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Riemann-Lebesgue lemma on LCA groups

Statement

Assume the Axiom of Choice (The Axiom of Choice) and Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $G$ be a locally compact Hausdorff abelian group with Haar measure $m_{G}$ and dual $\hat{G}$ . For every $f \in L^{1} (G, m_{G})$ its Fourier transform satisfies $\hat{f} \in C_{0} (\hat{G}), ∥ \hat{f} ∥_{\infty} \leq ∥ f ∥_{1} .$ Thus $f \mapsto \hat{f}$ maps $L^{1} (G, m_{G})$ into the Banach space $C_{0} (\hat{G})$ , and the vanishing at infinity is uniform, not merely along sequences.

Facts & Assumptions

Given: The Axiom of Choice and Dependent Choice, a locally compact Hausdorff abelian group $G$ with Haar measure $m_{G}$ , its dual $\hat{G}$ , and $f \in L^{1} (G, m_{G})$ .

[F1]

The Fourier transform is $\hat{f} (γ) = \int_{G} f (x) \overline{γ (x)} d m_{G} (x)$ , it is well defined on $L^{1} (G, m_{G})$ and linear, and $∣ \hat{f} (γ) ∣ \leq ∥ f ∥_{1}$ for every $γ$ (The Fourier transform on an LCA group).

[F2]

For $g \in C_{c} (G; C)$ the support of $g$ is compact with $m_{G} (supp g) < + \infty$ ; the evaluation pairing $(γ, x) \mapsto γ (x)$ is jointly continuous; and the compact-open topology of $\hat{G}$ is the topology of uniform convergence on compact subsets of $G$ (The Pontryagin dual with the compact-open topology, Evaluation of characters is jointly continuous, Compact support, $C_{c} (X)$ , and $C_{0} (X)$ , Haar measure is positive on nonempty open sets and finite on compact sets).

[F3]

Real $C_{c} (G)$ is dense in real $L^{1} (G, m_{G})$ . Applying this separately to $Re f$ and $Im f$ gives $a_{n}, b_{n} \in C_{c} (G; R)$ with both $L^{1}$ errors tending to zero; hence $g_{n} = a_{n} + i b_{n} \in C_{c} (G; C)$ satisfies $∥ f - g_{n} ∥_{1} \leq ∥ Re f - a_{n} ∥_{1} + ∥ Im f - b_{n} ∥_{1} \to 0$ (C_c(X) is dense in L^p(mu) for a Radon measure).

[F4]

Assume the Axiom of Choice and Dependent Choice. In $A^{+} = C \oplus A$ the characters are exactly $q (z, h) = z$ and $h_{γ}^{+} (z, h) = z + \hat{h} (γ)$ , the Gelfand transform $χ \mapsto χ (0, f)$ of $(0, f)$ is continuous on the compact Hausdorff space $Δ (A^{+})$ , $q (0, f) = 0$ and $h_{γ}^{+} (0, f) = \hat{f} (γ)$ , and $Δ (A^{+})$ is homeomorphic to $\hat{G} \cup {q}$ with $\hat{G}$ carrying its compact-open topology (Scalar unitisation of L^1 of an LCA group: characters, spectrum and identity criterion, Gelfand transform, Maximal ideal space is compact Hausdorff, The Axiom of Choice).

Proof

technique · direct

1.1F2

(Continuity on compactly supported functions.) Let $g \in C_{c} (G; C)$ and let $γ_{i} \to γ_{0}$ be a net in $\hat{G}$ . By [F2], $γ_{i} \to γ_{0}$ uniformly on the compact set $supp g$ , and $m_{G} (supp g) < + \infty$ , so $∣ \hat{g} (γ_{i}) - \hat{g} (γ_{0}) ∣ \leq \int_{supp g} ∣ g (x) ∣ ∣ γ_{i} (x) - γ_{0} (x) ∣ d m_{G} (x) \leq ∥ g ∥_{\infty} m_{G} (supp g) \sup_{x \in supp g} ∣ γ_{i} (x) - γ_{0} (x) ∣ \to 0.$ Hence $\hat{g}$ is continuous.

1.2F1

(Norm bound.) For every $γ \in \hat{G}$ , $∣ \hat{f} (γ) ∣ \leq \int_{G} ∣ f (x) ∣ ∣ γ (x) ∣ d m_{G} (x) = ∥ f ∥_{1}$ , so $∥ \hat{f} ∥_{\infty} \leq ∥ f ∥_{1}$ .

1.3F4

(Compact superlevel sets.) Let $F : Δ (A^{+}) \to C$ be the Gelfand transform $F (χ) : = χ (0, f)$ . By [F4], $F$ is continuous, $Δ (A^{+})$ is compact, $F (h_{γ}^{+}) = \hat{f} (γ)$ and $F (q) = 0$ . For $ε > 0$ the set $L_{ε} : = {χ \in Δ (A^{+}) : ∣ F (χ) ∣ \geq ε}$ is closed in $Δ (A^{+})$ , hence compact, and it does not contain $q$ ; therefore $L_{ε} = {γ \in \hat{G} : ∣ \hat{f} (γ) ∣ \geq ε}$ is a compact subset of $\hat{G}$ for its compact-open topology. Thus every superlevel set of $∣ \hat{f} ∣$ is compact.

2.1F1F3step 1.1

(Continuity in general.) Choose $g_{n} \in C_{c} (G; C)$ with $∥ f - g_{n} ∥_{1} \to 0$ by [F3]. Then $∥ {\hat{g}}_{n} - \hat{f} ∥_{\infty} \leq ∥ g_{n} - f ∥_{1} \to 0$ by [F1], so $\hat{f}$ is the uniform limit of the continuous functions ${\hat{g}}_{n}$ of step 1.1; hence $\hat{f}$ is continuous.

3.1F4step 1.3step 2.1

( $\hat{f} \in C_{0} (\hat{G})$ .) By step 2.1, $\hat{f}$ is continuous on $\hat{G}$ , and by step 1.3 every set ${γ : ∣ \hat{f} (γ) ∣ \geq ε}$ is compact; this is exactly the definition of $\hat{f} \in C_{0} (\hat{G})$ (Compact support, $C_{c} (X)$ , and $C_{0} (X)$ ).

4.1F1step 1.2step 1.3step 2.1step 3.1∎

Steps 2.1 and 3.1 show $\hat{f} \in C_{0} (\hat{G})$ , step 1.2 gives $∥ \hat{f} ∥_{\infty} \leq ∥ f ∥_{1}$ , and linearity of the transform makes $f \mapsto \hat{f}$ a map into $C_{0} (\hat{G})$ . For each $ε > 0$ , the transform has magnitude less than $ε$ outside a compact set, which is the stated uniform vanishing at infinity.

Depends on

Used by

Dependency tree · two levels

78 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lynn H. Loomis, Introduction to Abstract Harmonic Analysis, D. Van Nostrand, 1953 (Harvard-hosted full scan) (standard reference, not scraped)
Manfred Einsiedler and Thomas Ward, Ergodic Theory with a View Towards Number Theory, Appendix C.2-C.3 (course-hosted full text) (standard reference, not scraped)