Fourier's Theorem

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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 $\openint \alpha {\alpha + 2 \pi}$:

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


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:

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

such that:

$\ds a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$
$\ds b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$


Then for all $a \in \R$, $(1)$ converges to the sum:

$\ds \frac 1 2 \paren {\lim_{x \mathop \to a^+} \map f x + \lim_{x \mathop \to a^-} \map f x}$

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:

Let $f: \R \to \R$ be a real function which satisfies the Dirichlet conditions on $\R$.

Then:

$\dfrac {\map f {t^+} + \map f {t^-} } 2 = \ds \int_{-\infty}^\infty e^{2 \pi i t s} \paren {\int_{-\infty}^\infty e^{-2 \pi i t s} \map f t \rd t} \rd s$

where:

$\map f {t^+}$ and $\map f {t^-}$ denote the limit from above and the limit from below of $f$ at $t$.


Proof

Lemma 1

Let $\psi$ be a real function defined on a closed interval $\closedint a b$.

Let $\psi$ be piecewise continuous with one-sided limits on $\closedint a b$.

Then:

$\ds \lim_{N \mathop \to \infty} \int_a^b \map \psi u \sin N u \rd u = 0$


Lemma 2

Let $\psi$ be a real function defined on a half-open interval $\hointl 0 a$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\hointl 0 a$.

Let $\map \psi u$ have a right-hand derivative at $u = 0$.


Then:

$\ds \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$

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


Lemma 3

Let $\psi$ be a real function defined on an open interval $\openint a b$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\openint a b$.

Let $\map \psi u$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \openint a b$.


Then:

$\ds \lim_{N \mathop \to \infty} \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$

where:

$\map \psi {x^+}$ denotes the limit of $\psi$ at $x$ from the right
$\map \psi {x^-}$ 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:

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

where:

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


Substituting from $(3)$ and $(4)$ into $(2)$ and applying Integral of Integrable Function is Additive:

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

Now we have:

\(\ds \dfrac 1 2 + \ds \sum_{n \mathop = 1}^N \paren {\cos n x \cos n u + \sin n x \sin n u}\) \(=\) \(\ds \dfrac 1 2 + \ds \sum_{n \mathop = 1}^N \map \cos { n \paren {u - x} }\) Cosine of Difference
\(\ds \) \(=\) \(\ds \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } }\) Lagrange's Cosine Identity

Hence:

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

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



This article needs proofreading.
In particular: In the source book given, when initially stated, this theorem was as given here. When it is proved in the next chapter, it is given as being piecewise smooth. I believe one implies the other, but not sure. Initial thoughts are that piecewise smooth is stronger.
If you believe all issues are dealt with, please remove {{Proofread}} from the code.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Proofread}} from the code.



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:

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


Now:

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

and:

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

and so:

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

$\blacksquare$


Also known as


Source of Name

This entry was named for Joseph Fourier.


Sources

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