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This is Parseval's theorem from book PMA Rudin.

Let $\{\phi_n(x)\}$ is orthonormal system on $[a,b]$. Rudin defines Fourier coefficients as $c_n=\int_{a}^{b}f(t)\overline{\phi_n(t)}dt$. And to to every function we can correspond it's Fourier series: $f\sim \sum \limits_{n}c_n\phi_n(x)$.

Using that definition we get: System $\left\{\dfrac{e^{inx}}{\sqrt{2\pi}}\right\}_{n\in \mathbb{Z}}$ is orthonormal on $[-\pi,\pi]$ and Fourier coefficient in this case is $\int_{a}^{b}f(t)\overline{\phi_n(t)}dt=\frac{1}{\sqrt{2\pi}}\int_{a}^{b}f(t)e^{-int}dt$ and Fourier series in this case has form: $$\sum \limits_{n=-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\int_{a}^{b}f(t)e^{-int}dt\right)\dfrac{e^{inx}}{\sqrt{2\pi}}=\sum \limits_{n=-\infty}^{\infty}\underbrace{\left(\frac{1}{2\pi}\int_{a}^{b}f(t)e^{-int}dt\right)}_{c_n}e^{inx}$$

I think that above in picture where I marked with red line coefficient $c_n$ is equal to $\frac{1}{2\pi}\int_{a}^{b}f(t)e^{-int}dt$. Am I right?

Can anyone please help with this question?

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    $\begingroup$ yes you are right en.wikipedia.org/wiki/Fourier_series $\endgroup$ – reuns Jan 18 '16 at 6:24
  • $\begingroup$ @user1952009, Thanks! Сuriously but one my friend who also interests in mathematics said that coefficient there must be $\frac{1}{\sqrt{2\pi}}$ but I sure that he is wrong $\endgroup$ – ZFR Jan 18 '16 at 6:37
  • $\begingroup$ if $f(x+T) = f(x)$ and defining $c_n = \frac{1}{T} \int_0^T f(x) e^{-2 i \pi n x/T} dx$ then $f(x) = \sum_{n =-\infty}^\infty c_n e^{2 i \pi n x/T}$ if the last series converges (see for the convergence en.wikipedia.org/wiki/Dirichlet_kernel ) $\implies$ you split the normalization term $\frac{1}{T}$ as you want but for the Parseval theorem be careful to use the $\frac{\displaystyle e^{2 i \pi n x / T}}{\sqrt{T}}$ orthogonal basis $\endgroup$ – reuns Jan 18 '16 at 6:45

You're off on a small but important point. What Rudin is referring to as $c_n$ is indeed $$ c_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-int}~dt, $$ and with this definition we have $$ f(x) \sim \sum_{n=-\infty}^\infty c_ne^{inx}. $$ However, this definition of the Fourier coefficients depends on which particular inner product you use, and at this point Rudin is using one that isn't what you're thinking of. Typically when one refers to the square-integrable functions on $[-\pi,\pi]$, the inner product of choice is $$ \langle f,g\rangle = \int_{-\pi}^\pi f(t)\overline{g(t)}~dt. $$ Then the definition $$ \tilde{c}_n = \frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi f(t)e^{-int}~dt $$ and the series expression $$ f(x) \sim \sum_{n=-\infty}^\infty \tilde{c}_n\phi_n(x) $$ where $\phi_n(x) = e^{inx}/\sqrt{2\pi}$ is more appropriate, because the $\phi_n$ are orthonormal in this inner product. Then defining $\tilde{\gamma}_n$ similarly, Parseval reads $$ \int_{-\pi}^\pi f(t)\tilde{g(t)}~dt = \sum_{n=-\infty}^\infty \tilde{c}_n\overline{\tilde{\gamma}_n}, $$ $$ \int_{-\pi}^\pi |f(t)|^2~dt = \sum_{n=-\infty}^\infty |\tilde{c}_n|^2. $$ In particular notice how the factors of $1/2\pi$ are not there compared to Rudin's formulas.

Where did they go? Well, since Rudin wants a Fourier series expansion in the form $$ f(x) \sim \sum_{n=-\infty}^\infty c_ne^{inx}, $$ if he wants his version of Parseval to work then he needs to choose an inner product so that $\{e^{inx}\}$ is orthonormal. The way to do this is to define $$ (f,g) = \frac{1}{2\pi}\int_{-\pi}^\pi f(t)\overline{g(t)}~dt. $$ In this setting, the $c_n$ are indeed the Fourier coefficients with respect to the (now) orthonormal basis $\{e^{inx}\}$ and the inner product $(\cdot,\cdot)$. And Parseval reads $$ \frac{1}{2\pi}\int_{-\pi}^\pi f(t)\overline{g(t)}~dt = \sum_{n=-\infty}^\infty c_n\overline{\gamma}_n, $$ $$ \frac{1}{2\pi}\int_{-\pi}^\pi |f(t)|^2~dt = \sum_{n=-\infty}^\infty |c_n|^2. $$ So we've recovered Rudin's version of Parseval, but we had to change our inner product to accommodate.

Addendum: Of course, the formulas are equivalent to each other once you write them out, but proper representation is key here. This is basically equivalent to the idea that in a finite-dimensional vector space, the same vector can be represented in several different ways depending on the choice of basis. Rudin has made a change of basis by demanding that the $\{e^{inx}\}$ be orthonormal, and consequently the representation of a function $f$ as a sequence of Fourier coefficients has to change (namely, by a factor of $1/2\pi$ - it's a scaling change of variable).

  • $\begingroup$ +1. Very nice and extended answer! So Rudin defines inner product as $(f,g)=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt$. With this inner product system $\{e^{inx}\}$ is orthonormal and Fourier coefficients are defined by $c_n:=(f,e^{int})=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt$. Right? $\endgroup$ – ZFR Jan 18 '16 at 6:57
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    $\begingroup$ Yes, that's exactly right. $\endgroup$ – Gyu Eun Lee Jan 18 '16 at 6:59
  • $\begingroup$ Thank you very much for your help! I think that Fourier series are written better and detailed in Rudin's Real and Complex Analysis. $\endgroup$ – ZFR Jan 18 '16 at 7:00

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