Relative compactness with respect to an operator

Assume the Axiom of Choice. Let $A$ be a self-adjoint operator on a complex Hilbert space $H$, and let $B:D(A)\to H$ be linear and bounded for the graph norm of $A$ (Relative boundedness with respect to an operator). Then $B$ is $A$-compact, or relatively compact with respect to $A$, when $B{R}_A(z)\in \mathcal{K}(H)$ is a compact operator (Compact linear operator) for one, equivalently for every, $z\in \rho (A)$.

The resolvent set is nonempty: $i\in \rho (A)$ by Resolvent of a self-adjoint operator: nonreal resolvents and the estimate. If $H=\{0\}$, all operators here are the unique operator, compact with bound zero; the assertions hold directly. Below suppose $H\mathrm{\ne }\{0\}$.

Well-definedness, with proofs.

1. $B{R}_A(z)$ is everywhere defined and bounded. $R_A(z)$ maps $H$ into $D(A)$ and $B$ is graph-norm bounded there, so $\mathrm{\parallel }B{R}_A(z)y\mathrm{\parallel }\le a\mathrm{\parallel }A{R}_A(z)y\mathrm{\parallel }+b\mathrm{\parallel }{R}_A(z)y\mathrm{\parallel }\le (a(1+\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }{R}_A(z)\mathrm{\parallel })+b\mathrm{\parallel }{R}_A(z)\mathrm{\parallel })\mathrm{\parallel }y\mathrm{\parallel }$ using $A{R}_A(z)=z{R}_A(z)-I$, so that $\mathrm{\parallel }A{R}_A(z)\mathrm{\parallel }\le 1+\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }{R}_A(z)\mathrm{\parallel }$ (Resolvent and spectrum of an unbounded operator).

2. Independence of $z$. Write $R_z=(zI-A{)}^{-1}$. For $y\in H$, $R_{w}y\in D(A)$ and $(zI-A)R_{w}y=y+(z-w)R_{w}y$. Applying $R_z$ gives $R_{w}y=R_{z}y+(z-w)R_z{R}_{w}y,$ since $R_z(zI-A)$ is the identity on $D(A)$. Thus $R_z-R_w=(w-z)R_z{R}_w$. Interchanging $z,w$ also gives $R_z-R_w=(w-z)R_w{R}_z$. This derives both orders without an unproved resolvent identity; the two sides of the latter identity have values in $D(A)$, so applying the linear map $B$ gives $B{R}_z=B{R}_w+(w-z)(B{R}_w)R_z\mathrm{.}$ Compactness at $w$ implies compactness at $z$ by composition with the bounded $R_z$ and finite linear combinations Compositions with a compact operator are compact Linear combinations of compact operators are compact. Exchanging $z,w$ proves the converse, including the trivial case $z=w$.

3. Vector space. If $B_1,B_2$ are graph-norm bounded and $A$-compact, then $\alpha B_1+\beta B_2$ is graph-norm bounded and $(\alpha B_1+\beta B_2)R_A(z)=\alpha B_1{R}_A(z)+\beta B_2{R}_A(z)$ is compact, being a linear combination of compact operators (Linear combinations of compact operators are compact).

4. An $A$-compact $B$ has $A$-bound zero. For $z\in \rho (A)$ and $\psi \in D(A)$, the inverse identity gives $B\psi =B{R}_A(z)(zI-A)\psi$. Hence, with $a_z=\mathrm{\parallel }B{R}_A(z)\mathrm{\parallel }$, $\mathrm{\parallel }B\psi \mathrm{\parallel }\le a_z\mathrm{\parallel }A\psi \mathrm{\parallel }+a_z\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }\psi \mathrm{\parallel }\mathrm{.}$ It suffices to prove $a_{in}\to 0$ along positive integers $n$. The spectral theorem Spectral theorem for unbounded self-adjoint operators (PVM form) and product/domain rule Unbounded Borel functional calculus: domains, products, spectral mapping identify $R_A(in)$ with the bounded function $(in-\mu {)}^{-1}$ of $A$: multiplication by $in-\mu$ gives the two inverse identities, with range in $D(A)$ since both $(in-\mu {)}^{-1}$ and $\mu (in-\mu {)}^{-1}$ are bounded. Consequently $F_n:=(iI-A)R_A(in)$ is the bounded function $h_n(\mu )=(i-\mu )\mathrm{/}(in-\mu )$ of $A$. For real $\mu$ and $n\ge 1$, $\mathrm{\mid }{h}_n(\mu ){\mathrm{\mid }}^2=\frac{1+{\mu }^2}{n^2+{\mu }^2}\le 1,h_n(\mu )\to 0.$ The bounded PVM calculus and its adjoint rule Bounded borel pvm integral Pvm integral is a star homomorphism give $F_n^{\ast }=\overline{h_n}(A)$. For every $v\in H$ both squared norms $\mathrm{\parallel }{F}_{n}v{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }{F}_n^{\ast }v{\mathrm{\parallel }}^2$ equal $\int \mathrm{\mid }{h}_n{\mathrm{\mid }}^{2}d{E}_v$, which tends to zero by Dominated convergence, dominated by $1$ in the finite measure of mass $\mathrm{\parallel }v{\mathrm{\parallel }}^2$. In particular both families converge strongly to zero.

   Put $C=B{R}_A(i)$, compact by item 2. The inverse identity on $D(A)$ gives $C{F}_n=B{R}_A(in)$. Suppose its norm does not tend to zero. There exist $\delta >0$, a strictly increasing integer subsequence $n_k$, and, using the declared AC, vectors $y_k$ with $\mathrm{\parallel }{y}_k\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }C{F}_{n_k}{y}_k\mathrm{\parallel }>\delta$. For any $v\in H$, $\mathrm{\mid }⟨F_{n_k}{y}_k,v⟩\mathrm{\mid }=\mathrm{\mid }⟨y_k,F_{n_k}^{\ast }v⟩\mathrm{\mid }\le \mathrm{\parallel }{F}_{n_k}^{\ast }v\mathrm{\parallel }\to 0.$ Riesz representation Riesz representation for Hilbert spaces therefore proves $F_{n_k}{y}_k\rightharpoonup 0$. A compact operator sends a weakly null sequence to a norm-null sequence under AC Compact operator sends weakly convergent sequences to norm convergent sequences, contradicting the displayed lower bound. Thus $a_{in}\to 0$. Given any $\epsilon >0$, choose $n$ with $a_{in}<\epsilon$ in the first estimate: $b=n{a}_{in}$ is finite and the $A$ coefficient is below $\epsilon$. Its infimum is therefore zero. This does not assert that the zero coefficient itself is attained. The full AC assumption covers the spectral theorem and the compactness/sequence argument.