Kato-Rellich theorem

If H={0}, all domains and graphs are zero and all conclusions hold directly. Otherwise use [A4]. For z=plus or minus i mu, mu>0, put r_z(v)=(z-v)^{-1}. The bounds |r_z(v)|<=1/mu and |v r_z(v)|<=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>=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 (-infinity,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 (-infinity,c); their union has zero projection by countable subadditivity of each E_x. Therefore again T>=c. This does not require a finite spectral infimum or any resolvent-distance formula.

Put S=A+B on exactly D(A). For z=plus or minus 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>=gamma and let lambda+gamma=d>0. By step 1.2 E_A is carried by [gamma,infinity). Define r(v)=(-lambda-v)^{-1} on that half-line and zero outside. For v>=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>=0 the ratio v/(v+lambda) is monotone with its maximum at an endpoint or its limiting value 1; for gamma<=v<0, (-v)/(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>=|gamma| it equals a+b/d<1 (also when b=0); if d<|gamma| it equals (a|gamma|+b)/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>=gamma-C, the stated bound including its endpoint.

If A is essentially self-adjoint, put T=closure(A). For each x in D(T) choose x_n in D(A) with x_n to x and Ax_n to Tx, using [A5]. The inequality in [A1] applied to x_n-x_m makes Bx_n Cauchy. Define Btilde 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+Btilde self-adjoint. The inclusion A+B subset T+Btilde and closedness give closure(A+B) subset T+Btilde. Conversely the same approximating sequences satisfy (A+B)x_n to (T+Btilde)x, giving the reverse graph inclusion. This proves essential self-adjointness and the exact closure formula. If A>=gamma in this case, taking limits of its quadratic inequality gives T>=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/(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.