Prove that $C[a,b]$ is a complete metric space.

Question

Consider $C[a,b]$ - metric space with $d(f,g) = \int_{a}^{b} |f-g|dx$.

I consider a Cauchy-sequence $\{f_{n}\}$ then consider $f(x)=\lim_\limits{n \to \infty}f_{n}(x)$ then it's obviously that $f(x)$ is continuous. Now consider: $\forall \epsilon > 0 \exists N : \forall n,m > N : d(f_{n},f_{m}) < \epsilon$, so consider $\int_{a}^{b} |f_{n}-f_{m}| = \lim_\limits{m \to \infty} \int_{a}^{b} |f_{n}-f_{m}| < \int_{a}^{b} |f_{n}-f_{m}| < \epsilon$. Am I right?

How do you see that $f$ is continuous – Guy Fsone Sep 25 '17 at 13:38 — Guy Fsone, Sep 25 '17 at 13:38

score 1 · Answer 1 · 2017-09-25T14:07:53.650

proof-verification

I consider a Cauchy-sequence $\{f_{n}\}$ then consider $f(x)=\lim_\limits{n \to \infty}f_{n}(x)$ then it's obviously that $f(x)$ is continuous. (Instead of claiming so, one is supposed to show it.)

Now consider: $\forall \epsilon > 0 \exists N : \forall n,m > N : d(f_{n},f_{m}) < \epsilon$, so consider $$ \int_{a}^{b} |f_{n}-f_{m}| = \lim_\limits{m \to \infty} \int_{a}^{b} |f_{n}-f_{m}| < \int_{a}^{b} |f_{n}-f_{m}| < \epsilon.$$ (The first equality does not make sense and your proof breaks down here.)

Am I right? (No. Where do you show that the space is complete?)

There is actually nothing to prove because $C[a,b]$ with your metric $d$ is not a complete metric space. Take a look at this answer:

https://math.stackexchange.com/a/156905/9464

score 0 · Answer 2 · answered Sep 25 '17 at 13:37

0

No, not at all. How do you know that your definition of $f$ makes sense? It may be obvious for you that $f$ is continuous, but not for me. And what's the goal of the final string of inequalities?

answered Sep 25 '17 at 13:37

José Carlos Santos

427,504

At last string I tried to show that $f(x)$ applies our metric. – openspace Sep 25 '17 at 13:39
@openspace I don't know what that means. – José Carlos Santos Sep 25 '17 at 13:40
I tried to show that : $d(f,f_{n}) < \epsilon$ – openspace Sep 25 '17 at 13:40
@openspace Well, you didn't (and you couldn't, since it is not true). – José Carlos Santos Sep 25 '17 at 13:42
How can I change my proof? – openspace Sep 25 '17 at 13:43
for all $x \in [a,b]$ we have $f_{n}(x)$ converges for some $q_{x}$ now consider a function that for all $x$ : $f(x) = q_{x}$. And so we have some function in $C[a,b]$

openspace

Sep 25 '17 at 13:51

score 0 · Answer 3 · answered Sep 25 '17 at 13:42

Obvious is a dangerous word to use. I align my phiolosophy about that word based on two quotes by my professor of calculus:

"dear colleague, the word 'obvious' does not belong to your vocabulary. You are still a couple years off."
"The only things that you are allowed to call "obvious" are those that have a single line proof. And you should always have the time to write down single line proofs."

Your problems are:

1.I would not consider it obvious that $f$ is continuous.

You should provide a proof of why $f$ is continuous.

2.It is not clear why you define $f$ at all, since you don't use it anywhere else.

What you probably want is to prove that $f$ is the limit of $\{f_n\}$, but you haven't done that yet.

3.The final string of inequalities makes no sense. For example, it is not true that

$$\int_{a}^{b} |f_{n}-f_{m}| = \lim_\limits{m \to \infty} \int_{a}^{b} |f_{n}-f_{m}| $$

because the variable $m$, on the left, is unclear.

Prove that $C[a,b]$ is a complete metric space.

3 Answers3