Convex Open Neighborhood of Origin in Topological Vector Space contains Balanced Convex Open Neighborhood

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Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\Bbb F$.

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


Then there exists a balanced convex open neighborhood $V$ of ${\mathbf 0}_X$ with $V \subseteq U$.


Proof

Let:

$\ds A = \bigcap_{\cmod \alpha = 1} \alpha U$

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists an balanced open neighborhood $W$ of ${\mathbf 0}_X$ with $W \subseteq U$.

Since $W$ is balanced, we have:

$\alpha W \subseteq W$

for all $\alpha \in \Bbb F$ with $\cmod \alpha = 1$.

Note that if $\alpha \in \Bbb F$ has $\cmod \alpha = 1$, then we have $\cmod {\alpha^{-1} } = 1$, so we also have:

$\alpha^{-1} W \subseteq W$

giving:

$W \subseteq \alpha W$

and so:

$W = \alpha W$

Swapping $\alpha$ for $\alpha^{-1}$, we also get:

$W = \alpha^{-1} W$

So, for each $\alpha \in \Bbb F$ with $\cmod \alpha = 1$ we have:

$\alpha^{-1} W \subseteq U$

and so:

$W \subseteq \alpha U$

This gives, from Set Intersection Preserves Subsets:

$W \subseteq A$

Since $W$ is open and contains ${\mathbf 0}_X$, the interior $A^\circ$ is an open neighborhood of ${\mathbf 0}_X$.

Since:

$A \subseteq \alpha U$

for each $\alpha \in \Bbb F$ with $\cmod \alpha = 1$, we in particular have:

$A \subseteq U$

Since $A^\circ \subseteq A$, we have:

$A^\circ \subseteq U$

We show that $V = A^\circ$ is our desired balanced convex open neighborhood.

Since $A$ is the intersection of convex sets, it is convex by Intersection of Convex Sets is Convex Set (Vector Spaces).

From Interior of Convex Set in Topological Vector Space is Convex, $A^\circ$ is then convex.

It remains to show that $A^\circ$ is balanced.

From Interior of Balanced Set containing Origin in Topological Vector Space is Balanced, it suffices to show that $A$ is balanced.

Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.

Pick $r \in \closedint 0 1$ and $\beta \in \Bbb F$ with $\cmod \beta = 1$ such that $\lambda = r \beta$.

Then:

\(\ds \lambda A\) \(=\) \(\ds r \beta A\)
\(\ds \) \(=\) \(\ds r \beta \bigcap_{\cmod \alpha = 1} \alpha U\)
\(\ds \) \(=\) \(\ds \bigcap_{\cmod \alpha = 1} r \beta \alpha U\) Dilation of Intersection of Subsets of Vector Space

Note that for fixed $\beta \in \Bbb F$ with $\cmod \beta = 1$, $\beta \alpha$ runs through all $\gamma \in \Bbb F$ with $\cmod \gamma = 1$, setting $\alpha = \gamma/\beta$.

Conversely, $\cmod {\beta \alpha} = 1$ whenever $\cmod \alpha = 1$ and $\cmod \beta = 1$.

So, we have:

$\ds \bigcap_{\cmod \alpha = 1} r \beta \alpha U = \bigcap_{\cmod \gamma = 1} r \gamma U$

From Dilation of Convex Set in Vector Space is Convex, we have that:

$\gamma U$ is convex for each $\gamma \in \Bbb F$ with $\cmod \gamma = 1$.

So, we have:

$r \gamma U + \paren {1 - r} \gamma U \subseteq \gamma U$

and in particular:

$r \gamma U \subseteq \gamma U$

for each $\gamma \in \Bbb F$ with $\cmod \gamma = 1$.

From Set Intersection Preserves Subsets, we therefore have:

$\ds \bigcap_{\cmod \gamma = 1} r \gamma U \subseteq \bigcap_{\cmod \gamma = 1} \gamma U = A$

So we have:

$\lambda A \subseteq A$

for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$.

So $A$ is balanced.

So, from Interior of Balanced Set containing Origin in Topological Vector Space is Balanced, $A^\circ$ is balanced.

So $V = A^\circ$ is a balanced convex open neighborhood $V$ of ${\mathbf 0}_X$ with $V \subseteq U$.

$\blacksquare$


Sources

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