Sequential Characterization of von Neumann-Boundedness in Topological Vector Space

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Theorem

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

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

Let $E \subseteq X$.


Then $E$ is von Neumann-bounded if and only if:

for each sequence $\sequence {x_n}_{n \in \N}$ in $E$ and each sequence $\sequence {\alpha_n}_{n \in \N}$ in $\GF$ such that $\alpha_n \to 0$ as $n \to \infty$, we have $\alpha_n x_n \to {\mathbf 0}_X$ as $n \to \infty$.


Proof

Necessary Condition

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

Let $U$ be an open neighborhood of ${\mathbf 0}_X$.

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists a balanced open neighborhood $V$ of ${\mathbf 0}_X$ such that $V \subseteq X$.

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

$E \subseteq t V$

so that:

$\ds \frac 1 t E \subseteq V$

Then we have:

$\ds \frac {x_n} t \in V$

Since $\alpha_n \to 0$, there exists $N \in \N$ such that:

$\ds \cmod {\alpha_n} < \frac 1 t$ for $n \ge N$.

That is:

$\ds \cmod {\alpha_n} t = \cmod {\alpha_n t} < 1$ for $n \ge N$.

Since $V$ is balanced, we have:

$\alpha_n t V \subseteq V$

so that:

$\alpha_n x_n \in V$ for $n \ge N$

since $x_n/t \in V$.

So, we have:

$\alpha_n x_n \in V$ for $n \ge N$.

So $\alpha_n x_n \to \mathbf 0_X$ as $n \to \infty$.

$\Box$

Sufficient Condition

Let $E \subseteq X$ be such that:

for all sequences $\sequence {x_n}_{n \in \N}$ is a sequence in $E$ and $\sequence {\alpha_n}_{n \in \N}$ is a sequence in $\GF$ such that $\alpha_n \to 0$, we have $\alpha_n x_n \to {\mathbf 0}_X$.

Suppose that $E$ is not von Neumann-bounded.

Then for some open neighbourhood $V$ of ${\mathbf 0}_X$ and all $s > 0$, there exists $t > s$ such that $E \not \subseteq t V$.

That is, there exists a sequence $\sequence {r_n}_{n \in \N}$ of positive real numbers such that $r_n \to \infty$ and $E \not \subseteq r_n V$ for each $n \in \N$.

Then $r_n^{-1} \to 0$ as $n \to \infty$.

For each $n \in \N$, pick $x_n \in E \setminus r_n V$.

Then we have $r_n^{-1} x_n \not \in V$ for each $n \in \N$.

So $\sequence {r_n^{-1} x_n}_{n \in \N}$ cannot converge to $\mathbf 0_X$.

This is a contradiction, so $E$ is von Neumann-bounded.

$\blacksquare$


Sources

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