Fourier's Theorem/Lemma 1

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Lemma for Fourier's Theorem

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:

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


Proof

We are given that $\psi$ is piecewise continuous with one-sided limits on $\closedint a b$.

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

$\psi$ is continuous on $\openint {x_{i - 1} } {x_i}$
$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map \psi x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map \psi x$ exist.


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

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

Then:

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

The last integral is bounded.

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

Let $M = \max \set {\size {M_0}, \size {M_1}, \dotsc, \size {M_{m - 1} } }$.

Then:

$\displaystyle \size {\int_a^b \map \psi u \sin N u \rd u} < \dfrac {M m} N$

As $M$ and $m$ are finite:

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

Hence the result.

$\blacksquare$


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