Bochner Inversion and Plancherel on LCA Groups — Examples

For $b=(z,f)$ and $c=(v,g)$, the norm estimate in [F1] gives $\mathrm{\parallel }bc\mathrm{\parallel }\le \mathrm{\mid }zv\mathrm{\mid }+\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_1+\mathrm{\mid }v\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_1+\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1=(\mathrm{\mid }z\mathrm{\mid }+\mathrm{\parallel }f{\mathrm{\parallel }}_1)(\mathrm{\mid }v\mathrm{\mid }+\mathrm{\parallel }g{\mathrm{\parallel }}_1)$. A Cauchy sequence in $B$ has Cauchy scalar and $A$ coordinates; completeness of $\mathbb{C}$ from [F3] and of $A$ from [F1] makes it converge in the sum norm.

Bilinearity and commutativity follow from [F1], and $(1,0)$ is the identity. For $b=(z,f)$, $c=(v,g)$ and $d=(u,k)$, either bracketing of $bcd$ has scalar part $zvu$ and $A$ part $zvk+zug+vuf+z(g\ast k)+v(f\ast k)+u(f\ast g)+(f\ast g)\ast k$, by convolution associativity. Conjugate-linearity, involutivity and isometry of the star follow coordinatewise from [F1]; expanding the product and applying $(f\ast g{)}^{\ast }=g^{\ast }\ast f^{\ast }$ gives $(bc{)}^{\ast }=c^{\ast }{b}^{\ast }$.

In the conjugate-phase notation of [F1], set $\gamma =\bar{w}$. Then $h_{\gamma }^{+}(z,f)=z+{\int }_{G}f(t)\overline{\gamma (t)}dm(t)=h_w(z,f)$. The map $w\mapsto \bar{w}$ is its own inverse and preserves uniform convergence on each compact set, hence is a homeomorphism of the compact-open dual. Therefore [F1] makes every character of $B$ one of the displayed $h_w$ or $q$, with the asserted topology. Applying [F2] gives ${\mathrm{\Gamma }}_B(z,f)(h_w)=h_w(z,f)=z+\widehat{f}(w)$ and ${\mathrm{\Gamma }}_B(z,f)(q)=q(z,f)=z$. Taking $z=0$ and restricting to the dual gives the Fourier/Gelfand identity.

(The dual identifications.) A character $\chi :\mathbb{Z}\to \mathbb{T}$ is determined by $z:=\chi (1)$, and each $z\in \mathbb{T}$ gives ${\gamma }_z(k)=z^k$. This is a group isomorphism $\mathbb{T}\to \widehat{\mathbb{Z}}$. Its inverse is evaluation at $1$, while each evaluation $z\mapsto z^k$ is continuous; [F2] therefore makes this a homeomorphism. For a character $\chi :\mathbb{T}\to \mathbb{T}$, lift $t\mapsto \chi (e^{2\pi it})$ to $\mathbb{R}$ and apply [F1]; periodicity forces the resulting frequency to be an integer. Thus every character is $z\mapsto z^n$ for a unique $n\in \mathbb{Z}$. Since the compact-open topology on this dual is uniform on $\mathbb{T}$, and ${\sup }_{z\in \mathbb{T}}\mathrm{\mid }{z}^n-z^m\mathrm{\mid }=2$ for $n\mathrm{\ne }m$, $\widehat{\mathbb{T}}$ is discrete.

(Inversion on $\mathbb{T}$.) Let $f\in L^1(\mathbb{T})$ with $a_n:=\widehat{f}(n)$ satisfying ${\sum }_n\mathrm{\mid }{a}_n\mathrm{\mid }<\mathrm{\infty }$. The series $g(z):={\sum }_{n\in \mathbb{Z}}{a}_n{z}^n$ converges uniformly; [F3] shows its Fourier coefficients are $a_n$. Its Fejer means are ${\sigma }_{N}f(z)={\sum }_{\mathrm{\mid }n\mathrm{\mid }\le N}(1-\frac{\mathrm{\mid }n\mathrm{\mid }}{N+1})a_n{z}^n\mathrm{.}$ Absolute summability implies these weighted sums converge uniformly to $g$: first bound the tail by ${\sum }_{\mathrm{\mid }n\mathrm{\mid }>M}\mathrm{\mid }{a}_n\mathrm{\mid }$, then let $N\to \mathrm{\infty }$ on the finite central sum. By [F4], ${\sigma }_{N}f\to f$ in $L^1$; uniform convergence also gives ${\sigma }_{N}f\to g$ in $L^1$. Uniqueness of limits yields $f=g$ almost everywhere. Thus counting measure on $\mathbb{Z}$ gives the displayed dual inversion formula.

