Math 202B: Lecture 10

Let

G = C_{n_1} \times \dots \times C_{n_r}

be an abelian group, and let

\Lambda = \mathbb{Z}_{n_1} \times \dots \times \mathbb{Z}_{n_r}

be its dual group. In Lecture 8 we showed that the convolution algebra \mathcal{C}(G) admits a basis \{F^\lambda \colon \lambda \in \Lambda\} consisting of orthogonal projections, called its Fourier basis. Thus any function A \in \mathcal{C}(G) can be expanded in the group basis,

A = \sum\limits_{g \in G} A(g) E_g,

and in the Fourier basis

A = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda) F^\lambda.

The coefficients of the Fourier expansion of A \in \mathcal{C}(G) give a function \widehat{A} \in \mathcal{F}(\Lambda) defined by \lambda \mapsto \widehat{A}(\lambda).

Definition 9.1. The Fourier transform on \mathcal{C}(G) is the map \mathcal{C}(G) \to \mathcal{F}(\Lambda) defined by A \mapsto \widehat{A}.

In Lecture 3, we proved that any algebra which admits a basis of orthogonal projections is isomorphic to a function algebra, and the proof of this general fact given there applies here in particular, allowing us to conclude that the Fourier transform A \mapsto \widehat{A} is an algebra isomorphism \mathcal{C}(G) \to \mathcal{F}(G). Thus, we have

\widehat{A*B} = \widehat{A} \widehat{B},

where on the left hand side

[A*B](g) = \sum\limits_{h \in G} A(gh^{-1})B(h), \quad g \in G,

is the convolution product of two functions in \mathcal{C}(G), and on the right hand side

[\widehat{A}\widehat{B}](\lambda) = \widehat{A}(\lambda)\widehat{B}(\lambda), \quad \lambda \in \Lambda

is the pointwise product of two functions in \mathcal{F}(\Lambda).

By definition, the elements of the Fourier basis of \mathcal{C}(G) are

F^\lambda = \frac{1}{|G|} \sum\limits_{g \in G} \chi^\lambda(g) E_g,

where \{\chi^\lambda \colon \lambda \in \Lambda\} is the character basis of \mathcal{C}(G), which consists of all group homomorphisms G \to U(\mathbb{C}).

Theorem 9.1. For any A \in \mathcal{C}(G), we have

\widehat{A}(\lambda) = \frac{1}{|G|}\langle \chi^\lambda,A\rangle,

where the right hand side is an \ell^2-scalar product in \mathcal{C}(G).

Proof: Taking a slightly different normalization of the Fourier basis, we get an orthonormal basis of \mathcal{C}(G) consisting of the functions

E^\lambda = \sqrt{|G|}F^\lambda = \frac{1}{\sqrt{|G|}}\chi^\lambda, \quad \lambda \in \Lambda.

Because this basis is orthonormal, we have

A = \sum\limits_{\lambda \in \Lambda} \langle E^\lambda,A\rangle E^\lambda,

for any function A \in \mathcal{C}(G), and comparing with the Fourier expansion of A this gives

\widehat{A}(\lambda) = \frac{1}{\sqrt{|G|}}\langle E^\lambda,A\rangle E^\lambda=\frac{1}{|G|}\langle \chi^\lambda,A\rangle.

-QED

Theorem 9.1 is called the Fourier inversion formula, because

\widehat{A}(\lambda) = \frac{1}{|G|}\langle \chi^\lambda,A\rangle = \frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)}A(g)

recovers the values of the transformed function \widehat{A} \in \mathcal{F}(\Lambda) in terms of the values of the original function A \in \mathcal{C}(G). As an example, let us calculate the Fourier transform of the elementary function E_h \in \mathcal{C}(G) corresponding to a given group element h \in G. From the inversion formula, the Fourier transform of the indicator function of h is

\widehat{E_h}(\lambda) = \frac{1}{|G|}\langle \chi^\lambda,E_h\rangle = \frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)}E_h(g) = \frac{1}{|G|} \overline{\chi^\lambda(h)}.

Equivalently, we have

\widehat{E_h} = \frac{1}{|G|}\sum\limits_{\lambda \in \Lambda} \overline{\chi^\lambda(g)}E_\lambda,

where \{E_\lambda \colon \lambda \in \Lambda\} is the elementary basis of \mathcal{F}(\Lambda) consisting of indicator functions E_\lambda(\mu) = \delta_{\lambda\mu}.

