Characterization of von Neumann-Boundedness in Hausdorff Locally Convex Space

Theorem

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

Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$.

Let $U \subseteq X$.

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

for each $p \in \PP$ there exists $C_p > 0$ such that:

$\map p x < C_p$

for each $x \in U$.

Proof

For each $p \in \PP$, let:

$B_p = \set {y \in X : \map p y < 1}$

Note that by the definition of the standard topology, $B_p$ is an open neighborhood of $\mathbf 0_X$.

Let $r > 0$.

From Seminorm Axiom $\text N 2$: Positive Homogeneity, for $y \in X$ we have:

$\map p y < 1$

if and only if:

$\map p {r y} < r$

So, we have:

$r B_p = \set {y \in X : \map p y < r}$

Necessary Condition

Suppose that $U$ is von Neumann-bounded.

Then there exists $s > 0$ such that:

$U \subseteq t B_p = \set {y \in X : \map p y < t}$

for $t > s$

In particular:

$U \subseteq \set {y \in X : \map p y < s + 1}$

That is:

$\map p x < s + 1$

for all $x \in U$.

$\Box$

Sufficient Condition

Suppose that for each $p \in \PP$ there exists $C_p > 0$ such that:

$\map p x < C_p$

for each $x \in U$.

Let $V$ be an open neighborhood of $\mathbf 0_X$.

From Open Sets in Standard Topology of Locally Convex Space, there exists $p_1, \ldots, p_n \in \PP$ and $\epsilon > 0$ such that:

$\ds \bigcap_{k \mathop = 1}^n \epsilon B_{p_k} = \set {y \in X : \map {p_k} y < \epsilon \text { for each } 1 \le k \le n} \subseteq V$

Let:

$\ds r = \frac 1 \epsilon \max_{1 \le k \le n} C_{p_k}$

Then:

$\ds r \bigcap_{k \mathop = 1}^n \epsilon B_{p_k} = \bigcap_{k \mathop = 1}^n \paren {\max_{1 \le k \le n} C_{p_k} } B_{p_k} \subseteq r V$

and generally if $s > r$:

$\ds \bigcap_{k \mathop = 1}^n \paren {\max_{1 \le k \le n} C_{p_k} } B_{p_k} \subseteq s \bigcap_{k \mathop = 1}^n \epsilon B_{p_k} \subseteq s V$

Note that for $x \in U$, $1 \le k \le n$ and $s > r$ we then have:

$\ds \map {p_k} x < C_{p_k} \le \max_{1 \le k \le n} C_{p_k} < s \epsilon$

for $x \in U$, and hence:

$U \subseteq s V$

for $s > r$.

So $U$ is von Neumann-bounded.

$\blacksquare$