The measures in [F3] are Haar measures on the two groups. Under the identifications of step 1.1, the Fourier transforms are $\widehat{f}(z)={\sum }_{k\in \mathbb{Z}}f(k)z^{-k}$ for $f\in {\mathrm{\ell }}^1(\mathbb{Z})$ and $\widehat{f}(n)={\int }_{\mathbb{T}}f(z)z^{-n}d{m}_{\mathbb{T}}(z)$ for $f\in L^1(\mathbb{T})$.

(Inversion on $\mathbb{Z}$.) For $f\in {\mathrm{\ell }}^1(\mathbb{Z})$ the series for $\widehat{f}$ converges absolutely and uniformly, hence is integrable. Termwise integration and [F3] give, for each $j\in \mathbb{Z}$, ${\int }_{\mathbb{T}}\widehat{f}(z)z^{j}d{m}_{\mathbb{T}}(z)={\sum }_{k\in \mathbb{Z}}f(k){\int }_{\mathbb{T}}{z}^{j-k}d{m}_{\mathbb{T}}(z)=f(j)\mathrm{.}$ Thus normalized arc measure gives the displayed inversion formula for counting measure on $\mathbb{Z}$.

Steps 1.1 and 2.1 identify the dual groups and Haar measures, step 3.1 proves inversion for $\mathbb{Z}$, and step 1.2 proves Fourier-series inversion on $\mathbb{T}$ for summable Fourier coefficients. These explicit pairs give the usual normalization conventions.

(Order and topology of the dual.) If $G$ is trivial, take the empty product; otherwise [F3] writes $G\cong Z_1\oplus \cdots \oplus Z_k$ with $Z_j=\mathbb{Z}\mathrm{/}{N}_j\mathbb{Z}$. The local computation in [F3] shows $\widehat{Z_j}\cong \mathbb{Z}\mathrm{/}{N}_j\mathbb{Z}$, so by the finite-product duality $\widehat{G}\cong {\prod }_j\mathbb{Z}\mathrm{/}{N}_j\mathbb{Z}$ and $\mathrm{\mid }\widehat{G}\mathrm{\mid }={\prod }_j{N}_j=\mathrm{\mid }G\mathrm{\mid }$. The same statement holds for the trivial group, whose dual is trivial. Thus $\widehat{G}$ is finite and discrete, and by [F2] counting measure is a Haar measure of total mass $\mathrm{\mid }\widehat{G}\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }$.

(Orthogonality.) For $z\in G$, if $z=0$ then $\gamma (z)=1$ for every $\gamma$ and ${\sum }_{\gamma \in \widehat{G}}\gamma (z)=\mathrm{\mid }\widehat{G}\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }$. If $z\mathrm{\ne }0$, step 1.1 and [F4] provide ${\gamma }_0$ with ${\gamma }_0(z)\mathrm{\ne }1$; since $\gamma \mapsto {\gamma }_0\gamma$ is a bijection of $\widehat{G}$, ${\sum }_{\gamma }\gamma (z)={\sum }_{\gamma }({\gamma }_0\gamma )(z)={\gamma }_0(z){\sum }_{\gamma }\gamma (z)$, hence ${\sum }_{\gamma \in \widehat{G}}\gamma (z)=0$. Therefore ${\sum }_{\gamma \in \widehat{G}}\gamma (x-y)=\mathrm{\mid }G\mathrm{\mid }{\delta }_{x,y}$ for all $x,y\in G$, which is the displayed orthogonality relation.

