KindiakMath

Fourier Convergence

In a sense, I’m most excited to begin this segment in the discussion on complex analysis. Fourier analysis has been surprisingly fascinating for me to self-study, but has eluded my genuine study since I did not take complex analysis seriously as an undergraduate student.

To recap, for 2\pi-periodic functions f : \mathbb R \to \mathbb R, f = f(\cdot + 2\pi) that are integrable on [-\pi, \pi], we can write their Fourier series S(f) in the form

\displaystyle (S_n(f))(t) := \sum_{k=-n}^{n} \hat f(k) e^{i kt},\quad \hat f(k) := \frac 1{2\pi} \int_{-\pi}^{\pi} f(t) e^{-ikt}\, \mathrm dt,

where the right-hand side is the integral of the complex-valued function \mathbb R \to \mathbb C, t \mapsto f(t) e^{-ikt}, prompting the use of complex-analytic techniques in the future. The function \hat f : \mathbb R \to \mathbb C defined by

\displaystyle \hat f(\xi) := \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-i \xi t}\, \mathrm dt

also turns out to be the famous Fourier transform that we will explore later on as well (where f is suitably defined, of course).

However, most computational treatments assume, rather than prove, that S_n(f) \to f as n \to \infty via whatever meaningful mode of convergence. Having developed needful complex analytic machinery, we aim to establish said convergence under sufficiently mild conditions on f. We will follow the exposition written here.

Theorem 1. Let f : [-\pi, \pi] \to \mathbb R be continuously differentiable so that its periodic extension given by f = f(\cdot + 2\pi) is piecewise continuously differentiable. Define \tilde f = f on points of continuity and

\displaystyle \tilde f(t) := \frac{f(t^-) + f(t^+)}{2},\quad f(t^{\pm}) := \lim_{\delta \to 0^+} f(t \pm \delta),

whenever t is a point of discontinuity (i.e. t is an odd multiple of \pi). Then \displaystyle S_n(f) \to \tilde f pointwise.

Theorem 1 tells us that most of the various Fourier series problems students encounter are indeed point-wise equal to the functions they represent (in fact, many of these problems are designed to be continuous even at odd multiples of \pi), so that direct substitution (e.g. f(0) = (S_\infty(f))(0)) for problem solving is mathematically valid.

Lemma 1. For any t \in \mathbb R,

\displaystyle (S_n(f))(t) = \frac 1{\pi} \int_{-\pi}^{\pi} f(u) \cdot D_n(t-u)\, \mathrm du,

where the function D_n defined by

\displaystyle D_n(\theta) := \frac{ \sin((n+1/2) \theta) }{ 2 \sin( \theta /2) }

is called the Dirichlet kernel, and the integral (without the 1/\pi factor) is known as the convolution of f and D_n, sometimes abbreviated to f * D_n.

Proof. By expanding the definitions of all relevant terms (here, we regard \hat f := \hat f|_{[-\pi, \pi]}),

\begin{aligned} (S_n(f))(t) &= \sum_{k=-n}^{n} \hat f(k) e^{i kt} \\ &= \sum_{k=-n}^{n} \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} f(u) e^{-i k u}\, \mathrm du \right) e^{i kt} \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(u) \left( \sum_{k=-n}^{n} e^{i k (t-u)} \right)\, \mathrm du. \end{aligned}

To simplify the sum, we use a geometric series with common ratio r := e^{i(t-u)} and first term r^{-n}:

\begin{aligned} \sum_{k=-n}^{n} e^{i k (u-t)} &= \frac {r^{-n} \cdot (1 - r^{2n+1})}{1 - r} = \frac{ r^{-(n+1/2)} - r^{n+1/2} }{ r^{-1/2}-r^{1/2} }\end{aligned}

Denoting \theta := t-u, we observe that for any m, using Euler’s formula,

\begin{aligned} r^{-m} - r^m &= e^{-i m \theta} - e^{i m \theta} = -2i \cdot \frac{ e^{i m \theta} - e^{-i m \theta} }{ 2i } = -2i \cdot \sin(m \theta). \end{aligned}

Hence,

\begin{aligned} \sum_{k=-n}^{n} e^{-i k \theta} = \frac{ -2i \cdot \sin((n+1/2) \theta) }{ -2i \cdot \sin( \theta /2) } = \frac{ \sin((n+1/2) \theta) }{ \sin( \theta /2) }.\end{aligned}

Lemma 2. For any n,

\begin{aligned}(\tilde f - S_n(f))(t) &= \frac 1{2 \pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot (\sin(nv) \cot(v/2) + \cos(nv))\, \mathrm dv.\end{aligned}

