Relative boundedness with respect to an operator

Let $A$ be a linear operator on $H$ with domain $D(A)$. An operator $B$ is $A$-bounded, or relatively bounded with respect to $A$, when $D(A)\subseteq D(B)$ and there are finite constants $a,b\ge 0$ with $\mathrm{\parallel }Bx\mathrm{\parallel }\le a\mathrm{\parallel }Ax\mathrm{\parallel }+b\mathrm{\parallel }x\mathrm{\parallel }\text{for all }x\in D(A)\mathrm{.}$ The infimum of the admissible constants $a$ is the $A$-bound of $B$, and one says the $A$-bound is below one when some $a<1$ is admissible.

Equivalent form. $B$ is $A$-bounded exactly when $B$ is bounded on $D(A)$ for the graph norm $\mathrm{\parallel }x{\mathrm{\parallel }}_A=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Ax{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$ of Unbounded linear operators: domain, graph and extension: each estimate $\mathrm{\parallel }Bx\mathrm{\parallel }\le a\mathrm{\parallel }Ax\mathrm{\parallel }+b\mathrm{\parallel }x\mathrm{\parallel }$ gives $\mathrm{\parallel }Bx\mathrm{\parallel }\le \max (a,b)\sqrt{2}\mathrm{\parallel }x{\mathrm{\parallel }}_A$, and conversely a graph-norm bound $\mathrm{\parallel }Bx\mathrm{\parallel }\le C\mathrm{\parallel }x{\mathrm{\parallel }}_A$ gives the estimate with $a=b=C$. The constant $b$ is not intrinsic, and no closedness, density or resolvent hypothesis is needed for the definition. If $A$ is closed with nonempty resolvent set, then $A$-bounded operators are exactly those with $D(A)\subseteq D(B)$ for which $B{R}_A(z)$ is bounded for some, equivalently every, $z\in \rho (A)$ (Resolvent and spectrum of an unbounded operator). Indeed, an $A$-bound makes $B{R}_A(z)$ bounded because $A{R}_A(z)=z{R}_A(z)-I$; conversely, if $B{R}_A(z)$ is bounded, then $Bx=B{R}_A(z)(z-A)x$ gives an $A$-bound. Thus boundedness for one resolvent implies relative boundedness and hence boundedness for every resolvent. This is used in the Kato-Rellich theorem below and recorded here as an interface.