Showing composition of functions are uniformly continuous

[k3k3]

Showing the sum of functions are uniformly continuous

Homework Statement

Suppose f and g are uniformly continuous on an interval I. Prove f + g are uniformly continuous on I.

Homework Equations

The Attempt at a Solution

Let ε >0

By definition, since f and g are uniformly continuous 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?

 

[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

you could instead just write

|f(y)+g(y)-(f(x)+g(x))| ≤ |f(y)-f(x)|+|g(y)-g(x)| < ε/2+ε/2 =ε.

Nice work. And, by the way, you have shown the sum of two u.c. functions is u.c. If you really meant the composition, you did the wrong problem

 

[k3k3]

Oops! I did mean the sum. I am not sure why I wrote that... Thank you for the suggestions. I will rewrite it to make it look nicer!