8 Fourier Analysis
Organization:
- Fourier transform on a compact domain: the Poisson kernel, the continuous limit of the Poisson integral, orthonormal basis of \(L^2\), convergence conditions of the Fourier series, convolution and its norm properties.
- Haar measure: locally compact abelian groups and the construction of the Haar measure.
Key takeaways:
- The Fourier series of a continuous function
may fail to converge even on a compact domain,
but the following results hold:
- Strong convergence is guaranteed for twice continuously differentiable functions (proposition 8.6); the main idea here is to bound each term in the series using the derivative relation of Fourier coefficients.
- An Abel-sum like correction \(\mathcal P_{0\leq r<1} f\) (definition 8.3) corresponds to the Poisson integral, leading leading naturally to the definition 8.4 of the Poisson kernel, a disc version of the Dirac delta.
- The Fourier basis is an orthonormal basis of
\(L^2(\partial D)\) (proposition 8.7), so
every \(L^2\) function is the \(L^2\)-limit of its Fourier series.
- This holds by the special Hilbert space structure of \(L^2\).
- The definition of sums and a basis becomes nuanced in possibly uncountable-dimensional vector spaces (definition 8.7).
- Convolution is initially defined in \(L^1\) and satisfies a crucial norm property (proposition 8.8).
- The central convolution-multiplication duality 8.9 relies on a translation-invariant measure and the ability to exchange integrals.
Strengths of the topologies: convergence in \(L_p\) norm, pointwise convergence a.e, etc.
Compact domain: \(\partial D\)
Let \(D=\{z\in \mathbb C: |z|<1\}\) denote the open unit disk in the complex plane and \(\partial D\) its boundary consisting of unit-norm complex numbers.
Definition 8.1 (measurable subsets) Let \(\sigma\) be the measure on \(\partial E\) induced by the Lebesgue measure on \((-\pi, \pi]\) normalized so that \(\sigma(\partial D)=1\).
- \(E\subset \partial D\) is measurable if \(\{t\in (-\pi, \pi]: e^{it}\in E\}\) is a Borel subset.
- \(\sigma(E) = \mu(\{t\in (-\pi, \pi]: e^{it}\in E\})/2\pi\).
With this definition, we obtain \[ \int_{\partial D} F\, d\sigma = \dfrac 1 {2\pi} \int_{-\pi}^\pi f(e^{it})\, dt \]
For \(1\leq p\leq \infty\), the set \(L^p(\partial D)\) is defined w.r.t. the \(\sigma\) measure.
Proposition 8.1 (orthonormal family) \(\{z^n\}_{n\in \mathbb Z}\) is an orthonormal family in \(L^2(\partial D)\).
Note that, for \(z\in \partial D\), we have \(\bar z = 1/z\). Then for \(m\neq n\) \[\begin{align} \langle z^n, z^n\rangle &= \int_{\partial D} |z^n|^2\, d\sigma = 1 \\ \langle z^n, z^m\rangle &= \dfrac 1 {2\pi} \int_{-\pi}^\pi e^{imt} e^{-int}\, dt = \dfrac 1 {i(m-n)2\pi} e^{i(m-n)t} \big|_{t=-\pi}^\pi = 0 \end{align}\]
Definition 8.2 (fourier series) The Fourier series of \(f\in L^1(\partial D)\) is the formal sum \[ \sum_{n=-\infty}^\infty \hat f(n) z^n, \quad \hat f(n) = \int_{\partial D} f(z) \overline{z^n} = \dfrac 1 {2\pi} \int_{-\pi}^\pi f(e^{it}) e^{-int}\, dt \]
Note that we do need \(f\in L^1\) for the integral in \(\hat f(n)\) to make sense.
Lemma 8.1 (Riemann-Lebesgue lemma) For every \(f\in L^1(\partial D)\), we have \(\lim_{n\to \pm \infty} \hat f(n)=0\).
Proof
Since \((z^n)\) is an orthonormal family, we have \[ \sum_{n=-\infty}^\infty |\hat g(n)|^2 \leq \|g\|_2^2 < \infty \]The Poisson kernel
Note that there are continuous function in \(L^1(\partial D)\) whose Fourier series diverge. The next definition introduces a geometric “correction” of the Fourier series that converges; it also leads naturally to the definition of the Poisson kernel.
