Why does the inner product space $( C[0,1], \| \cdot \|_2$) have an orthonormal family $(e^{\color{red}{2\pi}inx})_{n\in \mathbb{N}}$ ?
An orthonormal family in an inner product space
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functional-analysis
hilbert-spaces
inner-product-space
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0This space has many collections of orthonormal elements. That is sort of the point of an inner product space. – 2012-04-07
1 Answers
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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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0I see: so if we add that the algebra is unital (which is the case here), we do not need this e$x$tra assumption. – 2013-03-24