TITLE: Fourier Analysis
QUESTION [7 upvotes]: I am interested in Fourier Analysis. But I don't get why the coefficients are choosen that way and why the Fourier series will converge to a given function?
Can someone provide me simple information or do you know an easy source for a beginner?
Regards,
Kevin

REPLY [3 votes]: It seems to me, that you are interested in the identity
\begin{equation}
f(x) = \sum_{k=-\infty}^\infty \frac{1}{2\pi} \int_{-\pi}^\pi f(t) e^{-ikt} dt e^{ikx} \ ,
\end{equation}
that is true for example if $f$ is $2\pi$-periodic and piecewise differentiable with finite number of pieces on the interval $[-\pi,\pi]$ and satisfies
\begin{equation}
f(x) = \frac{1}{2} (\lim_{t \rightarrow x^+} f(t) + \lim_{t \rightarrow x^-} f(t))
\end{equation}
for every $x \in \mathbb{R}$. To see this, we define
\begin{eqnarray}
f(x+) & = & \lim_{t \rightarrow x^+} f(t) \ , \\
f(x-) & = & \lim_{t \rightarrow x^-} f(t) \ .
\end{eqnarray}
We calculate for $x \neq 2\pi k$
\begin{eqnarray}
\sin(1/2 \ x) \sum_{k=-n}^n e^{ikx} & = & \sin(1/2 \ x) (e^{-ix})^n \frac{1-(e^{ix})^{2n+1}}{1-e^{ix}} = \sin((n+1/2)x) \ .
\end{eqnarray}
The left hand side of the first equation equals to the right hand side of the second equation also for $x = 2\pi k$. We define
\begin{eqnarray}
h_1(t) & = & \frac{f(x-t+)-f(x+)}{\sin(1/2 \ t)} \ , t \in (-\pi,0) \ , \\
h_2(t) & = & \frac{f(x-t-)-f(x-)}{\sin(1/2 \ t)} \ , t \in (0,\pi) \ .
\end{eqnarray}
Then
\begin{eqnarray}
\sin(1/2 \ t) h_1(t-) & = & f(x-t+)-f(x+) \ , t \in (-\pi,0] \ , \\
\sin(1/2 \ t) h_2(t+) & = & f(x-t-)-f(x-) \ , t \in [0,\pi) \ .
\end{eqnarray}
Note that $h_1(0-)$ equals to $-2$ times right derivative of $f(t+)$ at $x$ and $h_2(0+)$ $-2$ times left derivative of $f(t-)$ at $x$. This is obtained by dividing and multiplying the expression of $h_k$ by $1/2 \ t$ and taking limits when argument approaches to $0$ from left and right respectively. We calculate using the Riemann-Lebesgue lemma and obtain
\begin{eqnarray}
\sum_{k=-\infty}^\infty & \frac{1}{2\pi} & \int_{-\pi}^\pi f(t) e^{-ikt} dt e^{ikx} = \sum_{k=-\infty}^\infty \frac{1}{2\pi} \int_{x-\pi}^{x+\pi} f(t) e^{-ikt} dt e^{ikx} \\
& = & \lim_{n \rightarrow \infty} \sum_{k=-n}^n \frac{1}{2\pi} \int_{x-\pi}^{x+\pi} f(t) e^{-ikt} dt e^{ikx} = \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_{x-\pi}^{x+\pi} f(t) \sum_{k=-n}^n e^{ik(x-t)} dt \\
& = & \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_{-\pi}^\pi f(x-t) \sum_{k=-n}^n e^{ikt} dt \\
& = & \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_{-\pi}^0 f(x-t+) \sum_{k=-n}^n e^{ikt} dt + \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_0^\pi f(x-t-) \sum_{k=-n}^n e^{ikt} dt \\
& = & \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_{-\pi}^0 (f(x-t+) - f(x+)) \sum_{k=-n}^n e^{ikt} dt + \frac{1}{2} f(x+) \\
& & + \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_0^\pi (f(x-t-) - f(x-)) \sum_{k=-n}^n e^{ikt} dt +\frac{1}{2} f(x-) \\
& = & \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_{-\pi}^0 \sin(1/2 \ t) h_1(t-) \sum_{k=-n}^n e^{ikt} dt + \frac{1}{2} f(x+) \\
& & + \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_0^\pi \sin(1/2) h_2(t+) \sum_{k=-n}^n e^{ikt} dt +\frac{1}{2} f(x-) \\
& = & \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_{-\pi}^0 h_1(t-) \sin((n+1/2)t)  dt + \frac{1}{2} f(x+) \\
& & + \lim_{n \rightarrow \infty} \frac{1}{2\pi} \int_0^\pi h_2(t+) \sin((n+1/2)t) dt +\frac{1}{2} f(x-) \\
& = & \frac{1}{2} (f(x+) + f(x-)) = f(x) \ .
\end{eqnarray}
In the fifth equation integrands on the right hand side are changed to their right and left limits respectively. This doesn't influence to the value of the integral because the difference is nonzero on a finite set only. The function $h_1(t-)$ has a limit as $t \rightarrow 0^-$. Hence it is bounded in a boundary of $0$. The expression of $h_1$ is also bounded elsewhere. Similar reasoning shows that also $h_2$ is bounded. Hence they are absolutely integrable and we may apply the Riemann-Lebesgue lemma. This shows that Fourier series converges to the function, from which the coefficients were calculated.
Consider now two Fourier serieses of the same function $f$,
\begin{eqnarray}
f(x) & = & \sum_{k=-\infty}^\infty c_k^1 e^{ikx} \\
f(x) & = & \sum_{k=-\infty}^\infty c_k^2 e^{ikx} \ .
\end{eqnarray}
Sidewise substraction implies
\begin{equation}
0 = \sum_{k=0}^\infty (c_k^1-c_k^2) e^{ikx} \ .
\end{equation}
We also have
\begin{equation}
e^{inx} = \sum_{k=-\infty}^\infty \delta_{n,k} e^{ikx} \ .
\end{equation}
Parseval's theorem now implies
\begin{eqnarray}
\frac{1}{2\pi} \int_{-\pi}^\pi 0 \overline{e^{inx}} dx & = & \sum_{k=-\infty}^\infty (c_k^1-c_k^2)\overline{\delta_{n,k}} \\
0 & = & c_n^1-c_n^2 \ .
\end{eqnarray}
Hence $c_n^1=c_n^2$ for every $n \in \mathbb{Z}$. This shows that Fourier series is unique, that is there is only one way to choose the coefficients. It is
\begin{equation}
c_k = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) e^{-ikt} dt \ .
\end{equation}
Fourier inversion theorem is more difficult to prove.