Definition 8.3 (convergent Fourier series) Given \(f\in L^1(\partial D)\) and \(0\leq r<1\), define \(\mathcal P_r(f):\partial D\to \mathbb C\) as the extended Abel sum of the Fourier series \[ (\mathcal P_r f)(z) = \sum_{n=-\infty}^\infty r^{|n|} \hat f(n)z^n \]
Definition 8.4 (Poisson kernel) Given \(0\leq r<1\), the Poisson kernel \(P_r: \partial D\to \mathbb R_{>0}\) is \[ P_r(\xi) = \dfrac{1-r^2}{|1-r\xi|^2} = \sum_{n=-\infty}^\infty r^{|n|}\xi^n \]
Equality between the two representations
\[\begin{align} P_r(\xi) &= \sum_{n=-\infty}^\infty r^{|n|} \xi^n = \sum_{n=0}^\infty (r\xi)^n + \sum_{n=1}^\infty (r\bar \xi)^n \\ &= \dfrac 1 {1-r\xi} + \dfrac{r\bar \xi}{1 - r\bar \xi} = \dfrac{1-r^2}{|1-r\xi|^2} \end{align}\]Proposition 8.2 (Poisson kernel and Poisson-approximate Fourier series) Given \(f\in L^1(\partial D)\) and \(0\leq r<1\), we have \[ (\mathcal P_r f)(z) = \int_{\partial D} f(w) P_r(z\bar w)\, d\sigma(w) \]
Proof
The partial sums converge uniformly on \(\partial D\), thus \(\mathcal P_r f\) is a continuous function. Unrolling definitions yield \[\begin{align} (\mathcal P_r f)(z) &= \sum_{n=-\infty}^\infty r^{|n|} \int_{\partial D} f(w) \overline{w^n} z^n \, d\sigma(w) \\ &= \int_{\partial d} f(w) \left(\sum_{n=-\infty}^\infty r^{|n|} (z \bar w)^n \right)\, d\sigma(w) \end{align}\] Swapping the sum and integral is justified by uniform convergence.Proposition 8.3 (properties of the Poisson kernel) The Poisson kernel can be viewed as approximations to the Dirac delta function on \(\partial D\) (approximate identity) since it satisfies the following properties: for all \(0\leq r<1\):
- \(P_r(\forall \xi\in \partial D) > 0\).
- \(\int_{\partial D} P_r(\xi)\, d\sigma(\xi) = 1\).
- For all \(\delta>0\), \(\lim_{r\to 1^-} \int_{|1-\xi|<\delta} P_r(\xi)\, d\sigma(\xi) = 1\).
Proof
Property (1) follows from definition; property (2) follows from integrating the integral representation and noting that \(\forall n\in \mathbb Z - \{0\}\), we have \(\int_{\partial D} \xi^{n}\, d\sigma(\xi)=0\). To show (3), by properties (1), (2) it suffices to show that \[ \lim_{r\to 1^-} \int_{|1-\xi|\geq \delta} P_r(\xi)\, d\sigma(\xi) = 0 \] Consider the denominator \(|1-r\xi|^2\) in the Poisson kernel: for \(|1-\xi|\geq \delta\) and \(1-r<\delta/2\), we have \[\begin{align} |1-r\xi| &= |1-\xi - (r-1)\xi| \\ &\geq |1-\xi| - |r-1| |\xi| \\ &\geq \delta - \delta/2 = \delta/2 \end{align}\] Thus the denominator is uniformly (w.r.t. \(r\)) bounded away from \(0\) while the numerator approaches \(0\).Convergence results
Proposition 8.4 (continuous limit of the Poisson integral) If \(f\) is continuous, then \(\lim_{r\to 1^-} \|f - \mathcal P_r f\|_\infty = 0\). In other words, \(\mathcal P_r f\) converges uniformly to \(f\).
