# Strictly convex if and only if derivative strictly increasing?

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_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 $$f'(x_1). So $$f'$$ strictly increasing.

The other side is true by using Mean value theorem.

For $$x_1, $$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, $$f'(c_1), hence $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$$

• How does the mean value theorem argument go? – LucasSilva Nov 1 '14 at 18:36
• @LucasSilva I have added more details. – John Nov 2 '14 at 14:56
• 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 $x_1 < x_2 < x_3$? – LucasSilva Nov 3 '14 at 0:22
• @LucasSilva They are equivalent. See Proposition 1.1 for a proof of convex function. – John Nov 3 '14 at 4:32
• 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? – 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" .