# 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.
– user123641
Jul 5, 2017 at 2:27

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! Jan 8, 2016 at 13:51
• @Bernd $x^4$ satisfies also that definition :)
• $f'(0)=4\cdot 0^3=0 \ngtr 0$? Where am I wrong? Jan 8, 2016 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?
$\newcommand{\Reals}{\mathbf{R}}$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...).