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

Uniform convergence

  1. Aug 27, 2011 #1
    Is it true that a cauchy sequence of continuous functions defined on the whole real line converges uniformly to a continuous function?
    I thought this was only true for functions defined on a compact subset of the real line.
    Am I wrong?
  2. jcsd
  3. Aug 27, 2011 #2
    Yes, the uniform limit of a sequence of continuous functions is always continuous.

    Why we often work with compact sets in uniform convergence is because the uniform norm can be infinite. But on a compact interval, all functions are bounded. So this cannot happen there.
  4. Aug 27, 2011 #3
    My actual doubt was actually about the convergence. Why does it have to converge? The space of continuous functions defined on the real line is not a banach space as far as I know.
  5. Aug 27, 2011 #4
    Well, take a uniform Cauchy sequence [itex](f_n)_n[/itex]. By definition, it satisfies

    [tex]\forall \varepsilon>0:~\exists n_0:~\forall p,q\geq n_0:~\forall x\in \mathbb{R}:~d(f_p(x),f_q(x))<\varepsilon[/tex]

    It follows that every sequence [itex](f_n(x))_n[/itex] is Cauchy for all x, so it converges to a [itex]y_x[/itex]. This constructs a function

    [tex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow y_x[/tex]

    By taking limits, we get

    [tex]\forall \varepsilon>0:~\exists n_0:~\forall q\geq n_0:~\forall x\in \mathbb{R}:~d(f(x),f_q(x))=\lim_{p\rightarrow +\infty}{d(f_p(x),d_q(x))}\leq\varepsilon[/tex]

    This shows that [itex](f_n)_n[/itex] converges uniformly to f.

    Is this what you want?
  6. Aug 28, 2011 #5
    It seems to be. But doesn't this amount to saying that [tex] \mathcal{C}(\mathbb{R}) [/tex] is complete with respect to the uniform norm?
  7. Aug 28, 2011 #6
    No, because the uniform norm isn't well defined. For example, [itex]f(x)=x^2[/itex] has [itex]\|f\|_\infty=+\infty[/itex]. But infinity is not a valid value for a norm...
  8. Aug 28, 2011 #7
    Sure that's what I was missing, thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook