Inner Product vs. Integrals with Fourier Series, When to include 1/2pi?

Question

I am confused about when to include a prefactor of $\frac{1}{2\pi}$ when dealing with integrals of functions that are expressed as fourier series. This is what I understand (please correct me if I'm wrong). Assume I have a square integrable function, $f\left(x\right)$. I can express it as a fourier series according to:

$$f\left(x\right)=\sum_{k=-\infty}^{\infty} c_k e^{ikx}$$

Where the coefficients, $c_k$, are obtained via the following inner-product:

$$c_k=\frac{1}{2\pi}\int_{0}^{2\pi} e^{-ikx} f\left(x\right)\text{d}x$$

My understanding of the standard inner-product between functions is, that it is defined by:

$$\left\langle h\left(x\right),g\left(x\right)\right\rangle \equiv \int_{\Omega} h^*\left(x\right) g\left(x\right) \text{d}\Omega$$

Where the $^*$ denotes complex conjugation, $\Omega$ is the domain, and $\text{d}\Omega$ is the invariant measure.

So, using this definition, I would intuitively think that the coefficients for a fourier series ought to be computed from:

$$c_k=\int_{0}^{2\pi} e^{-ikx} f\left(x\right)\text{d}x$$

i.e. without the prefactor of $\frac{1}{2\pi}$.

Question 1: So, why is the prefactor of $\frac{1}{2\pi}$ included in the special case when the basis functions are complex exponentials, i.e. for Fourier series'? I know that the complex exponentials are not orthonormal, only orthogonal, but why should the normalization convention of the basis functions change the definition of the inner-product? I don't think it comes from the invariant measure, since for a circle the invariant measure should be $\text{d}x$, not $\frac{1}{2\pi}\text{d}x$

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Now, say that I have an equation that is defined using an integral whose arguments include functions expressed as Fourier series. Specifically:

$$\bar{S} = \int_{0}^{2\pi} f\left(x\right)S\left(x\right)\text{d}x$$

Where $\bar{S}$ represents the average value of $S\left(x\right)$ weighted by the probability density function $f\left(x\right)$. If I express $f\left(x\right)$ and $S\left(x\right)$ as Fourier series, and use the fact that $f\left(x\right)=f^*\left(x\right)$, then the equation should be:

$$\bar{S} = \sum_{k=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} c_k^* s_m\int_{0}^{2\pi} e^{-ikx} e^{imx} \text{d}x=2\pi \sum_{k=-\infty}^{\infty} c_k^* s_k$$

Question 2: In this case the integral amounts to doing an inner-product but without the prefactor of $\frac{1}{2\pi}$. So, when doing integrals of functions that are expressed as Fourier series do you need to always throw in that prefactor or not? When should you?

Answer 1

It depends on the inner product and the basis used.

Suppose you have an inner product $\langle \cdot, \cdot \rangle$ and an orthogonal (Schauder) basis $b_k$ for a Hilbert space $\mathbb{H}$.

If you represent some $f \in \mathbb{H}$ in terms of the basis, you will have $f = \sum_k \hat{f_k} b_k$, where $\hat{f_k}$ are the Fourier coefficients with respect to the basis $b_k$.

Then you have no choice over the $\hat{f_k}$, as they must satisfy $\langle b_k, f \rangle = \hat{f_k} \|b_k\|^2$, or $\hat{f_k} = \frac{\langle b_k, f \rangle}{\|b_k\|^2}$.

To answer your first question. Presumably you are dealing with $L^2[0,2 \pi]$, with $\langle f, g \rangle = \int_0^{2 \pi} \overline{f}(t)g(t) dt$, and the basis $b_k(t) = e^{i k t}$. Since $\|b_k\|^2 =\langle b_k, b_k \rangle = 2 \pi$, the coefficients are given by $\hat{f_k} = \frac{\langle b_k, f \rangle}{\|b_k\|^2} = \frac{1}{2\pi} \int_0^{2 \pi} e^{-ikt} f(t) dt$, whence the $2 \pi$.

For your second question, you have $\overline{S} = \int_{0}^{2\pi} f(x) S(x) dx$. In terms of the inner product from the previous question, this becomes $\overline{S} = \langle f, S \rangle$.

If $\hat{f_k}, \hat{S_k}$ are the Fourier coefficients (with respect to the basis and inner product above), then an easy computation (Parseval's theorem) shows that $\langle f, S \rangle = \sum_k \|b_k\|^2 \overline{\hat{f_k}} \hat{S_k}$, and so we obtain the result above (remember $\|b_k\|^2 = 2 \pi$).

Answer 2

The normalization doesn't change the inner product convention; what we are doing is projecting $f(x)$ onto the subspace spanned by $e^{ikx}$.

For instance, in a 'regular' vector space, if we wanted to expand a vector $v$ in terms of an orthogonal (not necessarily orthonormal) basis $\{e_k\}$, then we would write

$$ v = \sum_k c_k e_k; \quad c_k = \frac{\left<v,e_k\right>}{||e_k||^2} $$

In the same way, when we wish to expand a function in it's Fourier series, we have

$$ c_k = \frac{\left<f,e^{ikx}\right>}{||e^{ikx}||^2} = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ikx} dx. $$

If we wanted to, we could first normalize our basis functions, taking $e_k = \dfrac{e^{ikx}}{||e^{ikx}||} = \dfrac{1}{\sqrt{2\pi}} e^{ikx}$ so that $$ c_k = \frac{\left<f,e_k\right>}{||e_k||^2} = \left<f,e_k\right>= \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} f(x) e^{-ikx} dx. $$ but then when we write our function in terms of the Fourier basis $$ f = \sum_k c_k e_k = \sum_k c_k \frac{1}{\sqrt{2\pi}}e^{ikx} = \sum_k \left(\frac{1}{\sqrt{2\pi}}c_k\right) e^{ikx} $$ the missing factor of $\dfrac{1}{\sqrt{2\pi}}$ comes back anyway.

Your second question is essentially about Parseval's theorem. Again, it is a matter of applying the weight $2\pi$ somewhere to properly normalize everything. If you took the basis to be $\dfrac{1}{\sqrt{2\pi}} e^{ikx}$ so that $c_k = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} f(x) e^{-ikx} dx$ (and similarly for $s_k$). Then you would have

$$ \bar S = \sum_k c_k^* s_k. $$

All in all, this is a reflection of the fact that the Fourier series isn't quite unitary. You either have to (somewhat awkwardly) normalize the basis functions, absorb the extra $2\pi$'s into the measure, or just live with 'basically unitary' and accept (and then subsequently forget about) the $2\pi$'s.