Filtered derivative of an isolated finite-volume band

Let $u$ be the standard flat function of [F5] and put $q(\omega ):=1-(\omega \mathrm{/}\gamma {)}^2$ and $\chi (\omega ):=eu(q(\omega ))$. Since $\gamma >0$, $q$ is a smooth real polynomial. The chain and product rules in [F5], iterated at each derivative order, make $\chi$ smooth on all of $\mathbb{R}$: every positive-order derivative is a finite sum of products of a derivative $u^{(m)}\circ q$, with $m\ge 1$, and derivatives of $q$. At $\omega =\pm \gamma$, one has $q(\omega )=0$ and every such $u^{(m)}(0)=0$, so all these derivatives vanish and match the exterior zero branch. The value itself also vanishes there because $u(0)=0$. Since $q>0$ exactly when $\mathrm{\mid }\omega \mathrm{\mid }<\gamma$, this formula is exactly $e\exp (-1\mathrm{/}(1-(\omega \mathrm{/}\gamma {)}^2))$ inside that interval and zero outside. Thus $\chi$ is real, even, smooth, compactly supported in $[-\gamma ,\gamma ]$, and $\chi (0)=e\exp (-1)=1$. Define $w(t)=(2\pi {)}^{-1}\int \chi (\omega )e^{-it\omega }d\omega$. By [A1], [F3], and the sign/scale conversion stated in [F3], $w$ is Schwartz, real and even, its Fourier transform $\widehat{w}(\omega )=\int w(t)e^{it\omega }dt$ equals $\chi (\omega )$, and $\int w=1$. For $t>0$ put $W_{\gamma }(t)={\int }_t^{\mathrm{\infty }}w(v)dv$; for $t<0$ put $W_{\gamma }(t)=-{\int }_{-\mathrm{\infty }}^{t}w(v)dv$; set $W_{\gamma }(0)=0$. Evenness and unit integral make $W_{\gamma }$ odd with one-sided limits $W_{\gamma }(0^{\pm })=\pm \frac{1}{2}$. For every $N$, a Schwartz bound $\mathrm{\mid }w(v)\mathrm{\mid }\le c(1+\mathrm{\mid }v\mathrm{\mid }{)}^{-N-2}$ on either tail gives $\mathrm{\mid }{W}_{\gamma }(t)\mathrm{\mid }\le c^{\mathrm{\prime }}(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-N-1}$; hence $W_{\gamma }\in L^1$ and has the stated tail estimate. Integration by parts on the two half-lines, accounting for the jump of size $1$ at zero and using $W_{\gamma }^{\mathrm{\prime }}=-w$ away from zero, gives $-\widehat{w}(\omega )=-1-i\omega {\widehat{W}}_{\gamma }(\omega )$. Thus ${\widehat{W}}_{\gamma }(\omega )=i(1-\chi (\omega ))\mathrm{/}\omega$ for $\omega \mathrm{\ne }0$, so it equals $i\mathrm{/}\omega$ for $\mathrm{\mid }\omega \mathrm{\mid }\ge \gamma$.

Fix $s_0$. By the locally fixed spectral-cluster hypothesis and [F2], choose a finite union of positively oriented contours $\mathrm{\Gamma }$ surrounding exactly the selected eigenvalues and staying a positive distance from the remaining spectrum, with the same selected eigenvalues enclosed for nearby $s$. For $s$ near $s_0$, $\mathrm{\parallel }{H}_{\mathrm{\Lambda }}(s)-H_{\mathrm{\Lambda }}(s_0)\mathrm{\parallel }$ is small; factoring $zI-H_{\mathrm{\Lambda }}(s)$ through $zI-H_{\mathrm{\Lambda }}(s_0)$ and using the norm-convergent Neumann series keeps every $z\in \mathrm{\Gamma }$ in the resolvent set. The contour projection $P_{\mathrm{\Lambda }}(s)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}(zI-H_{\mathrm{\Lambda }}(s){)}^{-1}dz$ is therefore $C^1$, since the inverse matrix is $C^1$ in $s$.