(Inversion.) For $f:G\to \mathbb{C}$ and $\gamma \in \widehat{G}$, the transform is $\widehat{f}(\gamma )=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_{x\in G}f(x)\overline{\gamma (x)}$ by [F5]. Hence, using the orthogonality of step 2.1, ${\sum }_{\gamma \in \widehat{G}}\widehat{f}(\gamma )\gamma (x)=\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{y\in G}f(y){\sum }_{\gamma \in \widehat{G}}\gamma (x-y)=\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{y\in G}f(y)\mathrm{\mid }G\mathrm{\mid }{\delta }_{x,y}=f(x),$ which is the displayed inversion formula.

(The compatible measure is counting measure.) Counting measure on the finite group $\widehat{G}$ is Haar. Any Haar measure $\nu$ on this finite group assigns the same mass $c$ to every point by translation invariance, so $\nu =c\mathrm{\#}$. If $\nu$ is compatible with the transform, applying inversion to ${\delta }_0$ gives $1={\int }_{\widehat{G}}\widehat{{\delta }_0}(\gamma )d\nu (\gamma )=c\mathrm{\mid }\widehat{G}\mathrm{\mid }\mathrm{/}\mathrm{\mid }G\mathrm{\mid }=c$, using $\mathrm{\mid }\widehat{G}\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }$ from step 1.1. Thus counting measure is the unique compatible dual Haar measure.

(Unitary normalisation.) Set $N:=\mathrm{\mid }G\mathrm{\mid }$ and $Uf(\gamma ):=N^{-1\mathrm{/}2}{\sum }_{x}f(x)\overline{\gamma (x)}=N^{1\mathrm{/}2}\widehat{f}(\gamma )$. By step 3.1, $N^{-1\mathrm{/}2}{\sum }_{\gamma \in \widehat{G}}Uf(\gamma )\gamma (x)={\sum }_{\gamma \in \widehat{G}}\widehat{f}(\gamma )\gamma (x)=f(x)\mathrm{.}$ Expanding the finite sum and using step 2.1 gives ${\sum }_{\gamma \in \widehat{G}}\mathrm{\mid }Uf(\gamma ){\mathrm{\mid }}^2=N^{-1}{\sum }_{x,y\in G}f(x)\overline{f(y)}{\sum }_{\gamma \in \widehat{G}}\gamma (y-x)={\sum }_{x\in G}\mathrm{\mid }f(x){\mathrm{\mid }}^2\mathrm{.}$ Hence $U$ is a linear isometry for the counting-measure norms; the displayed inversion and $\mathrm{\mid }\widehat{G}\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }$ make it bijective, so it is unitary. Its relation to the probability-Haar transform is $Uf=N^{1\mathrm{/}2}\widehat{f}$.

Step 2.1 gives the orthogonality relation, step 3.1 gives the nonunitary inversion formula, step 4.1 identifies counting measure as the compatible dual Haar measure, and step 4.2 gives the unitary DFT, its inverse coefficient $\mathrm{\mid }G{\mathrm{\mid }}^{-1\mathrm{/}2}$ and the output conversion $Uf=\mathrm{\mid }G{\mathrm{\mid }}^{1\mathrm{/}2}\widehat{f}$.

(Rank-one positivity.) For every finite family $x_1,\dots ,x_n\in G$ and coefficients $c_1,\dots ,c_n\in \mathbb{C}$, put $u_j:=\gamma (x_j)$. Since $\gamma$ is a homomorphism into the unit circle, $\gamma (x_j-x_k)=\gamma (x_j)\gamma (-x_k)=\gamma (x_j)\overline{\gamma (x_k)}=u_j\overline{u_k}$; consequently ${\sum }_{j,k}{c}_j\overline{c_k}\gamma (x_j-x_k)={\sum }_{j,k}{c}_j\overline{c_k}{u}_j\overline{u_k}=\mid {\sum }_j{c}_j{u}_j{\mid }^2\ge 0.$ For $n\ge 1$ the vector $u$ is nonzero because every $u_j$ has modulus one, so the matrix $u{u}^{\ast }$ is positive semidefinite of rank one. For $n=0$ its rank and quadratic form are zero. Thus $\gamma$ is positive definite and $\gamma (0)=1$.

