Context: For a $2\pi$-periodic bounded function $f:\mathbb{R}\to\mathbb{C}$, we define the complex Fourier coefficients of $f$ by
$$
\hat{f_k}:=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-ikx}\,dx.
$$
We call Fourier series of $f$ the formal series
$$
\sum_{k=-\infty}^{\infty}\hat{f_k}e^{ikx}.\tag{1}
$$
Now, it is easily shown that
$$
\overline{\hat{f_k}}=\hat{\overline{f}}_{-k}.
$$
Hence if $f$ is real-valued then
$$
\overline{\hat{f_k}}=\hat{f}_{-k}.\tag{2}
$$
Reading some notes, it is said that if $f$ is a real-valued function, then we can write
$$
\hat{f_k}=\frac{1}{2\pi}\int_0^{2\pi}f(x)\cos(kx)\,dx-i\frac{1}{2\pi}\int_0^{2\pi}f(x)\sin(kx)\,dx.\tag{3}
$$
Putting
$$
a_k:=\frac{1}{\pi}\int_0^{2\pi}f(x)\cos(kx)\,dx,\\
b_k:=\frac{1}{\pi}\int_0^{2\pi}f(x)\sin(kx)\,dx,
$$
where $a_k,b_k$ are called the real Fourier coefficients of $f$, it is said that in view of $(2)$, we have the relations
\begin{align}
\hat{f_0}&=\frac{1}{2}a_0,\\
\hat{f_k}&=\frac{1}{2}(a_k-ib_k),\\
a_k&=\hat{f_k}+\hat{f}_{-k},\tag{4}\\
b_k&=i(\hat{f_k}-\hat{f}_{-k}),\\
a_{-k}&=a_k,\\
b_{-k}&=-b_k.
\end{align}
Finally it is said that if, once again, $f$ is real-valued, then we can write $(1)$ as the trigonometric series
$$
\frac{1}{2}a_0+\sum_{k=1}^{\infty}a_k\cos(kx)+b_k\sin(kx).\tag{5}
$$
Question: Is it really essential that $f$ be real-valued for $(3)$, $(4)$ and $(5)$ to hold? It seems to me that everything holds even if $f$ is complex-valued…
[Math] Fourier Series of Real-valued Functions
fourier analysisfourier series
Related Solutions
Yes! And these types of expansions occur in a variety of applications, e.g., solving the heat or wave equation on a rectangle with prescribed boundary and initial data.
As a specific example, we can think of the following expansion as a two dimensional Fourier sine series for $f(x,y)$ on $0<x<a$, $0<y<b$: $$ f(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty c_{nm}\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right), \quad 0<x<a,\ 0<y<b, $$ where the coefficients (obtained from the same type of orthogonality argument as in the 1D case) are given by \begin{align} c_{nm}&={\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy\over \int_0^b \int_0^a \sin^2\left({n\pi\, x\over a}\right)\sin^2\left({m\pi\, y\over b}\right)\,dx\,dy}\\ &={4\over a b}\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy, \quad n,m=1,2,3,\dots \end{align}
For example, the picture below shows (left) the surface $$f(x,y)=30x y^2 (1-x)(1-y)\cos(10x)\cos(10y), \quad 0<x<1,\ 0<y<1,$$ and a plot of the two dimensional Fourier sine series (right) of $f(x,y)$ for $n,m,=1,\dots,5$:
Finally, keep in mind that we are not limited just to double sums of the form sine-sine. We could have any combination we like so long as they form a complete orthogonal family on the domain under discussion.
A $1$-periodic function, discretized (sampled) at step $h=1/n$, is a collection of $n$ values $y_0,\dots,y_{n-1}$. The functions $f_k = \exp(2\pi i k/n)$, $k=0,\dots,n-1$, form an orthogonal basis among such functions, so there is an expansion $f=\sum c_k f_k$ where
$$
c_k = \frac{1}{n} \sum_{\ell} y_\ell \exp(-2\pi i k\ell/n) \tag{1}
$$
Except for $1/n$, this computation amounts to the multiplication of vector $y$ by the Fourier matrix
$W$ with the entries $\exp(-2\pi i kj)$ (if we index the rows and columns starting with 0). The Discrete Fourier transform is this multiplication (algorithmically, fft arranges it in a clever way, reusing previous intermediate results to speed it up.)
However, one should not jump to the conclusion that $$ y = \sum_{k=0}^{n-1} c_k \exp(2\pi i k t) \tag{2} $$ is the right way to interpolate the sampled values. Although the function on the right does interpolate $y_0,\dots, y_{n-1}$, it has unnatural oscillations in between. One should consider the aliasing of frequencies and fold them appropriately.
For example, take $n=5$, so that (2) says $$ c_0+ c_1e^{2\pi i t} + c_2e^{4\pi i t} + c_3e^{6\pi i t} + c_4e^{8\pi i t} \tag{3} $$ The aliasing of frequencies is the fact that $6 \pi t \equiv -4 \pi t \bmod 2\pi$ when $t$ is a grid point (one of $k/5$ points). Similarly, $8\pi t \equiv -2\pi t \bmod 2\pi $. So, we can get smoother curve by folding the frequencies, replacing (3) with $$ c_0+ c_1e^{2\pi i t} + c_2e^{4\pi i t} + c_3e^{ - 4\pi i t} + c_4e^{-2\pi i t} \tag{4} $$ The equation (4) is the correct interpolant, and it can be turned into sine-cosine Fourier series (rather, its partial sum) using Euler's formula.
Note that for real functions, the vector of coefficients is conjugate-symmetric: $c_{n-k}=\overline {c_k}$. So one can take the real part of the terms with $c_1, c_2$ and double them; this accounts for $c_3, c_4$. Separate treatment is required for the constant term $c_0$ (which has no counterpart) and, when $n$ is even, for the Nyquist frequency $k=n/2$, which also has no counterpart.
Best Answer
Yes they hold assuming $a_k,b_k$ are complex, but the point is that they are real valued and thus by using (2) we can show that $$ \hat f_k e^{ikx} + \hat f_{-k} e^{-ikx} = \hat f_k e^{ikx} + (\hat f_k e^{ikx})^* = 2 \mathrm{Re} \{ f_k e^{ikx} \} = 2 \mathrm{Re} \Big\{ \frac{1}{2} (a_k-ib_k)(\cos(kx)+ i\sin(kx)) \Big\} = a_k\cos(kx) + b_k \sin(kx) $$
We can also show that same thing holds if $a_k,b_k$ are complex:
\begin{align} \hat f_k e^{ikx} + \hat f_{-k} e^{-ikx} &= \frac{1}{2}(a_k - ib_k) e^{ikx} + \frac{1}{2}(a_{-k} - ib_{-k}) e^{-ikx}\\&= \frac{1}{2}(a_k - ib_k) e^{ikx} + \frac{1}{2}(a_{k} + ib_{k}) e^{-ikx} \\&= \frac{1}{2} a_k (e^{ikx}+e^{-ikx}) - \frac{1}{2}i b_{k}( e^{ikx} - e^{-ikx}) \\&= a_k \cos(kx) + b_{k}\sin(kx) \end{align}