If $K$ and $L$ are compact convex sets with smooth boundary, does their union have piecewise-smooth boundary?

Clarification: by "piecewise", I mean a finite number of pieces.

I'm sure this must be true, but my search for a citation was in vain (although I did learn the new term "polyconvex").

Thanks!

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I don't think this is true. Suppose one of the sets is essentially $\{(x,y):y\geq x^2\}$ in the plane (cut off in some smooth way at the top, to make it compact). And suppose the other one is the same except that the parabolic lower boundary has been replaced by the graph of something like $y=x^2+e^{-1/x^2}\sin (1/x)$. In other words add a fierce oscillation but damped so strongly that the region above the curve is still convex (i.e., $d^2y/dx^2$ remains positive). (I haven't done the arithmetic to make sure my $e^{-1/x^2}$ damping is sufficient; if it isn't, then replace it by a more vigorous damping.) The union of the two convex sets will have infinitely many corners, where $\sin (1/x)$ is 0.
Thanks! I thought assuming convexity would get rid of the $e^{-1/x^2} \sin(1/x)$ example but I guess not. I don't suppose you know if one can salvage it by assuming the boundaries of $K$ and $L$ are analytic, do you? – Ryan O'Donnell Nov 10 '10 at 17:35