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 $\mathcal{C}^2$?
~~~~~
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...