Fourier's Theorem/Lemma 1

From ProofWiki
Jump to navigation Jump to search

Lemma for Fourier's Theorem

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:

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


Proof

We are given that $\psi$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, b}\right]$.

Therefore, there exists a finite subdivision $\left\{{x_0, x_1, \ldots, x_m}\right\}$ of $\left[{a \,.\,.\, b}\right]$, where $x_0 = a$ and $x_m = b$, such that for all $i \in \left\{{1, 2, \ldots, m}\right\}$:

$\psi$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \psi \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \psi \left({x}\right)$ exist.


From the corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions:

$\displaystyle \int_a^b \psi \left({u}\right) \sin N u \rd u = \sum_{r \mathop = 0}^{m - 1} \int_{x_r}^{x_{r + 1} } \psi \left({u}\right) \sin N u \rd u$

Then:

\((1):\quad\) \(\displaystyle \int_{x_r}^{x_{r + 1} } \psi \left({u}\right) \sin N u \rd u\) \(=\) \(\displaystyle \left[{-\psi \left({u}\right) \frac {\cos N u} N}\right]_{x_r}^{x_{r + 1} }\) Integration by Parts
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac 1 N \int_{x_r}^{x_{r + 1} } \psi' \left({u}\right) \cos N u \rd u\)

The last integral is bounded.

Thus $(1)$ is less than $\dfrac {M_r} N$ for $M_r \in \R$.

Let $M = \max \left\{ {\left\lvert{M_0}\right\rvert, \left\lvert{M_1}\right\rvert, \ldots, \left\lvert{M_{m - 1} }\right\rvert}\right\}$.

Then:

$\displaystyle \left\lvert{\int_a^b \psi \left({u}\right) \sin N u \rd u}\right\rvert < \dfrac {M m} N$

As $M$ and $m$ are finite:

$\displaystyle \lim_{N \mathop \to \infty} \dfrac {M m} N = 0$

Hence the result.

$\blacksquare$


Sources