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I have the following problem:

Let $K$ and K' be two convex compact sets in $\mathbb{R}^n$, with $n \geq 2$. Assume that $K$ and K' both have a smooth ($\mathcal{C}^2$) boundary. Assume moreover that K \cup K' is convex. Then is the boundary of K \cup K' also $\mathcal{C}^2$?

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Edit : I'll add some context.

The motivation of the question is that I would like to prove that curvature integrals are additive over the space of $\mathcal{C}^2$ convex compact sets, which basically boils down to this problem.

My guess is that the answer is positive. Here are some elements in two dimensions. Let $K$ and K' be two convex compact sets in $\mathbb{R}^2$ with smooth boundary, and assume that K \cup K' is also convex. Let x \in \partial K \cap \partial K'. Then $\partial K$ and \partial K' have the same tangent space in $x$; otherwise, K \cup K' would not be convex. Now, let us look at the curvature at $x$. If the curvature of $\partial K$ is strictly smaller than the curvature of \partial K', then K' is locally included into $K$, so \partial K \cup K' is locally $\partial K$, which is $\mathcal{C}^2$. The same holds if we exchange $K$ and K'. The last possibility is that both curvatures are equal, which "trivially"* imply that the second derivative of any reasonable parametrization of \partial K \cup K' in a neighborhood of $x$ is $\mathcal{C}^2$. In all cases, \partial K \cup K' is a $\mathcal{C}^2$ curve.

This is sketchy, I am not really sure I can write it down neatly, let alone tackle higher dimensions...

$*$ this may be Jordan-theorem-like trivial: getting a rigorous proof does not look that easy...

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    @GregMartin: yes, this is equivalent and generalizes to the n-dimensional case. The boundary of a set is a manifold of class $C^2$ iff it is locally representable as the graph of a $C^2$-function.2012-03-29

1 Answers 1

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The problem is the following: if two $\mathcal{C}^2$ functions $f$ and $g$ from $]-\epsilon, \epsilon[$ to $\mathbb{R}$ are convex with $f(0)=g(0)=0$ and if $h=\min(f,g)$ is convex, is $h$ a $\mathcal{C}^2$ function too?

If f'(0)\neq g'(0), $h$ is not convex. So f'(0)=g'(0).

If f''(0)> g''(0), $f(x)\geq g(x)$ in a neighborhood of $0$. So $h=g$ in this neighborhood and $h$ is $\mathcal{C}^2$ in this neighborhood.

So we may suppose f''(0)=g''(0).

Let $I= \{x \in ]-\epsilon, \epsilon[ | f(x)=g(x)\}$,

$I^+= \{x \in ]-\epsilon, \epsilon[| f(x)>g(x)\}$,

$I^-= \{x \in ]-\epsilon, \epsilon[| f(x)

(1) If $x \in I^-$: $I^-$ is open so $h=f$ in a neighborhood of $x$, so h''(x)=f''(x) and h'(x)=f'(x).

(2) If $x \in I^+$: $I^+$ is open so $h=g$ in a neighborhood of $x$, so h''(x)=g''(x) and h'(x)=g'(x).

(3) If $x \in I$: $f(x)=g(x):=a$. $h$ is convex, so f'(x)=g'(x):=b.

(3.i) if f''(x)> g''(x), as f'(x)=g'(x) ,$f\geq g$ in a neighborhood of $x$, so h''(x)=g''(x) and h'(x)=g'(x).

(3.ii) if f''(x), as f'(x)=g'(x), $f \leq g$ in a neighborhood of $x$, so $h''(x)=f''(x)$and h'(x)=f'(x).

(3.iii) if f''(x)=g''(x):=c, for all $u$ in a neighborhood of $0$, $h(x+u)=\min(f(x+u),g(x+u))$ =\min(f(x)+f'(x)u+(c/2)u^2+ o_1(u^2), g(x)+g'(x)u+ (c/2)u^2+o_2(u^2))=a+bu+(c/2)u^2+o_3(u^2). where $\lim_{u \mapsto 0} o_i(u^2)/u^2=0.$ So h'(x)=b=f'(x)=g'(x).

So \forall x \in ]-\epsilon, \epsilon[, h'(x)=f'(x) or h'(x)=g'(x). And h'(0)=f'(0)=g'(0).

So \forall x \in ]-\epsilon,\epsilon[, \frac{h'(x)-h'(0)}{x} = \frac{f'(x)-f'(0)}{x} or \frac{h'(x)-h'(0)}{x} =\frac{g'(x)-g'(0)}{x}.

So the limit of \frac{h'(x)-h'(0)}{x} when $x \mapsto 0$ is f''(0)=g''(0).

So h''(0)=f''(0)=g''(0).

By the same way, we show that if $f(z)= g(z)$, then h''(z)=f''(z)=g''(z).

If $f(z) \neq g(z)$, then h''(z)=f''(z) or h''(z)=g''(z).

(1) if $f(z) < g(z)$, it's easy to see that $h(z)=f(z)$ so $h$is $\mathcal{C}^2$ in the neighborhood of $z$.

(2) idem if $f(z)>g(z)$.

(3) if $f(z)=g(z)$, then f'(z)=g'(z).

(3.i) if f''(z) < g''(z), $f \leq g$ in a neighborhood of $z$, so $h=f$ in this neighborhood an is $\mathcal{C}^2$ in this neighborhood.

(3.ii) idem if f''(z)> g''(z).

(3.iii) if f''(z)=g''(z), h''(z)=f''(z)=g''(z).And $\forall y$, h''(y)=f''(y) or h''(y)=g''(y). So \lim_{y \mapsto z} h''(y)=h''(z). So h'' is continuous at $z$.

So $h$ is a $\mathcal{C}^2$ function.

[Edit: I believed the problem was in $\mathbb{R}^2$] ]