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In the study of analysis and geometry of a bounded domain, its boundary regularity is important. For example, it is known that a bounded convex domain has Lipschitz bounday. This implies that a bounded convex domain in the complex Euclidean space $\mathbb C^n$ has to be hyperconvex, namely, it admits a bounded exhaustive plurisubharmonic function.

My question concerns stronger regularity of the boundary of a convex domain, which can be fomulated as: does a convex domain has one smooth boundary point? comparing to the whole boundary, how large is the set of smooth boundary points?

Thanks a lot!

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    $\begingroup$ Convex functions are twice differentiable almost everywhere, see e.g. people.math.sc.edu/howard/Notes/alex.pdf. $\endgroup$
    – Igor Belegradek
    Commented Aug 9, 2016 at 13:33
  • $\begingroup$ @Igor: Thanks! I am still not so clear. Try to represent a convex domain as $\{x|f(x)<0\}$, where $f$ is a convex function. How can we see that $D$ has smooth boundary points from the statement that convex functions are twice differentiable almost everywhere? Note that the boundary of $D$ has measure 0 in general. $\endgroup$
    – Entaou
    Commented Aug 9, 2016 at 14:29
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    $\begingroup$ The graph of a smooth function is smooth. I suggest you read some textbook on the subject, e.g. Schneider in "Convex bodies: Brunn-Minkowski theory", 2nd edition devotes a whole chapter to boundary regularity, and the result I mentioned above is Theorem 2.61. $\endgroup$
    – Igor Belegradek
    Commented Aug 9, 2016 at 15:11
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    $\begingroup$ To return to the original question, there need not be a smooth (as in $C^{\infty}$) boundary point, of course. You can just start with a negative nowhere differentiable (and, let's say, continuous) function and integrate it twice to produce a convex function that isn't smooth anywhere. $\endgroup$
    – Christian Remling
    Commented Aug 9, 2016 at 17:55
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    $\begingroup$ Take an atomic measure $\mu$ that has full support (for example a mass $1/q^{2+\delta}$ on every rational $p/q$) and integrate it up twice. This is convex, but nowhere $C^2$. (Not even $C^1$ in any interval...) Furthermore, the second derivative is zero wherever it exists, so it doesn't contain much information. $\endgroup$
    – Martin Hairer
    Commented Aug 10, 2016 at 12:22

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