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:
\(\ds T_\alpha \sqbrk E\) | \(\subseteq\) | \(\ds T_\alpha \sqbrk {t V}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds t T_\alpha \sqbrk V\) | Image of Dilation of Set under Linear Transformation is Dilation of Image | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds t W\) |
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.
$\blacksquare$
Sources
- 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $2.4$: Equicontinuity