# Strictly increasing, strictly convex function: is the second derivative positive?

Consider a twice continuously differentiable function $f \colon \mathbb{R} \to \mathbb{R}$. While $f''(x)>0\ \forall x$ implies strict convexity of $f$, the converse is not true (e.g. $f(x)=x^4$, strictly convex but $f''(0)=0$).

I was wondering whether the additional requirement of $f$ being strictly increasing can ensure $f''(x)>0$. It would at least rule out the example above.

From how I understand this answer, the requirement is sufficient for (once) differentiable functions, but for twice differentiable ones it says:

$f$ is strictly convex if and only if $f'' \geqslant 0$ everywhere and $f''$ does not vanish on any non-empty open interval $J \subset I$.

Is it correct that this is satisfied under strict increasingness and hence for $f$ as above, strict increasingness and strict convexity together imply $f''(x)>0\ \forall x$?

• The alteration that you would want to make to this conjecture to make it true is $f$ is strictly convex if and only if BOTH $f''\geq 0$ everywhere AND the set $\{x\in I: f''(x)=0\}$ has empty interior. – Robert Wolfe Jul 5 '17 at 2:27

## 2 Answers

$\newcommand{\Reals}{\mathbf{R}}$Let $h:\Reals \to \Reals$ be a continuous, non-negative function that does not vanish on any open interval, and whose integral over $\Reals$ exists. The function $$g(x) = \int_{0}^{x} h(t)\, dt$$ is strictly increasing and bounded, say $|g| < M$ for some $M > 0$, and the function $$f(x) = Mx + \int_{0}^{x} g(t)\, dt = Mx + \int_{0}^{x} (x - t) h(t)\, dt$$ is strictly increasing and strictly convex. By construction, $f'' = h$; the function $h$, however, can vanish at infinitely many points (the integers, the terms of a convergent sequence, a Cantor set...).

Take $f(x) = x^4$ on $[0,1]$. It is strictly increasing, strictly convex, but $f''(0) = 0$

• My bad. I took strict increasingness to be defined via $f'(x)>0$ rather via $x>y\Rightarrow f(x)>f(y)$. Thanks for the counter example! – Bernd Jan 8 '16 at 13:51
• @Bernd $x^4$ satisfies also that definition :) – Ant Jan 8 '16 at 13:59
• $f'(0)=4\cdot 0^3=0 \ngtr 0$? Where am I wrong? – Bernd Jan 8 '16 at 14:03
• @Bernd $x^4$ satisfies $x > y \implies f(x) > f(y)$, while it does not satisfy $f'(x) > 0$. So maybe you wanted to say the converse of what you said in your previous comment? – Ant Jan 8 '16 at 14:06