Let $f:\mathbb R\to\mathbb R$, maps a point $x \in \mathbb R$. $f$ is twice differentiable. Show that if second derivative is positive for all $x$ then $f$ is convex
Is there anyway to prove this without using MVT/definitions of derivatives?
Basically can this be proven with just the definition of convexity/concavity and maybe quasi-convexity and lower contours. The definition we learned use for convexity is: $f$ is convex if $$
f(\theta x_a + (1-\theta) x_b) \leq \theta f(x_a) + (1-\theta) f(x_b).
$$
Best Answer
We know that if $f$ is differentiable in some interval, say $I$, then $f$ is concave upward in $I$ iff $$f(x)<\frac{x_2-x}{x_2-x_1}f(x_1)+\frac{x-x_1}{x_2-x_1}f(x_2)$$ for all $x_1,x,x_2\in I$ such that $x_1<x<x_2$. Use this fact.