# Showing composition of functions are uniformly continuous

1. Sep 27, 2012

### k3k3

Showing the sum of functions are uniformly continuous

1. The problem statement, all variables and given/known data
Suppose f and g are uniformly continuous on an interval I. Prove f + g are uniformly continuous on I.

2. Relevant equations

3. The attempt at a solution

Let ε >0

By definition, since f and g are uniformly continous on I, there exists a $\delta$_1 such that |f(y)-f(x)| < ε for all x,y in I that satisfy |x-y| < $\delta$_1

Similarly, for g, there exists a $\delta$_2 such that |g(y)-g(x)| < ε for all |y-x| < $\delta$_2

Then, for all x,y in I, |f(y)+g(y)-(f(x)+g(x))| ≤ |f(y)-f(x)|+|g(y)-g(x)| by the triangle inequality. This implies |f(y)+g(y)-(f(x)+g(x))| < 2ε

Choose $\delta$=min{$\delta$_1,$\delta$_2} * 1/2

Then |f(y)+g(y)-(f(x)+g(x))| < 2ε for all x,y in I that satisfy |y-x| < $\delta$

∴f+g is uniformly continuous on I.

Is this correct?

Last edited: Sep 27, 2012
2. Sep 27, 2012

### LCKurtz

Yes, but a couple of minor changes would make it slightly nicer. For your first two inequalites use $\frac \epsilon 2$ so it will only add to $\epsilon$ at the end. And choosing $\delta$, there is no need to multiply by 1/2. Finally, where you have:

|f(y)+g(y)-(f(x)+g(x))| ≤ |f(y)-f(x)|+|g(y)-g(x)| by the triangle inequality. This implies |f(y)+g(y)-(f(x)+g(x))| < 2ε