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.

Solution 1:

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 $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$$

Solution 2:

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" .