Math 202B: Lecture 18

*** Problems in this lecture due March 1 ***

Any algebra \mathcal{A} which admits a basis \{F^\lambda \colon \lambda \in \Lambda\} of orthogonal projections is isomorphic to a function algebra, namely \mathcal{F}(\Lambda). The isomorphism \mathcal{A} \to \mathcal{F}(\Lambda) is called the Fourier transform on \mathcal{A}, and it sends A \in \mathcal{A} to the function \widehat{A} \in \mathcal{F}(\Lambda) uniquely determined by

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

We have implemented this abstract isomorphism concretely for the convolution algebra \mathcal{C}(G) of an abelian group G, which we take to be a product

G = G_1 \times \dots \times G_r

of cyclic groups. More precisely, G_i is a cyclic group of order n_i with generator g_i. The dual group is then

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

and we label elements of G with elements of \Lambda using the isomorphism \Lambda \to G which sends \alpha=(\alpha_1,\dots,\alpha_r) to

g_\alpha=(g_1^{\alpha_1},\dots,g_r^{\alpha_r}).

Last lecture we showed that the functions

\chi^\lambda(g_\alpha) = \omega_1^{\alpha_1\lambda_1} \dots \omega_r^{\alpha_r\lambda_r}, \quad \omega_k = \exp\left(\frac{2\pi i}{n_k}\right)

are precisely the multiplicative characters of G, and that \{\chi^\lambda \colon \lambda \in \Lambda\} is an orthogonal basis of \mathcal{C}(G),

\langle \chi^\lambda,\chi^\mu\rangle = \sum\limits_{g \in G} \overline{\chi^\lambda(g)}\chi^\mu(g)=\sum\limits_{g \in G} \chi^\lambda(g^{-1})\chi^\mu(g)=\delta_{\lambda\mu}|G|.

The normalized characters

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

form a Fourier basis \{F^\lambda \colon \lambda \in \Lambda\}. Thus for any function A \in \mathcal{C}(G) we have a corresponding Fourier expansion,

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

and this in turn defines the Fourier transform \mathcal{C}(G) \to \mathcal{F}(\Lambda), an algebra isomorphism.

Because we are working in a concrete setting, it is natural to ask for an explicit formula for the Fourier transform \widehat{A} \in \mathcal{F}(\Lambda) of a given function A \in \mathcal{C}(G).

Theorem 18.1. For any A \in \mathcal{C}(G), we have

\widehat{A}(\lambda) = \sum\limits_{g \in G} \overline{\chi^\lambda(g)}A(g), \quad \lambda \in \Lambda.

Problem 18.1. Prove Theorem 18.1.

For example, let us compute the Fourier transform of an elementary function E_g \in \mathcal{C}(G). Theorem 18.1 gives

\widehat{E}_g(\lambda) = \frac{1}{|G|}\sum\limits_{h \in G} \overline{\chi^\lambda(h)}E_g(h) = \frac{1}{|G|}\overline{\chi^\lambda(h)}, \quad \lambda \in \Lambda.

Notice that the function E_g is highly concentrated: it is nonzero at a single point, namely g \in G, where it has value 1. However, its Fourier transform \widehat{E}_g is very spread out: all of its values are complex numbers of modulus one, so it never equals zero. We will see below that this is a general phenomenon.

Because the Fourier transform is an isomorphism \mathcal{C}(G) \to \mathcal{F}(\Lambda), we can also ask for a formula which gives A in terms of \widehat{A}.

Theorem 18.2 (Fourier inversion). For any A \in \mathcal{C}(G), we have

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

Proof: The elementary expansion of A is

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

while its Fourier expansion is

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

Comparing the two yields the stated formula. \square

The Fourier inversion formula gives a useful extremal result: the maximum modulus of a function A \in \mathcal{C}(G) is bounded by the average modulus of its Fourier transform \widehat{A} \in \mathcal{C}(G).

Corollary 18.3 (Fourier bound). For all $g \in G,$ we have

|A(g)|\leq \frac{1}{|\Lambda|} \sum\limits_{\lambda \in \Lambda} |\widehat{A}(\lambda)|.

The above can also be written as the inequality

\|A\|_\infty \leq \frac{1}{|G|}\|A\|_1.

More generally, for any p >0 we have the corresponding p-norm on \mathcal{C}(G) defined by

\|A\|_p = \left( \sum_{g \in G} |A(g)|^p\right)^{1/p}.

The sup norm of a function A \in \mathcal{C}(G) is defined by

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

and heuristically it is the case p=\infty of the p-norm.

Theorem 18.4. (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

Proposition 18.5. 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 18.2. Prove Proposition 18.5.

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 18.6. (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

Published by Jonathan Novak

Mathematician at UC San Diego.

2 Comments

  1. Andrew Tan says:
    February 22, 2026 at 7:42 pm

    I think Theorem 18.1 as it is currently stated might be incorrect. I don’t think there should be a normalization by |G|, and the Fourier transform formula that was given in class doesn’t have it either.

    Reply
    1. Jonathan Novak says:
      February 23, 2026 at 7:37 am

      Corrected, thanks.

      Reply

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