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I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $\psi \in D(\sqrt{G})$ (that just means every $\psi$ in the Hilbert space if $\Lambda < +\infty$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle $$ where $x,x' \in [0, \delta\lambda]$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $ E^{(G)}(\delta, \psi)$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $G$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$ Let us fix the $1-\delta$ issue. I explicitly suppose that $||\psi||=1$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq 1\tag{M}$$ because the operator is unitary. As a consequence, from(E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))$$ where $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $\delta>0$, otherwise (M) is impossible. In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi) $$ where, if $0\leq \delta$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$

I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle+ \langle \psi |(e^{-i\delta G}-I) \psi\rangle =\langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $\psi \in D(\sqrt{G})$ (that just means every $\psi$ in the Hilbert space if $\Lambda < +\infty$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle $$ where $x,x' \in [0, \delta\lambda]$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $ E^{(G)}(\delta, \psi)$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $G$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$

Let us fix the $1-\delta$ issue. 

I explicitly suppose that $||\psi||=1$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq ||\psi||\: ||e^{-i\delta G} \psi|| = ||\psi||^2 =1\tag{M}$$ because the operator is unitary. As a consequence, from  (E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))$$ where $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $\delta>0$, otherwise (M) is impossible. In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi) $$ where, for $\delta\geq 0$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$

I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $\psi \in D(\sqrt{G})$ (that just means every $\psi$ in the Hilbert space if $\Lambda < +\infty$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle $$ where $x,x' \in [0, \delta\lambda]$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $ E^{(G)}(\delta, \psi)$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $G$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$ Let us fix the $1-\delta$ issue. I explicitly suppose that $||\psi||=1$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq 1\tag{M}$$ because the operator is unitary. As a consequence, from(E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\:.$$ Necessarily $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $\delta>0$, otherwise (M) is impossible. In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi) $$ where, if $0\leq \delta$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$

I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $\psi \in D(\sqrt{G})$ (that just means every $\psi$ in the Hilbert space if $\Lambda < +\infty$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle $$ where $x,x' \in [0, \delta\lambda]$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $ E^{(G)}(\delta, \psi)$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $G$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$ Let us fix the $1-\delta$ issue. I explicitly suppose that $||\psi||=1$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq 1\tag{M}$$ because the operator is unitary. As a consequence, from(E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))$$ where $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $\delta>0$, otherwise (M) is impossible. In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi) $$ where, if $0\leq \delta$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$

I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $\psi \in D(\sqrt{G})$ (that just means every $\psi$ in the Hilbert space if $\Lambda < +\infty$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle $$ where $x,x' \in [0, \delta\lambda]$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $ E^{(G)}(\delta, \psi)$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $G$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$ Let us fix the $1-\delta$ issue. I explicitly suppose that $||\psi||=1$ and that $||G|| \leq +\infty$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq 1\tag{M}$$ because the operator is unitary. As a consequence, from(E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\:.$$ Necessarily $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $\delta>0$, otherwise (M) is impossible.In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi) $$ where, if $0\leq \delta\leq \Delta <+\infty$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq K <+\infty\:.$$ With $K$ depending on $G$ and $\Delta$ only.

I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $\psi \in D(\sqrt{G})$ (that just means every $\psi$ in the Hilbert space if $\Lambda < +\infty$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle $$ where $x,x' \in [0, \delta\lambda]$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $ E^{(G)}(\delta, \psi)$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $G$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$ Let us fix the $1-\delta$ issue. I explicitly suppose that $||\psi||=1$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq 1\tag{M}$$ because the operator is unitary. As a consequence, from(E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\:.$$ Necessarily $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $\delta>0$, otherwise (M) is impossible. In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi) $$ where, if $0\leq \delta$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$