A First Theorem on Pointwise Convergence of Fourier Series
Given a Riemann-integrable function on \([0,L]\), let the \(N\)-th symmetric partial sum of the Fourier series of \(f\) be the quantity
\[ S_N f(x) := \sum_{n=-N}^N \widehat{f}(n) e^{\frac{2 \pi i x n}{L}}\]
where it should be recalled that
\[ \widehat{f}(n) := \frac{1}{L} \int_0^L e^\frac{-2\pi i t n}{L} f(t) dt. \]
1. The Dirichlet Kernel
Proposition
Let
\[ D_N (x) := \sum_{n=-N}^N e^\frac{2 \pi i n x}{L}. \]
This quantity is called the \(N\)-th Dirichlet kernel. Then
\[ S_N f(x) = \frac{1}{L} \int_0^L D_N(x-t) f(t) dt. \]
The \(N\)-th Dirichlet kernel satisfies the following properties:
- \(D_N\) is periodic of period \(L\).
- For each \(x \in {\mathbb R}\),\[ D_N(x) = \begin{cases} 2N+1 & x \in L {\mathbb Z} \\ \frac{\sin ((2N+1)\pi x/L)}{\sin (\pi x / L)} & x \not \in L {\mathbb Z} \end{cases}. \]
- For each integer \(n\),\[ \widehat{D_N}(n) = \begin{cases} 1 & |n| \leq N \\ 0 & \text{otherwise} \end{cases}. \]
In particular,
\[ \int_0^1 D_N(t) dt = 1. \]
Proof
2. Convolution
Proposition
Suppose that \(f\) and \(g\) are periodic of period \(L\) and are both Riemann integrable. Then
\[{}\frac{1}{L} \int_0^L f(t) g(x-t) dt{}\]
\[{}= \frac{1}{L} \int_0^L f(x-t) g(t) dt{}\]
\[{}= \frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f(t) g(x-t) dt{}\]
\[{}= \frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f(x-t) g(t) dt.{}\]
This expression is called the convolution (or, less commonly, the circular or periodic convolution) of \(f\) and \(g\), and is denoted \(f * g(x)\). In particular, \(f * g(x) = g * f (x)\).
Proof
3. A Pointwise Convergence Theorem
Using the fact that the integral of the Dirichlet kernel is \(1\) allows us to write a useful integral formula for the difference \(S_N f(x) - f(x)\):
\[{}S_N f(x) - f(x){}\]
\[{}= \frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f(x-t) D_N(t) dt - f(x){}\]
\[{}= \frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} f(x-t) D_N(t) dt{}\]
\[{}- f(x) \frac{1}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} D_N(t) dt{}\]
\[{}= \frac{1}{L}\int_{-\frac{L}{2}}^{\frac{L}{2}} (f(x-t) - f(x)) D_N(t) dt{}\]
When \(t \not \in L {\mathbb Z}\),
\[{}(f(x-t) - f(x)) D_N(t){}\]
\[{}= \frac{f(x-t)-f(x)}{\sin (\pi t / L)} \sin \frac{(2N+1)\pi t}{L}{}\]
\[{}= \frac{f(x-t)-f(x)}{2i \sin (\pi t / L)} e^{\frac{i \pi t}{L}} e^{\frac{2 \pi i N t}{L}}{}\]
\[{}- \frac{f(x-t)-f(x)}{2i \sin (\pi t / L)} e^{-\frac{i \pi t}{L}} e^{-\frac{2 \pi i N t}{L}}{}\]
If \(f\) happens to be differentiable at \(x\), then the ratio \((f(x-t)-f(x))/\sin (\pi t/L)\) will be bounded near \(t = 0\) and consequently will be Riemann integrable. It follows that for this fixed \(x\) and all \(t \in [-L/2,L/2]\),
\[{}(f(x-t) - f(x)) D_N(t){}\]
\[{}= h_1(t) e^{\frac{2 \pi i N t}{L}} + h_2(t) e^{-\frac{2 \pi i N t}{L}}{}\]
for some Riemann integrable functions \(h_1\) and \(h_2\). Thus
\[ S_N f(x) - f(x) = \widehat{h_1}(-N) + \widehat{h_2}(N). \]
By the Riemann-Lebesgue Lemma, the right-hand side tends to zero as \(N \rightarrow \infty\). We have proved the following theorem:
Theorem
Suppose that \(f\) is an \(L\)-periodic function which is Riemann integrable. Then at every point \(x \in {\mathbb R}\) for which \(f\) is differentiable, the symmetric partial sums of the Fourier series of \(f\) at \(x\) converge and converge to \(f(x)\), i.e.,
\[ \sum_{n=-\infty}^\infty \widehat{f}(n) e^{\frac{2 \pi i n x}{L}} = f(x). \]
Important
The criterion of differentiability of \(f\) is to be understood as applying to the periodic function. So, for example, taking
\[ \widehat{f}(n) := \int_0^1 x e^{-2 \pi i n x} dx, \]
one has
\[ \sum_{n=-\infty} \widehat{f}(n) e^{2 \pi i n x} \neq x \text{ when } x \not \in (0,1) \]
because the periodic function which agrees with \(f(x) :=x\) on the interval \([0,1]\) is not differentiable at \(x = 0,1\) and is not equal to \(x\) when \(x \not \in [0,1]\) (e.g., the periodic function equals \(x-1\) on \((1,2)\) and so on).