One of my analysis texts states this as an exercise
If $f$ is midpoint convex, continuous, and two times differentiable, then for any $a, b \in \mathbb{R}$, there exists $c \in [a, b]$ such that $f''(c) \geq 0$.
and says as a hint that I shouldn't have to prove that midpoint convexity and continuity together imply convexity because "that proof is difficult" (which I've seen after researching it a bit here )and that I "shouldn't assume the second derivative test."
I'm lost at how to prove this. My first thoughts were these:
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If $f$ is constant, it's convex, and $f'' = 0$ everywhere in the interval.
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If $f$ isn't constant, then because $[a, b]$ is compact and $f$ is continuous, $f$ achieves a minimum at some $c \in [a, b]$, so by the second derivative test, $f''(c) > 0$.
But, I'm told not to use the second derivative test, and furthermore, thus far in the book, it hasn't talked about derivatives at the end points of a closed interval, so I think it'll break down if $c = a$ or $c = b$.
Is this actually a simple proof, and I'm just missing something? Is there a way to prove this within the constraints the text lays down? What am I missing? I'm trying to work with the basic mean value theorem and intermediate value theorem to see where that gets me, but I haven't had much luck yet.
Best Answer
Let $m$ be the midpoint of $[a,b].$ We have $f(m)\le (f(a) + f(b))/2.$ This is the same as saying $(m,f(m))$ lies on or below the line through $(a,f(a)), (b,f(b)).$ Thus the slope from $(a,f(a))$ to $(m,f(m))$ is $\le$ the slope from $(m,f(m))$ to $(b,f(b)).$ Apply the MVT to see this implies $f'(c)\le f'(d)$ for some $c\in (a,m), d \in (m,b).$ Apply the MVT again: $ 0\le f'(d)-f'(c)= f''(z)(d-c)$ to see $f''(z)\ge 0.$