(The point mass represents the character.) Assume now that $G$ is locally compact Hausdorff abelian, so that Bochner's theorem applies. The point mass ${\delta }_{\gamma }$ at the point $\gamma \in \widehat{G}$ is a probability measure and, by [F3], a finite positive Radon measure on $\widehat{G}$. Its inverse (Fourier-Stieltjes) transform is ${\int }_{\widehat{G}}\eta (x)d{\delta }_{\gamma }(\eta )=\gamma (x)(x\in G),$ because the function $\eta \mapsto \eta (x)$ agrees with the constant $\gamma (x)$ off the ${\delta }_{\gamma }$-null set $\widehat{G}\setminus \{\gamma \}$ (evaluation formula of [F3]; the coordinate functions are measurable by joint continuity). By [F4] this transform is continuous and positive definite, so the computation identifies $\gamma$ as the Fourier-Stieltjes transform of the finite positive Radon measure ${\delta }_{\gamma }$; by uniqueness in Bochner's theorem [F2] the point mass is the representing measure of $\gamma$, and its total mass is ${\delta }_{\gamma }(\widehat{G})=1=\gamma (0)$.

Step 1.1 proves that a character is positive definite with $\gamma (0)=1$ and exhibits its rank-one nonempty test matrices and rank-zero empty matrix; step 1.2 identifies the representing probability measure as the point mass ${\delta }_{\gamma }$.

The values of $\varphi$ at the differences of $x_1=0,x_2=1,x_3=2$ are $\varphi (0)=1$, $\varphi (\pm 1)=1$ and $\varphi (\pm 2)=0$, so the Hermitian matrix of [F2] is $M=\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\\ 0& 1& 1\end{array}\right)$. Its determinant is $1\cdot (1\cdot 1-1\cdot 1)-1\cdot (1\cdot 1-1\cdot 0)+0\cdot (1\cdot 1-1\cdot 0)=-1<0$. The explicit negative quadratic form in the next step establishes the failure of positive semidefiniteness directly.

Explicitly, the coefficients $c_1=1,c_2=-2,c_3=1$ give ${\sum }_{j=1}^3{\sum }_{k=1}^3{c}_j{c}_k\varphi (x_j-x_k)=1+4+1+2\cdot (-2)+2\cdot (-2)=-2<0,$ the quadratic form of $M$ being $c_1^2+c_2^2+c_3^2+2c_1{c}_2+2c_2{c}_3$; hence the defining inequality of [F2] fails for this finite family, and $\varphi$ is not positive definite, even though it is continuous with $\varphi (0)=1$ and $\mathrm{\mid }\varphi (x)\mathrm{\mid }\le 1$ everywhere.

(Singletons are Haar-null.) Put $a:=m_G(\{0\})$, which is finite because $\{0\}$ is compact. If $a>0$, translation invariance gives $m_G(\{x\})=a$ for every $x\in G$. The open neighborhood $U$ in [F1] is infinite: if it were finite, then $U\setminus \{0\}$ would be closed in the Hausdorff space $G$, making $\{0\}=U\cap (G\setminus (U\setminus \{0\}))$ open and $G$ discrete. For every positive integer $N$, choose $N$ distinct points of $U$; their disjoint singletons lie in $K$, so finite additivity gives $Na\le m_G(K)$. Since this holds for all $N$ and $m_G(K)<\mathrm{\infty }$, $a=0$. Translation invariance then gives $m_G(\{x\})=0$ for every $x\in G$.

(A point change preserves the class.) Define $\tilde{u}$ to agree with $u$ off $\{x_0\}$ and choose its value at $x_0$ to be any complex number different from $u(x_0)$ and from $F_f(x_0)$. Such a value exists because $\mathbb{C}$ is infinite. By step 1.1, $u$ and $\tilde{u}$ agree almost everywhere, so [F2] gives $[u]=[\tilde{u}]=f$.

(The inverse integral cannot distinguish the representatives.) By [F3], $\widehat{u}=\widehat{\tilde{u}}=\widehat{f}$ on $\widehat{G}$. Therefore both representatives give the same absolutely convergent inverse integral $F_f(x_0)$, while $\tilde{u}(x_0)\mathrm{\ne }{F}_f(x_0)$ by construction. This is an explicit failure of pointwise recovery for an arbitrary representative.

The modification leaves the $L^1$ class and its Fourier transform unchanged but changes the value at $x_0$; hence no inverse formula determined by the transform can hold everywhere for every representative.