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Let A be a closed set in a normed space. Prove that A is convex $\iff$ $\forall{x, y \in A} $ we have $\frac{x+y}{2}\in A $

Well, my definition of a convex set A is that $\forall \lambda \in [0,1]; x,y \in A$ we have $\lambda x + (1-\lambda)y \in A$. I literally couldn't write anything else than the thesis when trying to prove both sides of the $\iff$. Is there some trick to be used? Why is the space being normed important?

I appreciate your help.

blahblah
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1 Answers1

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In general, this is not true. Take for instance $\mathbb{Q}$ in $\mathbb{R}$. It satisfies the property, but it is not convex.

However, if you assume the set to be closed, then it is true. Indeed, the $\Rightarrow$ way is easy (just take $\lambda=1/2$) in the definition of convexity. For the reverse way, prove inductively that for all $n \geq 1$, for all $k \in \lbrace 0, ..., 2^n \rbrace$, the definition of convexity is true for $\lambda=k/2^n$. And finally use a density argument to extand that to all $\lambda \in [0,1]$.

TheSilverDoe
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  • Whoops - my bad. The set is closed in the example. I have edited the question. – blahblah Oct 11 '20 at 07:58
  • what is the density argument? – blahblah Oct 11 '20 at 08:43
  • @НикитаВасильев You have to deduce the property for all $\lambda$ from the fact that it holds for all the $\lambda = k/2^n$... This can be done by taking a limit, since all the $\lambda \in [0,1]$ can be written as a limit of elements of the form $k/2^n$. – TheSilverDoe Oct 11 '20 at 09:30