Convex function implies max at endpoint for closed interval proof

Question

I need to prove that a function with a positive second derivative throughout interval [a,b] necessarily will have the maximum value at one of the endpoints. This seems pretty obvious for a concave up function, I'm just struggling to put it into the right terms that provide serious proof.

Answer 1

Let your function be $f\colon[a,b]\to\mathbb{R}$. Because it has a second derivative on $[a,b]$, it must also have a first derivative and must be continuous on $[a,b]$. Therefore, $f$ attains a maximum value on that interval at some point $c\in[a,b]$. Either $c$ is an endpoint (what you need to prove) or an internal point for that interval. Assume that it's an internal point. This means that at $x=c$, $f$ has a local maximum. Fermat's Theorem implies $f'(c)=0$. We also know the second derivative is positive, so $f''(c) > 0$. The second derivative test tells us that $x=c$ is a local minimum point. So, $x=c$ is both a local minimum and a local maximum point, which can only happen when $f$ is constant on some open interval containing $c$. But a constant has a second (and first) derivative of $0$, contradicting $f''(x)>0$ for all $x\in(a,b)$. Therefore, $c$ is an endpoint.