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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:

  1. If $f$ is constant, it's convex, and $f'' = 0$ everywhere in the interval.

  2. 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.

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M T
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2 Answers2

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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 $f''(z)\ge 0.$

zhw.
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  • Why is $f(m) \leq f(a)/2 + f(b)/2$ the same as saying that $(m,f(m))$ lies on or below the line through $(a,f(a)),(b,f(b))$? Intuitively it seems right, but I think you need to develop your argument to actually be certain of this statement. – Newb Apr 08 '15 at 06:15
  • It's clear from the picture (and the definition of midpoint convexity) as soon as you realize $(m, f(a)/2+f(b)/2)$ is on the line I described. You go down below, then back up to, the line. There's no way to do that unless the slope goes down, then up. – zhw. Apr 08 '15 at 06:57
  • @Newb It also becomes clear if you find the equation of the line using $(a, f(a))$ and $(b, f(b))$ in point-slope form. Once you have that equation, substituting $m = (a+b)/2$ into the equation for $x$ gives you the point zhw talked about. – M T Apr 08 '15 at 13:50
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Assuming midpoint convexity means $$ f(a/2+b/2)\leq f(a)/2+f(b)/2 \text{ for all }a,b, $$ I would try to approximate $f''(y)$ as with a two-sided second difference quotient, i.e., $$ (f(y+h)-2f(y)+f(y-h))/h^2\to f''(y),\text{ as }h\to0 $$ and then look at the implications of midpoint convexity on a relationship between the various summands in the numerator of the second difference quotient.

Oskar Limka
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  • Hmmm the book hasn't talked about two-sided difference quotients yet, so I'm not sure what that means (unless it's related to the basic definition of the derivative as a limit), but I'll look into it. – M T Apr 08 '15 at 00:25