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 $x \in X \setminus \set 0$:

where:
 * $\ds K' := \sup_{\alpha \mathop \in A} \norm {\T_\alpha x_0 }_Y$

is finite.

That is:
 * $\ds \norm {T_\alpha} = \sup_{x \mathop \in X \setminus \set 0} \frac {\norm {T_\alpha x}_Y}{\norm x_X} \le \frac {2 \paren {K + K'} } r$

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

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

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

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

Then there exists:
 * $\ds z := \lim_{k \mathop \to +\infty} r_k$

As $z \in \map {B_{r_k} } {x_k}$ for all $k \in \N_{>0}$, we have:
 * $\forall k \in \N_{>0} : \norm {T_{\alpha_k} z}_Y \ge k$

In particular:
 * $\ds \sup_{\alpha \mathop \in A} \norm {T z}_Y = + \infty$

This is a contradiction.