Prove f-g is uniformly continuous

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Homework Statement



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



Homework Equations



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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.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



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



Homework Equations



none

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

How do you know the same ##\delta## works for both f and g?
 
  • #3
Can't you just choose what ε and δ is?

It can have many values as long as its > 0. Right?
 
  • #4
LCKurtz
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Homework Statement



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



Homework Equations



none

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.

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

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

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

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
but first you have to figure out how you can get one δ that works for both f and g

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
LCKurtz
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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.
They aren't, so that is irrelevant.
2. Because they are both uniformly continuous.
No.
3. Because I am picking the same ε for each function.
No.
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.
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
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
LCKurtz
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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.

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.
 
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