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I know that, for any convex set $S$, there is at least one supporting hyperplane at every point in $B$, the boundary of $S$. Also, there can be more than one supporting hyperplane at the same point in $B$.

Let $S$ be an $n$-dimensional convex set in $\mathbb{R}^n$, $X=\{x \in B \mid S$ has more than one supporting hyperplane at $x\}$.

I am looking for some results on the size of $X$. My guess is that it must be a Lebesgue null set. Is that true?

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Yes, this set has zero $(n-1)$-dimensional Lebesgue measure. Since $S$ is $n$-dimensional, its interior is nonempty; after translation we may suppose $0$ is an interior point. Let $f:\mathbb{R}^n\to \mathbb{R}$ be the Minkowski gauge for $S$, that is $$f(x) = \inf\{t : t^{-1}x\in S\}$$ This is a convex Lipschitz function for which the level sets are rescaled copies of $\partial S$. If $x\in B$ admits multiple supporting hyperplanes, then $f$ is not differentiable at $t x$ for all $t>0$. But by Rademacher's theorem, a Lipschitz function is differentiable almost everywhere. The conclusion follows.

Actually, more can be said about the set of nondifferentiability of a convex function: see the paper On the points of non-differentiability of convex functions by Pavlica and reference [2] therein (1979 paper by Zajíček, On the differentiation of convex functions in finite and infinite dimensional spaces).

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