The shifted Hilbert-space normal operator is invertible with inverse norm at most $1\mathrm{/}\lambda$

Uniqueness in [F2] gives linearity of $A^{\ast }$: for scalars $a,b$ and $y_1,y_2\in Y$, conjugate-linearity of the second slot shows that $a{A}^{\ast }{y}_1+b{A}^{\ast }{y}_2$ satisfies the defining identity for $a{y}_1+b{y}_2$, so it equals $A^{\ast }(a{y}_1+b{y}_2)$. For $w=A^{\ast }y$, [F2], [F4] and [F7] give $\mathrm{\parallel }w{\mathrm{\parallel }}^2=⟨Aw,y⟩\le \mathrm{\mid }⟨Aw,y⟩\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$. Dividing when $w\mathrm{\ne }0$, and treating $w=0$ directly, proves $\mathrm{\parallel }{A}^{\ast }y\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$. Thus $A^{\ast }$ is bounded, and composition and addition make $B_{\lambda }$ linear and bounded with $\mathrm{\parallel }{B}_{\lambda }x\mathrm{\parallel }\le (\mathrm{\parallel }A{\mathrm{\parallel }}^2+\lambda )\mathrm{\parallel }x\mathrm{\parallel }$. The adjoint identity and conjugate symmetry [F6] give $⟨B_{\lambda }x,z⟩=⟨Ax,Az⟩+\lambda ⟨x,z⟩=⟨x,B_{\lambda }z⟩$ for all $x,z\in X$. In particular, $⟨B_{\lambda }x,x⟩=\mathrm{\parallel }Ax{\mathrm{\parallel }}^2+\lambda \mathrm{\parallel }x{\mathrm{\parallel }}^2\ge \lambda \mathrm{\parallel }x{\mathrm{\parallel }}^2$.

By [F4], $\mathrm{\parallel }{B}_{\lambda }x\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\ge \mathrm{\mid }⟨B_{\lambda }x,x⟩\mathrm{\mid }=⟨B_{\lambda }x,x⟩\ge \lambda \mathrm{\parallel }x{\mathrm{\parallel }}^2$. Thus $\mathrm{\parallel }{B}_{\lambda }x\mathrm{\parallel }\ge \lambda \mathrm{\parallel }x\mathrm{\parallel }$ for $x\mathrm{\ne }0$; the same inequality is trivial for $x=0$. Hence $B_{\lambda }$ is injective. If $B_{\lambda }{x}_n$ converges, this inequality applied to $x_n-x_m$ makes $(x_n)$ Cauchy; completeness of $X$ and boundedness of $B_{\lambda }$ show its limit lies in $\mathrm{ran}{B}_{\lambda }$. Therefore the range is closed.

If $z\in (\mathrm{ran}{B}_{\lambda }{)}^{\perp }$, [F8] gives $⟨z,B_{\lambda }x⟩=0$ for every $x\in X$. By conjugate symmetry [F6] this is $⟨B_{\lambda }x,z⟩=0$, and the symmetry identity from step 1.1 gives $⟨x,B_{\lambda }z⟩=0$ for every $x\in X$. Taking $x=B_{\lambda }z$ and using positive definiteness [F6] yields $B_{\lambda }z=0$; injectivity from step 2.1 then gives $z=0$. Thus the closed range has zero orthogonal complement.

By [F5] and the Countable Choice assumption [A1], a closed subspace with zero orthogonal complement is all of $X$; so $B_{\lambda }$ is surjective as well as injective. For $y=B_{\lambda }x$, step 2.1 gives $\mathrm{\parallel }{B}_{\lambda }^{-1}y\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}\lambda$, proving $\mathrm{\parallel }{B}_{\lambda }^{-1}\mathrm{\parallel }\le 1\mathrm{/}\lambda$.