Suppose $f$ is a real-valued function that is differentiable on an open interval $I$. It is well-known that $f^{\prime}$ is increasing on $I$ if and only if $f$ is convex on $I$.

Is the following true?

$f^{\prime}$ is strictly increasing on $I$ if and only if $f$ is strictly convex on $I$.

I'm pretty sure the $\Rightarrow$ direction is true. I'm less confident about the other direction.

Is it easier if we also assume $f^{\prime \prime}$ exists on $I$.

References or counterexamples greatly appreciated.


Suppose $f$ is strictly convex on $(a,b)$, let $x_1<x_2<x_3<x_4<x_5$, $x_i\in(a,b)$

By strictly convex, we have $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}<\frac{f(x_4)-f(x_3)}{x_4-x_3}<\frac{f(x_5)-f(x_4)}{x_5-x_4}$$

Let $x_2\to x_1^+, x_4\to x_5^-$, we have $f'(x_1)<f'(x_5)$. So $f'$ strictly increasing.

The other side is true by using Mean value theorem.

For $x_1<x_2<x_3$, $x_i\in(a,b)$ since $f$ is differentiable,

$\exists c_1\in(x_1,x_2)$, s.t. $\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c_1)$,

$\exists c_2\in(x_2,x_3)$, s.t. $\frac{f(x_3)-f(x_2)}{x_3-x_2}=f'(c_2)$,

Since $f'$ is strictly increasing, $f'(c_1)<f'(c_2)$, hence $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$$

  • $\begingroup$ How does the mean value theorem argument go? $\endgroup$ – LucasSilva Nov 1 '14 at 18:36
  • $\begingroup$ @LucasSilva I have added more details. $\endgroup$ – John Nov 2 '14 at 14:56
  • $\begingroup$ Don't you need to show that $\dfrac{f(y)-f(x)}{y-x}$ is strictly increasing in $y$ for fixed $x$? So we would also need to show an inequality like $\dfrac{f(x_3) - f(x_1)}{x_3 - x_1} < \dfrac{f(x_3) - f(x_2)}{x_3 - x_2}$ for $x_1 < x_2 < x_3$? $\endgroup$ – LucasSilva Nov 3 '14 at 0:22
  • $\begingroup$ @LucasSilva They are equivalent. See Proposition 1.1 for a proof of convex function. $\endgroup$ – John Nov 3 '14 at 4:32
  • $\begingroup$ I just tried this with $f(x)=x^2$ and got a contradiction. I chose $x_1=-2$, $x_2=-1$, $x_3=0$. Did I do something wrong? $\endgroup$ – MLK Jul 6 '18 at 12:37

Actually, we also have that if in addition $f$ is double differentiable everywhere, then $f'' \ge 0$ for a convex function $f$ (where the inequality is strict for strict convexity) as being the following (written in just another manner) "$f$ takes maximum values at some end point" .


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