Pointwise convergence of Fourier series

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Our main question has been: What type of functions can be represented as a Fourier series? We have seen two types of convergence of \(S[f]\) to \(f\). 

  1. Uniform convergence: If \(f\in C_{2\pi}^1\) (continuously differentiable), then \(S[f]\) converges uniformly to \(f\).
  2. \(L^2\) convergence: If \(f\in\mathcal{R}_{2\pi}^{2}\) (square-integrable), then \(S[f]\) converges to \(f\) in the \(L^2\) norm.
These are ``global'' convergence properties, so to speak, in the sense that they characterize the function \(f\) for the whole period \([-\pi, \pi]\).

Now, we consider ``local'' convergence properties such as the problem of pointwise convergence. 


To begin with, we have the following result:

Theorem (du Bois-Reymond)

Let \(f\) be a periodic continuous function. Then, the Fourier series of \(f\) may not converge pointwise.

But we don't really need the continuity of \(f\) for its Fourier series to converge to \(f(x_0)\) at \(x_0\). Thus, let's define the following.

Definition (piecewise continuous and smooth functions)

A function \(f\) with a period \(2\pi\) is said to be piecewise continuous and smooth if there exist finitely many points \(\gamma_1 < \gamma_2 < \cdots < \gamma_l\) in the interval \([-\pi, \pi)\) such that \(f\) and \(f'\) are continuous and bounded in each interval \((\gamma_{j}, \gamma_{j+1})\) and in \((\gamma_l - 2\pi, \gamma_1)\).

If \(f\) is piecewise continuous and smooth, the left limit \(f(\gamma_j-0) = \lim_{x\to \gamma_j - 0}f(x)\) and the right limit \(f(\gamma_j + 0) = \lim_{x\to \gamma_j + 0}f(x)\) exist at each \(\gamma_j\). In fact, if \(x_1, x_2 \in (\gamma_{j-1}, \gamma_{j})\), then
\[|f(x_1) - f(x_2)| = \left|\int_{x_1}^{x_2}f'(\xi)d\xi\right| \leq M_{j-1}|x_1 - x_2|\]
where \(M_{j-1} = \sup_{\xi\in(\gamma_{j-1},\gamma_{j})}|f'(\xi)|\). Thus,
\[\lim_{x_1,x_2\to \gamma_j - 0}|f(x_1) - f(x_2)| = 0.\]
Similarly, \(\lim_{x_1,x_2\to \gamma_{j-1} + 0}|f(x_1) - f(x_2)| = 0\).

In particular, \(f\) is uniformly continuous on each subinterval \((\gamma_{j-1}, \gamma_j)\).

See also: Uniform continuity (Wikipedia)

We use the following lemma without proof.

Lemma (Riemann-Lebesgue Lemma)

Let \(g(x)\) be a function on an arbitrary interval \(I = (a, b)\) that is continuous except for finitely many points and integrable on \(I\), that is,
\[\int_a^b|g(x)|\,dx < +\infty.\]
Then,
\[\int_a^be^{i\lambda x}g(x)dx \to 0 ~~~ (\lambda \to \pm\infty).\]
Remark. If we adopt the Lebesgue integral, the required conditions are that \(g\) is measurable and integrable on \(I\). □
Proof. Omitted. ■
See also: Riemann-Lebesgue Lemma (Wikipedia)

Theorem (Pointwise convergence of Fourier series)

If the function \(f\) is piecewise continuous and smooth, then, for each \(x_0\), the Fourier series of \(f\) converges to
\[\frac{1}{2}\{f(x_0 - 0) + f(x_0 + 0)\}.\]
Proof. Let \(S_n\) be the partial sum of the Fourier series of \(f\). Using the Dirichlet kernel (which is an even function),
\[\begin{eqnarray*} S_n(x_0) &=& \int_{x_0-\pi}^{x_0+\pi}D_n(y - x_0)f(y)dy\\ &=& \int_{-\pi}^{\pi}D_n(y)f(x_0 + y)dy\\ &=& \int_{-\pi}^{0}D_n(y)f(x_0 + y)dy + \int_0^{\pi}D_n(y)f(x_0 + y)dy\\ &=& \int_{0}^{\pi}D_n(y)\{f(x_0 + y) + f(x_0 - y)\}dy. \end{eqnarray*}\]
Since the Dirichlet kernel is an even function and normalized,
\[2\int_{0}^{\pi}D_n(y)dy = \int_{-\pi}^{\pi}D_n(y)dy = 1.\]
Therefore, it suffices to show (why?) that
\[\begin{eqnarray*} J_1 &\equiv& \int_0^{\pi}D_n(y)\{f(x_0 + y) - f(x_0 + 0)\}dy \to 0,\\ J_2 &\equiv& \int_0^{\pi}D_n(y)\{f(x_0 - y) - f(x_0 - 0)\}dy \to 0. \end{eqnarray*}\]
Since \(f\) is piecewise continuous and smooth (by assumption), there exists a \(\delta > 0\) such that \(f'(x)\) is continuous and bounded for all \(x \in (x_0, x_0 + \delta)\). Let us define
\[\begin{eqnarray*} M_{\delta} &=& \sup_{x \in (x_0, x_0 + \delta)}|f'(x)|,\\ g(y) &=& f(x_0 + y) - f(x_0 + 0). \end{eqnarray*}\]
Then, for \(0 < y < \delta\),
\[|g(y)| = \left|\int_{0}^{y}f'(x_0 + t)dt\right| \leq \int_{0}^{y}|f'(x_0 + t)|dt \leq M_{\delta}y.\]
Therefore,
\[\int_{0}^{\delta}\left|\frac{1}{\sin\frac{y}{2}}g(y)\right|dy \leq M_{\delta}\int_0^{\delta}\frac{y}{\sin\frac{y}{2}}dy < +\infty.\]
For \(\delta < y < \pi\),
\[\sin\frac{y}{2} \geq \sin\frac{\delta}{2} > 0\]
so that
\[\int_{\delta}^{\pi}\left|\frac{1}{\sin\frac{y}{2}}g(y)\right|dy \leq \frac{1}{\sin\frac{\delta}{2}}\int_{\delta}^{\pi}|g(y)|dy < +\infty.\]
Therefore, \(g(y)/\sin(y/2)\) is integrable on \((0, \pi)\). By the Riemann-Lebesgue lemma, 
\[J_1 = \frac{1}{2\pi}\int_0^{\pi}\left(\sin\left(n + \frac{1}{2}\right)y\right)\cdot\frac{g(y)}{\sin\frac{y}{2}}dy \to 0.\]
We can prove that \(J_2 \to 0\) in a similar manner (exercise!). ■

Remark. If \(f\) is continuous at \(x_0\), its Fourier series at \(x_0\) converges to \(f(x_0)\). If we define the function \(\tilde{f}\) by
\[\tilde{f}(x) = \begin{cases} f(x) & \text{if $f$ is continuous at $x$},\\ \frac{1}{2}\{f(x-0) + f(x+0)\} & \text{if $f$ is discontinuous at $x$}, \end{cases}\]
then, the Fourier series of \(\tilde{f}\) converges to \(\tilde{f}\) at each point. □

Remark. The above theorem assumes that \(f'\) is piecewise continuous and bounded. This condition can be relaxed in various ways. For example, we may assume instead that
  • \(f'\) is piecewise continuous and square-integrable.
□

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