Banach-Steinhaus Theorem/Normed Vector Space/Proof 2

Proof
From Banach Space is F-Space, $\struct {X, \norm {\, \cdot \,}_X}$ can be considered as an $F$-Space.

From Normed Vector Space is Topological Vector Space, $\struct {Y, \norm {\, \cdot \,}_Y}$ can be considered as a topological vector space.

Let $\Gamma = \set {T_\alpha : \alpha \in A}$ and:
 * $\map \Gamma x = \set {T_\alpha x : \alpha \in A}$

for each $x \in X$.

By hypothesis, we have:
 * $\ds \sup_{\alpha \in A} \norm {T_\alpha x}_Y < \infty$ for each $x \in X$.

From Characterization of von Neumann-Boundedness in Normed Vector Space, we have that $\map \Gamma x$ is von Neumann-bounded in $\struct {Y, \norm {\, \cdot \,}_Y}$ for each $x \in X$.

From Banach-Steinhaus Theorem: $F$-Space, $\Gamma$ is equicontinuous.

So there exists an open neighborhood $U$ of ${\mathbf 0}_X$ such that:
 * $T_\alpha \sqbrk U \subseteq \map {B_1^Y} { {\mathbf 0}_Y} \subseteq B^-_Y$ for each $\alpha \in A$.

where:
 * $\map {B_1^Y} { {\mathbf 0}_Y}$ is the open ball of center ${\mathbf 0}_Y$ and radius $1$ in $\struct {Y, \norm {\, \cdot \,}_Y}$
 * $B_Y^-$ is the closed unit ball of $\struct {Y, \norm {\, \cdot \,}_Y}$.

From the definition of an open set in a normed vector space, there exists $\delta > 0$ such that:
 * $\map {B_\delta^X} { {\mathbf 0}_X} \subseteq U$

and so:
 * $T_\alpha \sqbrk {\map {B_\epsilon^X} { {\mathbf 0}_X} } \subseteq \map {B_1^Y} { {\mathbf 0}_Y}$ for each $\alpha \in A$.

We then have:
 * $\paren {\delta/2} B_X^- \subseteq \map {B_\delta^X} { {\mathbf 0}_X}$

and:
 * $\map {B_1^Y} { {\mathbf 0}_Y} \subseteq B_Y^-$

where $B_X^-$ denotes the closed unit ball of $\struct {X, \norm {\, \cdot \,}_X}$.

Hence by Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:
 * $T_\alpha \sqbrk {B_X^-} \subseteq \paren {2/\delta} B_Y^-$ for each $\alpha \in A$.

From Norm of Bounded Linear Transformation in terms of Closed Unit Ball, we have:
 * $\ds \norm {T_\alpha} \le \frac 2 \delta$ for each $\alpha \in A$.

That is:
 * $\ds \sup_{\alpha \in A} \norm {T_\alpha} \le \frac 2 \delta$

as required.