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480 H. Schultz / Journal of Functional Analysis 236 (2006) 457–489<br />

Repeating this argument, we find that for arbitrary θ ∈[0, 2π[,<br />

i.e.<br />

Since θ was arbitrary, we conclu<strong>de</strong> that<br />

and this proves (6.7). ✷<br />

supp(μ e iθ (S+T) ) ⊆ z ∈ C Re z r ′ (S) + r ′ (T ) ,<br />

supp(μS+T ) ⊆ z ∈ C Re e −iθ z r ′ (S) + r ′ (T ) .<br />

supp(μS+T ) ⊆ B 0,r ′ (S) + r ′ (T ) ,<br />

Lemma 6.4. Let S,T ∈ M be commuting operators, and <strong>le</strong>t α, β ∈ C. Then μαS,βT is the pushforward<br />

mea<strong>sur</strong>e of μS,T via the map hα,β : C × C → C × C given by<br />

hα,β(z, w) = (αz, βw).<br />

Proof. Recall that μαS,βT is uniquely <strong>de</strong>termined by the property that for all B1,B2 ∈ B(C),<br />

Now, it is easily seen that for α = 0 and β = 0,<br />

Hence,<br />

μαS,βT (B1 × B2) = τ PαS(B1) ∧ PβT (B2) . (6.8)<br />

PαS(B1) = PS<br />

<br />

1<br />

α B1<br />

<br />

<br />

1<br />

μαS,βT (B1 × B2) = τ PS<br />

α B1<br />

<br />

1<br />

∧ PT<br />

and PβT (B2) = PT<br />

β B2<br />

<br />

1<br />

β B2<br />

<br />

.<br />

<br />

1<br />

= μS,T<br />

α B1 × 1<br />

β B2<br />

<br />

−1<br />

= μS,T hα,β (B1 × B2) . (6.9)<br />

If for instance α = 0, then PαS(B1) = 0if0/∈ B1 and PαS(B1) = 1 if 0 ∈ B1. It then follows that<br />

(6.9) holds in this case as well. Simi<strong>la</strong>r arguments apply if β = 0. ✷<br />

Proof of Theorem 6.1. As noted in Remark 6.2, it suffices to prove that for all α, β ∈ C, μαS+βT<br />

is the push-forward mea<strong>sur</strong>e of μS,T via the map (z, w) ↦→ αz + βw. At first we will consi<strong>de</strong>r<br />

the case α = β = 1. Define a : C × C → C by<br />

We are going to prove that for all B ∈ B(C),<br />

a(z,w) = z + w (z,w∈ C).<br />

−1<br />

μS+T (B) = μS,T a (B) . (6.10)

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