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I have the following task:

Prove that if $ f : I \rightarrow \mathbb{R} $ is continuous ($ I $ is a range) and $$ \forall {x,y \in I} \qquad f\left(\frac{x+y}{2}\right) \leq \frac{f(x) + f(y)}{2} $$ then $f$ is a convex function.

Can somebody give a hint what can I use to prove this? This is a homework assignment, so I'd like to try solve it myself.

Surb
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Krzysztof Antoniak
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  • Isn't that inequality the definition for a convex function? Meaning, if we know this inequality holds, doesn't that mean the function is convex just by definition? – peter.petrov Mar 30 '15 at 07:14
  • Maybe what the OP wants to prove is that $f''(x)>0$. – Arpan Mar 30 '15 at 07:17
  • @peter.petrov: The definition for convexity is that $f(tx + (1-t)y) \le tf(x) + (1-t)f(y)$ for all $x, y$ and $t\in [0,1]$. –  Mar 30 '15 at 07:17
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    Or this older thread: [Showing that $f$ is convex given that $f(\frac{x+y}2)\le\frac{f(x)+f(y)}2$](http://math.stackexchange.com/questions/83383/showing-that-f-is-convex-given-that-f-fracxy2-le-fracfxfy2). – Martin R Mar 30 '15 at 07:21
  • I think your condition is called "midpoint convex". – Martin R Mar 30 '15 at 07:30

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Take $x,y$ two points in $I$ and $\lambda \in (x,y)$.

You know that if $\lambda = (x+y)/2 := x'$ you're done. Otherwise, $\lambda$ is either in $(x,x')$ or in $(x',y)$.

Continue this way and you'll build two sequences converging to $\lambda$. Since $f$ is continuous, it preserves limits.

That should set you on the right path.

Alexandre Halm
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