Projection lemma for a large penalty

Set $K=\mathrm{\parallel }{H}_1\mathrm{\parallel }$. By [F1], the Hermitian compression $H_1{\mathrm{\mid }}_S$ has a unit eigenvector $u\in S$ with eigenvalue ${\lambda }_S$. Since $H_{2}u=0$, its Rayleigh quotient for $H_1+H_2$ is ${\lambda }_S$, so ${\lambda }_{\min }(H_1+H_2)\le {\lambda }_S$.

If $S^{\perp }=\{0\}$, then $S=H$ and $H_2=0$, so the conclusion is equality. Otherwise both $S$ and $S^{\perp }$ are nonzero. For a unit vector $x\in H$, write $x=\alpha u+\beta w$ with $u\in S$, $w\in S^{\perp }$ unit and $\alpha ,\beta \ge 0$, ${\alpha }^2+{\beta }^2=1$; absorb the phases into $u,w$. If either component of $x$ is zero, choose any unit vector in its nonzero subspace for the unused direction. Self-adjointness of $H_2$ and $H_2{\mathrm{\mid }}_S=0$ make its cross terms zero, so $⟨H_{2}x,x⟩\ge J{\beta }^2$. Also $⟨H_{1}u,u⟩\ge {\lambda }_S$, $⟨H_{1}w,w⟩\ge -K$, and $\mathrm{\mid }⟨H_{1}u,w⟩\mathrm{\mid }\le K$.

Expanding the quadratic form and using ${\lambda }_S\ge -K$ gives $⟨(H_1+H_2)x,x⟩\ge {\lambda }_S+(J-2K){\beta }^2-2K\beta \mathrm{.}$ Indeed the $S$-diagonal term is ${\alpha }^2⟨H_{1}u,u⟩\ge {\alpha }^2{\lambda }_S={\lambda }_S-{\beta }^2{\lambda }_S\ge {\lambda }_S-K{\beta }^2$, the $S^{\perp }$ term is at least $-K{\beta }^2$, and $\alpha \le 1$ bounds the cross term below by $-2K\beta$.

The scalar quadratic $(J-2K){\beta }^2-2K\beta$ has minimum $-K^2\mathrm{/}(J-2K)$ by completing the square. Thus every unit Rayleigh quotient is at least ${\lambda }_S-K^2\mathrm{/}(J-2K)$; [F1] identifies the least eigenvalue with the minimum Rayleigh quotient. This proves the lower bound.