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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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    @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)2012-10-31

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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.2012-11-07