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I know that convexity does not imply differentiability, for example f(x)=|x| is convex but not differentiable. However, |x| is not strictly convex. So I wonder whether strict convexity imply differentiability.

I did some search and found out the Wikipedia implicitly gives the negative answer: http://en.wikipedia.org/wiki/Convex_function#Strongly_convex_functions It says that "a strongly convex function is also strictly convex" and "a function doesn't have to be differentiable in order to be strongly convex".

Can anyone provide a concrete example? Thanks in advance.

3 Answers 3

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$f(x)=x^2+|x|$ is strictly convex because of the $x^2$ term but not differentiable at $0$ because of the $|x|$ term

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    Thanks! Now I feel stupid.. Haha2011-09-27
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Choose a strictly convex function $u$ and some sequences $(x_n)$ and $(a_n)$ such that every $a_n$ is positive and the series $\sum\limits_na_n(1+|x_n|)$ converges. Then the formula $ v(x)=\sum\limits_na_n|x-x_n| $ defines a proper function $v$ such that $u+v$ is strictly convex, and not differentiable at any $x_n$.

The countable set of points $X=\{x_n\}$ may be dense. The function $v$ is differentiable at every $x$ not in $X$, with v'(x)=\sum\limits_na_n\,\mathrm{sgn}(x-x_n).

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    @Did So you've proved that the set of non-diffrentiable points of a convex function can be at most countable and may well be dense. Is it only true for $\Bbb R$ or we can generalize it for general ${\Bbb R}^n$ or other metric spaces ?2018-01-05
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The function $f(x)=\max(e^x,e^{-x})$ is strictly convex but not differentiable at $0$.

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    Also your $f$ can be seen as $f(x)=\max(e^x,e^{-x})=e^{\max(x,-x)}=e^{|x|}$.2013-10-05