Fourier series of piece-wise smooth functions

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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 \(f'\in\mathcal{R}_{2\pi}^{2}\) (i.e., the derivative is square-integrable, not necessarily continuous).

In the following, we will consider the complex Fourier series:
\[f \sim \sum_{k=-\infty}^{\infty}c_ke^{ikx}\]
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
\[2\pi\sum_{k=-\infty}^{\infty}|c_k|^2 \leq \int_{-\pi}^{\pi}|f(x)|^2dx.\tag{eq:bessel}\]
Proof. Let \(S_n(x) = S_n[f](x) = \sum_{k=-n}^{n}c_ke^{ikx}\) where \(c_k\) are the complex Fourier coefficients of \(f\).
\[\begin{eqnarray*} \int_{-\pi}^{\pi}|f(x) - S_n(x)|^2dx &=& \int_{-\pi}^{\pi}(f(x) - S_n(x))(\overline{f(x)} - \overline{S_n(x)})dx\\ &=& \int_{-\pi}^{\pi}|f(x)|^2dx - \int_{-\pi}^{\pi}\left(\sum_{k=-n}^{n}c_ke^{ikx}\right)\overline{f(x)}dx \\ && - \int_{-\pi}^{\pi}{f(x)}\left(\sum_{k=-n}^{n}\overline{c}_ke^{-ikx}\right)dx \\ && + \int_{-\pi}^{\pi}\left(\sum_{k=-n}^{n}c_ke^{ikx}\right)\left(\sum_{k=-n}^{n}\overline{c}_ke^{-ikx}\right)dx. \end{eqnarray*}\]
Using the definition of \(c_k\),  and orthogonality of \(\{e^{ikx}\}\), we have
\[\begin{eqnarray} \int_{-\pi}^{\pi}|f(x) - S_n(x)|^2dx &=& \int_{-\pi}^{\pi}|f(x)|^2dx -2\pi\sum_{k=-n}^{n}|c_k|^2 -2\pi\sum_{k=-n}^{n}|c_k|^2 +2\pi\sum_{k=-n}^{n}|c_k|^2 \nonumber\\ &=& \int_{-\pi}^{\pi}|f(x)|^2dx - 2\pi\sum_{k=-n}^{n}|c_k|^2.\tag{eq:fSn} \end{eqnarray}\]
But the left-hand side is non-negative. Thus, 
\[\int_{-\pi}^{\pi}|f(x)|^2dx \geq 2\pi\sum_{k=-n}^{n}|c_k|^2.\]
The right-hand side is monotone increasing as \(n\to \infty\), and the left-hand side does not depend on \(n\). Therefore, (eq:bessel) holds. ■

Remark. Since \(\int_{-\pi}^{\pi}|S_n(x)|^2dx = 2\pi\sum_{k=-n}^{n}|c_k|^2\), we have
\[\int_{-\pi}^{\pi}|S_n(x)|^2dx  \leq \int_{-\pi}^{\pi}|f(x)|^2dx.\tag{eq:Snlef}\]
□

Theorem 

If \(f \in C_{2\pi}^{0}\) and \(f' \in \mathcal{R}_{2\pi}^{2}\), 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 \(f'\):
\[f' \sim \sum_{k=-\infty}^{\infty}\gamma_k e^{ikx}.\]
Since \(f'\in\mathcal{R}_{2\pi}^2\) by assumption, by applying the above lemma to \(f'\), we have
\[\sum_{k=-\infty}^{\infty}|\gamma_k|^2 < +\infty.\tag{eq:gammaineq}\]
Integrating by parts,
\[\begin{eqnarray*} 2\pi c_k &=& \int_{-\pi}^{\pi}f(x)e^{-ikx}dx = \frac{1}{ik}\int_{-\pi}^{\pi}f'(x)e^{-ikx}dx\\ &=& \frac{2\pi}{ik}\gamma_k ~~~ (k \neq 0). \end{eqnarray*}\]
That is,
\[c_k = \frac{1}{ik}\gamma_k ~~~ (k\neq 0, k \in \mathbb{Z}).\]
By the Cauchy-Schwarz inequality,
\[\sum_{k\neq 0}|c_k| = \sum_{k\neq 0}\frac{1}{|k|}|\gamma_k|\leq \sqrt{\sum_{k\neq 0}\frac{1}{k^2}}\sqrt{\sum_{k\neq 0}|\gamma_k|^2}.\]
Noting that \(\sum_{k=1}^{\infty}\frac{1}{k^2} < +\infty\) and (eq:gammaineq), the right-hand side is finite. Thus, \(\sum_{k\neq 0}|c_k| < +\infty\). 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 \(f' \in \mathcal{R}_{2\pi}^{2}\). ■

Corollary (Parseval's equality)

If \(f \in C_{2\pi}^{0}\) and \(f' \in \mathcal{R}_{2\pi}^{2}\), then
\[2\pi\sum_{k=-\infty}^{\infty}|c_k|^2 = \int_{-\pi}^{\pi}|f(x)|^2dx.\tag{eq:parseval}\]
Proof. Since \(S[f]\) converges uniformly to \(f\), the left-hand side of (eq:fSn) converges to 0. Thus, the above equality holds. ■



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