Well, I want to prove the following:
Let $f:(a,b)\to\mathbb R$ be double differentiable then $f$ is convex iff $\;\;f''(x)>0$ for all $x\in (a,b)$.
Then I tried te following:
$\Rightarrow]$ Lets suppose that $f:(a,b)\to\mathbb R$ is convex then if we define $h=-y+x$
we have that:
$$ f(x+h) \leq (1-t)f(x)+tf(x+h)$$
$$0 \leq (1-t)f(x)+tf(x+h)-f(x+h)$$
I was expecting to have something of this form
$$0 \leq f(x+h)-2f(x)+f(x-h)$$
and then take limit and have the result, I don't know how to define $h$ to have the above relation, Can someone help me ?
$\Leftarrow]$ I checked this, but I don't know if it is right, If it is not Can you help me to fix the mistakes please:
Second derivative positive $\implies$ convex
Thanks a lot in advance
Best Answer
From convexity, $f(ta+ (1-t)b) \leqslant tf(a) + (1-t)f(b)$.
Let $t = 1/2$, $a= x-h$, and $b = x+h$.
Then
$$f(x) \leqslant \frac1{2}f(x-h) + \frac1{2}f(x+h)\\ \implies f(x+h) - 2f(x) + f(x-h) \geqslant 0$$