Derivation of Complex Fourier Series coefficients through inner products (and swapping arugments)

eigenfunctionsfourier seriesfourier transforminner-productssturm-liouville

I am trying to derive the complex Fourier series coefficients given by:$$f(x)=\sum_{n=-\infty}^{\infty}{c_n}e^{inx}$$ with coefficients:$$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}{f(x)}e^{-inx}dx$$

I am trying to use the inner product defined as:
$$\lt f,g\gt\,\,=\int_a^b{f(x)\overline{g(x)}dx}$$
to find this. I know that $\lt f,g\gt \neq \lt g,f\gt$ if $f$ and $g$ are complex. I have no problem finding generalized Fourier series when my eigenfunctions are real like in trigonometric Fourier series. I am trying to apply to complex Fourier series the approach I learned from Sturm-Liouville theory where I wrote functions as an eigenfunction expansion.

I often see that from this inner product is used with the arguments in this order for the definition of the complex Fourier coefficient: $$c_n=\frac{\lt f,e^{inx} \gt}{\lt e^{inx},e^{inx} \gt}=\frac{1}{2\pi}\int_{-\pi}^{\pi}{f(x)}e^{-inx}dx$$.

It seems that you can instead take: $$a_n=\frac{\lt e^{inx},f \gt}{\lt e^{inx},e^{inx} \gt}=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{inx}\overline{f(x)}dx$$ (the inner product arguments flipped) and still create a "valid" complex fourier series. The choice to order the inner product with the exponential term being conjugated due to the inner product seems arbitrary.

Using the new definition for $c_n$ above (as $a_n$), it seems that this a valid series for $f$:

$$f(x)=\sum_{n=-\infty}^{\infty}{a_n}e^{-inx}$$

with coefficients:$$a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{inx}\overline{f(x)}dx$$

This appears to just flip the series upside down.

Now to my question: It seems arbitrary to me that $c_n$ corresponds to $e^{inx}$ whereas $a_n$ corresponds to $e^{-inx}$. Why can't it be the other way around? Why is the correspondence between $c_n$ and $e^{inx}$ drawn in the first place?

In essence, why would the series:$$f(x)=\sum_{n=-\infty}^{\infty}{a_n}e^{inx}$$

with coefficients:$$a_n=\frac{\lt e^{inx},f \gt}{\lt e^{inx},e^{inx} \gt}=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{inx}\overline{f(x)}dx$$

be invalid?

Best Answer

Note that the functions $\{e_n(x) := e^{inx}\}_{n\in\mathbb{Z}}$ form an orthogonal basis of $L^2([-\pi,\pi])$. In consequence, the Fourier series of a function $f$ is nothing else than the representation of $f$ in the said basis, i.e. $f(x) = \displaystyle \sum_{n\in\mathbb{Z}} c_n e_n(x)$, where the coefficients $c_n$ play the role of its components, which are themselves given by the orthogonal projection of $f$ on $e_n$. In other words, if $P_n := \frac{e_ne_n^\dagger}{e_n^\dagger e_n}$ represents the associated orthogonal projector, then one gets $$ c_n = \frac{e_n^\dagger f}{e_n^\dagger e_n} = \frac{\langle e_n,f \rangle}{\langle e_n,e_n \rangle} = \frac{1}{2\pi} \int_{-\pi}^\pi e_n(x)^*f(x) \,\mathrm{d}x = \frac{1}{2\pi} \int_{-\pi}^\pi f(x)\,e^{-inx} \,\mathrm{d}x, $$ and not the reverse formula you proposed.