Absolute gap equals one minus the largest nonconstant eigenvalue modulus

If $H_0=\{0\}$, the restriction has zero domain, so its operator norm is $0$ by [F1] and the gap is $1$. Otherwise, [F2] and [F4] give an orthonormal eigenbasis $e_1,\dots ,e_m$ of $H_0$. Each eigenvalue is real: if $P{e}_j={\lambda }_j{e}_j$, self-adjointness and [F7] give ${\lambda }_j\mathrm{\parallel }{e}_j{\mathrm{\parallel }}^2=⟨P{e}_j,e_j{⟩}_{\pi }=⟨e_j,P{e}_j{⟩}_{\pi }=\overline{{\lambda }_j}\mathrm{\parallel }{e}_j{\mathrm{\parallel }}^2$. [F5] identifies these eigenvalues as the spectrum of the restriction.

Write $f={\sum }_j{c}_j{e}_j$. Orthonormality gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{2,\pi }^2={\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$ and $\mathrm{\parallel }Pf{\mathrm{\parallel }}_{2,\pi }^2={\sum }_j\mathrm{\mid }{\lambda }_j{\mathrm{\mid }}^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le ({\max }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }{)}^2\mathrm{\parallel }f{\mathrm{\parallel }}_{2,\pi }^2$, with equality at an eigenvector whose eigenvalue has largest modulus. Therefore $\mathrm{\parallel }P{\mathrm{\mid }}_{H_0}{\mathrm{\parallel }}_{2,\pi }={\max }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }$.

By [F3] the restriction norm is at most $1$, and it is nonnegative by definition; substituting step 1.2 into [F1] proves both the eigenvalue formula and $0\le {\delta }_{\mathrm{a}\mathrm{b}\mathrm{s}}(P)\le 1$.

If $Pf=f$ and $f\mathrm{\notin }\mathrm{span}\{\mathbf{1}\}$, put $c={\sum }_x{\pi }_x{f}_x$ and $h=f-c\mathbf{1}$. The normalization ${\sum }_x{\pi }_x=1$ gives $h\in H_0$, while $h\mathrm{\ne }0$ because $f$ is not constant. Since $P\mathbf{1}=\mathbf{1}$, $Ph=Pf-cP\mathbf{1}=h$. Thus $1$ is an eigenvalue of $P{\mathrm{\mid }}_{H_0}$ and step 1.2 gives $\mathrm{\parallel }P{\mathrm{\mid }}_{H_0}\mathrm{\parallel }=1$, so [F1] gives ${\delta }_{\mathrm{a}\mathrm{b}\mathrm{s}}(P)=0$.