Theorem 9.2. (Plancherel and Parseval formulas) For any A,B \in \mathcal{C}(G), we have

\langle \widehat{A},\widehat{B}\rangle=|G|\langle A,B\rangle

and

\|\widehat{A}\|_2 = \sqrt{|G|}\|A\|_2.

Proof: From the definition of the Fourier transform, we have

\langle A,B \rangle = \left\langle \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)F^\lambda,\sum\limits_{\mu \in \Lambda} \widehat{B}(\mu)F^\mu\right\rangle=\sum\limits_{\lambda \in \Lambda}\sum\limits_{\mu \in \Lambda}\overline{\widehat{A}(\lambda)}\widehat{B}(\mu)\langle F^\lambda,F^\mu\rangle=\sum\limits_{\lambda \in \Lambda}\overline{\widehat{A}(\lambda)}\widehat{B}(\lambda)\frac{1}{|G|}.

This is the Plancherel formula, and the Parseval formula is the special case where A=B, together with the definition of the norm associated to a scalar product.

-QED

The \ell^2-norm on functions A \colon G\to \mathbb{C} the p=2 case of the \ell^p-norm, which for any p>0 is defined by

\|A\|_p =\left(\sum\limits_{g\in G}|A(g)|^p\right)^{\frac{1}{p}}.

Also familiar from analysis is the case p=\infty, which is defined by

\|A\|_\infty = \max\limits_{g \in g} |A(g)|.

The intuition behind this definition is that, as p \to \infty the sum \|A\|_p^p is approximately equal to its largest term, and this concentration becomes exact in the p=\infty limit. This intuition leads to the following inequality.

Proposition 11.3. For any p>0 and A \colon G\to \mathbb{C}, we have

\|A\|_p^p \leq \|A\|^p_\infty |\mathrm{supp}(A)|,

where by definition the support of A is the cardinality of the complement of its vanishing set,

\mathrm{supp}(A) = \{g\in G\colon A(g) \neq 0\}.

Problem 11.2. Prove Proposition 11.3.

We now prove a discrete version of the famous uncertainty principle, whose origins are in quantum mechanics; in Fourier analysis, this principle effectively says that if the support of a function is small than the support of its Fourier transform is large.

Theorem 11.4. (Uncertainty principle) For any nonzero function A \in \mathcal{C}(G), we have

|\mathrm{supp}(A)||\mathrm{supp}(\widehat{A}| \geq |\Lambda|.

Proof: Start from the Fourier expansion of A, which in terms of the character basis \{\chi^\lambda \colon \lambda \in \Lambda\} of \mathcal{C}(G) is

A = \frac{1}{|G|} \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)\chi^\lambda.

For any g \in G, we have

|A(g)| =\left|\frac{1}{|G|} \sum\limits_{\lambda \in \Lambda} \widehat{A}\chi^\lambda(g)\right| \leq \frac{1}{|G|}\sum\limits_{\lambda \in \Lambda} |\widehat{A}(\lambda)| =\frac{1}{|G|}\sum\limits_{\lambda \in \mathrm{supp}(\widehat{A})} |\widehat{A}(\lambda)|,

where we used the triangle inequality and the fact that |\chi^\lambda(g)|=1. Since g\in G was arbitrary, the above shows that

\|A\|_\infty \leq \frac{1}{|G|}\sum\limits_{\lambda \in \mathrm{supp}(\widehat{A})} |\widehat{A}(\lambda)|.

Squaring both sides, we thus have

\|A\|_\infty^2 \leq \frac{1}{|G|^2}\left(\sum\limits_{\lambda \in \Lambda} \delta_{\lambda \in \mathrm{supp}(A)} |\widehat{A}(\lambda)|\right)^2\leq \frac{1}{|G|^2} |\mathrm{supp}(\widehat{A})| \|\widehat{A}\|_2^2=\frac{1}{|G|} |\mathrm{supp}(\widehat{A})| \|A\|_2^2\leq \frac{1}{|G|} |\mathrm{supp}(\widehat{A})| \|A\|_\infty^2|\mathrm{supp}(A)|,

where we applied the Cauchy-Schwarz inequality, the Parseval formula, and Proposition 11.3. Since \|A\|_\infty^2 \neq 0, it can be cancelled from both sides of the above inequality, leaving

1 \leq \frac{1}{|\Lambda|} |\mathrm{supp}(\widehat{A})||\mathrm{supp}(A)|.

-QED

Problem 11.3. Prove that the uncertainty principle is sharp by exhibiting a function A \colon G\to \mathbb{C} for which equality holds in Theorem 11.4.

Published by Jonathan Novak

Mathematician at UC San Diego.

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