Banach-Steinhaus Theorem/Normed Vector Space/Proof 3

Proof
It suffices to show that there exist an $x_0 \in X$ and an $r \in \R_{>0}$ such that:
 * $\ds K : = \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite

where $\map {B_r} {x_0}$ is the open $r$-ball of $x_0$.

Indeed, then we have for all $x \in X \setminus \set 0$:

That is:
 * $\ds \norm {T_\alpha} = \sup_{x \mathop \in X \setminus \set 0} \frac {\norm {T_\alpha x}_Y}{\norm x_X} \le \frac {4 K} r$

that for all $x_0 \in X$ and $r \in \R_{>0}$:
 * $\ds \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y = + \infty$

Then we can choose:
 * $\map {B_{r_1} } {x_1} \supseteq \map {B_{r_2} } {x_2} \supseteq \cdots$

and:
 * $\alpha_1, \alpha_2, \ldots \in A$

such that:
 * $\ds \inf _{x \mathop \in \map {B_{r_k} } {x_k} } \norm {T_{\alpha_k} x}_Y \ge k$

and:
 * $0 < r_k < \frac 1 k$

Indeed, let $r_0 \in \R_{>0}$ and $x_0 \in X$ be fixed.

Then, for $k = 1, 2, \ldots$, we have:
 * $\exists x_k \in \map {B_{r_{k-1} } } {x_{k-1} } \exists \alpha_k \in A : \norm {T_{\alpha_k} {x_k} }_Y > k$

Furthermore, as $T_{\alpha_k}$ is continuous, $\exists r_k \in \openint 0 {\frac 1 k}$ such that:
 * $\map {B_{r_{k-1} } } {x_{k-1} } \supseteq \map {B_{r_k} } {x_k}$

and:
 * $\ds \inf _{x \mathop \in \map {B_{r_k} } {x_k} } \norm {T_{\alpha_k} x}_Y \ge k$

Then $\sequence {x_k}$ is a Cauchy sequence in $X$, since for each $N \in \N_{>0}$:
 * $k, l \ge N \implies \norm {x_k - x_l}_X \le \norm {\underbrace {x_k - x_N}_{\in \map {B_{r_N} } {x_N} } }_X + \norm {\underbrace {x_l - x_N}_{\in \map {B_{r_N} } {x_N} } }_X\le 2 \, r_N \le \frac 2 N$

As $X$ is a Banach space, there exists:
 * $\ds z := \lim_{k \mathop \to +\infty} x_k$

Observe that:
 * $\ds z \in \bigcap_{k \in \N_{>0} } \map {\overline {B_{r_k} } } {x_k}$

where $\map {\overline {B_{r_k} } } {x_k}$ is the $r_k$-closed ball of $x_k$.

In particular:
 * $\forall k \in \N_{>0} : \norm {T_{\alpha_k} z}_Y \ge k$

Thus:
 * $\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha z}_Y = + \infty$

This is a contradiction.