# 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 $\closedint \alpha {\alpha + 2 \pi}$.

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

$(\mathrm D 1): \quad f$ is absolutely integrable.
$(\mathrm D 2): \quad f$ has a finite number of local maxima and local minima.
$(\mathrm D 3): \quad f$ has a finite number of discontinuities, all of them finite.

Outside the interval $\closedint \alpha {\alpha + 2 \pi}$, 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.

## Proof

### Lemma 1

Let $\psi$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$.

Let $\psi$ be piecewise continuous with one-sided limits on $\left[{a \,.\,.\, b}\right]$.

Then:

$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \psi \left({u}\right) \sin N u \rd u = 0$

### Lemma 2

Let $\psi$ be a real function defined on a half-open interval $\left({0 \,.\,.\, a}\right]$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\left({0 \,.\,.\, a}\right]$.

Let $\psi \left({u}\right)$ have a right-hand derivative at $u = 0$.

Then:

$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \psi \left({u}\right) \frac {\sin N u} u \rd u = \frac \pi 2 \psi \left({0^+}\right)$

where $\psi \left({0^+}\right)$ denotes the limit of $\psi$ at $0$ from the right.

### Lemma 3

Let $\psi$ be a real function defined on an open interval $\left({a \,.\,.\, b}\right)$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\left({a \,.\,.\, b}\right)$.

Let $\psi \left({u}\right)$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \left({a \,.\,.\, b}\right)$.

Then:

$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u = \frac \pi 2 \left({\psi \left({x^+}\right) + \psi \left({x^-}\right)}\right)$

where:

$\psi \left({x^+}\right)$ denotes the limit of $\psi$ at $x$ from the right
$\psi \left({x^-}\right)$ denotes the limit of $\psi$ at $x$ from the left.

### 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:

$\displaystyle\frac 1 2 + \sum_{n \mathop = 1}^N \paren {\cos n x \cos n u + \sin n x \sin n u} = \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } }$

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 $\closedint \alpha {\alpha + 2 \pi}$.

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

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

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

$\displaystyle \lim_{n \mathop \to N} \map {S_N} x = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$

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:

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

$\blacksquare$

## Source of Name

This entry was named for Joseph Fourier.