Finding the Extrema Non-differentiable Functions. Are there any examples of solving for the global maximum of a non-differentiable function where you:


*

*Construct a series of differentiable functions that approach the non-differentiable function in the limit

*Show the maximum of each differentiable function converges to some value, which is thus your answer.


For all I know, the procedure above is fatally flawed (or there are trivial examples, I would be most interested in non-trivial examples) in some way, if that is the case let me know.
I am specifically interested in examples involving absolute values.
 A: The approach you outlined is commonly used in practice.  If your original problem has some nice properties, such as convexity, the approach will work well.  For example, the soft maximum is a common way to construct a series of smooth, convex approximations to the maximum function. 
A: A simple example:
Let $F_n(x) = \sqrt{x^2+2^{-n}}$. It is not hard to show that $F_n(x) \to \sqrt{x^2} = |x|$.
Every $F_n$ is differentiable and has a local minimum at 0, and indeed so does |x|.
Let me know if this is what you're looking for.
A: Here's a stab at why the problem is hard for highly non-differentiable but continuous functions. So say $f$ is nowhere differentiable on $[a,b]$. I claim that there are either infinitely or no local extrema of $f$. (By local extrema, I mean extrema that occur in the interior of the interval $[a,b]$.)
Indeed, suppose there were finitely many, say $c_1 < c_2<\dots f(c_2)$. Then we can take the global maximum of $f$ on $[c_1, c_2]$, which occurs at an interior point; it is thus a local extremum of $f$.
If $f$ is a monotone function, then it is a theorem of Lebesgue that it is differentiable a.e. In particular, the above reasoning shows that the existence of finitely many local extrema implies that $f$ is differentiable a.e.
