Is it true that If $A + A = 2A$ in a TVS then $A$ is convex? what if we assume A is complete?

Asked Mar 19 '23 at 02:56

Active Mar 19 '23 at 04:58

Viewed 202 times

We know that if $A$ is convex then it implies that $ A + A = 2A$ by simple arguments.

$2A \subset A + A $ this always holds. Now Let $ u+v \in A + A $ for some $u,v \in A$ Then $ (u+v)/2 \in A $ since A is convex. Therefore $ 2×(u+v)/2 = u+v \in 2A$

Now Does converse hold? If not then does converse hold if we assume tvs is complete?

My try: Let $u,v \in A$ Then $u + v \in 2A \implies (u+v)/2 \in A$.

Now repeating this arguments for diff pairs of $u/2^n$ and $v/2^m$ .

We get all dyadic coefficients of $u$ and $v$. So if we assume A to be complete we can get $u+v$ as well. That is my idea but I don't know how to write it formally?

Can you provide counter eg or proof if we don't assume completeness?

edited Mar 19 '23 at 04:40

asked Mar 19 '23 at 02:56

Mr.Multitalented

Assume it as a vector space over $R $or $C$ – Mr.Multitalented Mar 19 '23 at 04:37
But $Q$ is not complete. Assume $A$ to be complete – Mr.Multitalented Mar 19 '23 at 04:41
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why the downvote? – Snared Mar 19 '23 at 04:57
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$A+A\subset 2A$ implies $A$ is midpoint convex $[\frac{1}{2}x+\frac{1}{2}y\in A]$ $$\text{Closed+midpoint convex}\implies \text{convex}$$ – SoG Mar 19 '23 at 07:07
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Does this answer your question? If $C$ is midpoint-convex and closed then it is a convex set – SoG Mar 19 '23 at 07:08

Is it true that If $A + A = 2A$ in a TVS then $A$ is convex? what if we assume A is complete?

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