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In my lecture, I need prove the existence of the anyone Cauchy sequence $(f_n)_{n\in \mathbb{N}}$ belonging $C(K)$ with the norm ${\Vert \cdot \Vert}_{\infty}=\sup_{x \in K}|f(x)|$ where $K$ is a compact space. How I can do it? I'm trying that below

Fixing $\varepsilon>0$, there exist any $x_0$ such that

${\Vert f_n - f_m \Vert}_{\infty} < |(f_n - f_m)x_0| + \varepsilon/2$. Then for $n,m$ such that

$|(f_n - f_m)x_0|<\varepsilon/2$. Then $(f_n)_{n\in \mathbb{N}}$ is a Cauchy sequence.

Is this correct?

Rudy the Reindeer
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juaninf
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    (I fixed a bit your English :-) ) – Mariano Suárez-Álvarez Aug 17 '12 at 18:11
  • Do you have to show that any Cauchy Sequence converges in that space with that norm? Please correct the formulation of the question, since in it's current shape it is incomprehensible. –  Aug 17 '12 at 18:12
  • (Notice that all your questions have titles which begin with «Question abput...»: there is no need to include that, as it is assumed that what you are writing is a question!) – Mariano Suárez-Álvarez Aug 17 '12 at 18:12
  • Are you trying to prove that there is some Cauchy sequence? A constant sequence is already Cauchy... – tomasz Aug 17 '12 at 18:14
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    I have no idea what's going on here... – David Mitra Aug 17 '12 at 18:14
  • sorry I write again the question – juaninf Aug 17 '12 at 18:18
  • Are you supposed to show that every Cauchy sequence in $C(K)$ is convergent? – Martin Sleziak Aug 17 '12 at 18:20
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    well, it looks like you're trying to prove the completeness of $C(K)$. If that is so, observe first that for $(f_n)$ a Cauchy sequence in $C(K)$ you have for each $x \in K$ that $|f_n(x) - f_m(x)| \leq |f_n - f_m|\infty$, so $(f_n(x)){n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Put $f(x) = \lim_{n\to\infty} f_n(x)$. Prove first that $\sup_{x \in K} |f(x) - f_n(x)| \xrightarrow{n\to\infty} 0$ and deduce that $f$ is in $C(K)$. – t.b. Aug 17 '12 at 18:20
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    The question still is unclear to me. Are you trying to show the existence of some Cauchy sequence in that space? If your mother tongue is spanish you can use it here: I and several others have that language as mother tongue. Puedes escribir en español: varios aquí lo hablamos y podemos ayudar en la traducción. – DonAntonio Aug 17 '12 at 18:38
  • @t.b. but if $f_n$ is not a sequence Cauchy then $|f_n(x) - f_m(x)| \leq |f_n - f_m|_\infty$ yet is true, Why I have taking $f_n$ as Cauchy sequence? – juaninf Aug 17 '12 at 18:39
  • Of course it is still true, but the conclusion "so $(f_n(x))_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$." no longer holds. – t.b. Aug 17 '12 at 18:40
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    Related: Is the space $C[0,1]$ complete? – Martin Sleziak Aug 17 '12 at 18:44
  • @t.b. Why ...?, – juaninf Aug 17 '12 at 18:47
  • @t.b. is $\sup_{x \in K} |f(x) - f_n(x)| \xrightarrow{n\to\infty} 0$ or $\sup_{x \in K} |f - f_n| \xrightarrow{n\to\infty} 0$? – juaninf Aug 17 '12 at 19:07
  • @Matt the norm is ${\Vert \cdot \Vert}{\infty}=\sup{x \in K}|f(x)|$, i don't understand why you write the norm $|\cdot|_\infty$ is independent of $x$. – juaninf Aug 17 '12 at 19:17
  • @Juan I meant that $|f|\infty$ is an expression independent of $x$ and meant to say that that is the reason why we write $|f|\infty$ instead of $|f(x)|_\infty$. But then I thought my comment could be more confusing than helpful and also that I was interrupting your conversation with tb so I deleted my comment. – Rudy the Reindeer Aug 17 '12 at 19:21

1 Answers1

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To have an answer, I'll expand on the comments.

You claim that $C(K)$ is complete with respect to $\|\cdot\|_\infty$. What this means is that every Cauchy sequence $f_k$ has its limit in $C(K)$.

So let $f_k$ be a Cauchy sequence in $C(K)$, that is, for $\varepsilon > 0$ there is $N$ such that for $n,m > N$ you have $\|f_n - f_m \|_\infty < \varepsilon$. You want to show that there is an $f$ in $C(K)$ such that $\|f-f_k\|_\infty \to 0$. Your guess for $f$ is that $f$ is the pointwise limit of $f_k$. For this observe first that the pointwise limit exists: for every $x \in K$ fixed, $f_k (x)$ is a Cauchy sequence in $\mathbb R$, $\mathbb R $ is complete hence its limit exists.

Now that we have a candidate for the limit let's denote it by $f(x) = \lim_{n \to \infty} f_n (x)$.

To finish the proof you need to show that $f$ is in $C(K)$ that is, that $f$ is continuous, and also that $f_k$ converges to $f$ in norm.

If you show that $f_k$ converge to $f$ in norm you will get continuity for free by the uniform limit theorem. So let's show that $f_k \to f$ in $\|\cdot\|_\infty$:

Let $N$ be such that for $k \geq N$ you have $\|f_k - f_N \|_\infty < \varepsilon / 2$ (by Cauchyness of $f_k$). Then $$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)|$$

The second term is $< \varepsilon / 2$. For the first term observe that $$ |f(x) - f_N (x)| = |\lim_{n \to \infty} f_n (x) - f_N(x)|$$

So $|f(x) - f_N (x)| \leq \varepsilon /2$, so that $$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)| < \varepsilon$$

Rudy the Reindeer
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