Put \(\varepsilon =1/(4p)\). Bernoulli’s inequality gives \(|a|^p\ge (1-\varepsilon )^p\ge 3/4\) and \((1+\varepsilon )^p\le 4/3\), and the geometric sum gives \(|a^p-1|\le p\, \varepsilon \, (1+\varepsilon )^p\le 1/3\). Hence \(|a^{-p}-1|\le 4/9\). For the second claim write \(1-z^2=w\, a\) with \(a-1=(u-z)(u+z)/w\), so that \(|a-1|\le 3|z-u|/w\le 1/(4p)\).

The first bound is Cauchy’s estimate on the circle of radius \(w/(12p)\) about \(u\), where \(|g|\le e\, e^{-w^{-p}/2}\) by Lemma 20.1.1. For the second, put \(s=w^{-p}\), so that \(w^{-n}=w^{pk-n}s^k\), and use \(s^ke^{-s/2}\le 2^k k!\), which follows from the exponential series.

Let \(D_n\) be the \(n\)-th derivative of \(g\) on \((-1,1)\), extended by zero. By Theorem 20.1.2 with \(k=n+1\), \(|D_n(u)|\le C_n|1-u^2|\), so each \(D_n\) is continuous at \(\pm 1\). On \((-1,1)\) the derivative of \(D_n\) is \(D_{n+1}\), outside \([-1,1]\) both vanish, and at \(\pm 1\) a function continuous at a point whose derivative extends continuously there is differentiable there with that derivative. Hence \(D_0\) is infinitely differentiable with \(n\)-th derivative \(D_n\), and \(\chi (\omega )=D_0(\omega /\delta )\). The bound is the Gevrey estimate \(\| \chi ^{(m)}\| _\infty \le C_1^{m+1}(m!)^{1+1/p}\) of  [ Ope26b , Lemma 4.2 ] , written with integer exponents.

Evenness of \(\chi \) removes the sine part of \(e^{-it\omega }\). Integrating by parts \(n\) times, with no boundary terms, gives \((it)^n\int \chi e^{-it\omega }=\int \chi ^{(n)}e^{-it\omega }\); the integrand is supported in \([-\delta ,\delta ]\) and bounded by Theorem 20.2.2.

Put \(T=\delta |t|\) and \(B_p=2(12p^2)^p\). With \(n=pk\), Lemma 20.3.2, \((pk)!\le (pk)^{pk}\) and \(k!\le k^k\) give \(T^{pk}|f(t)|\le (\delta e/\pi )(B_pk^{p+1})^k\). Choosing \(k=\lfloor \kappa _pT^\alpha \rfloor \) makes \(B_pk^{p+1}\le T^p/2\), hence \(|f(t)|\le (\delta e/\pi )2^{-k}\le Ce^{-c|t|^\alpha }\); the case \(k=0\) covers bounded \(t\). The moment bounds give \((1+|t|)^N|f(t)|\le A_N\) for every \(N\), whence integrability.

The Fourier transform of \(\chi \) at \(\xi \) is \(2\pi f(2\pi \xi )\), which is integrable by Theorem 20.3.3. Fourier inversion for a continuous integrable function with integrable transform, followed by the substitution \(t=2\pi \xi \), gives the first identity; \(\omega =0\) gives \(\int f=\chi (0)=1\).

Since \(e^{itH}\) is unitary, \(\| e^{itH}Ae^{-itH}\| =\| A\| \), and \(f\) is integrable by Theorem 20.3.3. The adjoint passes through the integral because \(f\) is real. Finally \(e^{itH}\) commutes with \(H\), so \(\mathcal F_H(H)=(\int f)\, H=H\).

The exponential series gives \(e^{-itH}w=e^{-itE'}w\) and \(\langle v,e^{itH}y\rangle =e^{itE}\langle v,y\rangle \), so the integrand has matrix element \(f(t)e^{it(E-E')}\langle v,Aw\rangle \), and Fourier inversion (Theorem 20.3.4) evaluates the integral. For the last claim, \(\chi (0)=1\) and \(\chi \) vanishes outside \((-\delta ,\delta )\), so the coefficients of \(\mathcal F_H(A)w\) in the eigenbasis are \(\langle u_k,Aw\rangle \) when \(\lambda _k=E'\) and zero otherwise.

