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Let $A$ be a locally convex algebra. By $\mathcal{B}_1$, we denote the collection of all subsets $B$ of $A$ such that:

  1. $B$ is absolutely convex and $B^2\subseteq B$
  2. $B$ is closed and bounded in $A$.

Now let $S$ be a closed subalgebra of $A$. Let $\mathcal{B}_1$ have its usual meaning and let $\mathcal{B}_2$ be the corresponding collection for the subalgebra $S$.

My question: Is $\mathcal{B_2} \subseteq \mathcal{B}_1?$

I am thinking yes.

  1. Clearly if $B \in \mathcal{B}_2$, then $B\subseteq S \subseteq A$, i.e. $B$ is a subset of $A$ that is absolutely convex and satisfies $B^2 \subseteq B$.

However, since $B \in \mathcal{B}_2$ (the collection corresponding to $S$) we only know that $B$ is closed and bounded in $S$, i.e. the in the relative topology on $S$. How do we know that $B$ is also closed and bounded in the topology on $A$?

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If $B\subset S$ is closed in the relative topology of $S$, then since $S$ is closed in $A$ it follows that $B$ is closed in $A$ (see for instance this question).

If now $B$ is bounded in $S$, then for every neighborhood $U$ of $0$ in $A$, $U\cap S$ is an open neighborhood of $0$ in $S$, so there is some $t_0>0$ such that $B\subset t(U\cap S)$ for all $t> t_0$. But then for all $t>t_0$ we have $B\subset tU$, and thus $B$ is bounded in $A$.

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  • $\begingroup$ By the way, I am using the definition of a bounded set found in Rudin's Functional Analysis. If there is another definition you would prefer, and cannot translate the proof given here, let me know and I will edit. $\endgroup$ – Aweygan Apr 15 '18 at 1:00

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