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As seen by this question, $C([0,1])$ is not complete in the $L^p$ norm ($1 \le p < \infty$) because one can choose a Cauchy sequence $$f_n(x) := \begin{cases} 0 & \text{if $0 \le x \le \frac1{2}$}\\ n(x-\frac{1}{2}) & \text{if $\frac {1}{2} < x \le \frac {1}{2} + \frac {1}{n}$}\\ 1 & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $} \end{cases}$$ and find that, as $f_n \to f$, we have $$f(x)=\begin{cases} 0 &\text{if } 0 \le x < \frac 12 \\ 1 & \text{if } \frac 12 \le x \le 1 \end{cases},$$ which means $f \not\in C[0,1]$. Therefore $C[0,1]$ is not complete in the $L^p$ norm.

But what about the supremum norm $L^\infty$? I am not sure if $C([0,1])$ is complete in the $L^\infty$ norm, because would we not still have the counterexample of $f \not\in C[0,1]$?

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Cookie
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  • The $L^\infty$ norm is just the good old maximum norm on $C([0,1])$, it is complete in that norm. – Daniel Fischer Jun 16 '15 at 22:49
  • @DanielFischer I'm aware of that part; $|f|{L^\infty} = \max{x \in [0,1]} |f(x)|$. But why does the "counterexample" I posted not contradict the infinity norm, even though it shows that all the other $L^p$'s are not complete? – Cookie Jun 16 '15 at 22:51
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    The maximum norm is the norm of uniform convergence, the "counterexample" doesn't converge uniformly. – Daniel Fischer Jun 16 '15 at 22:52
  • Note that $| f_n-f|_\infty \not\to 0 $ as $n\to \infty $ – Alonso Delfín Jun 16 '15 at 22:57

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I think you're forgetting the fact that completeness requires Cauchy sequences. Specifically you need $\|f_n-f_m\|_\infty<\epsilon$ for all $n,m\geq N$. However, $\lim_{m\rightarrow\infty}\|f_n-f_m\|_\infty\rightarrow 1$ for any fixed $n$. So your example is not a Cauchy sequence in $L^\infty$. Double check that it is Cauchy in $L^p$.

Alex R.
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