# real analysis lipschitz continuous

• Mar 23rd 2009, 06:10 AM
varkoume.com
real analysis lipschitz continuous
Show that f: R-->R defined as f(x)=(x^2) / (1+x^2) is Lipschitz continuous

Can anybody help me with this please??
• Mar 23rd 2009, 06:50 AM
HallsofIvy
I presume that by "Lipschitz continuous" you mean what I would call just "Lipschitz": A function, f, is Lipschitz on a set A if and only if there exist a constant C such that, for all x, y in the set, |f(x)- f(y)|< C|x- y|.

The problem with that is that it requires a "set A". On what set do you want to prove this function is Lipschitz? The set of all real numbers?

It is relatively easy to prove, using the mean value theorem, that if a function is differentiable on a set then it is Lipschitz on that set. And this function is clearly differentiable for all x.
• Mar 23rd 2009, 06:52 AM
varkoume.com
It is for all x,y on R

Thanks
• Mar 26th 2009, 06:38 AM
varkoume.com
I have already proof that |f(x)-f(y)|= |x-y| |x+y| / |(1+x^2) (1+y^2)|

I also know that |f(x)-f(y)| < c by uniform continuity.

Moreover by MVT f(c)= f(b)-f(a)/b-a

But I cannot see what I have to do next ;/
• Mar 26th 2009, 07:56 AM
Plato
Have you actually looked carefully at the derivative of this function?
Have you graphed the function and its derivative?
Is the derivative bounded?
• Mar 26th 2009, 08:01 AM
varkoume.com
Yes I noticed that it is bounded.. If it is bounded then the function is uniform continuous..right? and then |f(x)-f(y)|< c

after that?.