Kato-Rellich theorem

If $H=\{0\}$, all domains and graphs are zero and all conclusions hold directly. Otherwise use [A4]. For $z=\pm i\mu$, $\mu >0$, put $r_z(v)=(z-v{)}^{-1}$. The bounds $\mathrm{\mid }{r}_z(v)\mathrm{\mid }\le 1\mathrm{/}\mu$ and $\mathrm{\mid }v{r}_z(v)\mathrm{\mid }\le 1$ show that $r_z(A)$ maps $H$ into $D(A)$, that $\mathrm{\parallel }{r}_z(A)\mathrm{\parallel }\le 1\mathrm{/}\mu$, and that $\mathrm{\parallel }A{r}_z(A)\mathrm{\parallel }\le 1$. The product rule gives $(z-A)r_z(A)=I$ on $H$ and $r_z(A)(z-A)=I$ on $D(A)$, so this is $R_A(z)$. Thus $\mathrm{\parallel }B{R}_A(z)\mathrm{\parallel }\le a+b\mathrm{/}\mu <1$ when, for example, $\mu =1+2b\mathrm{/}(1-a)$.

For any nonzero-space self-adjoint $T$ and real $c$, the spectral-measure equivalence $T\ge c$ if and only if $E((-\mathrm{\infty },c))=0$ follows directly. If $T\ge c$ and $J_n=[-(n+1),c-1\mathrm{/}(n+1)]$ has a nonzero projection, a nonzero $x$ in its range belongs to $D(T)$, has $E_x$ carried by $J_n$, and satisfies $⟨Tx,x⟩\le (c-1\mathrm{/}(n+1))\mathrm{\parallel }x{\mathrm{\parallel }}^2$, a contradiction. These increasing sets exhaust $(-\mathrm{\infty },c)$, so countable additivity gives zero projection. Conversely zero projection below $c$ gives $⟨Tx,x⟩=\int vd{E}_x\ge c\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $x$ in $D(T)$. Also if every real $t<c$ is a resolvent point, the essential-range characterization in [A4] supplies a zero-projection open neighborhood of each $t$. The rational intervals contained in such neighborhoods form a countable cover of $(-\mathrm{\infty },c)$; their union has zero projection by countable subadditivity of each $E_x$. Therefore again $T\ge c$. This does not require a finite spectral infimum or any resolvent-distance formula.

Put $S=A+B$ on exactly $D(A)$. For $z=\pm i\mu$ the identity $z-S=(I-B{R}_A(z))(z-A)$ holds on $D(A)$, since $R_A(z)(z-A)x=x$ there. The first factor is boundedly invertible by [A3], and $z-A$ is bijective $D(A)$ to $H$. Hence both nonreal shifts of $S$ are onto. $S$ is symmetric by summing the two symmetry identities and densely defined because $D(A)$ is dense. The range criterion makes $S$ self-adjoint. No second-resolvent identity with a previously closed $S$ is assumed.

Suppose $A\ge \gamma$ and let $\lambda +\gamma =d>0$. By step 1.2 $E_A$ is carried by $[\gamma ,\mathrm{\infty })$. Define $r(v)=(-\lambda -v{)}^{-1}$ on that half-line and zero outside. For $v\ge \gamma$, $\mathrm{\mid }r(v)\mathrm{\mid }\le 1\mathrm{/}d$ and $\mathrm{\mid }vr(v)\mathrm{\mid }\le \max (1,\mathrm{\mid }\gamma \mathrm{\mid }\mathrm{/}d)$: for $v\ge 0$ the ratio $v\mathrm{/}(v+\lambda )$ is monotone with its maximum at an endpoint or its limiting value $1$; for $\gamma \le v<0$, $(-v)\mathrm{/}(v+\lambda )$ decreases with $v$ since $\lambda >0$ in that case. The spectral product rule shows $r(A)=R_A(-\lambda )$, exactly as in step 1.1, and hence $\mathrm{\parallel }B{R}_A(-\lambda )\mathrm{\parallel }\le a\max (1,\mathrm{\mid }\gamma \mathrm{\mid }\mathrm{/}d)+b\mathrm{/}d$. Put $C=\max \{a\mathrm{\mid }\gamma \mathrm{\mid }+b,b\mathrm{/}(1-a)\}$. For every $d>C$ this last bound is strictly below one: if $d\ge \mathrm{\mid }\gamma \mathrm{\mid }$ it equals $a+b\mathrm{/}d<1$ (also when $b=0$); if $d<\mathrm{\mid }\gamma \mathrm{\mid }$ it equals $(a\mathrm{\mid }\gamma \mathrm{\mid }+b)\mathrm{/}d<1$. The factorization of step 2.1 therefore proves that every real number $t<\gamma -C$ is in $\rho (S)$, with bounded inverse $R_A(t)(I-B{R}_A(t){)}^{-1}$. Apply step 1.2 to the already self-adjoint $S$ to obtain $S\ge \gamma -C$, the stated bound including its endpoint.

If $A$ is essentially self-adjoint, put $T=\bar{A}$. For each $x$ in $D(T)$ choose $x_n$ in $D(A)$ with $x_n\to x$ and $A{x}_n\to Tx$, using [A5]. The inequality in [A1] applied to $x_n-x_m$ makes $B{x}_n$ Cauchy. Define $\tilde{B}x$ as its limit. Two such approximations give the same limit by the same inequality applied to their difference. Approximating $x$ and $y$ and their linear combinations proves linearity, $\mathrm{\parallel }\tilde{B}x\mathrm{\parallel }\le a\mathrm{\parallel }Tx\mathrm{\parallel }+b\mathrm{\parallel }x\mathrm{\parallel }$, and symmetry by passing to the limit in $⟨B{x}_n,y_n⟩=⟨x_n,B{y}_n⟩$. It extends $B$ restricted to $D(A)$; no extension of $B$'s possibly larger domain is claimed. Step 2.1 makes $T+\tilde{B}$ self-adjoint. The inclusion $A+B\subseteq T+\tilde{B}$ and closedness give $\overline{A+B}\subseteq T+\tilde{B}$. Conversely the same approximating sequences satisfy $(A+B)x_n\to (T+\tilde{B})x$, giving the reverse graph inclusion. This proves essential self-adjointness and the exact closure formula. If $A\ge \gamma$ in this case, taking limits of its quadratic inequality gives $T\ge \gamma$; step 3.1 then gives the bound for the closure and its restriction $A+B$.

The choice use is exactly [A5]. The cases $B=0$ or $a=b=0$ are admitted by the same estimates and return the original lower bound. Dimension one requires no change. The strict hypothesis $a<1$ is used in the positive choice of $\mu$ and in $b\mathrm{/}(1-a)$; no conclusion at $a=1$ is asserted. The endpoint $\gamma -C$ is included by the zero-projection argument, without asserting that the spectrum is nonempty on the zero space.