# Finding the Extrema Non-differentiable Functions.

Exist any kind of instances of addressing for the international maximum of a non-differentiable function where you :

- Construct a collection of differentiable features that come close to the non-differentiable function in the restriction
- Show the maximum of each differentiable function merges to some value, which is hence your solution.

For all I recognize, the procedure over is fatally flawed (or there are unimportant instances, I would certainly be most curious about non-trivial instances ) somehow, if that holds true allow me recognize.

I am especially curious about instances entailing outright values.

A straightforward instance:

Let $F_n(x) = \sqrt{x^2+2^{-n}}$. It is not tough to show that $F_n(x) \to \sqrt{x^2} = |x|$. Every $F_n$ is differentiable and also has a neighborhood minimum at 0, and also without a doubt so does |x |.

Allow me recognize if this is what you're seeking.

The strategy you laid out is generally made use of in technique. If your initial trouble has some wonderful buildings, such as convexity, the strategy will certainly function well. As an example, the soft maximum is an usual means to construct a collection of smooth, convex estimates to the maximum function.

Here's a stab at why the trouble is tough for very non - differentiable yet continual features. So claim $f$ is no place differentiable on $[a,b]$. I assert that there are either definitely or no neighborhood extrema of $f$. (By neighborhood extrema, I suggest extrema that take place in the inside of the interval $[a,b]$.)

Without a doubt, intend there were finitely several, claim $c_1 < c_2<\dots f(c_2)$. After that we can take the international maximum of $f$ on $[c_1, c_2]$, which takes place at an *indoor * factor ; it is hence a neighborhood extremum of $f$.

If $f$ is a monotone function, after that it is a theory of Lebesgue that it is differentiable a.e. Specifically, the above thinking reveals that the presence of finitely several neighborhood extrema indicates that $f$ is differentiable a.e.

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