📂Fourier Analysis

The Fourier Series of a Riemann Integrable Function Converges

Theorem[^1]

Let’s assume that the function $f$ is Riemann integrable on the interval $[-L,\ L)$. Then, for any point of continuity $t$, the Fourier series $\lim \limits_{N \to \infty }S^{f}_{N}(t)$ of $f$ converges to $f(t)$.

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

Where

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

Proof

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

Relationship 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 equation.

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

Integration 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 $f(t)$ gives the following equation.

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

Substituting this into $(1)$ and arranging gives the following result.

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

Substituting $x-t=\lambda$ here yields the following result.

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

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

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

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

And due to the above fact, for a fixed positive number $\delta>0$, there exists a natural number $n$ such that $|x|>\delta$ and when $N \gt n$, $\left| D_{N}\left( \dfrac{\pi x}{L} \right) \right| \lt \varepsilon$.

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

Moreover, since $f(x)$ is assumed to be Riemann integrable, it is bounded. Therefore, a real number $M$ exists that satisfies $|f(t)| < M$.

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

Now, dividing the range of integration for $(2)$ 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 the $(3)$ condition to the first term and the $(4)$, $(5)$ conditions to the second term yields the following.

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

Since the above equation must hold for all $\varepsilon >0$, it must consequently hold for all $\varepsilon^{\prime}>0$. Therefore, we obtain the following equation.

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

Hence, the Fourier series of $f$ converges at the continuous point $t$ to $f$.

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

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