Continuous Linear Transformation between Topological Vector Spaces is Bounded

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

Let $T : X \to Y$ be a continuous linear transformation.


Then $T$ is bounded.


Proof

Let $E$ be a von Neumann-bounded subset of $X$.

We want to show that $T \sqbrk E$ is a von Neumann-bounded subset of $Y$.

Let $W$ be a open neighborhood of ${\mathbf 0}_Y$ in $Y$.

Since $T$ is continuous, it is continuous at ${\mathbf 0}_X$.

Since $\map T { {\mathbf 0}_X} = {\mathbf 0}_Y$, there exists an open neighborhood $V$ of ${\mathbf 0}_X$ in $X$ such that:

$T \sqbrk V \subseteq W$

Since $E$ is von Neumann-bounded subset, there exists $s > 0$ such that:

$E \subseteq t V$ for $t > s$.

Then we have:

\(\ds T \sqbrk E\) \(\subseteq\) \(\ds T \sqbrk {t V}\) Image of Subset under Mapping is Subset of Image
\(\ds \) \(=\) \(\ds t T \sqbrk V\) Image of Dilation of Set under Linear Transformation is Dilation of Image
\(\ds \) \(\subseteq\) \(\ds t W\)

for $t > s$.

Since $W$ was an arbitrary open neighborhood of ${\mathbf 0}_Y$ in $Y$, we have that $T \sqbrk E$ is von Neumann-bounded.

So $T$ is bounded.

$\blacksquare$


Sources