Lipschitz continuity

[zoopreme]

Hello everyone,
this is my very first post and I hope you may help me! I got a huge problem with the lipschitz continuity...
______

is called lipschitz continuous if there exists a real constant

such that

for all

. That is the definition we had in our lecture.
______
My question is now. How can i show that there's always a smallest

, if

is lipschitz continuous?

Thanks for any help.

Sincerely,
zoopreme

 

[pka]

My question is now. How can i show that there's always a smallest

, if

is lipschitz continuous?

Consider the set of all such L's.
Could that set be empty?
Is the set bounded below?
Can you use the greatest lower property?

 

[zoopreme]

Thanks for the quick answer.
I thought about that, too. If f is Lipschitz continuous, the set of all L's is never empty and always bounded. But how do I know that the infimum of the set of all Ls is in that set as well? I mean if the set could be open to the bounded side like (1,

), there is no smallest element...

let's say f: (1,

) -> R, x -> f(x)=sqrt(x). Therefore the infimum of A := { L ; |f(x)-f(y)| ≤ L* |x-y|, x,y e (1,

)} is 1, which is not the smallest L because it is not in A.

i hope you see what i am trying to say...

 

[zoopreme]

Consider the set of all such L's.
Could that set be empty?
Is the set bounded below?
Can you use the greatest lower property?

thanks for the quick answer.thought about that, too. If f is Lipschitz continuous, the set of all L's is never empty and always bounded. But how do I know that the infimum of the set of all Ls is in that set as well? I mean if the set could be open to the bounded side like (1, infinity), there is no smallest element...

 

[pka]

But how do I know that the infimum of the set of all Ls is in that set as well?

You don't know that. But can you show that it does belong to the set of all lower bounds?

 

[zoopreme]

I figured it out by myself, but thanks for the help anyways!