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I am currently working through some lecture notes on the Geometry of Hilbert spaces, and I am stuck with the following comment:

If we are given the inner product space $C([0,1])$ of continuous functions $f : [0,1] \to \mathbb{F}$, where $\mathbb{F} = \mathbb{C}$ or $\mathbb{R}$ with the inner product \begin{equation} (f,g) = \int^1_0 f(t)\overline{g(t)} \, dt \quad f,g, \in C([0,1]) \end{equation} then the set \begin{equation}\{ e^{i2\pi nt} \}_{n \in \mathbb{Z}} \end{equation} forms a complete orthonormal set, that is, \begin{equation} (f,e^{i2\pi nt}) = 0 \quad \forall n \in \mathbb{Z} \quad \Rightarrow \quad f \equiv0 \end{equation} I tried to find an argument on my own, but somehow I get stuck each time, I try to estimate the magnitude of $f$ and derive it is zer0, but each time I get caught up in a double integral which I am sure is wrong. The proofs in notes that I found on the web all prove statements that are equivalent to the one above - I would really like to understand how to derive the completeness using only the above definition - if somebody could give me a small hint how to go about this that would be great, many thanks !

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$C[0,1]$ is not complete for the norm given by the inner product, so it's not a Hilbert space. If you know Stone-Weierstass theorem, you can show that the vector space generated by the $f_n$, $f_n(t)=e^{i2\pi nt}$, is dense for the supremum norm. –  Davide Giraudo Apr 13 '12 at 17:30
    
@DavideGiraudo: my fault, it says inner product space in the notes, sorry for this ! I shall look up the Stone Weierstrass theorem now, thanks for your comment ! –  harlekin Apr 13 '12 at 17:34

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The key is that we can approximate a continuous function by a sequence of elements of the set $$\mathcal A:=\{\sum_{k=-n}^na_ke^{2i\pi kx},n\in\Bbb N, a_k\in\Bbb C\},$$ with the uniform norm. Indeed, by Stone-Weierstrass theorem, since $\mathcal A$ is an algebra of continuous function on $[0,1]$, non-vanishing at any point, and separates the points, it's dense in $C[0,1]$ endowed with the supremum norm.

We apply this to our problem: We have $(f,P)=0$ for each element $P$ of $\mathcal A$. Take $\{P_n\}\subset\mathcal A$ such that $\sup_{0\leq t\leq 1}|f(t)-P_n(t)|\leq n^{—1}$. Then for each integer $n$, $$|(f,f)|\leq \lVert f\rVert_{\infty}n^{-1}$$ which gives that $\int_0^1f(t)^2dt=0$. This implies that $f\equiv 0$.

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