Fourier's Theorem

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

Let $\map f x$ be a real function which is defined and bounded on the interval $\openint \alpha {\alpha + 2 \pi}$.

Let $f$ satisfy the Dirichlet conditions on $\openint \alpha {\alpha + 2 \pi}$:

Outside the interval $\openint \alpha {\alpha + 2 \pi}$, let $f$ be periodic and defined such that:
 * $\map f x = \map f {x + 2 \pi}$

Let $f$ be defined by the Fourier series:


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

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

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

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

Integral Form
This theorem can often be seen presented in the form:

Main Theorem
Let $\map {S_N} x$ denote the first $N$ terms of the Fourier series:


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

where:


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

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


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

Now we have:
 * $\dfrac 1 2 + \displaystyle\sum_{n \mathop = 1}^N \paren {\cos n x \cos n u + \sin n x \sin n u} = \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } }$

Hence:
 * $\displaystyle \map {S_N} x = \int_\alpha^{\alpha + 2 \pi} \map \psi u \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } } \rd u$

where:
 * $\map \psi u = \dfrac 1 \pi \map f u \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} }$

We have that $\map f u$ satisfies the Dirichlet conditions on $\openint \alpha {\alpha + 2 \pi}$.

Hence $f$ is piecewise smooth on $\openint \alpha {\alpha + 2 \pi}$.

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

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

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

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

and:


 * $\map \psi {x^-} = \displaystyle \frac 1 \pi \map f {x^-} \lim_{u \mathop \to x} \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} } = \frac 1 \pi \map f {x^-}$

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

Also known as

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