# Lipschitz continuous

• Nov 13th 2007, 02:24 AM
dopi
Lipschitz continuous
Let f : R ->R be a function.

We say that f is Lipschitz continuous if there is some
L > 0 such that |f(x) − f(y)| < L|x − y| for all x, y in R.

I want to show that f : R ->R defined as f(x) = x^2 / (1+ x^2)is Lipschitz continuous.

I know I should use the Mean Value Theorem but i cant seem to apply it can some one please help thanks
• Nov 13th 2007, 03:04 AM
Plato
This is the absolute value of the derivative:
$\displaystyle \left| {\frac{{2x}}{{\left( {1 + x^2 } \right)^2 }}} \right|$.
Graph it to see that it is bounded.

From the mean value theorem:
$\displaystyle x < c < y\;\& \;\left| {\frac{{2c}}{{\left( {1 + c^2 } \right)^2 }}} \right| = \left| {\frac{{f(y) - f(x)}}{{x - y}}} \right|$.
• Nov 13th 2007, 03:14 AM
dopi
not Lipschitz continuous.
Then i was wondering how do you Show that f : R -> R defined as f(x) = x^2 is not Lipschitz continuous.
• Nov 13th 2007, 06:55 AM
ThePerfectHacker
Quote:

Originally Posted by dopi
Then i was wondering how do you Show that f : R -> R defined as f(x) = x^2 is not Lipschitz continuous.

Assume that there is a number $\displaystyle A$ so that:
$\displaystyle |x^2-y^2|\leq A|x-y|$ for all $\displaystyle x,y\in \mathbb{R}$.
So in particular for $\displaystyle x\not = y$ this inequality should work.
But then,
$\displaystyle |x-y||x+y|\leq A |x-y| \implies |x+y|\leq A$.
Which is impossible because $\displaystyle |x+y|$ can be made larger than $\displaystyle A$, i.e. $\displaystyle y=A+1,x=0$.