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Why does the inner product space $( C[0,1], \| \cdot \|_2$) have an orthonormal family $(e^{\color{red}{2\pi}inx})_{n\in \mathbb{N}}$ ?

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    This question makes no sense. What is ∥⋅∥, is it the sup norm on continuous functions? Then what does orthonormality have to do with this? There's no inner product! Are you asking why the family is orthonormal under the usual integral inner product? They are on $[-\pi,\pi]$. For density, check that the family satisfies the criteria for the Stone-Weierstrass theorem for C[0,1].2012-04-06
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    Sorry, It makes sense with the 2-norm2012-04-06
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    Your family should probably be $(e^{2\pi i n x})_{n \in \mathbb{Z}}$2012-04-06
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    I think your right, I can see it working in this case, with the normal inner product defined on $\|\cdot\|_2$, it must be a typo in the lecture notes..2012-04-06
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    Nooo! I edited the title but introduced another typo...2012-04-07
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    @Tyler: fixed, no worries :)2012-04-07
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    This space has many collections of orthonormal elements. That is sort of the point of an inner product space.2012-04-07

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First, we have to consider the family $f_n(x):=e^{2i\pi nx}$; as written in the OP and pointed out in the comments, this won't be an orthonormal basis.

If $j\neq k$, using an anti-derivative of $e^{iax}$ for $a\neq 0$ and the periodicity of $x\mapsto e^{2\pi ix}$, we can see the family is orthogonal.

Furthermore, we can use Stone-Weierstrass theorem to see that the vector space $V$ generated by the family $\{f_n,n\in\Bbb Z\}$ is dense in $C[0,1]$ for the $L^2$ norm. Indeed, it's enough to do it for the uniform norm. We can see that $V$ is an algebra, which separates points, it's stable taking the conjugate and doesn't vanishes everywhere.

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    What do you mean by: doesn't vanish everywhere? I'm not used to such a requirement for Stone-Weierstrass, so I might have something to learn here.2013-03-24
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    I mean that we cannot find a $x$ such that $f(x)=0$ for all $f\in A$, where $A$ is the considered algebra.2013-03-24
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    I see: so if we add that the algebra is unital (which is the case here), we do not need this extra assumption.2013-03-24