Strictly convex function: how often can its second derivative be zero?

It’s a basic fact that a twice-differentiable function from $\mathbb{R}$ to $\mathbb{R}$ is strictly convex if its derivative is positive everywhere.

The converse is not true: consider, e.g., $f(x) = x^4$, which is strictly converse, with $f ”(0)=0$.

Is there a partial converse, however?

Is it true, e.g., that a strictly convex twice-differentiable function from $\mathbb{R}$ to $\mathbb{R}$ can have zero second derivative at at most one point?

Thanks for your help!

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We have a partial converse. It’s not quite as strong as allowing only finitely many or only isolated zeros of the second derivative, but it’s strong enough.

For a function $f \colon I \to \mathbb{R}$, where $I \subset \mathbb{R}$ is an open interval, possibly $I = \mathbb{R}$, convexity can be formulated as

$$\bigl(\forall u,v,w \in I\bigr)\left(u < v < w \Rightarrow \frac{f(v)-f(u)}{v-u} \leqslant \frac{f(w)-f(v)}{w-v}\right)$$

and strict convexity with the strict inequality.

For a differentiable function $f$, these conditions can be seen to be equivalent to

  • $f$ is convex if and only if the derivative $f’$ is non-decreasing,
  • $f$ is strictly convex if and only if $f’$ is strictly increasing.

For a twice continuously differentiable function $f$, the above can be easily seen to be equivalent to

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