In an orthonormal eigenbasis of $H_{\mathrm{\Lambda }}(s)$, write its eigenvalues as $E_m$ and the band indicator as $p_m\in \{0,1\}$. Differentiating $P^2=P$ gives $P_{mn}^{\mathrm{\prime }}=0$ when $p_m=p_n$. Differentiating $HP=PH$ gives $P_{mn}^{\mathrm{\prime }}=(p_m-p_n)H_{mn}^{\mathrm{\prime }}\mathrm{/}(E_m-E_n)$ when $p_m\mathrm{\ne }{p}_n$; the denominator has magnitude at least $\gamma$. The integral defining $D_{\mathrm{\Lambda }}$ converges in norm because $W_{\gamma }\in L^1$; it is self-adjoint because $W_{\gamma }$ is real and each conjugated $H_{\mathrm{\Lambda }}^{\mathrm{\prime }}$ is self-adjoint. For each fixed $t$, the matrix exponential series for $e^{\pm it{H}_{\mathrm{\Lambda }}(s)}$ converges uniformly in $s\in [0,1]$: if $M={\sup }_s\mathrm{\parallel }{H}_{\mathrm{\Lambda }}(s)\mathrm{\parallel }<\mathrm{\infty }$, its order-$r$ term has norm at most $(\mathrm{\mid }t\mathrm{\mid }M{)}^r\mathrm{/}r!$. The terms are continuous in $s$, so both exponentials, and hence their product with the continuous $H_{\mathrm{\Lambda }}^{\mathrm{\prime }}(s)$, are norm-continuous in $s$. For any sequence $s_m\to s$ in $[0,1]$, each matrix entry of the difference of the filtered integrands tends pointwise to zero and has modulus at most $2{\sup }_v\mathrm{\parallel }{H}_{\mathrm{\Lambda }}^{\mathrm{\prime }}(v)\mathrm{\parallel }\mathrm{\mid }{W}_{\gamma }(t)\mathrm{\mid }$, an integrable majorant by step 1.1 and compact-interval boundedness of $H_{\mathrm{\Lambda }}^{\mathrm{\prime }}$. The entries are measurable in $t$, since the exponentials are continuous in $t$ and $W_{\gamma }$ is measurable. Applying [F6] entrywise gives convergence of every entry of $D_{\mathrm{\Lambda }}(s_m)-D_{\mathrm{\Lambda }}(s)$. If the matrix dimension is $m_{\mathrm{\Lambda }}$, then $\mathrm{\parallel }R\mathrm{\parallel }\le m_{\mathrm{\Lambda }}{\max }_{i,j}\mathrm{\mid }{R}_{ij}\mathrm{\mid }$: for a unit vector $x$, $\mathrm{\mid }(Rx{)}_i\mathrm{\mid }\le {\max }_{i,j}\mathrm{\mid }{R}_{ij}\mathrm{\mid }{\sum }_j\mathrm{\mid }{x}_j\mathrm{\mid }\le \sqrt{m_{\mathrm{\Lambda }}}{\max }_{i,j}\mathrm{\mid }{R}_{ij}\mathrm{\mid }$, and summing squares over $i$ proves the bound. Thus entrywise convergence gives norm convergence, proving norm-continuity of $D_{\mathrm{\Lambda }}$ without an undeclared norm-equivalence theorem. Its matrix entries are $(D_{\mathrm{\Lambda }}{)}_{mn}=H_{mn}^{\mathrm{\prime }}{\widehat{W}}_{\gamma }(E_m-E_n)=i{H}_{mn}^{\mathrm{\prime }}\mathrm{/}(E_m-E_n)$ across the band gap. Hence $i[D_{\mathrm{\Lambda }},P_{\mathrm{\Lambda }}{]}_{mn}=(p_n-p_m)i(D_{\mathrm{\Lambda }}{)}_{mn}=(p_m-p_n)H_{mn}^{\mathrm{\prime }}\mathrm{/}(E_m-E_n)=P_{mn}^{\mathrm{\prime }}$ across the gap, and both sides vanish within either band block. This proves the identity in every matrix entry.