Prove or disprove the following:

For every strictly convex $f : \Bbb R^n \to \Bbb R$, for almost every $x \in \Bbb R^n$ there exists a neighborhood $N(x)$ of $x$ such that $f$ is strongly convex in $N(x)$.

The usual examples of strictly convex functions that are not strongly convex, for instance $x \mapsto x^4$, do satisfy this property.


1 Answer 1


A counterexample for $n=1$: Let $h: \Bbb R \to \Bbb R$ be a function with the following properties:

  • $h$ is continuous,
  • $h$ is strictly increasing,
  • $h'(x) = 0$ almost everywhere.

Such functions do exist, see for example Showing the existence of a continuous, strictly increasing function $f$ on $\mathbb{R}$ such that $f'(x) = 0$ almost everywhere or Existence of a Strictly Increasing, Continuous Function whose Derivative is 0 a.e. on $\mathbb{R}$.

Then $f: \Bbb R \to \Bbb R$ defined by $f(x) = \int_0^x h(t) \, dt$ is strictly convex, because $f'=h$ is strictly increasing.

But $f$ is not strongly convex on any non-empty open interval $I$: Assume that there is a constant $m > 0$ such that $$ h(y) - h(x) = f'(y) - f'(x) \ge m (y-x) $$ for all $x, y \in I$ with $x< y$. Then $h'(x) \ge m > 0$ a.e. in $I$, in contradiction to the construction of $h$.

Therefore no $x \in \Bbb R$ has a neighborhood on which $f$ is strongly convex.

This can be extended to a counterexample $F: \Bbb R^n \to \Bbb R$ for any $n$ by defining $$ F(x_1, \ldots, x_n) = f(x_1) + \cdots + f(x_n) \, . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.