Cauchy Sequence of Continuous Fns: Uniform Convergence?

[Kalidor]

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?

 

[micromass]

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.

 

[Kalidor]

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.

 

[micromass]

Well, take a uniform Cauchy sequence (f_n)_n. By definition, it satisfies

\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

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

f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow y_x

By taking limits, we get

\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

This shows that (f_n)_n converges uniformly to f.

Is this what you want?

 

[Kalidor]

It seems to be. But doesn't this amount to saying that \mathcal{C}(\mathbb{R}) is complete with respect to the uniform norm?

 

[micromass]

No, because the uniform norm isn't well defined. For example, f(x)=x^2 has \|f\|_\infty=+\infty. But infinity is not a valid value for a norm...

 

[Kalidor]

Sure that's what I was missing, thanks.