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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!

math student
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    See Theorem 25.1 (on p. 242 in my copy) in Rockafeller's Convex Analysis. The notion of subgradient he developed in section 23 is essentially the same as supporting hyperplanes. – Willie Wong Oct 28 '14 at 09:40
  • Hi Willie Wong, thanks for your attention. You helped me a lot with your commentary. I studied what you said, and I can obtain from what you said the following affirmation: "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 the graph (locally )of a differentiable function in $x_0$. Then $x_0$ has at least two supporting hyperplanes " . Do you know how can proceed in the $C^1$ case? – math student Oct 28 '14 at 22:18
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    Can you take this condition " one supporting hyperplane " as the definition of "$C^1$ in $x_0$ "? Then the problem is solved. – orangeskid Oct 29 '14 at 16:04
  • because of th counterexample of Willie Wong, I believe that the authors did what you said ... – math student Oct 29 '14 at 20:42

1 Answers1

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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).

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.

Willie Wong
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