F,g uni. cont. => max(f,g) uni. cont.?

[mattmns]

Here is the question.
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Prove or give a counterexample that if $f:X\to \mathbb{R}$ and $g:X\to \mathbb{R}$ are uniformly continuous functions then $\max(f,g)$ is uniformly continuous.
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Some definitions.

If $x,y \in \mathbb{R}, \ \text{then} \ \max(x,y) = x$ if $x\geq y$ and $\max(x,y) = y$ if $y\geq x$.Let $f:X \to Y$ Then $f$ is uniformly continuous if given $\epsilon > 0$ there is $\delta > 0$ such that if $d_X(x,x_0) < \delta$ then $d_Y(f(x),f(x_0)) < \epsilon'$ for all $x\in X$.-----------

I think that the max(f,g) will be uniformly continuous, and so I have been trying to prove it.

First we are given that f,g are uniformly continuous. So we know that given $\epsilon' > 0$ there is $\delta' > 0$ such that if $d_X(x,x_0) < \delta'$ then $|f(x)-f(x_0)| < \epsilon'$ and $|g(x)-g(x_0)| < \epsilon'$ for all $x\in X$. (which we can do if we just take the smaller of the two deltas that we would get from the original definition).

We want to show that the max(f,g) is uniformly continuous. That is, given $\epsilon > 0$ there is $\delta > 0$ such that if $d_X(x,x_0) < \delta$ then $|\max(f(x),g(x)) - \max(f(x_0),g(x_0))| < \epsilon$ for all $x\in X$.

The way I am trying to do this is by cases. Namely the following cases:

Case 1. $\max(f(x),g(x)) = f(x)$, and $\max(f(x_0),g(x_0)) = f(x_0)$.

This is obvious.

Case 2. $\max(f(x),g(x)) = g(x)$, and $\max(f(x_0),g(x_0)) = f(x_0)$

This is what I am having trouble with. We get that $|\max(f(x),g(x)) - \max(f(x_0),g(x_0)| = |g(x)-f(x_0)|$.

Now it looks like we have to add a mixed term, but then it seems like we are going to have to choose epsilon to depend on x or x_0 here, which would mess up the uniform continuity.

Am I missing something here (or perhaps doing the problem incorrectly), is there a way around this? Thanks!

 

[HallsofIvy]

Here is the question.
------
Prove or give a counterexample that if $f:X\to \mathbb{R}$ and $g:X\to \mathbb{R}$ are uniformly continuous functions then $\max(f,g)$ is uniformly continuous.
------

Some definitions.

If $x,y \in \mathbb{R}, \ \text{then} \ \max(x,y) = x$ if $x\geq y$ and $\max(x,y) = y$ if $y\geq x$.


Let $f:X \to Y$ Then $f$ is uniformly continuous if given $\epsilon > 0$ there is $\delta > 0$ such that if $d_X(x,x_0) < \delta$ then $d_Y(f(x),f(x_0)) < \epsilon'$ for all $x\in X$.

Strictly speaking we don't know that! At least from the definition of uniformly connected we know that for each $\epsilon> 0$, there exist $\delta_1$ such that is $|x-a|< \delta_1$ then |f(x)- f(a)|< \epsilon[/itex] and there exist $\delta_2$ such that is |x-a|< \delta_2[/itex] then $|g(x)-g(a)|< \epsilon$. Of course, if we take $\delta$ to be the smaller of $\delta_1$ and $\delta_2$ then what you say is true. Now you need to look at |(f(x)+g(x)- (f(a)-g(a)| which is "less than or equal to" |f(x)-f(a)|+ |g(x)- g(a)|.

-----------

I think that the max(f,g) will be uniformly continuous, and so I have been trying to prove it.

First we are given that f,g are uniformly continuous. So we know that given $\epsilon' > 0$ there is $\delta' > 0$ such that if $d_X(x,x_0) < \delta'$ then $|f(x)-f(x_0)| < \epsilon'$ and $|g(x)-g(x_0)| < \epsilon'$ for all $x\in X$.

We want to show that the max(f,g) is uniformly continuous. That is, given $\epsilon > 0$ there is $\delta > 0$ such that if $d_X(x,x_0) < \delta$ then $|\max(f(x),g(x)) - \max(f(x_0),g(x_0))| < \epsilon$ for all $x\in X$.

The way I am trying to do this is by cases. Namely the following cases:

Case 1. $\max(f(x),g(x)) = f(x)$, and $\max(f(x_0),g(x_0)) = f(x_0)$.

This is obvious.

Case 2. $\max(f(x),g(x)) = g(x)$, and $\max(f(x_0),g(x_0)) = f(x_0)$

This is what I am having trouble with. We get that $|\max(f(x),g(x)) - \max(f(x_0),g(x_0)| = |g(x)-f(x_0)|$.

Now it looks like we have to add a mixed term, but then it seems like we are going to have to choose epsilon to depend on x or x_0 here, which would mess up the uniform continuity.

Am I missing something here (or perhaps doing the problem incorrectly), is there a way around this? Thanks!

 

[mattmns]

Strictly speaking we don't know that! At least from the definition of uniformly connected we know that for each $\epsilon> 0$, there exist $\delta_1$ such that is $|x-a|< \delta_1$ then |f(x)- f(a)|< \epsilon[/itex] and there exist $\delta_2$ such that is |x-a|< \delta_2[/itex] then $|g(x)-g(a)|< \epsilon$. Of course, if we take $\delta$ to be the smaller of $\delta_1$ and $\delta_2$ then what you say is true.

That is what I was doing, I should have noted that in the post.

Now you need to look at |(f(x)+g(x)- (f(a)-g(a)| which is "less than or equal to" |f(x)-f(a)|+ |g(x)- g(a)|.

I am not sure if I understand what you mean here. A previous part of the exercise was to prove that f+g, and f-g are uniformly continuous if f,g are uniformly continuous. (Is this what you were getting at?)

 

[Mystic998]

Isn't there a definition of the max function that involves just the absolute value and sums, differences, and multiplication by constants of the arguments?

 

[mattmns]

I don't think I have ever seen such a definition.

Anybody have any ideas on this exercise? Thanks!

 

[matt grime]

Yes. Post 4. Thereis such an expression. Find it.

 

[mattmns]

I can't figure out the expression nor have I been able to find it online. However, I talked to my professor and she gave me a hint on how to finish the mixed case (using the intermediate value theorem to find a point z between x and x_0 such that f(z)=g(z), and then using the triangle inequality with that point).

 

[ZioX]

Well, suppose we know the absolute difference between x and y. Then x+y+diff(x,y) is what? What is x+y-diff(x,y)?

 

[mattmns]

Interesting, the first seems to be twice the max and the second twice the min, and it is easy to show as well. Thanks.