My task is as follows:
Let $f:\mathbb{R}\to\mathbb{R}$ be a twice-differentiable function,
and let $f$'s second derivative be continuous. Let $f$ be convex with
the following definition of convexity: for any $a<b \in \mathbb{R}$:
$$f\left(\frac{a+b}{2}\right) \leq \frac{f(a)+f(b)}{2}$$ Prove that
$f'' \geq 0$ everywhere.
I've thought of trying to show that there exists a $c$ in every $[a,b] \subset \mathbb{R}$ such that $f''(c) \geq 0$, and then just generalizing that, but I haven't been able to actually do it — I don't know how to approach this. I'm thinking that I should use the mean-value theorem. I've also thought about picking $a < v < w < b$ and then using the MVT on $[a,v]$ and $[w,b]$ to identify points in these intervals and then to take the second derivative between them, and showing that it's nonnegative.
However I'm really having trouble even formalizing any of these thoughts: I can't even get to a any statements about $f'$. I've looked at a few proofs of similar statements, but they used different definitions of convexity, and I haven't really been able to bend them to my situation.
I'd appreciate any help/hints/sketches of proofs or directions.
Best Answer
Given $f$ is a continuous and using the results from this answer, $f$ can be proven to satisfy: $f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda)f(x_2)\ \forall \ \lambda \in [0,1]$
Now, by using Taylor's expansion, $f''(x)$ can be written as: $$ f''(x) = \lim_{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} $$
$f(\frac{1}{2}(x+h) + \frac{1}{2}(x-h)) \leq \frac{1}{2}f(x+h) + \frac{1}{2}f(x-h) \implies 2f(x) \leq f(x+h)+f(x-h)$ or $f(x+h)+f(x-h)-2f(x) \geq 0$.
Since $h^2 \geq 0$ and $f$ being twice differentiable, $f''(x) \geq 0$ follows.