Pointwise convergence of Fourier series

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 $f^{\prime }$ 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 $f({\gamma }_j-0)=\underset{x\to {\gamma }_j-0}{lim}f(x)$ and the right limit $f({\gamma }_j+0)=\underset{x\to {\gamma }_j+0}{lim}f(x)$ exist at each ${\gamma }_j$. In fact, if $x_1,x_2\in ({\gamma }_{j-1},{\gamma }_j)$, then

$$|f(x_1)-f(x_2)|=|{\int }_{x_1}^{x_2}{f}^{\prime }(\xi )d\xi |\le M_{j-1}|x_1-x_2|$$

where $M_{j-1}=\underset{\xi \in ({\gamma }_{j-1},{\gamma }_j)}{sup}|f^{\prime }(\xi )|$. Thus,

$$\underset{x_1,x_2\to {\gamma }_j-0}{lim}|f(x_1)-f(x_2)|=0.$$

Similarly, $\underset{x_1,x_2\to {\gamma }_{j-1}+0}{lim}|f(x_1)-f(x_2)|=0$.

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,

$${\int }_a^b|g(x)|dx<+\mathrm{\infty }.$$

Then,

$${\int }_a^b{e}^{i\lambda x}g(x)dx\to 0(\lambda \to \pm \mathrm{\infty }).$$

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),

$$\begin{array}{rcl}{S}_n(x_0)& =& {\int }_{x_0-\pi }^{x_0+\pi }{D}_n(y-x_0)f(y)dy\\ & =& {\int }_{-\pi }^{\pi }{D}_n(y)f(x_0+y)dy\\ & =& {\int }_{-\pi }^0{D}_n(y)f(x_0+y)dy+{\int }_0^{\pi }{D}_n(y)f(x_0+y)dy\\ & =& {\int }_0^{\pi }{D}_n(y)\{f(x_0+y)+f(x_0-y)\}dy.\end{array}$$

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

$$\begin{array}{rcl}{J}_1& \equiv & {\int }_0^{\pi }{D}_n(y)\{f(x_0+y)-f(x_0+0)\}dy\to 0,\\ J_2& \equiv & {\int }_0^{\pi }{D}_n(y)\{f(x_0-y)-f(x_0-0)\}dy\to 0.\end{array}$$

Since $f$ is piecewise continuous and smooth (by assumption), there exists a $\delta >0$ such that $f^{\prime }(x)$ is continuous and bounded for all $x\in (x_0,x_0+\delta )$. Let us define

$$\begin{array}{rcl}{M}_{\delta }& =& \underset{x\in (x_0,x_0+\delta )}{sup}|f^{\prime }(x)|,\\ g(y)& =& f(x_0+y)-f(x_0+0).\end{array}$$

Then, for $0<y<\delta$,

$$|g(y)|=|{\int }_0^y{f}^{\prime }(x_0+t)dt|\le {\int }_0^y|f^{\prime }(x_0+t)|dt\le M_{\delta }y.$$

Therefore,

$${\int }_0^{\delta }|\frac{1}{\sin \frac{y}{2}}g(y)|dy\le M_{\delta }{\int }_0^{\delta }\frac{y}{\sin \frac{y}{2}}dy<+\mathrm{\infty }.$$

For $\delta <y<\pi$,

$$\sin \frac{y}{2}\ge \sin \frac{\delta }{2}>0$$

so that

$${\int }_{\delta }^{\pi }|\frac{1}{\sin \frac{y}{2}}g(y)|dy\le \frac{1}{\sin \frac{\delta }{2}}{\int }_{\delta }^{\pi }|g(y)|dy<+\mathrm{\infty }.$$

Therefore, $g(y)/\sin (y/2)$ is integrable on $(0,\pi )$. By the Riemann-Lebesgue lemma, 

$$J_1=\frac{1}{2\pi }{\int }_0^{\pi }(\sin (n+\frac{1}{2})y)\cdot \frac{g(y)}{\sin \frac{y}{2}}dy\to 0.$$

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

$$\tilde{f}(x)=\{\begin{array}{ll}f(x)& \text{if }f\text{ is continuous at }x,\\ \frac{1}{2}\{f(x-0)+f(x+0)\}& \text{if }f\text{ is discontinuous at }x,\end{array}$$

then, the Fourier series of $\tilde{f}$ converges to $\tilde{f}$ at each point. □

Remark. The above theorem assumes that $f^{\prime }$ is piecewise continuous and bounded. This condition can be relaxed in various ways. For example, we may assume instead that

• $f^{\prime }$ is piecewise continuous and square-integrable.

□