# Uniform Continuity

• Nov 9th 2011, 02:14 AM
H12504106
Uniform Continuity
Qn: Show that the function f(x) = 1/(1+x^2) is uniformly continuous on R.

I tried to show that the function satifies the Lipschitz condition

|1/(1+x^2) - 1/(1+y^2)| = |(x^2 - y^2)/(1+x^2)(1+y^2)| = |(x+y)/(1+x^2)(1+y^2)||x-y| --- (*)

Then I observe that since (x-1)^2 >= 0. Hence 1+x^2 >= 2x and 1+y^2 >= 2y.

Thus (*) <= |(x+y)/4xy||x-y| = 1/4 |1/x + 1/y||x-y|

However, I am unable to proceed after that. How should i do it? I have also tried to prove it by using the epsilon-delta definition of uniform continuity but the argument is very similar to the one above.

Thanks.
• Nov 9th 2011, 03:28 AM
girdav
Re: Uniform Continuity
You only have to show that $\displaystyle g(x,y):=\frac{x+y}{(1+x^2)(1+y^2)}$ is bounded independently of $\displaystyle x$ and $\displaystyle y$. Write $\displaystyle \left|g(x,y)\right| =\left|\frac{x}{(1+x^2)(1+y^2)}\right|+\left|\frac {y}{(1+x^2)(1+y^2)}\right|\leq \frac{|x|}{1+x^2}+\frac{|y|}{1+y^2}$, and $\displaystyle x\mapsto \frac{|x|}{1+x^2}$ is bounded by $\displaystyle \frac 12$.