Lipschitz function and Baire Category Theorem

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Discussion Overview

The discussion revolves around demonstrating the existence of a continuous function from the interval [0,1] to the real numbers that is not Lipschitz on any subinterval [r,s] for 0 ≤ r < s ≤ 1, using the Baire Category Theorem. Participants explore the properties of the set of Lipschitz continuous functions and the construction of specific functions to illustrate the argument.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant defines the set A(r,s) as the collection of continuous functions that are Lipschitz on the interval [r,s] and notes that this set is closed.
  • Another participant suggests that to show A(r,s) has empty interior, one must find a function g that is arbitrarily close to a function f in A(r,s) but not in A(r,s).
  • A participant proposes the function sqrt(x) as an example of a continuous function that is not Lipschitz on [0,1] and questions how to extend this idea to the interval [r,s].
  • It is noted that sqrt(x) is not Lipschitz because its first derivative is unbounded near 0.
  • A construction of a new function g is presented, defined piecewise, which is not Lipschitz on [r,s] and is used to create a function h that is also not Lipschitz.

Areas of Agreement / Disagreement

Participants are engaged in a collaborative exploration of the problem, with no explicit consensus reached on the final argument or construction. Multiple approaches and ideas are being discussed without resolution.

Contextual Notes

Participants express uncertainty about the continuity and Lipschitz properties of the functions being discussed, and the discussion includes attempts to refine the definitions and constructions involved.

itzik26
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hey,
I need to show, using Baire Category Theorem, that there exits a continuous function
f: [0,1] to R , that isn't Lipschitz on the interval [r,s] for every 0<=r<s<=1 .

I defined the set A(r,s) to be all the continuous functions that are lipschitz on the interval [r,s]. I showed that A(r,s) is closed , but I'm having trouble showing it's nowhere dense.

help please! :)
 
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itzik26 said:
I defined the set A(r,s) to be all the continuous functions that are lipschitz on the interval [r,s]. I showed that A(r,s) is closed , but I'm having trouble showing it's nowhere dense.

So, you must show that A(r,s) has empty interior. So take an f in A(r,s). Try to find a g arbitrary close to f such that g\notin A(r,s). This will show that f is not in the interior of A(r,s).
Note that our g will always be uniform continuous (since it is continuous on a compact domain). So you must somehow use a function that is uniform continuous but not Lipschitz. Do you know such a function?
 
sqrt(x), for example, is continuous but not lipschitz on [0,1], but i don't know how to continue from here.
 
itzik26 said:
sqrt(x), for example, is continuous but not lipschitz on [0,1], but i don't know how to continue from here.

Do you see why it is not Lipschitz?? Can you do something similar on [r,s]??
And can you modify the function such that it is close to f?
 
it's not Lipschitz because it's first derivative isn't bounded close to 0.
o.k., so given f and a>0, define g as follows: g=sqrt[x-r] in the interval [r,a^2/4+r], g=0 for x<=r and g=a/2 for x>=a^2/4+r. g is not Lipschitz on [r,s]. define h=f+g.
h isn't Lipschitz because g isn't Lipschitz and d(f,h)<a.
 

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