Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems

Contents
1.   Introduction 1  
2.   Boundary  triplets and self-adjoint  extensions 5   
3.   Gelfand  triples 9   
4.   Abstract  boundary  problems 12   
5.   Families of  abstract  boundary  problems 22   
6.   Differential  boundary  problems of  order one 28   
7.   Dirac-like  boundary  problems 33   
8.   Comparing  two boundary conditions 38   
9.   Rellich  example 43   
References 44

1. Introduction

The  Lagrange  identity  (Green  formula).   Let  $X$  be a compact  manifold  with  the boundary  $Y\hskip 3.99994pt=\hskip 3.99994pt\partial\hskip 1.00006ptX$.   Let  $D$ be a differential  operator acting on section of  a  Hermitian  bundle $E$ over $X$,   and  $D^{\prime}$  be  the operator  formally adjoint  to $D$.   The  theory of  boundary  problems for $D$ crucially depends on an  identity of  the form

(1)

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptD\hskip 0.50003ptu\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.00003pt-\hskip 3.00003pt\langle\hskip 1.49994pt\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997ptD^{\prime}\hskip 0.50003ptv\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\gamma_{1}\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997pt\gamma_{0}\hskip 1.00006ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004pt\partial}\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\gamma_{0}\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997pt\gamma_{1}\hskip 1.00006ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004pt\partial}\hskip 3.00003pt,$$

where $u\hskip 0.50003pt,\hskip 1.99997ptv$ are sections of  $E$.   It  is  known as  the  Lagrange  identity  or  the  Green  formula.   The scalar  products  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle$  and  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004pt\partial}$  are obtained  by  taking  the scalar product  in $E$ and  integrating over $X$ and  $Y$ respectively.   The operators $\gamma_{0}\hskip 1.00006pt,\hskip 3.00003pt\gamma_{1}$ are  the  boundary operators  taking  sections of  $E$ over $X$  to sections of  the restriction  $E\hskip 1.49994pt|\hskip 1.49994ptY$ of  $E$  to $Y$.   The identity  (1)  is  established  first  for smooth sections $u\hskip 0.50003pt,\hskip 1.99997ptv$  and  then extended  to sections  in  Sobolev  spaces.

When $D$  is  a differential  operator of  order $2$,   it  is  only  natural  to take as  $\gamma_{0}$  the restriction of  sections  to $Y$ and as  $\gamma_{1}$  the normal  derivative along  $Y$.   M.I.  Vishik  [V]  discovered  that  this  “naïve”  choice of  $\gamma_{0}\hskip 1.00006pt,\hskip 3.00003pt\gamma_{1}$  is  not  the most  efficient  one.   Namely,   it  is  advantageous  to adjust  the operator $\gamma_{1}$ by  replacing  it  by $\gamma_{1}\hskip 1.99997pt-\hskip 1.99997ptP$,   where  $P$,   nowadays  known as  the  Dirichlet–to–Neumann  operator,   takes a section  $f$ over  $Y$  to  $\gamma_{1}\hskip 1.00006ptu$,   where  $u$  is  the solution of  boundary  problem  $D\hskip 1.00006ptu\hskip 3.99994pt=\hskip 3.99994pt0$,  $\gamma_{0}\hskip 1.00006ptu\hskip 3.99994pt=\hskip 3.99994ptf$.   Then  the identity  (1)  still  holds,   and,   moreover,   is  valid  for $u\hskip 0.50003pt,\hskip 1.99997ptv$  belonging  to  the maximal  domains of  definition of  $D\hskip 0.50003pt,\hskip 1.99997ptD^{\prime}$ respectively.

G.  Grubb  [$G_{\hskip 0.70004pt1}$]  extended  Vishik’s  theory  to operators of  even  order $2m$  (and strengthened  it  in some  respects).   Both  Vishik  and  Grubb  considered only operators acting  on  functions,   but  the generalization  to sections of  bundles  is  routine.   At  the same  time  the assumption  of  the even order was used  throughout  and was even  build  into  the notations.   There are $2m$  boundary operators and  they are split  into  to  two groups of  $m$  operators each,   one leading  to $\gamma_{0}$ and  the other  to $\gamma_{1}$.   Much  later  B.M.  Brown,   G.  Grubb  and  I.G.  Wood  [BGW]  indicated  that  Grubb’s  theory can  be adapted  to some matrix operators of  order $1$.

