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Let $C[0,1]$ be the set of continuous real valued functions on $C[0,1]$. Show that $(C[0,1],\rho_\infty)$ is complete. Is $(C[0,1],\rho_1)$ complete? Justify your answer. Here,

$\rho_\infty(f,g)=\sup_{x\epsilon[0,1]}|f(x)-g(x)|$

$\rho_1(f,g)=\int_0^1|f-g|dx$

Any help is much appreciated.

Martin Sleziak
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pseudogeek
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2 Answers2

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Let $K=[0,1]$. Suppose $\langle f_n\rangle$ is a Cauchy sequence. This means that for each $\varepsilon >0$ there exist $N$ such that whenever $n,m>N$, $|f_n(x)-f_m(x)|<\varepsilon$ for each $x\in K$. In particular, for each fixed $x\in K$ $$x_n=f_n(x)$$

is Cauchy... so?

Pedro
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  • if $<f_n>$ is Cauchy doesn't it mean that sup|f(x)-g(x)| is less than epsilon? – pseudogeek Sep 24 '13 at 04:34
  • @pseudogeek You're mixing things up, it seems. We have a sequence $f_n$ of elements of $C[0,1]$. They are Cauchy iff $d(f_m,f_n)<\epsilon$ provided we take $n,m$ sufficiently large, which is what I wrote above. – Pedro Sep 24 '13 at 04:36
  • Don't we have to take the distance with respect to the metric that we have? – pseudogeek Sep 24 '13 at 04:39
  • @pseudogeek Yes, and our metric is the $\sup$ norm, yes? But what I said is the same as saying $\sup |f_n-f_m|\leqslant \varepsilon $ – Pedro Sep 24 '13 at 04:40
  • ohhh...right. So it converges to a point. I think I can take it from here. Thanks! – pseudogeek Sep 24 '13 at 04:43
  • okay I'm confused again. Why does $sup|f_n-f_m|<\varepsilon$ imply $|f_n-f_m|<\varepsilon$? – pseudogeek Sep 24 '13 at 05:21
  • @pseudogeek I mean to say that $|f_n(x)-f_m(x)|<\epsilon$ for each $x\in K$ means $\sup_{x\in K}|f_n(x)-f_m(x)|<\epsilon$. – Pedro Sep 24 '13 at 05:27
  • okay. now I got it. – pseudogeek Sep 24 '13 at 06:51
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okay I'm confused again. If $<f_n>$ is a Cauchy sequence, for each $\varepsilon>0$ there exists $n_\varepsilon$ such that $sup|f_n(x)-f_m(y)|<\varepsilon$ for each $m,n>n_\varepsilon$. where do I go from this? Why does it mean that $|f_n(x)-f_m(y)|<\varepsilon$ for each $x \ \epsilon \,K$?

pseudogeek
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