Uniform Convergence: Showing Continuous & Uniformly Continuous

In summary, the conversation is discussing the definitions of continuity and uniform continuity, as well as the relationship between the two. The definition of uniform continuity is provided as well as a definition for uniform convergence. The participants in the conversation also share their thoughts and ideas on an exercise related to uniform continuity.
  • #1
mattmns
1,128
6
Here is the question from the book:
----------
Let [itex]f: \mathbb{R} \to \mathbb{R}[/itex] be a function. For any [itex]a\in \mathbb{R}[/itex], let [itex]f_a :\mathbb{R}\to \mathbb{R}[/itex] be the shifted function [itex]f_a(x):=f(x-a)[/itex].(a) Show that [itex]f[/itex] is continuous if and only if, whenever [itex](a_n)_{n=0}^{\infty}[/itex] is a sequence of real numbers which converges to zero, the shifted functions [itex]f_{a_n}[/itex] converge pointwise to [itex]f[/itex].

(b) Show that [itex]f[/itex] is uniformly continuous if and only if, whenever [itex](a_n)_{n=0}^{\infty}[/itex] is a sequence of real numbers which converges to zero, the shifted functions [itex]f_{a_n}[/itex] converge uniformly to [itex]f[/itex].---------
[itex](\Rightarrow)[/itex] Let [itex]x_0 \in \mathbb{R}[/itex]. Suppose [itex]f[/itex] is continuous. That is, given [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]|x-x_0| < \delta[/itex] then [itex]|f(x)-f(x_0)| < \epsilon[/itex].Let [itex]x_n = x_0 - a_n[/itex]. So given [itex]\delta > 0[/itex] there exists [itex]N' > 0[/itex] such that [itex]|x_n - x_0|< \delta[/itex] for all [itex]n > N'[/itex].

Given [itex]\epsilon' > 0[/itex] take [itex]\epsilon = \epsilon'[/itex], and take [itex]N = N'[/itex].

So by the continuity of [itex]f[/itex] we get that [itex]|x_n - x_0| < \delta[/itex] for all [itex]n>N' = N[/itex].

Thus, [itex]|f(x_n) - f(x_0)| < \epsilon = \epsilon'[/itex].

But, [itex]f(x_n) = f(x_0-a_n) = f_{a_n}(x_0)[/itex] for all [itex]n>N[/itex].

So we have [itex]|f_{a_n}(x_0) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].Thus [itex]f_{a_n}[/itex] converges pointwise to [itex]f[/itex].
[itex](\Leftarrow)[/itex] Suppose given [itex]a_n \to 0[/itex] that [itex]f_{a_n}[/itex] converges pointwise to [itex]f[/itex]. Given [itex]x_0[/itex] take a sequence [itex]x_n[/itex] which converges to [itex]x_0[/itex]. That is, [itex]a_n = x_0 - x_n \to 0[/itex].

Since [itex]f_n[/itex] converges pointwise to [itex]f[/itex], given [itex]\epsilon > 0[/itex] there is some [itex]N>0[/itex] such that [itex]|f_{a_n}(x_0) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].

But, [itex]|f_{a_n}(x_0) - f(x_0)| = |f(x_0 - a_n) - f(x_0)| = |f(x_n) - f(x_0)|[/itex].

Hence, [itex]|f(x_n) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].

That is, [itex]f(x_n)[/itex] converges to [itex]f(x_0)[/itex]. Thus, [itex]f[/itex] is continuous.----------

I don't think there is anything wrong with any of that (if there is, or you have any comments please say something).

However, I am not too sure about part (b).

The first direction is the same as in part (a), but the other direction I am not sure about.

I have been trying just the basic epsilon-delta definition way (similar to what I did in part (a) with this direction) and not really getting anywhere (which I guess could be expected since we don't have a uniform continuity definition with sequences [at least in our book]).

Any ideas? Thanks!edit... I am converting the $ signs to [itex]'s. Is there an easy (fast and painless) way to do this?
 
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  • #2
Copy into notepad, search and replace, using pre-spaces to distinguish opening from closing ones. If it's not too late.
 
  • #3
What is your definition of uniform continuity? (Please always post relevant definitions)
 
  • #4
Cool, that worked nicely! Any ideas about the exercise now :smile:

Sorry, I thought that was a standard definition. Here it is:

Uniform Continuity: Let [itex]f:X\to Y[/itex] be a map from one metric space [itex](X,d_X)[/itex] to another [itex](Y,d_Y).[/itex] Then [itex]f[/itex] is uniformly continuous if, for every [itex]\epsilon > 0[/itex], there exists a [itex]\delta > 0[/itex] such that [itex]d_Y(f(x),f(x')) < \epsilon[/itex] whenever [itex]x,x'\in X[/itex] are such that [itex]d_X(x,x')<\delta[/itex].

Here is our book's definition of uniform convergence:

Let [itex](f^{(n)})_{n=1}^{\infty}[/itex] be a sequence of functions from one metric space [itex](X,d_X)[/itex] to another [itex](Y,d_Y)[/itex], and let [itex]f: X\to Y[/itex] be another function. We say that [itex](f^{(n)})_{n=1}^{\infty}[/itex] converges uniformly to [itex]f[/itex] on [itex]X[/itex] if for every [itex]\epsilon > 0[/itex] there exists [itex]N>0[/itex] such that [itex]d_Y(f^{(n)}(x),f(x))<\epsilon[/itex] for every [itex]n>N[/itex] and [itex]x\in X[/itex].
 
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  • #5
It is standard, it's just a lot easier to refer to a definition that's here. Remeber that the definition of continuity in terms of sequences was just a result of the more general definition of limits in terms of sequences, and you should still be able to apply that.
 

1. What is uniform convergence?

Uniform convergence is a type of convergence in which a sequence of functions converges to a single function in such a way that the rate of convergence is the same at every point in the domain. In other words, the convergence is not only pointwise, but also uniform.

2. How is uniform convergence different from pointwise convergence?

Pointwise convergence only requires that a sequence of functions converges to a single function at every point in the domain, while uniform convergence also requires that the rate of convergence is the same at every point. This means that the sequence of functions must get closer to the limit function at the same rate everywhere, not just at individual points.

3. How do you show that a sequence of functions is uniformly continuous?

To show that a sequence of functions is uniformly continuous, you must prove that the rate of convergence is the same at every point in the domain. This can be done by using the definition of uniform continuity and showing that for any given epsilon, there exists a delta that works for all points in the domain.

4. What is the importance of uniform convergence in analysis?

Uniform convergence is important in analysis because it allows us to extend certain properties of individual functions to an entire sequence of functions. This can be useful in proving the continuity of limit functions and in proving the interchangeability of limits and integrals.

5. Can a sequence of continuous functions converge to a non-continuous function uniformly?

No, a sequence of continuous functions cannot converge to a non-continuous function uniformly. This is because the uniform limit of continuous functions must also be continuous. If a sequence of functions converges uniformly, then the limit function must have the same continuity properties as the functions in the sequence.

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