Give an example or prove that it is impossible:

[davitykale]

Homework Statement

A sequence of Lipschitz functions f_n: [0,1] --> R which converges uniformly to a non-Lipschitz function

Homework Equations

a function f: A --> R is Lipschitz if there exists a constant M \in R such that |f(x)-f(y)|<=M|x-y|

The Attempt at a Solution

I don't think it's possible but I'm not sure how to prove that this is the case

 

[LCKurtz]

Think about a polygonal approximation to the upper half of a circle.

 

[davitykale]

Sorry, I'm really confused :/

 

[LCKurtz]

Think about a polygonal approximation to the upper half of a circle.

Sorry, I'm really confused :/

Look at the top half of x²+y²=1. Mark the points on that semicircle that correspond to x = -1,-1/2,0,1/2, and 1. Join these points with straight line segments. That would give what is called a polygonal approximation to the curve with 4 segments. You might call that function f₄(x). Think about f_n(x).

 

[davitykale]

How is f not Lipschitz if f_n(x) is? Doesn't f_n --> f where f is the semicircle?

 

[micromass]

How is f not Lipschitz if f_n(x) is? Doesn't f_n --> f where f is the semicircle?

There is no reason for f to be Lipschitz even if the f_n are. Doesn't the (beautiful) example of LCKurtz show this?

 

[davitykale]

Maybe I'm just confused...f is the semicircle, correct?

 

[micromass]

Maybe I'm just confused...f is the semicircle, correct?

Yes, but any non-Lipschitz function will work...

 

[Redbelly98]

Thread locked.