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Let $J=[a,b] \subset \mathbb{R}$, how does one then see that if $C(J)=C(J,\mathbb{C}^n)$ with the norm $$||u||_J = \sum_{i=1}^{n} ||u_i||_\infty = \sum _{i=1} ^{n} \max _{x\in J} |u_i (x) |$$

is a Banach space?

What have I tried:

One has to show completeness by showing every cauchy sequence $(u_n)_\mathbb{N}$ converges in $C(J,\mathbb{C}^n)$. Let $(u_n)_n$ be a Cauchy sequence, with n,k>N : $$|u_n - u_k|< \epsilon $$

Can one use that J is compact so there is at least one Cauchy subsequence which converges, (and then use that to make an epsilon/2 proof)?

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Be careful not to use $n$ for your sequences as it is the dimension of $\mathbb{C}^n$. – 1015 Feb 3 '13 at 21:02
up vote 2 down vote accepted

I will assume that you know $C(J,\mathbb{C})$ is a Banach space when equipped with the uniform norm. Then $C(J,\mathbb{C}^n)$ is just a $\ell^1$ sum of $n$ copies of the latter. So it is a Banach space.

First you should prove that $\|u\|_J$ is indeed a norm. This is not difficult.

Now you need to show it is complete. Let $u^m=(u^m_1,\ldots,u^m_n)$ be a Cauchy sequence. Then observe that for all index $i$ $$ \max|u^m_i-u^k_i|\leq \|u^m-u^k\|_J. $$ So each component $(u^m_i)$ is Cauchy in $C(J,\mathbb{C})$. So each one converges to some $u_i$ uniformly.

It remains to check that $(u^m)$ converges to $u=(u_1,\ldots,u_n)$ for the norm under consideration.

Take $\epsilon>0$. For each $i$, find $N_i$ such that $\max|u_i^m-u_i|<\epsilon/n$ for all $m\geq N_i$. Set $N:=\max(N_1,\ldots,N_n)$. Then for all $m\geq N$, you have $$ \|u^m-u\|_J=\sum_{i=1}^n \max|u^m_i-u_i|< n\frac{\epsilon}{n}=\epsilon. $$

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Thank you, julien. – bakabakabaka Feb 3 '13 at 21:15
You're welcome, $\mbox{baka}^3$. – 1015 Feb 3 '13 at 21:17

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