📂Fourier Analysis

Convergence of Fourier Series at Discontinuities

Theorem¹

Let’s say the function $f(t)$, defined in the interval $[-L,\ L)$, is piecewise continuous. Denoting the points of discontinuity as $t_{i}\ (i=1,\ \cdots m )$ and assuming at each point of discontinuity, there exist left-hand derivative $f(a-)$ and right-hand derivative $f(a+)$. Then, the Fourier series of $f(t)$ converges to the midpoint of the left-hand and right-hand limits at the point of discontinuity $t_{i}$.

$$ \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty}\left( a_{n} \cos \dfrac{n \pi t_{i} }{L} +b_{n}\sin\dfrac{n\pi t_{i}}{L} \right) = \dfrac{f(t_{i}+)+f(t_{i}-)}{2} $$

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If $f$ is Riemann integrable, its Fourier series converges to $f$ at the continuous point $t$. At points of discontinuity, it can be understood from the above theorem that it converges to the midpoint of the left and right derivatives.

Proof

Consider an arbitrary point of discontinuity $t_{i}=t$.

The relationship between the Fourier series and the Dirichlet kernel

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

Through the above relation, the following equation is obtained.

$$ \begin{align} & \lim_{N \rightarrow \infty} S_{N} (t) \nonumber \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(x)D_{N}\left( \dfrac{\pi (x-t)}{L} \right) dx \nonumber \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L-t}^{L-t} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L} \right) d\lambda \nonumber \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L} \right) d\lambda \nonumber \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0} f(\lambda + t)D_{N}\left(\dfrac{\pi \lambda }{L}\right) d\lambda +\lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda +t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda \end{align} $$

The second equality holds by substituting with $x-t=\lambda$. The third equality holds because it only matters that the integration is over a period $(2L)$.

Integration of the Dirichlet kernel

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

Since the Dirichlet kernel is an even function, the following equation is derived from the above.

$$ \dfrac{1}{L} \int_{0}^{L} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{L} \int_{-L}^{0} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{2} $$

Therefore, the following equation holds.

$$ \begin{equation} \begin{aligned} \dfrac{1}{2}f(t+) &= f(t+)\dfrac{1}{L} \int_{0}^{L} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{L} \int_{0}^{L} f(t+)D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda \\ \dfrac{1}{2}f(t-) &= f(t-)\dfrac{1}{L} \int_{-L}^{0} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{L} \int_{-L}^{0} f(t-)D_{N} \left( \dfrac{\pi \lambda }{L} \right) d\lambda \end{aligned} \end{equation} $$

Aligning with the integration range of the two terms of $(1)$ and subtracting the two equations of $(2)$ respectively, we get the following.

$$ \begin{align*} & \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda - \dfrac{1}{2}f(t+) \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}\Big( f(\lambda + t) -f(t+) \Big) D_{N}\left(\dfrac{\pi \lambda }{L}\right) d\lambda \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}\Big( f(\lambda + t) -f(t+)) \Big)\dfrac{\sin\left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L}}{2\sin \dfrac{\pi \lambda}{2L}} d\lambda \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}g_+(\lambda) \sin\left[ \left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L}\right] d\lambda \end{align*} $$

Where, $g_+(\lambda) = \dfrac{ f(\lambda + t) -f(t+) }{\lambda}\dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}}$ holds. The second equality is due to the equation below.

The relationship between the Fourier series and the Dirichlet kernel

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

Calculating in the same way, we get the following.

$$ \begin{align*} & \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda - \dfrac{1}{2}f(t-) \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0}g_-(\lambda) \sin\left[ \left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L}\right] d\lambda \end{align*} $$

Here, $g_-(\lambda) = \dfrac{ f(\lambda + t) -f(t-) }{\lambda}\dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}}$ holds. Now we aim to show that $g_\pm(\lambda)$ is piecewise continuous. Due to the properties of the limit of the sine function,

$$ \lim \limits_{\lambda \rightarrow 0} \dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}} =\lim \limits_{\lambda \rightarrow 0} \dfrac{\dfrac{\pi \lambda}{2L}}{\sin \dfrac{\pi \lambda}{2L}} \dfrac{L}{\pi}=\dfrac{L}{\pi} $$

there exists $M_{1}>0$ satisfying the equation below at any given $\lambda \in [-L,\ L)$.

$$ \left| \dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}}\right| \le M_{1} <\infty $$

It’s certain not to diverge except at $0$, and it’s shown not to diverge even at $0$. This means it is bounded. That is, there are a finite number of points of discontinuity within the interval, and both left/right limits exist at the points of discontinuity. Hence, it is piecewise continuous within the interval. If continuous, then it is Riemann integrable, and being Riemann integrable, it is bounded, and $f$ has a right-hand derivative at $t$, so there exists $M_2$ satisfying the equation below at any given $\lambda \in (0,\ L)$.

$$ \left| \dfrac{f(\lambda+t) -f(t+)}{\lambda} \right| \le M_2 <\infty $$

Similarly, it is piecewise continuous within the interval. By the two facts above, $g_+(\lambda)$ is piecewise continuous and therefore Riemann integrable at $[0,\ L)$. Therefore,

$$ \begin{align*} & \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda - \dfrac{1}{2}f(t+) \\ =& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}g_+(\lambda) \sin\left[ \left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L} \right]d\lambda \\ =&\ 0 \end{align*} $$

As $g_+(\lambda)$ is piecewise continuous within the interval, the second equality holds due to the Riemann-Lebesgue lemma.

Riemann-Lebesgue lemma

If the function $f(t)$ is piecewise continuous within the interval $[-L,\ L)$, the following equation holds:

$$ \lim \limits_{n \rightarrow \infty} a_{n} = \lim \limits_{n \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(t) \cos \dfrac{n \pi}{L}t dt=0 $$

$$ \lim \limits_{n \rightarrow \infty} b_{n} = \lim \limits_{n \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(t) \sin \dfrac{n \pi}{L}t dt=0 $$

Therefore,

$$ \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda=\dfrac{1}{2}f(t+) $$

In the same way, the following equation can be obtained.

$$ \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda=\dfrac{1}{2}f(t-) $$

Combining the two equations,

$$ \lim \limits_{N \rightarrow \infty} S_{N}(t) = \dfrac{1}{2}\big(f(t+)+f(t-)\big) $$

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