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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.

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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 $$ F(x_1, \ldots, x_n) = f(x_1) + \cdots + f(x_n) \, . $$

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