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Let $f$ be a sequentially continuous real function defined on $(a,b)$. Suppose $\exists \lambda \in (0,1)$ such that $\forall x,y \in (a,b)$, $f(\lambda x + (1-\lambda) y) ≦ \lambda f(x) + (1-\lambda) f(y)$.

Then, how do I prove $f$ is convex?

I proved that $f$ is convex when $\lambda=1/2$, but i think it would work for arbitrary $\lambda \in (0,1)$.

Let $A=\{m\in [0,1]|\forall x,y \in (a,b), f(mx+(1-m)y)≦m f(x) + (1-m) f(y)\}$.

It can be easily seen that "$j,k\in A \Rightarrow \lambda j + (1-\lambda) k \in A$".

The problem is i don't know how to show that $A$ is dense in $[0,1]$. ($A$ is dense when $\lambda=1/2$)

Help!

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Just to clarify: you use the term "sequentially continuous", which, under normal circumstances, is the same as "continuous". Does this mean that you work in an unusual setting, e.g. without the Axiom of Choice? If so, then this is a bit out of my league, but if you are in the usual ZFC, then I could probably answer your question. –  Dan Shved Oct 31 '12 at 7:33
    
@Dan Shved Choice need to be used to show that it is convex when $\lambda$ is irrational. However, i think we don't need choice to show that $A$ is dense. Would you show me your argument using choice to show that $A$ is dense? (i.e. in ZFC) –  Katlus Oct 31 '12 at 9:51

1 Answer 1

up vote 1 down vote accepted

OK, here is an argument to show that $A$ is dense in $[0,1]$. I'm not sure if it does or does not use Choice somewhere implicitly.

We have a set $A \subset [0,1]$, we know that it contains $0$ and $1$. We also know that it has the property that was stated in the question: $j,k \in A \Rightarrow \lambda j + (1-\lambda) k \in A$. We will only use this knowledge to prove that $A$ is dense.

Suppose it isn't. Then there exist $a, b$ such that $0 < a < b < 1$ and $A \cap (a, b) = \emptyset$. Define $a' = \sup (A \cap [0, a])$ and $b' = \inf(A \cap [b, 1])$. It is clear that $0 \leqslant a' < b' \leqslant 1$ and $A \cap (a', b') = \emptyset$.

By definition of supremum, there exists an $x \in A$ such that $a' - \lambda(b'-a') < x \leqslant a'$. By definition of infimum, there exists an $y \in A$ such that $b' \leqslant y < b' + (1 - \lambda)(b'-a')$.

Consider $z = \lambda y + (1 - \lambda)x \in A$. Observe that $$ z - x = \lambda(y-x) \geqslant \lambda (b'-a') > a'-x, $$ therefore $z > a'$. By a similar reason, $z < b'$. But then $z$ belongs to $A \cap (a', b')$, which is empty. So we have arrived to a contradiction, therefore $A$ is dense in $[0,1]$.

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I'm sorry that i took so long time to accept your answer, and your argument works fine without choice. Thank you. –  Katlus Nov 7 '12 at 11:12

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