Lipschitz function and Baire Category Theorem

[itzik26]

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! :)

 

[micromass]

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?

 

[itzik26]

sqrt(x), for example, is continuous but not lipschitz on [0,1], but i don't know how to continue from here.

 

[micromass]

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?

 

[itzik26]

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.