Continuous Linear Transformation between Topological Vector Spaces is Bounded
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
- 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.32$: Theorem