Advanced Analysis

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
Meta (Key Ideas)
Substitute the definition of the Fourier coefficients into the formula for \(S_N\) and then simplify exponentials and pull the sum inside the integral.

To express \(D_N\) exactly as a ratio of sines, observe that \(D_N\) is already given as a finite geometric series with common ratio equal to \(e^{2 \pi i x/L}\). Careful simplification gives the stated formula.

To find the Fourier coefficients of \(D_N\), use the formula for \(D_N\) as a sum of exponentials and use the orthogonality properties of those exponentials.

Below is a plot of the Diriclet kernels \(D_4, D_8\), and \(D_{16}\) on the interval \([-\frac{3}{2},\frac{3}{2}]\) (here \(L = 1\) for simplicity). Notice that the kernels become more peaked near the integers as the parameter increases. Notice also that the kernels are highly oscillatory and are not particularly small at points inbetween the integers.
Figure. Plot of \(D_4, D_8\), and \(D_{16}\)

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
Meta (Key Ideas)
For any fixed \(x\), \(f(t) g(x-t)\) is periodic of period \(L\) as a function of \(t\). In particular, for any real \(a\) and any \(b \in (a,a+L)\), this means that
\[{}\int_{a}^{a +L} f(t) g(x-t) dt{}\]
\[{}= \left(\int_{a}^b + \int_b^{a+L} \right) f(t) g(x-t) dt{}\]
\[{}= \int_a^b f(t) g(x-t) dt{}\]
\[{}+ \int_b^{a+L} f(t-L) g(x-t+L) dt{}\]
\[{}= \int_a^b f(t) g(x-t) dt{}\]
\[{}+ \int_{b-L}^{a} f(t+L) g(x-t-L) dt{}\]
\[{}= \int_{b-L}^{b} f(t)g(x-t) dt{}\]
by virtue of the \(L\)-periodicity of the integrand and a change of variables \(t \mapsto t + L\) in the second integral in the third equality. The case \(a = 0\) and \(b = L/2\) gives one of the asserted equalities. In the remaining cases, apply a change of variables \(t \mapsto x - t\):
\[{}\int_{0}^{L} f(t) g(x-t) dt{}\]
\[{}= \int_{x-L}^{x} f(x-t)g(t) dt.{}\]
If \(x \in (0,L)\), fixing \(a := x-L\) and \(b = 0\) achieves the desired equality. If \(x = L\), there is nothing to prove; and in all remaining cases, periodicity of \(f(t) g(x-t)\) and a change of variables shows that \(f*g\) must also be periodic of period \(L\).

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).