Non-Zero Subspace of Topological Vector Space is not von Neumann-Bounded

Theorem

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

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

Let $Y \ne \set {\mathbf 0_X}$ be a non-trivial subspace of $X$.

Then $Y$ is not von Neumann-bounded.

Proof

Let $x \in Y \setminus \set {\mathbf 0_X}$.

Then $t x \in Y$ for $t > 0$.

Since $X$ is Hausdorff, there exists an open neighborhood $V$ of $x$ that does not contain $\mathbf 0_X$.

Then $t x \not \in t V$ for each $t > 0$.

So we do not have $Y \subseteq t V$ for any $t > 0$.

So $Y$ is not von Neumann-bounded.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.29$: Bounded sets