Image of von Neumann-Bounded Set under Equicontinuous Family of Linear Transformations is Contained in von Neumann-Bounded Set

Theorem
Let $\GF \in \set {\R, \C}$.

Let $X$ and $Y$ be a topological vector space over $\GF$.

Let $\family {T_\alpha}_{\alpha \in I}$ be an equicontinuous family of linear transformations.

Let $E \subseteq X$ be von Neumann-bounded.

Then there exists a von Neumann-bounded set $F \subseteq Y$ such that:


 * $T_\alpha \sqbrk E \subseteq F$ for each $\alpha \in I$.

Proof
Let:


 * $\ds F = \bigcup_{\alpha \mathop \in I} T_\alpha \sqbrk E$

Then $T_\alpha \sqbrk E \subseteq F$ for each $\alpha \in I$.

It is enough to show that $F$ is von Neumann-bounded.

Let $W$ be an open neighborhood of $\mathbf 0_Y$.

Since $\family {T_\alpha}_{\alpha \in I}$ is equicontinuous, there exists an open neighborhood $V$ of $\mathbf 0_X$ such that:


 * $T_\alpha \sqbrk V \subseteq W$ for each $\alpha \in I$.

Since $E$ is von Neumann-bounded, there exists $s > 0$ such that for all $t > s$ we have $E \subseteq t W$.

Then for each $t > s$ and $\alpha \in I$, we have:

So, from Union of Subsets is Subset:


 * $\ds F = \bigcup_{\alpha \in I} T_\alpha \sqbrk E \subseteq t W$

for $t > s$.

So $F$ is von Neumann-bounded.