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We know that the Fourier system is complete, i.e. that $\lbrace e_n: ~ n \in \mathbb{N} \rbrace$ defined by \begin{equation} e_n(x)=\frac{1}{\sqrt{2 \pi}}\exp(inx), ~~~ n \in \mathbb{Z} \end{equation} is an orthonormal basis of $L^2(-\pi,\pi)$.

My question is:

How do you proof, that \begin{equation} e_n(x)=\frac{1}{\sqrt{2N}}\exp\left( \frac{i \pi n}{N}x \right), ~~~ n \in \mathbb{Z} \end{equation} defines an orthonormal basis on $L^2(-N,N)$ (if it does)?

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2 Answers 2

up vote 2 down vote accepted

If your first formula is correct, your second can't be. Taking $N=\pi$ you don't have normed vectors. After scaling them the proof should be the same with a little substituion in the Integrals. Edit: you corrected the formel, now just using the subsitution rules for Integrals will bring you the result.

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Thank you. So I will have to do the proof again. There is no way to proof this when the basis for $L^2(-\pi,\pi)$ is already known? –  mjb Feb 15 '13 at 13:51
    
With the subsitution in the Integral, you have the orthonormality, and in addition the basis. –  Dominic Michaelis Feb 15 '13 at 13:54

In general, for a proposed family $\{e_n(x)\}$, to determine if it is orthonormal, just check whether $$\langle e_n(x),e_m(x)\rangle=\begin{cases}1, &n=m,\\ 0, &n\not=m,\end{cases}$$ where here the inner product is $$\langle u,v\rangle=\int_a^b u(x)\overline{v(x)}\,dx.$$

To show that an orthonormal family forms an orthonormal basis, you need to show that the family is complete. Two criteria which are equivalent in your setting are:

  1. $\langle f,e_n\rangle=0$ for all $n$ implies $f$ is the zero function.
  2. Parseval's equation is true for all $f\in L^2[a,b]$: $\|f\|^2=\sum_n c_n^2$ where $c_n$'s are the coefficients in the expansion of $f$ in this orthogonal family.
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Yes, but this is just a proof for orthonormality and not for the basis-assumption –  mjb Feb 15 '13 at 13:50
    
Thank you! But this how do I actually do that? I know the equivalent characterizations of bases, but which one to use and how? –  mjb Feb 20 '13 at 6:50

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