Banach-Steinhaus Theorem/Normed Vector Space/Proof 1

Proof
For each $n \in \N$, define:


 * $F_n = \set {x \in X : \norm {T_\alpha x} \le n \text { for each } \alpha \in A}$

We have:

where $\map {\overline {B_Y} } {0, n}$ is the closed ball in $\struct {Y, \norm \cdot_Y}$ of radius $n$ centred at $0$.

From Closed Ball is Closed in Normed Vector Space, we have:


 * $\map {\overline {B_Y} } {0, n}$ is closed.

From Mapping is Continuous iff Inverse Images of Open Sets are Open: Corollary, we have:


 * ${T_\alpha}^{-1} \sqbrk {\map {\overline {B_Y} } {0, n} }$ is closed for each $\alpha \in A$.

From Intersection of Closed Sets is Closed in Normed Vector Space, we have:


 * $F_n$ is closed for each $n \in \N$.

Clearly we have:


 * $\ds \bigcup_{n \mathop = 1}^\infty F_n \subseteq X$

Recall that by hypothesis, for every $x \in X$, we have:


 * $\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite.

So in particular there exists $N \in \N$ such that:


 * $\ds \norm {T_\alpha x}_Y \le N$

so that:


 * $x \in F_N$

That is:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty F_n$

So:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty F_n$

Since $\struct {X, \norm \cdot_X}$ is a Banach space, from the Baire Category Theorem we have that:


 * $\struct {X, \norm \cdot_X}$ is a Baire space.

From Baire Space is Non-Meager, we therefore have:


 * $X$ is non-meager.

So:


 * $X$ is not the countable union of nowhere dense subsets of $X$.

So, we must have:


 * $F_n$ is not nowhere dense for some $n \in \N$.

Fix such an $n$.

Recall that $F_n$ is closed.

From Set is Closed iff Equals Topological Closure, we therefore have:


 * $F_n^- = F_n$

where $F_n^-$ denotes the closure of $F_n$.

So, since $F_n$ is not nowhere dense, we have that:


 * there exists a non-empty open $U \subseteq X$ such that $U \subseteq F_n$.

Pick $u \in U$.

Since $U$ is open, there exists $y \in X$ and $r > 0$ such that:


 * $\map {B_X} {y, r} \subseteq F_n$

where $\map {B_X} {y, r}$ denotes the open ball in $\struct {X, \norm \cdot_X}$ of radius $r$, centre $y$.

Now let $\alpha \in A$ and $x \ne 0$.

Since:


 * $\ds \norm {\frac {r x} {2 \norm x} }_X = \frac r 2 < r$

we have:


 * $\ds y + \frac {r x} {2 \norm x_X} \in \map {B_X} {y, r}$

So, we have:


 * $\ds \norm {\map {T_\alpha} {y + \frac r {2 \norm x_X} } }_Y \le n$

So, from linearity, we have:


 * $\ds \norm {T_\alpha y + \frac r {2 \norm x_X} T_\alpha x}_Y \le n$

From Reverse Triangle Inequality: Normed Vector Space and positive homogeneity of the norm, we therefore have:


 * $\ds \size {\norm {T_\alpha y}_Y - \frac r {2 \norm x_X} \norm {T_\alpha x}_Y} \le n$

So that:


 * $\ds \frac r {2 \norm x_X} \norm {T_\alpha x}_Y \le \norm {T_\alpha y}_Y + n$

Since $y \in F_n$, we have:


 * $\ds \norm {T_\alpha y}_Y \le n$

So:


 * $\ds \frac r {2 \norm x_X} \norm {T_\alpha x}_Y \le 2 n$

That is:


 * $\ds \norm {T_\alpha x}_Y \le \frac {4 n} r \norm x_X$

for all $\alpha \in A$ and $x \ne 0$.

This inequality also clearly holds for $x = 0$.

So:


 * $\ds \norm {T_\alpha} \le \frac {4 n} r$

for all $\alpha \in A$ from the definition of the norm on bounded linear transformations.

From the Continuum Property, we therefore have:


 * $\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha}$ exists as a real number

as required.