📂Fourier Analysis

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Theorem¹

Let the function $f$ be Riemann integrable on the interval $[-L,\ L)$. Then for a point $t$ at which the function is continuous, the Fourier series $\lim \limits_{N \to \infty }S^{f}_{N}(t)$ of $f$ converges to $f(t)$.

$$ \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t)=f(t) $$

In this case

$$ \begin{align*} S^{f}_{N}(t)&=\dfrac{a_{0}}{2}+\sum \limits_{n=1}^{N} \left( a_{n} \cos \dfrac{n\pi t}{L} + b_{n}\sin\dfrac{n\pi t} {L} \right) \\ a_{0} &=\dfrac{1}{L}\int_{-L}^{L}f(t)dt \\ a_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{n\pi t}{L} dt \\ b_{n}&=\dfrac{1}{L}\int_{-L}^{L}f(t)\sin\dfrac{n\pi t}{L}dt \end{align*} $$

Proof

Strategy: We complete the proof by showing $\left| \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t)-f(t) \right|=0$.

Relation between Fourier series and the Dirichlet kernel

$$ S^{f}_{N}(t)=\dfrac{1}{L}\int_{-L}^{L}f(x)D_{N}\left(\dfrac{\pi (x-t)}{L}\right)dx $$

From the above fact we obtain the following expression.

$$ \begin{equation} \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t)-f(t) =\lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(x)D_{N}\left( \dfrac{\pi (x-t)}{L}\right)dx -f(t) \end{equation} $$

Integral of the Dirichlet kernel

$$ \dfrac{1}{L}\int_{-L}^{L}D_{N}\left( \dfrac{\pi (x-t)}{L} \right)dx=1 $$

Multiplying both sides of the above equation by $f(t)$ yields the following.

$$ \dfrac{1}{L}\int_{-L}^{L} f(t) D_{N}\left( \dfrac{\pi (x-t)}{L} \right)dx = f(t) $$

Substituting this into $(1)$ and simplifying gives:

$$ \begin{align*} &\lim \limits_{N \rightarrow \infty} S^{f}_{N}(t)-f(t) \\ &= \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(x)D_{N}\left( \dfrac{\pi (x-t)}{L}\right)dx -f(t) \\ &= \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(x)D_{N}\left( \dfrac{\pi (x-t)}{L}\right)dx -\dfrac{1}{L}\int_{-L}^{L} f(t) D_{N}\left( \dfrac{\pi (x-t)}{L}\right)dx \\ &= \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} \Big[ f(x)-f(t) \Big] D_{N}\left( \dfrac{\pi (x-t)}{L}\right)dx \end{align*} $$

Introducing the substitution $x-t=\lambda$ gives:

$$ \begin{equation} \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L-t}^{L-t} \Big[ f(\lambda+t)-f(t) \Big] D_{N}\left( \dfrac{\pi \lambda}{L} \right)d\lambda \end{equation} $$

Since $f(x)$ is continuous at $t$, by the definition there exists $\delta>0$ such that for $s,t\in [-L,\ L)$ and $\varepsilon >0$ the following holds.

$$ \begin{equation} \exists \delta>0\quad \text{s.t. } \left| s-t \right|<\delta \implies \left| f(s)-f(t) \right| <\varepsilon \end{equation} $$

The Dirichlet kernel converges to the Dirac delta function $$ \lim \limits_{n \to \infty} D_{n}(t)=\delta (t) $$

Moreover, from the above fact, for the fixed positive $\delta>0$ there exists a natural number $n$ such that whenever $|x|>\delta$ and $N \gt n$ hold, then $\left| D_{N}\left( \dfrac{\pi x}{L} \right) \right| \lt \varepsilon$ holds.

$$ \begin{equation} \exists n\in \mathbb{N}\quad \text{s.t. } \left| x \right| > \delta,\ N>n \implies \left| D_{N} \left( \frac{\pi x}{L} \right) \right| < \varepsilon \end{equation} $$

For the fixed $t \in [-L, L)$ there exists a real number $M$ satisfying $|f(t)| < M$.

$$ \begin{equation} \exists M\quad \text{s.t. } \left| f(t) \right| <M \end{equation} $$

Now splitting the integration range of $(2)$ yields the following inequality.

$$ \begin{align*} &| \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t)-f(t)| \\ &= \lim \limits_{N \rightarrow \infty} \left| \dfrac{1}{L} \int_{-L-t}^{L-t} \left[ f(\lambda+t)-f(t) \right] D_{N}\left( \dfrac{\pi \lambda}{L}\right)d\lambda \right| \\ &\le \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \left[ \left|\int_{-\delta}^{\delta} \left[ f(\lambda+t)-f(t) \right] D_{N}\left( \dfrac{\pi \lambda}{L}\right) d\lambda \right| +\left| \int_{\lambda \notin[-\delta,\delta]} \left[ f(\lambda+t)-f(t) \right] D_{N}\left( \dfrac{\pi \lambda}{L}\right) d\lambda \right| \right] \end{align*} $$

Applying condition $(3)$ to the first term, and conditions $(4)$, $(5)$ to the second term gives:

$$ \begin{align*} &\lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \left[ \left|\int_{-\delta}^{\delta} {\color{red} \left[ f(\lambda+t)-f(t) \right] } D_{N}\left( \dfrac{\pi \lambda}{L}\right) d\lambda \right| +\left| \int_{\lambda \notin[-\delta,\delta]} {\color{blue} \left[ f(\lambda+t)-f(t) \right]} {\color{orange}D_{N}\left( \dfrac{\pi \lambda}{L}\right) } d\lambda \right| \right] \\ &\le \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \left[ \left|\int_{-\delta}^{\delta} {\color{red} \epsilon} D_{N}\left( \dfrac{\pi \lambda}{L}\right) d\lambda \right| + {\color{blue}2M} {\color{orange}\epsilon} \right] \\ &= \lim \limits_{N \rightarrow \infty} \dfrac{\epsilon}{L} \left( \left|\int_{-\delta}^{\delta} D_{N}\left( \dfrac{\pi \lambda}{L}\right) d\lambda \right| +2M \right) \\ &= \varepsilon^{\prime} \end{align*} $$

Since the above expression must hold for all $\varepsilon >0$, it is equivalent to holding for all $\varepsilon^{\prime}>0$. Therefore we obtain:

$$ \left| \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t)-f(t) \right|=0 $$

Thus the Fourier series of $f$ converges to $f$ at the point $t$ where it is continuous.

$$ \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t) = \dfrac{1}{2}a_{0}+\sum \limits_{n=1}^{\infty}\left( a_{n}\cos\dfrac{n\pi t}{L}+b_{n}\sin\dfrac{n\pi t}{L} \right) = f(t) $$

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