# Prove f-g is uniformly continuous

1. Mar 27, 2013

### Fire flame

1. The problem statement, all variables and given/known data

Let f, g : D→R be uniformly continuous. Prove that f-g: D→R is uniformly continuous aswell

2. Relevant equations

none

3. The attempt at a solution

Okay, I am posting this question because I want to make sure that my solution is correct and if it isn't I would really be thankful if someone pointed out its flaws.

My solution:

For all ε >0 there exists δ >0 s.t.

abs( x - y ) < δ => abs( f(x) - f(y)) <ε

then

abs( f(x) - f(y)) <ε/2

abs( g(x) - g(y)) <ε/2

now its f - g so

abs( f(x) - f(y) - g(x) + g(y)) < ε

abs( f(x) - f(y)) + abs( - g(x) + g(y)) < ε

abs( f(x) - f(y)) + abs(-1)*abs( +g(x) -g(y)) < ε

abs( f(x) - f(y)) + abs( +g(x) -g(y)) < ε

ε/2 + ε/2 <ε

ε ≤ ε

Also can anyone describe to me what uniformly continuous means besides from the actual definition. I have a hard time understanding what uniformly continuous means.

Thanks.

2. Mar 27, 2013

### LCKurtz

How do you know the same $\delta$ works for both f and g?

3. Mar 27, 2013

### Fire flame

Can't you just choose what ε and δ is?

It can have many values as long as its > 0. Right?

4. Mar 27, 2013

### LCKurtz

No. $\epsilon$ is a given, presumably small, number. If you have a uniformly continuous function $f$, that means that given any $\epsilon > 0$ there exists a $\delta$, depending only on $\epsilon$ and not $x$ or $y$ such that if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$. You don't get to choose $\delta$ but you are guaranteed that there is one. A different function $g$ would normally need a different $\delta$ for the same $\epsilon$.

There are other issues with your writeup, but first you have to figure out how you can get one $\delta$ that works for both $f$ and $g$.

5. Mar 27, 2013

### Fire flame

I'm going to take a few guesses.

1. What if both f and g are the same function then they would have the same δ and ε values.

2. Because they are both uniformly continuous.

3. Because I am picking the same ε for each function.

I have tried to understand the whole ε δ thing many times and although I think I kind of get it I don't think that I have a fundamental understanding of it.

Also, my professor just picked the same δ like I did for a similar problem. I do know that they don't have to be the same.

6. Mar 27, 2013

### LCKurtz

They aren't, so that is irrelevant.
No.
No.
Yes, I don't think you quite have it yet either. Think about my question like this. Suppose you have a $\delta_1$ that makes $|f(x)-f(y)|< \epsilon$ when $|x-y|<\delta_1$ and you also have a $\delta_2$ that makes $|g(x)-g(y)|< \epsilon$ when $|x-y|<\delta_2$. Can you see how to make a $\delta$ from that that works for both?

7. Mar 28, 2013

### Fire flame

Actually now that I'm looking at my professors notes more I do see that he indeed labeled δ1 and δ2 separately even though they are in the end the same.

So is there anything else wrong with my write up? I just turned in the homework yesterday and I will see what I get on this problem.

8. Mar 28, 2013

### LCKurtz

No, they aren't the same. What you want is to to pick $\delta_1$ for $f$ and $\delta_2$ for $g$ so $|f(x) - f(y)|< \frac \epsilon 2$ when $|x-y|<\delta_1$ and $|g(x) - g(y)|< \frac \epsilon 2$ when $|x-y|<\delta_2$. Then choose $\delta = \text{min}\{\delta_1,\delta_2\}$ so that when $|x-y|<\delta$ they both work.

Then in your writeup, you don't want to end up with $\epsilon\le \epsilon$. After the above paragraph you want to say that if $|x-y|<\delta$ then$$|f(x)+g(x) - (f(y)+g(y))| \le ... < \epsilon$$where you fill in the dots with a string of inequalities similar to what you did write.

Last edited: Mar 28, 2013