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Let $f: [a,b] \to \mathbb R$ twice differentiable, and $f''(x) \ge 0$, $\forall x \in [a,b]$.Prove that $f(\frac{x_1 + x_2} {2}) \le \frac{1}{2}[f(x_1) + f(x_2)], \forall x_1, x_2 \in [a,b].$

My attempt:

As a hypothesis we assume that $f''(x) \ge 0$ and we know that .If $f: (a,b) \to \mathbb R$ has second order derivatives on $(a, b)$ the function $f$ is convex if and only if $f''(x) \ge 0$. If $f’’(x) \gt 0$, the function $f$ is strictly convex. So $f$ is convex. And by the definition of convex, A function $f: (a,b) \to \mathbb R$ is calling convex on (a,b) if for any $x_1, x_2 \in (a,b)$ and for any pair of real numbers $\alpha_1 \ge 0, \alpha_2 \ge 0$ such that $\alpha_1 + \alpha_2 = 1$ the following inequality $f(\alpha_1x_1 + \alpha_2x_2) \le \alpha_1f(x_1) + \alpha_2f( x_2)$.

So we have:

$f(\frac{x_1 + x_2}{2}) \le \frac{1}{2}[f(x_1) + f(x_2)] = f(\frac{1}{2}[x_1]+\frac{1}{2}[x_2]) \le \frac{1}{2}f(x_1) + \frac{1}{2}f(x_2) \to f(\alpha_1x_1 + \alpha_2x_2) \le \alpha_1f(x_1) + \alpha_2f(x_2)$

How is my answer?

Thank's for any help

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    $\begingroup$ I don't think you are allowed to use the fact that $f''(x) \ge 0$ implies convexity of $f$; this is essentially what the problem is asking you to prove. $\endgroup$
    – angryavian
    May 14, 2022 at 17:40

2 Answers 2

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Definition. a function $f$ on an interval $(a,b)$ is said to be midpoint convex if $$f\left(\frac{x_1 + x_2} {2}\right) \le \frac{1}{2}[f(x_1) + f(x_2)] \tag{1}$$ for all $x_1, x_2 \in (a,b)$.

Equivalently you can replace (1) with (2): $$f(x+h) +f(x-h) - 2f(x) \geq 0 \tag{2}$$ for all $x,x+h,x-h\in (a,b)$.

As the OP correctly notes, every convex function is midpoint convex, so it is sufficient to rely on any theorem that asserts that a function is convex.

Or...one might prove directly since (as observed in a comment) that defeats the spirit of the problem.

Problem. Show [directly] that a differentiable function $f$ with a nondecreasing derivative $f'$ is midpoint convex.

The simplest proof maybe. Consider the function $$t \to \frac{f(x+t) + f(x-t) -2 f(x)}{t}$$ and apply the Cauchy Mean Value theorem [see below] on the interval $[0,h]$ to obtain $\tau \in (0,h)$ with $$ \frac{f(x+h) + f(x-h) -2 f(x)}{h} = \frac{f'(x+\tau) - f'(x-\tau)}{1} \geq 0.$$ QED.

But I jumped in here with a totally different motive.

What about this notion of midpoint convexity. Is that a thing? If every convex function is midpoint convex, then is every midpoint convex function really just convex?

  1. No. Not every midpoint convex function is convex.

  2. But every continuous midpoint convex function is convex.

  3. If a midpoint convex function is not convex then it is pretty weird. Blumberg (1919) and Sierpiński (1920) independently proved that every measurable midpoint convex function must be convex.

  4. But there are plenty of nonmeasurable functions that are midpoint convex. All of them are unbounded in every open subinterval, so not your ordinary everyday function.

  5. There is a big literature on midpoint convex functions which I can only encourage interested parties to consult.

Notes:

Calculus students are not provided with many tools. About the only reliable and often-used one is the mean-value theorem. I would suggest that you memorize a small upgrade. The weak version is easy to remember; this is almost as easy.

Cauchy's Mean Value Theorem: Let $F,G:R→R$ be continuous on $[a, b] $ and differentiable on $(a, b)$. Suppose that $G(b)≠G(a)$. Then there exists $c∈(a, b)$ such that $G′(c)≠0$ and such that $$\frac{F(b) - F(a)}{G(b) - G(a)} = \frac{F'(c)}{G'(c)}.$$

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Let $x_0 := (x_1+x_2)/2$ and let $h = (x_2-x_1)/2$. The desired inequality is $f(x_0) \le \frac{1}{2}(f(x_0-h) + f(x_0+h))$, which can be rearranged as $$f(x_0) - f(x_0-h) \le f(x_0 + h) - f(x_0).\tag{$*$}$$ By Taylor's theorem, there exist $\xi_-$ and $\xi_+$ in $[0, h]$ such that $$f(x_0-h) = f(x_0) - h f'(x_0) + \frac{h^2}{2} f''(x_0-\xi_-)$$ and $$f(x_0+h) = f(x_0) + h f'(x_0) + \frac{h^2}{2} f''(x_0+\xi_+).$$ Can use these two equations to prove the above inequality ($*$)?

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