Haar averaging of bounded operators as a weak operator integral

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff group with normalized Haar probability measure $\mu$ (Normalized Haar probability on a compact group). Under the assumed Axiom of Choice the Axiom of Countable Choice holds as well (AC supplies the countable and dependent choices used in Banach integration).

Let $\pi :K\to U(H)$ and $\sigma :K\to U(J)$ be strongly continuous unitary representations of $K$ on complex Hilbert spaces $H$ and $J$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space), and let $T\in \mathcal{B}(H,J)$ be a bounded linear operator (A bounded linear operator between normed spaces). The pairing below is linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

The scalar integrals. Fix $v\in H$ and $w\in J$ and put

$b_v(w):={\int }_K⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩d\mu (k)\mathrm{.}$

The integrand is continuous on $K$. Indeed, $k\mapsto T\pi (k{)}^{-1}v$ is continuous because $\pi$ is strongly continuous and $T$ is bounded, and for a continuous map $\psi$ into $J$ the map $k\mapsto \sigma (k)\psi (k)$ is continuous because every $\sigma (k)$ is an isometry: for fixed $k_0$,

$\mathrm{\parallel }\sigma (k)\psi (k)-\sigma (k_0)\psi (k_0)\mathrm{\parallel }\le \mathrm{\parallel }\psi (k)-\psi (k_0)\mathrm{\parallel }+\mathrm{\parallel }(\sigma (k)-\sigma (k_0))\psi (k_0)\mathrm{\parallel },$

and both terms tend to zero as $k\to k_0$. A continuous complex function on the compact space $K$ is bounded and Borel, so the displayed integral is a finite complex number; and $\mid ⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩\mid \le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }$ for every $k$ by the Cauchy–Schwarz inequality (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs), so $\mathrm{\mid }{b}_v(w)\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }$. For fixed $v$ the assignment $w\mapsto b_v(w)$ is conjugate-linear and $\mathrm{\parallel }w\mathrm{\parallel }\mapsto \mathrm{\parallel }{b}_v(w)\mathrm{\parallel }$ is bounded by $\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$; hence

${\phi }_v(w):=\overline{b_v(w)}={\int }_K⟨w,\sigma (k)T\pi (k{)}^{-1}v⟩d\mu (k)$

is a bounded linear functional on $J$ of norm at most $\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$.

The definition. Under Countable Choice the Hilbert space $J$ has the Riesz representation property (Riesz representation for Hilbert spaces): there is a unique vector $A(T)v\in J$ with ${\phi }_v(w)=⟨w,A(T)v⟩$ for every $w\in J$, equivalently

$⟨A(T)v,w⟩=b_v(w)={\int }_K⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩d\mu (k)(v\in H,w\in J)\mathrm{.}$

This determines $A(T)$ as a map $H\to J$, the Haar average of $T$ with respect to the pair $(\sigma ,\pi )$.

Well-definedness. For each $v$ the vector $A(T)v$ is unique, so $A(T)$ is a well-defined function. It is linear: if $v=a{v}_1+b{v}_2$ then, by the conjugate-linearity in $f$ of the representing vector recorded in Riesz representation for Hilbert spaces, the vector representing ${\phi }_{a{v}_1+b{v}_2}=\bar{a}{\phi }_{v_1}+\bar{b}{\phi }_{v_2}$ is $aA(T)v_1+bA(T)v_2$; and $\mathrm{\parallel }A(T)v\mathrm{\parallel }=\mathrm{\parallel }{\phi }_v\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ because the Riesz representation is isometric, so $A(T)$ is a bounded linear operator with $\mathrm{\parallel }A(T)\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$. The assignment $A$ is itself linear in $T$: for fixed $v,w$ the integrand is linear in $T$, so the scalar integral is, and equality of the weak matrix elements together with uniqueness of the Riesz vector gives $A(aT+b{T}^{\mathrm{\prime }})=aA(T)+bA(T^{\mathrm{\prime }})$. Thus $A:\mathcal{B}(H,J)\to \mathcal{B}(H,J)$ is a well-defined map.

This is a weak operator integral. Only scalar functions are integrated in the definition: for fixed $v$ and $w$, the continuous function $k\mapsto ⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩$ is integrated against the scalar measure $\mu$, and the vector $A(T)v$ is then produced by the Riesz representation theorem. No operator-norm continuity of the orbit $k\mapsto \sigma (k)T\pi (k{)}^{-1}$ of an arbitrary bounded operator $T$ is assumed or asserted. For finite-rank $T:H\to J$, it follows from Finite-rank conjugation orbits are operator-norm continuous as follows. Equip $H\oplus J$ with the sum inner product; it is a Hilbert space because Cauchy sequences converge in each coordinate. The representation $\tau (k)(v,w)=(\pi (k)v,\sigma (k)w)$ is unitary and strongly continuous. The bounded finite-rank operator $S(v,w)=(0,Tv)$ satisfies $\tau (k)S\tau (k{)}^{-1}(v,w)=(0,\sigma (k)T\pi (k{)}^{-1}v)$. The cited lemma makes this conjugation orbit norm continuous, and restriction to $H\oplus \{0\}$ followed by projection onto $J$ does not increase operator norm, proving the claimed continuity for $T$. Only the Axiom of Choice is used, through normalized Haar measure and Countable Choice. That $A$ is an idempotent contraction onto the bounded intertwiners, with operator norm one exactly when ${\mathrm{Hom}}_K(H,J)\mathrm{\ne }\{0\}$, is the content of Haar averaging projects contractively onto the bounded intertwiners ↗, which justifies the present definition.