Mid-Point Convexity Implies Convexity in Banach Spaces?

Asked Jan 25 '13 at 18:32

Active Jan 25 '13 at 18:32

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Showing that $f$ is convex

Let $X$ be a real Banach space and $f:X\rightarrow \mathbb{R}$ a continuous function. We say that $f$ is Mid-Point convex if for all $x,y\in X$ $$\tag{1}f\Big(\frac{x+y}{2}\Big)\leq \frac{f(x)}{2}+\frac{f(y)}{2}$$

Can I conclude from (1) that $f$ is convex?

edited Apr 13 '17 at 12:20

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asked Jan 25 '13 at 18:32

Tomás

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1

I know that if I can prove $$f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$$ for all $x,y\in X$ and $\lambda \in (0,1)$ rational, then by continuity the result follows. – Tomás Jan 25 '13 at 18:39
1

Tomás: Thank you. Perhaps it would be easier to work with the (still dense) set of rationals of the form $a/2^n$. – Jonas Meyer Jan 25 '13 at 18:41
It is easy to show that it holds for all dyadic-rational $\lambda$. – Arin Chaudhuri Jan 25 '13 at 18:41
@5PM is right, please disregard my qeustion. – Tomás Jan 25 '13 at 19:28

Mid-Point Convexity Implies Convexity in Banach Spaces?

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