Large clock-penalty perturbation

Write $H_J$ in blocks as $\left(\begin{array}{cc}A& B\\ B^{\ast }& D\end{array}\right)$ for $S\oplus S^{\perp }$. Then $\mathrm{\parallel }B\mathrm{\parallel }\le K$ and $D=Q_S{H}_1{Q}_S+J{Q}_S{H}_2{Q}_S\ge (J-K)I$ by [F1]–[F2]. For real $\lambda \le K$, $D-\lambda I\ge (J-2K)I>0$, so the Schur complement ${\displaystyle S_{\lambda }=A-\lambda I-B(D-\lambda I{)}^{-1}{B}^{\ast }}$ is defined and $0\le B(D-\lambda I{)}^{-1}{B}^{\ast }\le {ϵ}_{J}I$.

The invertible block triangular change of variables completing the square in $⟨(H_J-\lambda )x,x⟩$ gives the direct sum quadratic form of $S_{\lambda }$ and $D-\lambda I$. Since the latter is positive definite, the largest dimension of a subspace on which $H_J-\lambda I$ is negative definite equals the corresponding dimension for $S_{\lambda }$. This follows directly because the triangular change is an invertible linear map and hence preserves subspace dimension and signs of the quadratic form.

For $k=1,2$, the span of the first $k$ eigenvectors of $A$ is a $k$-dimensional subspace of $S$ on which the Rayleigh quotient of $H_J$ is at most ${\mu }_k$. If $H_J$ had fewer than $k$ eigenvalues at most ${\mu }_k$, that subspace would contain a nonzero vector orthogonal to its first $k-1$ eigenvectors, whose Rayleigh quotient would be strictly larger than ${\mu }_k$, a contradiction. Thus ${\lambda }_k(H_J)\le {\mu }_k\le K$.

Fix $k=1$ or $2$ and suppose $\lambda <{\mu }_k-{ϵ}_J$. Then $\lambda <K$, and step 1.1 gives $S_{\lambda }\ge A-(\lambda +{ϵ}_J)I$. Since $\lambda +{ϵ}_J<{\mu }_k$, at most $k-1$ eigenvalues of $S_{\lambda }$ are negative. By step 2.1, at most $k-1$ eigenvalues of $H_J-\lambda I$ are negative. Letting $\lambda$ increase to ${\mu }_k-{ϵ}_J$ proves ${\lambda }_k(H_J)\ge {\mu }_k-{ϵ}_J$. The gap estimate follows by subtracting the bounds for $k=1,2$.

If $x=s+t$ is a unit eigenvector of eigenvalue $\lambda \le K$, its $S^{\perp }$ block equation is $B^{\ast }s+(D-\lambda I)t=0$. By step 1.1, $\mathrm{\parallel }(D-\lambda I{)}^{-1}\mathrm{\parallel }\le 1\mathrm{/}(J-2K)$; then [F2] yields $\mathrm{\parallel }t\mathrm{\parallel }\le K\mathrm{\parallel }s\mathrm{\parallel }\mathrm{/}(J-2K)$. No lower bound on the compressed gap was used in this leakage estimate. For a counterexample to any gap conclusion from $J>2K$ alone, take $\dim S=2$, $H_1=0$, and $H_2=0$ on $S$ and $I$ on $S^{\perp }$; then $K=0$, every $J>0$ satisfies $J>2K$, and the lowest eigenvalue of $H_J$ has multiplicity two. Hence the separate compressed-gap condition is necessary.