- #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]
(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]