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I am so new to functional analysis so I am looking for an answer of a confusion I am having right now in my mind because I have seen many different answers for the question I am gonna ask below. I hope you will reply as simple as possible because I am not a Mathematician. Thanks for the help in advance...

Is the space of continuous functions $C^{0}$ a Cauchy complete? Therefore is it a Hilbert space or not?

There's a thesis online, which says that this space is not Cauchy complete and is therefore not a Hilbert space. $L^2$ square integrable functions space is the Cauchy completion of the function space $C^0$ and in other words, contnuous functions on domain $X$ are dense in $L^{2}(X)$.

However, I run into some documents which support that the space of continuous functions is a Cauchy complete.

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  • $\begingroup$ "Some documents" may be talking about a metric other than "uniform convergence" ... Or they may be talking about continuous functions defined on some space other than a compact interval. Who knows? $\endgroup$ – GEdgar Mar 24 '14 at 13:57
  • $\begingroup$ I think I understand you. $d_{\infty }(f,g)=max|f(x)-g(x)|$ is one of the metrics, with this metric, $C^{0}[a,b]$ is complete. But, with these two other metrics, $d_{1}(f,g)=\int_{a}^{b} |f(x)-g(x)|dx$ and $d_{2}(f,g)=\int_{a}^{b} (f(x)-g(x))^{2}dx , $C[a,b]$ is not complete with d1 (or d2) as the metric. This is what you say right? $\endgroup$ – ARAT Mar 25 '14 at 14:57
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This question has been addressed ad infinitum.

For proof that $C^0$ is complete with the supremum norm (hence is a Banach space) see here: How to show that $C=C[0,1]$ is a Banach space

For proof that the supremum norm arises from no inner product (and hence is not a Hilbert space) see here: $C[0,1]$ is not Hilbert space

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    $\begingroup$ but this says that the space $C[0,1]$ of continuous functions from $[0,1]$, it's a closed interval, right? What about continuous functions defined on other closed intervals? By the way, I am not asking for any proof. It's just a matter of a confusion. $\endgroup$ – ARAT Mar 24 '14 at 13:39
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    $\begingroup$ $[0,1]$ is homeomorphic to any other interval $[a,b]$ with $a<b$. They are topologically the same. Therefore, so are $C[0,1]$ and $C[a,b]$. If you dont want proof, the short answer to your question is: it's Banach, not Hilbert. $\endgroup$ – Frank Mar 24 '14 at 13:47
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    $\begingroup$ Does it change if you are looking at an open interval instead of a closed interval? $\endgroup$ – ruler501 Mar 24 '14 at 13:55
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    $\begingroup$ Yes. Then (0,1) is no longer compact and the result doesn't hold. The more general result is: if $K \subset \mathbb{R}$ is compact, then $C(K)$ is a Banach space with the supremum norm. $\endgroup$ – Frank Mar 24 '14 at 13:56
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    $\begingroup$ I use the term 'complete' for what you call 'Cauchy complete', i.e. a metric space in which any Cauchy sequence also converges. If we now have a normed vector space which happens to be complete, we call it a Banach space. If we have an inner product space which happens to be complete, we call it a Hilbert space. Let $C$ stand for $C[a,b]$ with the supremum norm. Then this space is complete. It is thus a complete normed vector space, i.e. a Banach space. However, there exists no inner product whose corresponding norm coincides with the supremum norm. $C$ is therefore not Hilbert. $\endgroup$ – Frank Mar 25 '14 at 15:31

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