Proof. By Lemma 1, since D_n is an even function,

\begin{aligned} (S_n(f))(t) &= \frac 1{\pi} \int_{-\pi}^{\pi} f(u) \cdot D_n(t-u)\, \mathrm du \\ &= \frac 1{\pi} \int_{-\pi-t}^{\pi-t} f(t + v) \cdot D_n (-v) \, \mathrm dv \\ &= \frac 1{\pi} \int_{-\pi-t}^{\pi-t} f(t + v) \cdot D_n (v) \, \mathrm dv \\ &= \frac 1{\pi} \left( \int_{-\pi-t}^{-\pi} + \int_{-\pi}^{\pi} + \int_{\pi}^{\pi - t} \right)f(t + v) \cdot D_n (v) \, \mathrm dv \\ &= \frac 1{\pi} \left( \int_{-\pi-t}^{-\pi} + \int_{-\pi}^{\pi} - \int_{\pi-t}^{\pi } \right)f(t + v) \cdot D_n (v) \, \mathrm dv \\ &= \frac 1{\pi}\int_{-\pi}^{\pi} f(t + v) \cdot D_n (v) \, \mathrm dv \end{aligned}

where \int_{-\pi-t}^{-\pi} = \int_{\pi-t}^{\pi} by 2\pi-periodicity. Now

\begin{aligned} \int_{-\pi}^{\pi} D_n(v)\, \mathrm dv &= \frac 12 \cdot \sum_{k=-n}^{n} \int_{-\pi}^{\pi} e^{-i k v}\, \mathrm dv \\ &= \frac 12 \cdot \int_{-\pi}^{\pi} e^{-i 0 v}\, \mathrm dv \\ &= \frac 12 \cdot 2\pi = \pi, \end{aligned}

so that

\displaystyle \frac 1{\pi} \int_{-\pi}^{\pi} \tilde f(t) \cdot D_n(v)\, \mathrm dv = \tilde f(t)

implies that

\begin{aligned}(\tilde f - S_n(f))(t) &= \frac 1{\pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot D_n(v)\, \mathrm dv \\ &= \frac 1{2 \pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot (\sin(nv) \cot(v/2) + \cos(nv))\, \mathrm dv. \end{aligned}

Lemma 3. For f continuously differentiable on [-\pi, \pi], \hat f(\xi) \to 0 for natural numbers \xi \to \infty. This result can be further generalised into the Riemann-Lebesgue lemma.

Proof. Denote M := \max_{t \in [-\pi, \pi]} |f'(t)|, which exists since f' is continuous. Integrating by parts,

\begin{aligned}|\hat f(\xi)| &= \left| \int_{-\pi}^{\pi} f(t) e^{-i \xi t}\, \mathrm dt \right| \\ &= \left| \left[ \frac{ f(t) e^{-i \xi t} }{ -i \xi } \right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} \frac{ f'(t) e^{-i \xi t} }{ -i \xi }\, \mathrm dt \right| \\ &\leq \frac 1{\xi} \left[ |f(\pi)| + | f(-\pi ) | + \int_{-\pi}^{\pi} |f'(t)|\, \mathrm dt \right] \\ &\leq \frac 1{\xi} \cdot [ |f(\pi)| + | f(-\pi ) | + 2\pi M]. \end{aligned}

Taking \xi \to \infty, |\hat f(\xi)| \to 0, which implies \hat f(\xi) \to 0.

Proof of Theorem 1. We pick up from Lemma 2:

\begin{aligned}(\tilde f - S_n(f))(t) &= \frac 1{2 \pi} \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot (\sin(nv) \cot(v/2) + \cos(nv))\, \mathrm dv.\end{aligned}

We claim that the two integrals I_1(n), I_2(n) defined by

\begin{aligned} I_1(n) &= \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot \sin(nv) \cot(v/2) \, \mathrm dv,\\ I_2(n) &= \int_{-\pi}^{\pi} (\tilde f(t) - f(t+v)) \cdot \cos(nv)\, \mathrm dv,\end{aligned}

tend to 0 as n \to \infty. For the first integral, we use the classic tangent limit

\begin{aligned} \lim_{v \to 0} \frac{ \tilde f(t) - f(t+v) }{ \tan(v/2) } &= 2 \cdot \lim_{v \to 0} \frac{f(t) - f(t+v)}{v} \cdot \lim_{v \to 0} \frac{v/2}{\tan(v/2)} = 2 f'(t), \end{aligned}

so that the function g(t):= (\tilde f(t) - f(t+v)) \cdot \cot(v/2) is bounded, and hence piecewise continuously differentiable on [-\pi, \pi]. By Lemma 3,

\displaystyle |I_1(n)| \leq |\hat g(n)| \to 0.

Since h(t):= \tilde f(t) - f(t+v) is continuously differentiable on [-\pi, \pi], by Lemma 3, |I_2(n)| \leq |\hat h(n)| \to 0.

Therefore, for any t \in \mathbb R

\begin{aligned} |(\tilde f - S_n(f))(t)| &\leq |I_1(n)| + |I_2(n)| \\ &\to 0 + 0 = 0, \end{aligned}

yielding S_n(f) \to \tilde f point-wise, as required.

Couple point-wise convergence with an integrable upper-bound, we can use the dominated convergence theorem to resolve most integration-related problems. For differentiation, however, we would require the stronger technology of uniform convergence. Whether we need such technology or not, however, is up for grabs.

For now, we want to develop the notion of the Fourier transform more carefully. This we will do in the next post.

—Joel Kindiak, 25 Aug 25, 2244H

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