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

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In summary, the conversation revolves around proving or disproving the uniform continuity of the function max(f,g) given that f and g are uniformly continuous functions. The conversation includes some definitions and a discussion of different cases to consider. However, the exact approach to solving the problem is still unclear and the participants are seeking input from others.
  • #1
mattmns
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Here is the question.
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Prove or give a counterexample that if [itex]f:X\to \mathbb{R}[/itex] and [itex]g:X\to \mathbb{R}[/itex] are uniformly continuous functions then [itex]\max(f,g)[/itex] is uniformly continuous.
------

Some definitions.

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

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 [itex]\epsilon' > 0[/itex] there is [itex]\delta' > 0[/itex] such that if [itex]d_X(x,x_0) < \delta'[/itex] then [itex]|f(x)-f(x_0)| < \epsilon'[/itex] and [itex]|g(x)-g(x_0)| < \epsilon'[/itex] for all [itex]x\in X[/itex]. (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 [itex]\epsilon > 0[/itex] there is [itex]\delta > 0[/itex] such that if [itex]d_X(x,x_0) < \delta[/itex] then [itex]|\max(f(x),g(x)) - \max(f(x_0),g(x_0))| < \epsilon[/itex] for all [itex]x\in X[/itex].

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

Case 1. [itex]\max(f(x),g(x)) = f(x)[/itex], and [itex]\max(f(x_0),g(x_0)) = f(x_0)[/itex].

This is obvious.

Case 2. [itex]\max(f(x),g(x)) = g(x)[/itex], and [itex]\max(f(x_0),g(x_0)) = f(x_0)[/itex]

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

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!
 
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  • #2
mattmns said:
Here is the question.
------
Prove or give a counterexample that if [itex]f:X\to \mathbb{R}[/itex] and [itex]g:X\to \mathbb{R}[/itex] are uniformly continuous functions then [itex]\max(f,g)[/itex] is uniformly continuous.
------

Some definitions.

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


Let [itex]f:X \to Y[/itex] Then [itex]f[/itex] is uniformly continuous if given [itex]\epsilon > 0[/itex] there is [itex]\delta > 0[/itex] such that if [itex]d_X(x,x_0) < \delta[/itex] then [itex]d_Y(f(x),f(x_0)) < \epsilon'[/itex] for all [itex]x\in X[/itex].
Strictly speaking we don't know that! At least from the definition of uniformly connected we know that for each [itex]\epsilon> 0[/itex], there exist [itex]\delta_1[/itex] such that is [itex]|x-a|< \delta_1[/itex] then |f(x)- f(a)|< \epsilon[/itex] and there exist [itex]\delta_2[/itex] such that is |x-a|< \delta_2[/itex] then [itex]|g(x)-g(a)|< \epsilon[/itex]. Of course, if we take [itex]\delta[/itex] to be the smaller of [itex]\delta_1[/itex] and [itex]\delta_2[/itex] 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 [itex]\epsilon' > 0[/itex] there is [itex]\delta' > 0[/itex] such that if [itex]d_X(x,x_0) < \delta'[/itex] then [itex]|f(x)-f(x_0)| < \epsilon'[/itex] and [itex]|g(x)-g(x_0)| < \epsilon'[/itex] for all [itex]x\in X[/itex].

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

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

Case 1. [itex]\max(f(x),g(x)) = f(x)[/itex], and [itex]\max(f(x_0),g(x_0)) = f(x_0)[/itex].

This is obvious.

Case 2. [itex]\max(f(x),g(x)) = g(x)[/itex], and [itex]\max(f(x_0),g(x_0)) = f(x_0)[/itex]

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

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!
 
  • #3
HallsofIvy said:
Strictly speaking we don't know that! At least from the definition of uniformly connected we know that for each [itex]\epsilon> 0[/itex], there exist [itex]\delta_1[/itex] such that is [itex]|x-a|< \delta_1[/itex] then |f(x)- f(a)|< \epsilon[/itex] and there exist [itex]\delta_2[/itex] such that is |x-a|< \delta_2[/itex] then [itex]|g(x)-g(a)|< \epsilon[/itex]. Of course, if we take [itex]\delta[/itex] to be the smaller of [itex]\delta_1[/itex] and [itex]\delta_2[/itex] 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?)
 
  • #4
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?
 
  • #5
I don't think I have ever seen such a definition.

Anybody have any ideas on this exercise? Thanks!
 
  • #6
Yes. Post 4. Thereis such an expression. Find it.
 
  • #7
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).
 
  • #8
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)?
 
  • #9
Interesting, the first seems to be twice the max and the second twice the min, and it is easy to show as well. Thanks.
 

1. What does "F,g uni. cont." stand for?

"F,g" refers to two functions, "F" and "g", while "uni. cont." stands for uniform continuity.

2. What does the arrow (=>) represent in "F,g uni. cont. => max(f,g) uni. cont."?

The arrow represents a logical implication, meaning that if the functions "F" and "g" are both uniformly continuous, then the maximum of these two functions, denoted as "max(f,g)", will also be uniformly continuous.

3. Why is uniform continuity important in this context?

Uniform continuity is important because it ensures that the functions "F" and "g" have a consistent rate of change over their entire domain, making it possible to compare and combine them in a meaningful way.

4. How is the maximum of two functions calculated?

The maximum of two functions, denoted as "max(f,g)", is the larger value of the two functions at any given point. In other words, if "f(x)" is the maximum of "F(x)" and "g(x)" at a certain point "x", then "f(x) = max(F(x), g(x))".

5. Can you provide an example of "F,g uni. cont. => max(f,g) uni. cont." in action?

One example of this concept in action is the proof that the maximum of two uniformly continuous functions is also uniformly continuous. This can be shown by using the definition of uniform continuity and the epsilon-delta criteria to prove that "max(f,g)" also satisfies these conditions.

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