Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uniform continuity and the sup norm

  1. Jan 6, 2014 #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]
     
  2. jcsd
  3. Jan 6, 2014 #2
    I believe I have found the answer to (1) here, and it is yes, but the proof isn't exactly trivial.
     
  4. Jan 6, 2014 #3

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook