Let φ∈ Pos(G). The function φ is bounded on G and one has ∥φ∥∞=φ(e). Moreover, one has
∣φ(g−1h)−φ(h)∣⩽√2φ(e)(φ(e)−R(φ(g)))(1)
for every (g,h)∈G×G.
Let (ϱ,x) be a Hilbert realization of φ. One has φ(e)=∥x∥2. For every g∈G, one therefore has ∣φ(g)∣=∣⟨x∣ϱ(g)x⟩∣⩽φ(e) by the Cauchy-Schwarz inequality. This proves the first assertion.
For every (g,h)∈G×G, one has
φ(g−1h)−φ(h)=⟨x∣ϱ(g−1h)x⟩−⟨x∣ϱ(h)x⟩=⟨ϱ(g)x−x∣ϱ(h)x⟩
since ϱ is unitary, whence
∣φ(g−1h)−φ(h)∣⩽∥ϱ(g)x−x∥∥ϱ(h)x∥⩽φ(e)1/2∥ϱ(g)x−x∥
We conclude by observing that
∥ϱ(g)x−x∥2=2∥x∥2−2R(⟨x∣ϱ(g)x⟩)=2(φ(e)−R(φ(g)))
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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