Give an example or prove that it is impossible:

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Homework Help Overview

The discussion revolves around the properties of Lipschitz functions and their convergence to a non-Lipschitz function. The original poster questions the possibility of a sequence of Lipschitz functions converging uniformly to a function that is not Lipschitz.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the idea of polygonal approximations to curves, specifically the upper half of a circle, as a potential example. Questions arise regarding the nature of convergence and the properties of the limit function compared to the sequence of functions.

Discussion Status

There is an ongoing exploration of examples and counterexamples, with some participants expressing confusion about the definitions and implications of Lipschitz continuity. Guidance has been offered regarding specific examples, but no consensus has been reached on the original question.

Contextual Notes

Participants are navigating the definitions of Lipschitz functions and the implications of uniform convergence, with some uncertainty about the characteristics of the limit function in relation to the sequence.

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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
 
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Think about a polygonal approximation to the upper half of a circle.
 
Sorry, I'm really confused :/
 
LCKurtz said:
Think about a polygonal approximation to the upper half of a circle.

davitykale said:
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 fn(x).
 
How is f not Lipschitz if f_n(x) is? Doesn't f_n --> f where f is the semicircle?
 
davitykale said:
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 fn are. Doesn't the (beautiful) example of LCKurtz show this?
 
Maybe I'm just confused...f is the semicircle, correct?
 
davitykale said:
Maybe I'm just confused...f is the semicircle, correct?

Yes, but any non-Lipschitz function will work...
 
Thread locked.
 
Last edited:

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