TITLE: Does strict convexity imply differentiability?
QUESTION [25 upvotes]: 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.

REPLY [30 votes]: $$f(x)=x^2+|x|$$ is strictly convex because of the $x^2$ term but not differentiable at $0$ because of the $|x|$ term

REPLY [20 votes]: 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
$$
v'(x)=\sum\limits_na_n\,\mathrm{sgn}(x-x_n).
$$

REPLY [11 votes]: The function
$$f(x)=\max(e^x,e^{-x})$$
is strictly convex but not differentiable at $0$.