# Uniform convergence

1. Aug 27, 2011

### 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?

2. Aug 27, 2011

### micromass

Staff Emeritus
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.

3. Aug 27, 2011

### 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.

4. Aug 27, 2011

### micromass

Staff Emeritus
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?

5. Aug 28, 2011

### 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?

6. Aug 28, 2011

### micromass

Staff Emeritus
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...

7. Aug 28, 2011

### Kalidor

Sure that's what I was missing, thanks.