Let $\mathcal{H}$ be a complex Hilbert space, $A$ a self-adjoint unbounded operator on $\mathcal{H}$ with domain $\mathcal{D}(A)$, and $g:\mathcal{D}(A)\rightarrow\mathbb{C}$ be an unbounded linear functional on $\mathcal{H}$ which is continuous with respect to the graph norm associated with $A$. I will denote its action on $\varphi\in\mathcal{D}(A)$ by $\langle g,\varphi\rangle$. The operator $A_0$ defined by \begin{equation} A_0=A\restriction\ker g \end{equation} is, as long as $g$ is unbounded, a densely defined symmetric operator on $\mathcal{H}$. Furthermore, its deficiency (or defect) indices, defined as usual by \begin{equation} d_\pm=\dim\mathrm{ran}(A_0\pm i)^\perp=\dim\ker(A_0^*\mp i) \end{equation} should be both equal to $1$. This is claimed in the classic reference Singular perturbations of differential operators by S. Albeverio and P. Kurasov, and it is the starting point to define rigorously rank-one singular perturbations of $A$: that is, operators corresponding to the formal expression "$A+\alpha\langle g,\cdot\rangle g$".

Now, unless I am losing something, in the book the authors do not prove explicitly that the deficiency indices of $A_0$ are in fact equal to $d_\pm=1$. They do show that the vectors $(A\pm i)^{-1}g$, to be intended as the vectors uniquely associated with the (bounded) forms \begin{equation} \varphi\in\mathcal{H}\mapsto\left\langle g,(A\pm i)^{-1}\varphi\right\rangle \end{equation} are indeed deficiency elements for $A_0$. This is easy: given $\varphi\in\mathrm{ran}(A_0\pm i)$, that is, $\varphi=(A\pm i)\psi$ for some $\psi\in\mathcal{D}(A_0)$ (i.e. $\psi\in\mathcal{D}(A)$ and $\langle g,\psi\rangle=0$), then \begin{equation} \left\langle g,(A\pm i)^{-1}\varphi\right\rangle=\left\langle g,\psi\right\rangle=0. \end{equation} However, this merely shows that $(A\pm i)^{-1}g$ are nonzero elements of the deficiency subspaces (whence $d_\pm\geq1$); one should show that any other element of these spaces is proportional to them, and the referenced book does not seem to provide a proof for that, merely stating that "the deficiency element is unique (up to multiplication by complex numbers)".

I am aware this is likely a very easy thing to prove. A possibility I was considering is the following: any element $\theta_{\pm i}\in\ker(A_0^*\mp i)$ must agree with $(A\pm i)^{-1}g$, as a bounded functional on $\mathcal{H}$, on the dense subspace $\mathrm{ker}\,g\subset\mathcal{H}$ (since both functionals must be identically zero on it), and this may somehow imply the desired relation. However, this does not seem to be enough.


1 Answer 1


The proof should be the following one, along the lines of Lemma II.2 in M. A. Astaburuaga et. al, J. Math. Phys. 63, 023502 (2022).

Since $g$ is continuous with respect to the graph norm on $\mathcal{D}(A)$, which is equivalent (see Section 1.2.2 of Albeverio and Kurasov's book referenced in the question) to the norms \begin{equation} \phi\mapsto\left\|(A\pm i)\phi\right\|, \end{equation} the functional $(A\pm i)^{-1}g$, defined by $\psi\in\mathcal{H}\mapsto\left\langle (A\pm i)^{-1}g,\psi\right\rangle:=\left\langle g,(A\mp i)^{-1}\psi\right\rangle$, is continuous on $\mathcal{H}$. The Riesz representation theorem implies that there exist two vectors $h_{\pm i}\in\mathcal{H}$ such that $h_{\pm i}=(A\mp i)^{-1}g$.

Now, let $\psi\in\mathcal{H}$. One has \begin{eqnarray} \langle h_{\pm i},\psi\rangle =0 &\iff& \left\langle g,(A\mp i)^{-1}\psi\right\rangle=0\\ &\iff&(A\mp i)^{-1}\psi\in\ker \langle g,\cdot\rangle\\ &\iff&\exists\phi\in\mathrm{dom}\,A,\,\langle g,\phi\rangle=0,\text{ s.t. } \psi=(A\mp i)\phi\\ &\iff&\exists\phi\in\mathrm{dom}\,A_0\text{ s.t. } \psi=(A\mp i)\phi\\ &\iff&\psi\in\mathrm{ran}(A_0\mp i). \end{eqnarray} Therefore, $\{h_{\pm i}\}^\perp = \mathrm{ran}(A_0\mp i)$ and thus, taking orthogonals, $\overline{\{h_{\pm i}\}}=\mathrm{span}\{h_{\pm i}\}=\mathrm{ran}(A_0\mp i)^\perp$. The linear span of a vector is one-dimensional, thus concluding the proof.


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