Uniform continuity and the sup norm

[AxiomOfChoice]

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]

 

[AxiomOfChoice]

I believe I have found the answer to (1) here, and it is yes, but the proof isn't exactly trivial.

 

[jbunniii]

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##.