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...