33.1 Pontryagin Duality

Proof, [Theorem 4.3, Fol16]. (Continuous): Let $\seqf{f_j}\subset L^{1}(G; \complex)$ and $\eps > 0$. By inner regularity of the Haar measure, there exists $K \subset G$ compact such that $\int_{G\setminus K}|f_{j}| < \eps$ for all $1 \le j \le n$. In which case, for any $\phi, \psi \in \wh G$ with $\sup_{x \in K}|\phi(x) - \psi(x)| < \eps$,

(Injective): By (3) of Definition 32.2.1, $\phi$ is uniquely determined by $\pi_{\phi}$.

(Surjective): Let $\pi \in \Omega(L^{1}(G; \complex))$. Since $G$ is Abelian, $L^{1}(G; \complex)$ is also Abelian by Proposition 31.4.2. As such, for any $x \in G$ and $f, g \in L^{1}(G; \complex)$, $f * (L_{x}g) = L_{x}(f * g) = (L_{x}f) * g$ and $\pi(L_{x}f)\pi(g) = \pi(f)\pi(L_{x}g)$. In particular, if $\pi(f), \pi(g) \ne 0$, then $\pi(L_{x}f)/\pi(f) = \pi(L_{x}g)/\pi(g)$. Let^[1]

then $\phi$ is defined independently of the choice of $f$. By continuity of translation, $\phi$ is a continuous homomorphism from $G$ to $\complex \setminus \bracsn{0}$. Moreover, by Proposition 38.7.3, $\norm{\pi}_{L^1(G; \complex)^*}\le 1$, and $\phi$ is bounded. Given that $\phi$ is a homomorphism, $\phi(G) \subset \partial B_{\complex}(0, 1)$, and $\phi \in \wh G$.

Since $\pi \ne 0$, there exists $\psi \in L^{1}(G; \complex)$ with $\pi(\psi) = 1$. Using Riemann sums,

(Embedding): By Proposition 38.7.3, every multiplicative linear functional on $L^{1}(G; \complex)$ is contractive. Hence $\Omega(L^{1}(G; \complex))$ is equicontinuous. By the Arzelà-Ascoli Theorem, the topology of compact convergence on $\Omega(L^{1}(G; \complex))$ coincides with the weak* topology.

Let $\phi, \psi \in \wh G$. Since $\phi \ne 0$, there exists $f \in L^{1}(G; \complex)$ such that $\pi_{\phi}(f) \ne 0$. If $|\pi_{\phi}(f) - \pi_{\psi}(f)|$ is sufficiently small, then $\pi_{\psi}(f) \ne 0$ as well. Let $K \subset G$ be compact, then $\bracsn{L_xf|x \in K}$ is compact in $L^{1}(G; \complex)$ by continuity of translation. In which case,

Thus for any $\eps > 0$, $\bracsn{\pi_\psi|\psi \in \wh G, \sup_{x \in K}|\phi(x) - \psi(x)| \le \eps}$ is a neighbourhood of $\pi_{\phi}$ in the weak* topology, and the mapping $\psi \mapsto \pi_{\psi}$ is an embedding.$\square$

Proof, [Proposition 4.4, Fol16]. Let $\phi, \psi \in \wh G$ be distinct, then there exists $x_{0} \in G$ with $\dpn{x_0, \phi}{G}\dpn{x_0, \psi}{G}^{-1}\ne 1$. In which case,

so $\phi \perp \psi$.$\square$

Proof, [Proposition 4.5, Fol16]. If $G$ is discrete, then $L^{1}(G; \complex)$ is a unital Banach algebra. By Definition 38.7.4, $\wh G \iso \Omega(L^{1}(G; \complex))$ is compact.

If $G$ is compact, then $\one \in L^{1}(G; \complex)$, and $\bracsn{f \in L^\infty(G; \complex)|\int f| > 1/2}$ is weak*-open. By Proposition 33.1.3, for any $\phi \in \wh G \setminus \bracsn{\one}$, $\int \phi = 0$. Therefore $\bracsn{1}$ is open in $\wh G$, and $\wh G$ is discrete.$\square$

Proof, [Proposition 4.6, Fol16]. (1): Let $\phi \in \wh \real$, then $\dpn{0, \phi}{\real}= 1$. Thus there exists $\alpha > 0$ with $A := \int_{0}^{\alpha} \dpn{x, \phi}{\real}dx \ne 0$. In which case,

so $\phi \in C^{1}(\real)$ and there exists $c \in \complex$ with $\phi' = c\phi$. Therefore $\dpn{x, \phi}{\real}= e^{cx}$ for all $x \in \real$.$\square$