Fourier's Theorem

Theorem
Let $\alpha \in \R$ be a real number.

Let $f \left({x}\right)$ be a real function which is defined and bounded on the interval $\left[{\alpha \,.\,.\, \alpha + 2 \pi}\right]$.

Let $f$ satisfy the Dirichlet conditions on $\left[{\alpha \,.\,.\, \alpha + 2 \pi}\right]$:

Outside the interval $\left[{\alpha \,.\,.\, \alpha + 2 \pi}\right]$, let $f$ be periodic and defined such that:
 * $f \left({x}\right) = f \left({x + 2 \pi}\right)$

Let $f$ be defined by the Fourier series:


 * $(1): \quad \displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \left({a_n \cos n x + b_n \sin n x}\right)$

such that:
 * $\displaystyle a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} f \left({x}\right) \cos n x \rd x$
 * $\displaystyle b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} f \left({x}\right) \sin n x \rd x$

Then for all $a \in \R$, $(1)$ converges to the sum:
 * $\displaystyle \frac 1 2 \left({\lim_{x \mathop \to a^+} f \left({x}\right) + \lim_{x \mathop \to a^-} f \left({x}\right)}\right)$

where the $\lim$ symbols denote the limit from the right and the limit from the left.

Main Theorem
Let $S_N \left({x}\right)$ denote the first $N$ terms of the Fourier series:


 * $(2): \quad S_N \left({x}\right) = \displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^N \left({a_n \cos n x + b_n \sin n x}\right)$

where:


 * $(3): \quad \displaystyle a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} f \left({x}\right) \cos n x \rd x$
 * $(4): \quad \displaystyle b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} f \left({x}\right) \sin n x \rd x$

Substituting from $(3)$ and $(4)$ into $(2)$ and rearranging:


 * $S_N \left({x}\right) = \displaystyle \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} f \left({u}\right) \left({\frac 1 2 + \sum_{n \mathop = 1}^N \left({\cos n x \cos n u + \sin n x \sin n u}\right)}\right) \rd u$

Now we have:
 * $\displaystyle\frac 1 2 + \sum_{n \mathop = 1}^N \left({\cos n x \cos n u + \sin n x \sin n u}\right) = \frac {\sin \left({\left({N + \frac 1 2}\right) \left({u - x}\right)}\right)} {2 \sin \left({\frac 1 2 \left({u - x}\right)}\right)}$

Hence:
 * $\displaystyle S_N \left({x}\right) = \int_\alpha^{\alpha + 2 \pi} \psi \left({u}\right) \frac {\sin \left({\left({N + \frac 1 2}\right) \left({u - x}\right)}\right)} {2 \sin \left({\frac 1 2 \left({u - x}\right)}\right)} \rd u$

where:
 * $\psi \left({u}\right) = \dfrac 1 \pi f \left({u}\right) \dfrac {\frac 1 2 \left({u - x}\right)} {\sin \frac 1 2 \left({u - x}\right)}$

We have that $f \left({u}\right)$ satisfies the Dirichlet conditions on $\left[{\alpha \,.\,.\, \alpha + 2 \pi}\right]$.

Hence $f$ is piecewise smooth on $\left[{\alpha \,.\,.\, \alpha + 2 \pi}\right]$.

That is, $f$ has right-hand derivative and left-hand derivative at all $x$ in $\left[{\alpha \,.\,.\, \alpha + 2 \pi}\right]$.

Thus at the point $u = x$, $f$ has right-hand derivative and left-hand derivative, and so does $\psi \left({u}\right)$.

So by Fourier's Theorem: Lemma 3:
 * $\displaystyle \lim_{n \mathop \to N} S_N \left({x}\right) = \frac \pi 2 \left({\psi \left({x^+}\right) + \psi \left({x^-}\right)}\right)$

Now:
 * $\psi \left({x^+}\right) = \displaystyle \frac 1 \pi f \left({x^+}\right) \lim_{u \mathop \to x} \dfrac {\frac 1 2 \left({u - x}\right)} {\sin \frac 1 2 \left({u - x}\right)} = \frac 1 \pi f \left({x^+}\right)$

and:


 * $\psi \left({x^-}\right) = \displaystyle \frac 1 \pi f \left({x^-}\right) \lim_{u \mathop \to x} \dfrac {\frac 1 2 \left({u - x}\right)} {\sin \frac 1 2 \left({u - x}\right)} = \frac 1 \pi f \left({x^-}\right)$

and so:
 * $\displaystyle \lim_{n \mathop \to N} S_N \left({x}\right) = \frac 1 2 \left({\lim_{x \mathop \to a^+} f \left({x}\right) + \lim_{x \mathop \to a^-} f \left({x}\right)}\right)$

Also known as

 * Fourier's Theorem is also known as Dirichlet's Theorem for 1-Dimensional Fourier Series'''.