Let $H$ be a countable subset of $[0,1]$. Construct a convex function $f:[0,1]\rightarrow\mathbb{R}$ such that $f$ is nondifferentiable on $H$ and differentiable in the rest.
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6$\begingroup$ Why do you think such a function might exist? Why do you need to know? $\endgroup$– Yemon ChoiCommented Sep 8, 2013 at 20:40
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1$\begingroup$ Hm, I'd rather not. $\endgroup$– Noah SchweberCommented Sep 8, 2013 at 20:53
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1$\begingroup$ Based on the fact that the set of nondifferentiablity points for a convex function is countable, the reverse problem, which I stated, arises quite natural. I've got some ideas, but they seem to work only for some particular sets. $\endgroup$– mihainexCommented Sep 8, 2013 at 20:56
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2$\begingroup$ The second antiderivative of a finite discrete measure concentrated in $H$. $\endgroup$– Pietro MajerCommented Sep 8, 2013 at 22:13
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2 Answers
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If you think in terms of constructing the derivative of $f$ rather than $f$ itself, you're looking for an increasing function that has an arbitrary countable set $H$ of discontinuities. There's a standard trick to do this: choose an injection $i:H\to\mathbb{N}$, and define $$g(x)=\sum_{h<x, h\in H} 2^{-i(h)}.$$ If you then define $f$ by integrating $g$, it will be convex and differentiable everywhere except on $H$.
answered Sep 8, 2013 at 21:32
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$\begingroup$ Like shooting fish in a barrel. :-) $\endgroup$ Commented Sep 8, 2013 at 22:00
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Enumerate $H$ as $h_1, h_2, \ldots$, and take $f(x) = \sum_{j=1}^\infty 2^{-j} |x - h_j|$.
answered Sep 8, 2013 at 21:39