in $X$. Let $V_n =\{x \in X : \underset {T \in \mathcal A} {\sup} \|Tx\| > n \}$, $n \geq 1$. Then $\bigcap\limits_{n=1}^{\infty} V_n = \{x \in X : \underset {T \in \mathcal A} {\sup} \|Tx\| = \infty \}$. If $\exists$ $n_0 \in \Bbb N$ such that $V_{n_0}$ is not dense in $X$ then $E_{n_0} = X - V_{n_0}$ is closed and contains a non-empty open set. So it's interior is non-empty. Hence $\exists$ $x_0 \in E_{n_0}$ and an $\epsilon > 0$ such that the closed ball $\overline B(x_0;\epsilon) \subset E_{n_0} = \{x \in X : \underset {T \in \mathcal A} {\sup} \|Tx\| \leq n_0 \}$. Now take $T \in \mathcal A$ and $y \in \overline B(0 ;1)$ then $x_0 + \epsilon y \in B(x_0;\epsilon)$. So $\epsilon \|Ty \| = \|T(\epsilon y)\| = \|T(x_0+\epsilon y - x_0)\| = \|T(x_0+\epsilon y) - T(x_0)\| \leq \|T(x_0+\epsilon y) \| + \|T(x_0)\| \leq n_0 + n_0 = 2n_0$ $\implies$ $\|T(y)\| \leq \frac {2n_0} {\epsilon}$. Since $y \in \overline B(0 ;1)$ is arbitrary so we have $\|T\| \leq \frac {2n_0} {\epsilon}$. Also $T \in \mathcal A$ was arbitrary and hence $\underset {T \in \mathcal A} {\sup}\|T\| \leq \frac {2n_0} {\epsilon} = M$ (say).

This completes the proof.

I don't understand that what the proof actually is. What does it try to prove? Also as a corollary he (my instructor) stated that if $X$ is a Banach space and $Y$ is a normed linear space then either $\bigcap\limits_{n=1}^{\infty} V_n$ is dense or an empty set.

Why is it so? Please help me in this regard.

Thank you very much.

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  • $\begingroup$ I think your intersection being empty or dense comes from an appeal to the Baire category theorem. $\endgroup$ – Lord Shark the Unknown Sep 1 '18 at 15:41
  • $\begingroup$ Here the family is given to be pointwise bounded hence for all $x \in X$, $\underset {T \in \mathcal A} {\sup} \|Tx\| < \infty$. So in our case $\bigcap\limits_{n=1}^{\infty} V_n = \emptyset$. Right? $\endgroup$ – Dbchatto67 Sep 1 '18 at 15:57
  • $\begingroup$ Hence there should exist some $n_0 \in \Bbb N$ such that $V_{n_0}$ is not dense by Baire's category theorem. So we are done with the proof. Right? $\endgroup$ – Dbchatto67 Sep 1 '18 at 16:00
  • $\begingroup$ According to Baire's category theorem we know that countable intersection of dense open sets is dense. $\endgroup$ – Dbchatto67 Sep 1 '18 at 16:02
  • $\begingroup$ So if each $V_n$ was dense; since they are all open so their intersection $\bigcap\limits_{n=1}^{\infty} V_n$ is also dense by Baire's category theorem which is not true in our case. Which gives rise to some $n_0 \in \Bbb N$ such that $V_{n_0}$ is not dense in $X$. Right or not? $\endgroup$ – Dbchatto67 Sep 1 '18 at 16:06

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