Proof for convexity of a function

Asked Dec 15 '16 at 18:11

Active Dec 15 '16 at 18:16

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A continuous function is called convex on an interval $I\subseteq\mathbb R$ iff $$\forall x,y\in I\;\forall t\in[0,1]:f\big((1-t)x+ty\big)\leq (1-t)f(x)+tf(y).$$ I want to prove that this is equivalent to $$\forall x,y\in I:f\left(\frac{x+y}2\right)\leq \frac{f(x)+f(y)}2.$$ One direction is clear as the second inequality follows from the first one if you choose $t=\frac12$.

However, I am getting stuck proving the reverse direction. How should I proceed here?

By the way, it seems like the second inequality is just a special case of the first one for $t=\frac12$ and is therefore weaker. How can both definitions be equivalent then?

edited Dec 15 '16 at 18:16

asked Dec 15 '16 at 18:11

Mophotla

5

The reverse implication only works for continuous functions, see for example Midpoint-Convex and Continuous Implies Convex. – dxiv Dec 15 '16 at 18:13
Do you know that $\mathbb{R}$ is Cauchy complete? – Nik Bren Dec 15 '16 at 18:13
2

It is not equivalent unless you add some other conditions. Continuity is a straightforward one. – copper.hat Dec 15 '16 at 18:14
A little work shows that if the second holds the the first formula is true whenever $t$ is of the form ${k \over 2^n}$ for $k \in {0,...,2^n}$. Then you can use continuity to extend to all of $[0,1]$. – copper.hat Dec 15 '16 at 18:16
@dxiv, copper.hat: You are right, the function should be continuous. I added that in my question. – Mophotla Dec 15 '16 at 18:17
@Mophotla The question is fully answered in the post I linked to in my first comment. – dxiv Dec 15 '16 at 18:20

Proof for convexity of a function

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