Proof
\(f\) is uniformly continuous in \(\partial D\), and apply the approximate identity properties of \(P_r\) above.This does not contradict the fact that the Fourier series of some continuous functions in \(L^1(\partial D)\) diverge: in these cases, the \(\lim_{r\to 1^-}\) cannot be exchanged with the series sum.
Definition 8.5 (differentiation) Given \(f:\partial D\to \mathbb C\), we define \(f^{[k]}\) to be the \(k\)-th derivative \(\tilde f^{(k)}\) of \(f\) in the chart \([-\pi, \pi)\) with periodic boundary conditions, then expressed back in the \(\partial D\subset \mathbb C\) chart.
Proposition 8.5 (Fourier coefficients of derivatives) Given \(k\in \mathbb N\) and \(f:\partial D\to \mathbb C\) being \(k\)-times continuously differentiable, then \[ \widehat{f^{[k]}}(n) = (in)^k \hat f(n) \]
Proof
For \(n=0\), the integral vanishes due to periodic boundary conditions: \[ \widehat{f^{[k]}}(0) = \dfrac 1 {2\pi} \int_{-\pi}^\pi \tilde f^{(k)}(t) \, dt = \tilde f^{(k-1})(t) \big|_{-\pi}^\pi \] Next choose \(n\neq 0\) and apply integration by parts \[\begin{align} \widehat{f^{[k]}}(n) &= \dfrac 1 {2\pi} \int_{-\pi}^\pi \tilde f^{(k)}(t)e^{-int} \, dt \\ &= \dfrac 1 {2\pi} f^{(k-1)}(t) e^{-int} \big|_{-\pi}^\pi + \dfrac{in} {2\pi} \int_{-\pi}^\pi \tilde f^{(k-1)}(t) e^{-int}\, dt \\ &= in \widehat{f^{[k-1]}}(n) \end{align}\]Proposition 8.6 (convergence of twice continuously differentiable functions) Given \(f:\partial \mathbb C\) twice continuously differentiable,
then the following sum converges uniformly
\[
f(\forall z\in \partial D) = \sum_{n=-\infty}^\infty \hat f(n) z^n
\]
Proof
Proposition 8.5 implies that \[ |\hat f(n\neq 0)| = \dfrac{|\hat f(n)|}{n^2} \leq \dfrac{\|f^{[2]}\|_1}{n^2} \] The inequality follows from \(|z^n|\leq 1\) on \(\partial D|\) and Cauchy-Schwarz. Applying the dominated convergence theorem to the sum proves that the partial sums converge continuously (thus uniformly continuously) on \(\partial D\).Convolution and \(L^p\)
An orthonormal family \(\{e_k\in V\}_{k\in \Gamma}\) in a Hilbert space (complete inner product space) satisfies \(\langle e_j, e_k\rangle= \delta_{jk}\); note that \(\Gamma\) may be uncountable! Also recall the following:
Definition 8.6 (unordered sum) Given an unordered index set \(\Gamma\) and \(\{f_k\in V\}_{k\in \Gamma}\), where \(V\) is a normed vector space, the sum \(\sum_{k\in \Gamma}f_k\) is said to converge if there exists \(g\in V\) such that \(\forall \epsilon>0\), there exists a finite subset \(\Omega\subset \Gamma\) such that \[ \left\|g - \sum_{j\in \Omega'} f_j \right\| < \epsilon \] for all finite \(\Omega'\) such that \(\Omega\subset \Omega' \subset \Gamma\).
Definition 8.7 (orthonormal basis) An orthonormal family \(\{e_k\}_{k\in \Gamma}\) is an orthonormal basis of \(V\) if every element in \(V\) has a decomposition \[ f = \sum_{k\in \Gamma} \langle f, e_k\rangle e_k, \quad \sum_{k\in \Gamma} |\langle f, e_k\rangle|^2 < \infty \]
Proposition 8.7 (orthonormal Fourier basis) The Fourier basis \(\{z^n\}_{n\in \mathbb Z}\) is an orthonormal basis of \(L^2(\partial D)\).