Test against an orthonormal eigenbasis \((u_k)\) of \(H\). If \(\lambda _k=E_0\), the gap inequality at \(u_k\) gives \(\Delta (1-|\langle \Omega ,u_k\rangle |^2)\le 0\), hence \(\| u_k-\langle \Omega ,u_k\rangle \Omega \| ^2=1-|\langle \Omega ,u_k\rangle |^2=0\), and \(\chi (0)=1\). If \(\lambda _k\ne E_0\), then \(u_k\perp \Omega \), so the gap inequality gives \(\lambda _k-E_0\ge \Delta \ge \delta \) and \(\chi \) vanishes.

Lemma 20.5.1 gives \(M_i\Omega =0\). Linearity and \(\mathcal F_H(H)=H\) give the sum, since \(\sum _i\langle \Omega ,h_i\Omega \rangle =\langle \Omega ,H\Omega \rangle =E_0\). The norm bound combines \(\| \mathcal F_H(h_i)\| \le \| f\| _1\| h_i\| \) with \(|\langle \Omega ,h_i\Omega \rangle |\le \| h_i\| \).

With \(\epsilon =\| a-b\| \), operator monotonicity of the square root gives \(a^{1/2}\le (b+\epsilon )^{1/2}\le b^{1/2}+\epsilon ^{1/2}\), the last step because \(b+\epsilon \le (b^{1/2}+\epsilon ^{1/2})^2\). Exchanging \(a\) and \(b\) bounds the norm. For the absolute values apply this to \(a^2,b^2\) and use \(a^2-b^2=a(a-b)+(a-b)b\).

Theorem 20.5.2 gives \(\| M_i\| \le c_*\), hence \(\| k_i\| \le 1\). Since \(|M_i|^2=M_i^2\), the kernel of \(M_i\) lies in that of \(|M_i|\), so \(k_i\Omega =0\). Finally \(M_i\le |M_i|\) and \(\sum _iM_i=H-E_0I\).

The map \(v\mapsto Gv\oplus Kv\) is an isometry, and \(\mathcal E(B)\) is the compression of \(B\oplus B\) by it; tensoring with the identity preserves \(G^2+K^2=I\). The difference bound follows from \(ABA-A'BA'=(A-A')BA+A'B(A-A')\) and \(\| A\otimes I\| \le \| A\| \). The root tails are Lemma 20.5.3. A matrix commuting with \(k\) commutes with both square roots, so its image is itself times \(G^2+K^2=I\).

The identities \(K_i\Omega =0\) and \(G_i\Omega =\Omega \) give \(G_iBG_i\Omega +K_iBK_i\Omega =G_iB\Omega \). Iterating this equality in the order of the word proves the assertion; each later contraction acts on the left.

Multiply the bounds \(|f(t)|\le A(1+|t|)^{-4}\) and \(|f(t)|\le Ke^{-c|t|^\alpha }\) and take square roots.

The linear map \(L\) passes through the integral. For \(|t|\le cl/(2v)\) the hypothesis gives the factor \(Ce^{-cl/2}\); for larger \(|t|\) use the bound \(2\) and Lemma 20.7.1. Finally \(e^{-cl/2}\le e^{c/2}e^{-(c/2)l^\alpha }\).

Write \(|M|-L|M|=(|M|-|LM|)+L(|LM|-|M|)\) and apply Lemma 20.5.3. Unitality gives \(M-LM=\mathcal F_H(h)-L\mathcal F_H(h)\).

With \(X=P-Q\) and \(s=|\langle \psi ,\phi \rangle |^2\) one has \(PQP=sP\) and \(QPQ=sQ\), hence \(X^3=(1-s)X\); so \(|X|=X^2/(1-s)^{1/2}\) and \(\operatorname {tr}X^2=2(1-s)\).

A minimizer of the quadratic form on the unit sphere is an eigenvector. For \(\psi \perp \Omega _0\), a combination of \(\psi \) and \(\Omega _0\) orthogonal to \(\Omega \) has energy at least \((g-\epsilon )\) times its squared norm, while the \(\Omega _0\) component carries energy \(e_0\le g-\epsilon \); hence \(\langle \psi ,H'\psi \rangle \ge (g-\epsilon )\| \psi \| ^2\). Testing on \(\Omega \) gives \(e_0\le \epsilon \), and the gap of \(H_F\) at \(\Omega _0\) gives \(g(1-|\langle \Omega ,\Omega _0\rangle |^2)\le 2\epsilon \).