About hyperplanes on the boundary (with no $C^1$ regularity ) of compact convex sets

Question

I am reading a paper and the authors use the following property:

"Let $K$ a compact and convex set in $R^n$ with nonempty interior. Let $x_0 \in \partial K$ and suppose that the boundary is not $C^1$ in $x_0$. Then $x_0$ has at least two supporting hyperplanes "

I have no idea to how to prove this. My geometry is not good ... =\

Someone please could help me ? (or point a reference with a proof)

Thanks in advance!

Answer 1

You should check your paper to see what they actually mean by $C^1$: under the usual interpretation of the phrase "continuously differentiable", the claim is false.

The usual definition: Let $f:(a,b) \to \mathbb{R}$ be a function. We say that $f$ is continuously differentiable on the interval $(a,b)$ iff

• $f$ is differentiable at every point in $(a,b)$
• the derivative $f'$ is a continuous function in $(a,b)$.

Now, I have never, ever, seen a definition for a function being $C^1$ at a point. But based on the above definition I would surmise that

Possible definition: Let $f:(a,b) \to \mathbb{R}$ be a function and $x\in (a,b)$ a point, we say that $f$ is continuously differentiable at $x$ iff there exists $\epsilon > 0$ such that the restriction $f|_{(x-\epsilon,x+\epsilon)}$ is continuously differentiable.

But clearly with this definition the claim is false. Let $f: [-1,1]\to\mathbb{R}$ be the function that satisfies

• $f(\pm 2^{-n}) = 2^{-2n}$, for every $n\in \mathbb{N}$
• $f(0) = 0$.
• $f$ restricted to intervals of the form $(2^{-n-1},2^{-n})$ and $(-2^{-n}, -2^{-n-1})$ is linear.

We see that away from the points of the form $\pm 2^{-n}$, the function $f$ is linear, and hence differentiable. At the points of the form $\pm 2^{-n}$ the left and right derivatives disagree, so we have a jump discontinuity. That is to say that $f$ is not differentiable there. You easily can check, however, that $f$ is differentiable at $0$, with derivative $0$.

Lastly, you can also easily check that the function $f$ is convex.

So we have here a convex function, a point $x$ such that $f$ is differentiable and hence also has only one subgradient, but such that by the definitions above cannot be considered as $C^1$ (this is due to in any interval $(-\epsilon,\epsilon)$ there exist points on which $f$ is not differentiable).

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Now, given a convex function in a finite dimensional linear space, the set on which it is not differentiable is well-known to be both Lebesgue measure 0 and meager. So maybe the authors mean something like $f$ is $C^1$ on the set where the derivative is defined under the induced topology? But you will have to check the paper and find out (possible infer) what the authors actually meant.