Proof
Consider \(f\) not in the span of \(\{z^n\}\), we have \(\langle f, z^n\rangle= 0\). Next fix \(\epsilon>0\) and choose a twice continuously differentiable function (dense in \(L^2\) by approximating step functions) \(g\) such that \(\|f - g\|_2 < \epsilon\); note that \[\begin{align} \|g\|_2^2 &= \sum_{n\in \mathbb Z} |\hat g(n)|^2 = \sum_{n\in \mathbb Z} |\widehat{(g-f)}(n)|^2 \leq \|g-f\|_2^2 \end{align}\] The second equation holds by \(\hat f(n)=0\). Then \(\|f\|_2 \leq \|f-g\|_2 + \|g_2\| \leq 2\|f-g\|_2 = 2\epsilon\).Corollary 8.1 (convergence in L2) Given \(f\in L^2(\partial D)\), the Fourier series converges to \(f\) in the norm of \(L^2(\partial D)\).
Definition 8.8 (convolution on the disc) Given \(f, g\in L^1(\partial D)\), their convolution \(f\ast g\) is \[ (f\ast g)(z) = \int_{\partial D} f(w) g(z\bar w)\, d\sigma(w) \] for those \(z\) for which the integral makes sense.
Proposition 8.8 (norm property of convolution) Given \(f\in L^1(\partial D)\) and \(g\in L^{1\leq p\leq \infty}(\partial D)\), we have \[ \|f\ast g\|_p \leq \|f\|_1 \|g\|_p \]
Proof
First derive the following inequality for \(\|h\|_{\bar p}\leq 1\): distribute \(|\cdot|\), apply Tonelli, then apply Holder’s inequality. \[\begin{align} \|(f\ast g)h\|_1 &= \int_{\partial D} \left|\int_{\partial D} f(w) g(z\bar w)\, d\sigma(w)\, h(z)\right|\, d\sigma(z) \\ &\leq \int_{\partial D} |f(w)| \int_{\partial D} |h(z) g(z\bar w)|\, d\sigma(z) \, d\sigma(w) \\ &\leq \|h\|_{\bar p} \|g\|_p \int_{\partial D} |f(w)|\, d\sigma(w) \leq \|g\|_p \|f\|_1 \end{align}\] Next recall the variational \(p\)-norm formula (proposition 7.6) with suitable generalizations to edge cases to obtain \[ \|f\ast g\|_p = \sup_{\|h\in L^{\bar p}(\partial D)\|_{\bar p} = 1} \int_{\partial D} |(f\ast g) h|\, d\sigma \] Take \(\sup_h\) to obtain the desired bound.As corollaries of the following proposition, we obtain the commutativity and associativity of convolution.
Proposition 8.9 (convolution-multiplication duality) Given \(f, g\in L^1(\partial D)\) we have \[ \widehat{(f\ast g)}(n) = \hat f(n)\hat g(n) \]
Proof
First note that, by rotation invariance of the measure (generalized by the Haar measure), substitute \(z\bar w=\xi\iff z=\xi w\) to obtain \[\begin{align} \int_{\partial D} g(z\bar w) \overline{z^n}\, d\sigma(z) = \int_{\partial D} g(\xi) \overline{\xi^n w^n}\, d\sigma(\xi) = \bar w^n \hat g(n) \end{align}\] Next expand the definitions, apply Fubini, then recombine the integral to obtain \[\begin{align} \widehat{(f\ast g)}(n) &= \int_{\partial D} (f\ast g)(z)\overline{z^n}\, d\sigma(z) \\ &= \int_{\partial D} \bar z^n \int_{\partial D} f(w)g(z\bar w)\, d\sigma(w)\, d\sigma(z) \\ &= \int_{\partial D} f(w) \int_{\partial D} g(z\bar w)\bar z^n\, d\sigma(z)\, d\sigma(w) \\ &= \hat g(n) \int_{\partial D} f(w) \bar w^n \, d\sigma(w) = \hat f(n)\hat g(n) \end{align}\]Proposition 8.10 (convergence of the Poisson integral) Given \(f\in L^p(\partial D)\) for \(1\leq p<\infty\) \[ \lim_{r\to 1^-} \|f - \mathcal P_r f\|_p = 0 \]