Riemann–Lebesgue lemma for abelian groups

Remarks

Proof idea: by the description of the Gelfand spectrum of L1(G), $\widehat{f}$ is the Gelfand transform of $f$ in the commutative Banach algebra L1(G). Gelfand transforms are continuous and vanish at infinity on the character space, which is $\widehat{G}$. A direct proof approximates $f$ in $L^1$ by functions whose transforms are known to vanish at infinity, and uses $\abs{\widehat{f}(\chi) - \widehat{g}(\chi)} \le \norm{f - g}_1$.

Example: for $G = \R$ this is the classical Riemann–Lebesgue lemma. For $G = \mathbb{T}$ it says that the Fourier coefficients of an integrable function on the circle tend to $0$. For $G = \Z$ the dual $\mathbb{T}$ is compact, and the statement only says that $\widehat{f}$ is continuous.

The hypothesis $f \in L^1(G)$ cannot be replaced by a finite measure. The Dirac measure at $0$ has Fourier–Stieltjes transform equal to $1$, which does not vanish at infinity on a non-compact dual.

Going deeper

For a student

The lemma says that the Fourier transform maps $L^1(G)$ into $C_0(\widehat{G})$, the continuous functions vanishing at infinity. This map is injective, but it is not onto when $\widehat{G}$ is infinite. Together with the bound $\norm{\widehat{f}}_\infty \le \norm{f}_1$, it makes the Fourier transform a norm-decreasing algebra homomorphism $L^1(G) \to C_0(\widehat{G})$.