# An orthonormal family in an inner product space

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]. – Alex R. Apr 6 '12 at 23:33
Sorry, It makes sense with the 2-norm – rk101 Apr 6 '12 at 23:43
Your family should probably be $(e^{2\pi i n x})_{n \in \mathbb{Z}}$ – t.b. Apr 6 '12 at 23:46
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.. – rk101 Apr 6 '12 at 23:56
Nooo! I edited the title but introduced another typo... – Tyler Apr 7 '12 at 1:08

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.
I mean that we cannot find a $x$ such that $f(x)=0$ for all $f\in A$, where $A$ is the considered algebra. – Davide Giraudo Mar 24 '13 at 19:29