0

Prove that the set of bounded continuous functions on $\Bbb R$ is complete in the sup norm.

I know the necessary definitions, but not how to combine them to get the appropriate result.

Brian Fitzpatrick
  • 26,933
kiwifruit
  • 737

1 Answers1

2

A metric spaced X is said to be complete if every Cauchy sequence converges in X

Consider a Cauchy sequence of bounded continuous functions $(f_n)_{n\in\mathbb{N}}$ on $\mathbb{R}$,

let $x$ be a real number, one can note that the sequence $(f_n(x))_{n\in\mathbb{N}}$ is a Cauchy sequence in the complete metric space $\mathbb{R}$, and therefore converges towards a real that we shall note $f(x)$.

All we need to show is that $(f_n)$ converges towards $f$ and that $f$ is a bounded continuous function.

$f_n$ being a Cauchy sequence one has: $\forall \varepsilon>0, \exists N \in\mathbb{N}, \forall n,p \geq N, ||f_n-f_p||_{\infty}<\varepsilon$

i.e.$\forall \varepsilon>0, \exists N \in\mathbb{N}, \forall n,p \geq N, \forall x \in \mathbb{R}, |f_n(x)-f_p(x)|<\varepsilon$

With $x$ a real number, when $p\to\infty$,$|f_n(x)-f(x)|<\varepsilon$ and this is true for all $x$, so $||f_n-f||_{\infty}<\varepsilon$ Which implies:

1) that $f_n$ converges uniformly towards $f$ ($f$ is therefore continuous)

2) that $f_n-f$ is bounded and therefore $f$ is a bounded continuous function as $f_n$ converges uniformly towards $f$.

Jack
  • 526
  • why the down vote ? – Jack Mar 15 '14 at 22:37
  • I did not put the down vote. But I am wondering why is is important that f is bounded? – kiwifruit Mar 15 '14 at 22:38
  • 1
    the limit has to be an element of the metric space. (sorry I forgot to mention that in my original post, I've added it now) – Jack Mar 15 '14 at 22:41
  • Ok, and how does $||f_n - f_p|| \leq \varepsilon$ imply that $||f_n - f|| \leq \varepsilon$? – kiwifruit Mar 15 '14 at 22:44
  • 1
    one has: $\forall x \in \mathbb{R}, |f_n(x)-f_p(x)| < \varepsilon$, so if we choose $x$ and let $p\to\infty$ , we have $|f_n(x)-f(x)| < \varepsilon$ because the sequence $(f_n)$ converges towards $f(x)$ by definition, this is true for all $x$, and so $||f_n-f||<\varepsilon$ – Jack Mar 15 '14 at 22:49
  • Ok, thank you, but is this special case OK if we want any Cauchy sequence to have these properties, not just when p is infinity? Wouldn't this be a special case? – kiwifruit Mar 15 '14 at 22:51
  • The Cauchy sequence is $(f_n)_{n\in\mathbb{N}}$, $f_n$ and $f_p$ $n,p \in \mathbb{N}$ are terms of the sequence. In my earlier comment I meant to say that the sequence of real numbers $(f_n(x))$ converges towards $f(x)$ – Jack Mar 15 '14 at 22:56
  • Right, but if we are restricting p to be infinity, does this necessarily satisfy the "every Cauchy sequence" part? – kiwifruit Mar 15 '14 at 22:59
  • 1
    To show that $f_n$ converges towards $f$ we need to show that: $\forall \varepsilon>0, \exists N \in \mathbb{N}, \forall n \geq N, ||f_n-f||< \varepsilon$. We've chosen epsilon and we know that there is an $N$ such that for all natural integers n,p greater than N $||f_n-f_p||< \epsilon$. The first paragraph in my original post consists in choosing $x$ and using the fact that $\mathbb{R}$ is complete we show that the sequence of real numbers (f_n(x)) has a limit in R that we call f(x). We can repeat this for all $x$ and thus define a point wise limit of the sequence $f$ – Jack Mar 15 '14 at 23:06
  • the rest of the proof consists in showing that this convergence is not simply point-wise, and that the sequence of functions $(f_n)$ converges towards $f$ for the sup norm. We use the fact that the sequence $(f_p(x))\longrightarrow f(x)$ when $p\to\infty$ in order to compare f_n and f. We don't need both n and p. – Jack Mar 15 '14 at 23:11
  • let us continue this discussion in chat – Jack Mar 15 '14 at 23:15