# Does strict convexity everywhere implies strong convexity on a neighborhood of almost every point?

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.

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.

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