- #1

- 1,085

- 6

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

-----------

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!

------

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!

Last edited: