Uniform continuity and the sup norm

In summary: I want to show that ##f## is uniformly continuous on the whole real line.This is easily done. Suppose we have a function ##g: \mathbb R^n \rightarrow \mathbb R## such that for all ##x \in \mathbb R^n## and all ##y \in \mathbb R##, we have ##g(x, y) \in \mathbb R##. Then, by the continuity of ##g##, there exists a neighborhood ##V = \{x \in \mathbb R^n : g(x, y) \in V}\}## such that for all ##z \in V##, we have ##
  • #1
AxiomOfChoice
533
1
Suppose I have a function [itex]f(x) \in C_0^\infty(\mathbb R)[/itex], the real-valued, infinitely differentiable functions with compact support. Here are a few questions:

(1) The function [itex]f[/itex] is trivially uniformly continuous on its support, but is it necessarily uniformly continuous on [itex]\mathbb R[/itex]?
(2) I want
[tex]
\left\| f(x-t) - f \left(\frac{x}{1+\epsilon} - t \right) \right\|_\infty \to 0
[/tex]
as [itex]\epsilon \to 0[/itex]. Is that so? It seems so, intuitively, if [itex]f[/itex] is uniformly continuous.
(3) Also, is it true that
[tex]
\left\| f(x-t) - f \left(\frac{x}{1+\epsilon} - t \right) \right\|_\infty = \left\| f(x) - f \left(\frac{x}{1+\epsilon} \right) \right\|_\infty
[/tex]
 
Physics news on Phys.org
  • #2
I believe I have found the answer to (1) here, and it is yes, but the proof isn't exactly trivial.
 
  • #3
AxiomOfChoice said:
Suppose I have a function [itex]f(x) \in C_0^\infty(\mathbb R)[/itex], the real-valued, infinitely differentiable functions with compact support. Here are a few questions:

(1) The function [itex]f[/itex] is trivially uniformly continuous on its support, but is it necessarily uniformly continuous on [itex]\mathbb R[/itex]?
Yes. Suppose the support of ##f## is ##K##. Since ##K## is compact, it is bounded, so it is contained within some closed interval ##I## with radius ##r##, centered at the origin. Now expand this to a closed interval ##J## with radius ##r+1##, centered at the origin. Since ##f## is continuous on the compact set ##J##, it is uniformly continuous on that set. Let ##\epsilon > 0##. There exists ##\delta > 0## such that whenever ##x,y \in J## with ##|x-y| < \delta##, we have ##|f(x)-f(y)| < \epsilon##.

Now let ##\delta_0 = \min(\delta, 1)##, and consider any points ##x,y \in \mathbb{R}## with ##|x-y| < \delta_0##. If ##x,y\in J## then ##|f(x) - f(y)| < \epsilon## because of the uniform continuity. If ##x,y\in J^c## then ##f(x) = f(y) = 0## so surely ##|f(x)-f(y)| < \epsilon##.

The only other possibility is that one of the points (say ##x##) is in ##J## and the other (##y##) is in ##J^c##. Once again this gives us ##f(x) = f(y) = 0##, because ##|x-y| < 1## and ##y \in J^c## imply that ##x \in J \setminus I##, so ##x## is outside the support of ##f##. Thus in this case we also have ##|f(x) - f(y)| < \epsilon##.
 

FAQ: Uniform continuity and the sup norm

What is uniform continuity?

Uniform continuity is a property of a function that describes how it behaves over the entire domain of the function. In simple terms, a function is uniformly continuous if small changes in the input result in small changes in the output, regardless of where the input is in the domain.

2. How is uniform continuity different from ordinary continuity?

Uniform continuity is a stronger condition than ordinary continuity. While ordinary continuity only requires the function to be continuous at each individual point in the domain, uniform continuity requires the function to be continuous over the entire domain.

3. What is the sup norm?

The sup norm, also known as the supremum norm, is a way of measuring the distance between two functions. In particular, it measures the maximum difference between the values of two functions over a given interval. In the context of uniform continuity, the sup norm is often used to determine if a function is uniformly continuous or not.

4. How is uniform continuity related to the sup norm?

In order for a function to be uniformly continuous, the sup norm of the function must be finite. This means that the function cannot have any large spikes or abrupt changes that would result in an infinite sup norm. Conversely, if the sup norm of a function is finite, it is possible for the function to be uniformly continuous.

5. Can a function be uniformly continuous but not have a finite sup norm?

No, a function cannot be uniformly continuous if its sup norm is infinite. This is because an infinite sup norm would indicate that the function has large spikes or abrupt changes, which would violate the definition of uniform continuity. Therefore, a function must have a finite sup norm in order to be uniformly continuous.

Similar threads

Back
Top