Fourier series of piece-wise smooth functions

Our goal in this post is to prove the following:

• If $f$ is ``piece-wise smooth,'' then the (complex) Fourier series of $f$ converges uniformly to $f$.

Note that the condition in this theorem does not involve the Fourier coefficients of $f$; hence, it is more direct. That is, we can tell if the Fourier series of a function converges uniformly from the property of the function alone.

Let $C_{2\pi }^m$ be the collection of all functions of class $C^m$ such that all derivatives up to the $m$-th order have period $2\pi$. 

We say a function $f$ is piece-wise smooth if $f\in C_{2\pi }^0$ (i.e., continuous) and $f^{\prime }\in {\mathcal{R}}_{2\pi }^2$ (i.e., the derivative is square-integrable, not necessarily continuous).

In the following, we will consider the complex Fourier series:

$$f\sim \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{e}^{ikx}$$

where the Fourier coefficients are

$$c_k=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)e^{-ikx}dx.$$

Lemma (Bessel's inequality)

If $f\in {\mathcal{R}}_{2\pi }^2$, then

$$\begin{array}{}\text{(eq:bessel)}& 2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}|c_k{|}^2\le {\int }_{-\pi }^{\pi }|f(x){|}^{2}dx.\end{array}$$

Proof. Let $S_n(x)=S_n[f](x)=\sum _{k=-n}^n{c}_k{e}^{ikx}$ where $c_k$ are the complex Fourier coefficients of $f$.

$$\begin{array}{rcl}{\int }_{-\pi }^{\pi }|f(x)-S_n(x){|}^{2}dx& =& {\int }_{-\pi }^{\pi }(f(x)-S_n(x))(\overline{f(x)}-\overline{S_n(x)})dx\\ & =& {\int }_{-\pi }^{\pi }|f(x){|}^{2}dx-{\int }_{-\pi }^{\pi }\left(\sum _{k=-n}^n{c}_k{e}^{ikx}\right)\overline{f(x)}dx\\ & & -{\int }_{-\pi }^{\pi }f(x)\left(\sum _{k=-n}^n{\bar{c}}_k{e}^{-ikx}\right)dx\\ & & +{\int }_{-\pi }^{\pi }\left(\sum _{k=-n}^n{c}_k{e}^{ikx}\right)\left(\sum _{k=-n}^n{\bar{c}}_k{e}^{-ikx}\right)dx.\end{array}$$

Using the definition of $c_k$,  and orthogonality of $\{e^{ikx}\}$, we have

$$\begin{array}{rcl}{\int }_{-\pi }^{\pi }|f(x)-S_n(x){|}^{2}dx& =& {\int }_{-\pi }^{\pi }|f(x){|}^{2}dx-2\pi \sum _{k=-n}^n|c_k{|}^2-2\pi \sum _{k=-n}^n|c_k{|}^2+2\pi \sum _{k=-n}^n|c_k{|}^2\\ \text{(eq:fSn)}& & =& {\int }_{-\pi }^{\pi }|f(x){|}^{2}dx-2\pi \sum _{k=-n}^n|c_k{|}^2.\end{array}$$

But the left-hand side is non-negative. Thus, 

$${\int }_{-\pi }^{\pi }|f(x){|}^{2}dx\ge 2\pi \sum _{k=-n}^n|c_k{|}^2.$$

The right-hand side is monotone increasing as $n\to \mathrm{\infty }$, and the left-hand side does not depend on $n$. Therefore, (eq:bessel) holds. ■

Remark. Since ${\int }_{-\pi }^{\pi }|S_n(x){|}^{2}dx=2\pi \sum _{k=-n}^n|c_k{|}^2$, we have

$$\begin{array}{}\text{(eq:Snlef)}& {\int }_{-\pi }^{\pi }|S_n(x){|}^{2}dx\le {\int }_{-\pi }^{\pi }|f(x){|}^{2}dx.\end{array}$$

□

Theorem 

If $f\in C_{2\pi }^0$ and $f^{\prime }\in {\mathcal{R}}_{2\pi }^2$, then the (complex) Fourier series of $f$ converges uniformly to $f$.

Proof. We use the following theorem (see Complex Fourier series):

• (*) If the series of Fourier coefficients of $f$ converges absolutely, then the Fourier series $S[f]$ converges uniformly to $f$ at continuous points. 

Let ${\gamma }_k$ denote the complex Fourier coefficients of $f^{\prime }$:

$$f^{\prime }\sim \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\gamma }_k{e}^{ikx}.$$

Since $f^{\prime }\in {\mathcal{R}}_{2\pi }^2$ by assumption, by applying the above lemma to $f^{\prime }$, we have

$$\begin{array}{}\text{(eq:gammaineq)}& \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}|{\gamma }_k{|}^2<+\mathrm{\infty }.\end{array}$$

Integrating by parts,

$$\begin{array}{rcl}2\pi c_k& =& {\int }_{-\pi }^{\pi }f(x)e^{-ikx}dx=\frac{1}{ik}{\int }_{-\pi }^{\pi }{f}^{\prime }(x)e^{-ikx}dx\\ & =& \frac{2\pi }{ik}{\gamma }_k(k\ne 0).\end{array}$$

That is,

$$c_k=\frac{1}{ik}{\gamma }_k(k\ne 0,k\in \mathbb{Z}).$$

By the Cauchy-Schwarz inequality,

$$\sum _{k\ne 0}|c_k|=\sum _{k\ne 0}\frac{1}{|k|}|{\gamma }_k|\le \sqrt{\sum _{k\ne 0}\frac{1}{k^2}}\sqrt{\sum _{k\ne 0}|{\gamma }_k{|}^2}.$$

Noting that $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}<+\mathrm{\infty }$ and (eq:gammaineq), the right-hand side is finite. Thus, $\sum _{k\ne 0}|c_k|<+\mathrm{\infty }$. Therefore, by the above Theorem (*), $S[f]$ converges uniformly to $f$. ■

The following corollary is the main result of this post:

Corollary

If $f\in C_{2\pi }^1$, then the (complex) Fourier series of $f$ converges uniformly to $f$.

Proof. $f\in C_{2\pi }^1$ implies $f\in C_{2\pi }^0$ and $f^{\prime }\in {\mathcal{R}}_{2\pi }^2$. ■

Corollary (Parseval's equality)

If $f\in C_{2\pi }^0$ and $f^{\prime }\in {\mathcal{R}}_{2\pi }^2$, then

$$\begin{array}{}\text{(eq:parseval)}& 2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}|c_k{|}^2={\int }_{-\pi }^{\pi }|f(x){|}^{2}dx.\end{array}$$

Proof. Since $S[f]$ converges uniformly to $f$, the left-hand side of (eq:fSn) converges to 0. Thus, the above equality holds. ■