Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 9 / 3 / Solution

Use additive notation and fix Haar measure $m$ on $G$. A continuous unitary character is a continuous homomorphism $\chi :G\to \mathbb{T}$. With the conventionthe Fubini theorem and the character identity imply $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Also $|\hat{f}(\chi )|\le \Vert f{\Vert }_1$, and choosing $f=\chi a$ for a nonzero nonnegative $a\in C_c(G)$ shows that evaluation at $\chi$ is a nonzero functional. Thus each character gives a character of an algebra on the L1 convolution algebra. Here a multiplicative linear functional means a nonzero multiplicative linear functional; the zero functional must be excluded for the claimed bijection.

Conversely, let $\phi$ be such a functional. It is automatically bounded with $|\phi (f)|\le \Vert f{\Vert }_1$: extend it to the unitization of an algebra by $\tilde{\phi }(f+z1)=\phi (f)+z$. If $|z|>\Vert f{\Vert }_1$, the Neumann series makes $z1-f$ invertible, so its image under the unital homomorphism cannot be zero. Hence $\phi (f)$ cannot lie outside the norm disc.

Choose $a\in L^1(G)$ with $\phi (a)\ne 0$, and let $T_{x}a(t)=a(t-x)$. The Abelian translation identities $(T_{x}a)\ast b=a\ast (T_{x}b)$ giveTherefore, defining $c(x)=\phi (T_{x}a)/\phi (a)$, one has $\phi (T_{x}b)=c(x)\phi (b)$ for every $b$. Applying this to two successive translations gives $c(x+y)=c(x)c(y)$ and $c(0)=1$. By translation continuity in Lp on a locally compact group, $c$ is continuous. It is uniformly bounded by $\Vert a{\Vert }_1/|\phi (a)|$; applying this bound to $c(nx)=c(x{)}^n$ for positive and negative integers forces $|c(x)|=1$. Thus $\chi (x)=\overline{c(x)}$ is a continuous unitary character.

The Bochner integral identity $a\ast b={\int }_{G}b(y)T_{y}adm(y)$ holds in $L^1$, since $\Vert T_{y}a{\Vert }_1=\Vert a{\Vert }_1$. Applying the bounded functional yieldsCancel $\phi (a)\ne 0$ to obtain the required correspondenceThe recovery formula $\overline{\chi (x)}=\phi (T_{x}a)/\phi (a)$ proves uniqueness. These functionals have norm exactly one, since $b=\chi a$ with $a\ge 0$ and $\int a=1$ has $\Vert b{\Vert }_1=\phi (b)=1$.

For the topology, identify the Pontryagin dual group with these functionals and give it the Gelfand topology, namely pointwise convergence of $\hat{f}(\chi )$ for every $f\in L^1$. If a net ${\chi }_{\lambda }$ converges to $\gamma$ uniformly on every compact subset of $G$, then for $f\in C_c(G)$,Density of $C_c$ in $L^1$ and the uniform functional norm bound one extend this convergence to every $f\in L^1$.

Conversely, suppose convergence in the Gelfand topology and choose $a$ with $\hat{a}(\gamma )\ne 0$. For compact $K\subseteq G$, the set $\{T_{x}a:x\in K\}$ is compact in $L^1$ by translation continuity in Lp on a locally compact group. Evaluation convergence is uniform on this compact set: a finite $\delta$-net reduces its supremum to finitely many convergent evaluations plus $2\delta$, using the common functional norm one. ThusNow $\widehat{T_{x}a}(\chi )=\overline{\chi (x)}\hat{a}(\chi )$, and $\hat{a}({\chi }_{\lambda })\to \hat{a}(\gamma )\ne 0$. Dividing this identity proves ${\sup }_{x\in K}|{\chi }_{\lambda }(x)-\gamma (x)|\to 0$. Hence the Gelfand topology is exactly the compact-open topology, with the requested neighbourhood basisFor compact $K$, the continuous difference attains its maximum, so this strict pointwise inequality on $K$ is equivalent to a strict uniform bound. Finite intersections contain another such set by taking the union of the compact sets and the minimum of the tolerances.