Lemma 32.4.6.label Let $G$ be a locally compact group and $\phi \in S(G)$, then for each $x, y \in G$,
\[|\phi(x) - \phi(y)|^{2} \le 2 - 2\text{Re}\phi(yx^{-1})\]
Proof, [Lemma 3.30, Fol16]. Let $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ be the GNS triple associated with $\phi$, then by Corollary 32.3.4, $\phi(x) = \dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H_\phi}$ for all $x \in G$. Since $\norm{\phi}_{u} = \phi(1_{G}) = 1$, $\norm{\xi_\phi}_{H_\phi}= 1$. As such,
\begin{align*}|\phi(x) - \phi(y)|^{2}&= |\dpn{[\pi_\phi(x) - \pi_\phi(y)]\xi_\phi, \xi_\phi}{H_\phi}|^{2} \\&= |\dpn{\xi_\phi, [\pi_\phi(x^{-1}) - \pi_\phi(y^{-1})]\xi_\phi}{H_\phi}|^{2} \\&\le \normn{[\pi_\phi(x^{-1}) - \pi_\phi(y^{-1})]\xi_\phi}_{H_\phi}^{2} \\&= 2 - 2\text{Re}\dpn{\pi_\phi(x^{-1})\xi_\phi, \pi_\phi(y^{-1})\xi_\phi}{H_\phi}\\&= 2 - 2\text{Re}\dpn{\pi_\phi(yx^{-1})\xi_\phi, \xi_\phi}{H_\phi}= 2 - 2\text{Re}\phi(yx^{-1})\end{align*}
$\square$
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