Fourier Series Convergence for Piecewise Smooth Functions

0. Preliminaries: A Broader Class of Functions.

Definition. A function $f$ is piecewise continuous on a finite interval $[a,b]$ if it is continuous at every point in $[a,b]$ except for a finite number of points $x_i$, at each of which the left-hand and right-hand limits, denoted $f(x_i^{-})={\lim }_{t\to x_i^{-}}f(t)$ and $f(x_i^{+})={\lim }_{t\to x_i^{+}}f(t)$, exist and are finite.

Definition. A function $f$ is piecewise $C^1$ on an interval if both the function $f$ and its derivative $f^{\prime }$ are piecewise continuous on that interval. A periodic function is piecewise $C^1$ if it is piecewise $C^1$ on any finite interval of one period.

1. Pointwise Convergence Theorem for Piecewise Smooth Functions.

The following theorem provides a definitive statement on the pointwise convergence behavior of a Fourier series for a function that is well-behaved between a finite number of jump discontinuities.

Theorem 1 (Dirichlet’s Convergence Theorem). Let $f:\mathbb{R}\to \mathbb{C}$ be a $2\pi$-periodic and piecewise $C^1$ function. The Fourier series of $f$, denoted $S[f](x)$, converges for all $x\in \mathbb{R}$. The limit of the series is as follows:

1. At any point $x$ where $f$ is continuous, the series converges to the value of the function: $S[f](x)=f(x)$.
2. At any point of discontinuity $x_0$, the series converges to the arithmetic mean of the left-hand and right-hand limits:

$$S[f](x_0)=\frac{f(x_0^{+})+f(x_0^{-})}{2}$$

Proof. The $N$-th symmetric partial sum of the Fourier series, $S_N[f](x)$, is the convolution of $f$ with the Dirichlet kernel $D_N(t)={\sum }_{k=-N}^N{e}^{ikt}=\frac{\sin ((N+1/2)t)}{\sin (t/2)}$. Let $L(x)$ be the asserted limit.

$$S_N[f](x)-L(x)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }(f(x-t)-L(x))D_N(t)dt$$

The integral is split at $t=0$, and a change of variable $u=-t$ is applied to the integral over $[-\pi ,0]$.

$${\int }_{-\pi }^0\frac{f(x-t)-L(x)}{\sin (t/2)}\sin \left(\left(N+\frac{1}{2}\right)t\right)dt={\int }_0^{\pi }\frac{f(x+u)-L(x)}{\sin (u/2)}\sin \left(\left(N+\frac{1}{2}\right)u\right)du$$

Combining this with the integral over $[0,\pi ]$ yields:

$$S_N[f](x)-L(x)=\frac{1}{2\pi }{\int }_0^{\pi }\frac{f(x+t)+f(x-t)-2L(x)}{\sin (t/2)}\sin \left(\left(N+\frac{1}{2}\right)t\right)dt$$

Let $h(t)=\frac{f(x+t)+f(x-t)-2L(x)}{\sin (t/2)}$. For the Riemann-Lebesgue Lemma to apply, $h(t)$ must be integrable on $[0,\pi ]$. The only point of concern is the behavior as $t\to 0^{+}$. The limit is an indeterminate form $\frac{0}{0}$ because ${\lim }_{t\to 0^{+}}(f(x+t)+f(x-t)-2L(x))=f(x^{+})+f(x^{-})-2L(x)=0$. By L’Hôpital’s rule:

$$\underset{t\to 0^{+}}{\lim }h(t)=\underset{t\to 0^{+}}{\lim }\frac{f^{\prime }(x+t)-f^{\prime }(x-t)}{\frac{1}{2}\cos (t/2)}=\frac{f^{\prime }(x^{+})-f^{\prime }(x^{-})}{1/2}=2(f^{\prime }(x^{+})-f^{\prime }(x^{-}))$$

Since $f$ is piecewise $C^1$, $f^{\prime }$ is piecewise continuous. Thus, the one-sided limits $f^{\prime }(x^{+})$ and $f^{\prime }(x^{-})$ are finite, which implies ${\lim }_{t\to 0^{+}}h(t)$ is finite. Therefore, $h(t)$ is integrable on $[0,\pi ]$. The Riemann-Lebesgue Lemma states that ${\lim }_{N\to \infty }{\int }_0^{\pi }h(t)\sin (((N+1/2))t)dt=0$. Consequently, ${\lim }_{N\to \infty }(S_N[f](x)-L(x))=0$. $\square$

2. Example: The Sawtooth Wave.

Consider the function $f(x)=x$ on the interval $[0,2\pi )$, extended periodically to all of $\mathbb{R}$. This function has jump discontinuities at every $x=2k\pi$ for $k\in \mathbb{Z}$. The Fourier coefficients are computed on $[0,2\pi )$. The constant term is $c_0=\pi$. For $n\ne 0$, integration by parts gives:

$$\begin{array}{rl}{c}_n[f]& =\frac{1}{2\pi }{\int }_0^{2\pi }x{e}^{-inx}dx\\ & =\frac{1}{2\pi }\left({\left[x\frac{e^{-inx}}{-in}\right]}_0^{2\pi }-{\int }_0^{2\pi }\frac{e^{-inx}}{-in}dx\right)\\ & =\frac{1}{2\pi }\left(\frac{2\pi }{-in}-{\left[\frac{e^{-inx}}{(-in{)}^2}\right]}_0^{2\pi }\right)\\ & =\frac{1}{-in}-0=\frac{i}{n}\end{array}$$

The Fourier series is $S[f](x)=\pi +{\sum }_{n\in \mathbb{Z},n\ne 0}\frac{i}{n}{e}^{inx}$. The function $f$ is piecewise $C^1$. For any $x\in (0,2\pi )$, $f$ is continuous, so $S[f](x)=f(x)=x$. At the discontinuity $x=0$, the left and right limits are $f(0^{-})=2\pi$ and $f(0^{+})=0$. By Dirichlet’s Theorem, the series converges to the midpoint:

$$S[f](0)=\frac{f(0^{+})+f(0^{-})}{2}=\frac{0+2\pi }{2}=\pi$$

This value matches the constant term of the series. $\square$