Translation estimates for continuous positive type functions

For every $g\in G$ one has $\mathrm{\parallel }{\pi }_{\phi }(g)\xi -\xi {\mathrm{\parallel }}^2=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2-2\mathrm{Re}⟨{\pi }_{\phi }(g)\xi ,\xi ⟩+\mathrm{\parallel }\xi {\mathrm{\parallel }}^2=2\phi (e)-2\mathrm{Re}\phi (g)$: expanding the squared norm with [A1], unitarity gives $\mathrm{\parallel }{\pi }_{\phi }(g)\xi {\mathrm{\parallel }}^2=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$, and $⟨{\pi }_{\phi }(g)\xi ,\xi ⟩=\phi (g)$. This is claim 3 with $g=y^{-1}x$.

For claim 1, unitarity gives $\phi (y^{-1}x)=⟨{\pi }_{\phi }(y^{-1}x)\xi ,\xi ⟩=⟨{\pi }_{\phi }(y{)}^{\ast }{\pi }_{\phi }(x)\xi ,\xi ⟩=⟨{\pi }_{\phi }(x)\xi ,{\pi }_{\phi }(y)\xi ⟩$, so $\phi (y^{-1}x)-\phi (x)=⟨{\pi }_{\phi }(x)\xi ,{\pi }_{\phi }(y)\xi -\xi ⟩$; Cauchy–Schwarz and $\mathrm{\parallel }{\pi }_{\phi }(x)\xi \mathrm{\parallel }=\mathrm{\parallel }\xi \mathrm{\parallel }$ give $\mathrm{\mid }\phi (y^{-1}x)-\phi (x)\mathrm{\mid }\le \mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }{\pi }_{\phi }(y)\xi -\xi \mathrm{\parallel }$, and step 1.1 with $g=y$ turns $\mathrm{\parallel }{\pi }_{\phi }(y)\xi -\xi \mathrm{\parallel }$ into $(2(\phi (e)-\mathrm{Re}\phi (y)){)}^{1\mathrm{/}2}$, hence the second bound with $\phi (e)$ in place of $\mathrm{\parallel }\xi \mathrm{\parallel }$.

For claim 2, $\phi (x)-\phi (y)=⟨({\pi }_{\phi }(x)-{\pi }_{\phi }(y))\xi ,\xi ⟩=⟨{\pi }_{\phi }(y)({\pi }_{\phi }(y^{-1}x)-I)\xi ,\xi ⟩=⟨({\pi }_{\phi }(y^{-1}x)-I)\xi ,{\pi }_{\phi }(y{)}^{\ast }\xi ⟩$, so Cauchy–Schwarz gives $\mathrm{\mid }\phi (x)-\phi (y)\mathrm{\mid }\le \mathrm{\parallel }{\pi }_{\phi }(y^{-1}x)\xi -\xi \mathrm{\parallel }\mathrm{\parallel }\xi \mathrm{\parallel }$; squaring and using step 1.1 with $g=y^{-1}x$ and $\mathrm{\parallel }\xi {\mathrm{\parallel }}^2=\phi (e)$ gives $\mathrm{\mid }\phi (x)-\phi (y){\mathrm{\mid }}^2\le 2\phi (e)(\phi (e)-\mathrm{Re}\phi (y^{-1}x))$.

For claim 4 assume $\phi (e)=1$. Since ${\pi }_{\phi }(xy)\xi -\xi ={\pi }_{\phi }(x)({\pi }_{\phi }(y)\xi -\xi )+({\pi }_{\phi }(x)\xi -\xi )$, the triangle inequality and $\mathrm{\parallel }a+b{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }a{\mathrm{\parallel }}^2+2\mathrm{\parallel }b{\mathrm{\parallel }}^2$ give $\mathrm{\parallel }{\pi }_{\phi }(xy)\xi -\xi {\mathrm{\parallel }}^2\le 2\mathrm{\parallel }{\pi }_{\phi }(y)\xi -\xi {\mathrm{\parallel }}^2+2\mathrm{\parallel }{\pi }_{\phi }(x)\xi -\xi {\mathrm{\parallel }}^2$; substituting $\frac{1}{2}\mathrm{\parallel }{\pi }_{\phi }(g)\xi -\xi {\mathrm{\parallel }}^2=1-\mathrm{Re}\phi (g)$ from step 1.1 and $\phi (e)=1$ yields $1-\mathrm{Re}\phi (xy)\le 2(1-\mathrm{Re}\phi (y))+2(1-\mathrm{Re}\phi (x))$, which is claim 4.