Self-adjoint  operators and  boundary  triplets.   In  the present  paper we are interested only  in  formally self-adjoint  operators,   and  their realizations by self-adjoint  operators  in  Hilbert  spaces.   The  theory of  boundary  triplets provides an abstract  axiomatic framework for using  the  Lagrange  identity  to construct  self-adjoint  operators in  Hilbert  spaces.   Let  $H_{\hskip 0.70004pt0}$  and  $K^{\hskip 0.70004pt\partial}$  be Hilbert spaces and $T$ be a densely defined  symmetric operator  in $H_{\hskip 0.70004pt0}$,   $T^{\hskip 0.70004pt*}$ be its adjoint  operator,  and  $\mathcal{D}\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$ be  its domain.   Let  $\gamma_{0}\hskip 1.00006pt,\hskip 3.00003pt\gamma_{1}\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{D}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK^{\hskip 0.70004pt\partial}$  be  two  linear maps.   The  triple $(\hskip 1.49994pt\mathcal{D}\hskip 0.50003pt,\hskip 3.00003pt\gamma_{0}\hskip 1.00006pt,\hskip 3.00003pt\gamma_{1}\hskip 1.49994pt)$  is  a  boundary  triplet  for $T^{\hskip 0.70004pt*}$  if  $\gamma_{0}\hskip 1.00006pt\oplus\hskip 1.00006pt\gamma_{1}\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{D}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK^{\hskip 0.70004pt\partial}\hskip 1.00006pt\oplus\hskip 1.00006ptK^{\hskip 0.70004pt\partial}$  is  surjective and

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.00003pt-\hskip 3.00003pt\langle\hskip 1.49994pt\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\gamma_{1}\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997pt\gamma_{0}\hskip 1.00006ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004pt\partial}\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\gamma_{0}\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997pt\gamma_{1}\hskip 1.00006ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004pt\partial}$$

for every  $u\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}$.   Here  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle$  is  the scalar product  in $H_{\hskip 0.70004pt0}$ and  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004pt\partial}$  is  the scalar product  in $K^{\hskip 0.70004pt\partial}$.   A boundary  triplet  for $T^{\hskip 0.70004pt*}$ exists if  and  only  if  $T$ admits a self-adjoint  extension,   and  then  the self-adjoint  extensions of  $T$ are in a natural  one-to-one correspondence with  self-adjoint  relations  $\mathcal{B}\hskip 1.99997pt\subset\hskip 1.99997ptK^{\hskip 0.70004pt\partial}\hskip 1.00006pt\oplus\hskip 1.00006ptK^{\hskip 0.70004pt\partial}$,   which should  be understood as boundary conditions for $T$ or $T^{\hskip 0.70004pt*}$.   Such self-adjoint  relations are a minor  but  essential  generalization  of  self-adjoint  operators  $K^{\hskip 0.70004pt\partial}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK^{\hskip 0.70004pt\partial}$.   Also,   if  a boundary  triplet  exists,   then  $\mathcal{D}\hskip 1.00006pt(\hskip 1.99997pt\overline{T}\hskip 1.99997pt)\hskip 3.99994pt=\hskip 3.99994pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003pt\gamma_{0}\hskip 1.00006pt\oplus\hskip 1.00006pt\gamma_{1}$.   We refer  to  the book of  K.  Schmüdgen  [Schm],   Chapter  14,   for  the details.   An outline of  this  theory,   sufficient  for our  purposes,   is  contained  in  [$I_{\hskip 1.04996pt2}$],   Sections  11  and  12.

Let  $D$  be a formally self-adjoint  differential  operator.   In  this case  the  Lagrange  identity  (1)  leads  to a description of  some self-adjoint  boundary conditions,   i.e.   of  relations between  $\gamma_{0}\hskip 1.00006ptu$  and  $\gamma_{1}\hskip 1.00006ptu$  leading  to self-adjoint  realisations of  $D$  in  a  Hilbert  space,   usually  in  the  Sobolev  space  $H_{\hskip 0.70004pt0}\hskip 1.00006pt(\hskip 1.49994ptX^{\hskip 0.70004pt\circ}\hskip 0.50003pt,\hskip 1.99997ptE\hskip 1.49994pt)$,   where  $X^{\hskip 0.70004pt\circ}\hskip 3.99994pt=\hskip 3.99994ptX\hskip 1.99997pt\smallsetminus\hskip 1.99997ptY$.   In  fact,   the most  interesting  self-adjoint  boundary conditions are defined  in  terms of  “naïve”  boundary operators,   not  the ones adjusted according  to  Vishik and  Grubb.   But  the  “naïve”  boundary operators do not  form a boundary  triplet  and by  this reason are not  good enough.   The results of  Vishik  [V]  and  Grubb  [$G_{\hskip 0.70004pt1}$]  concerned  with  adjusting  the  “naïve”  boundary operators for $D$ can  be understood  as a construction of  a boundary  triplet  for  $D^{\hskip 0.35002pt*}$ starting  with  the  “naïve”  Lagrange  identity  (1).   But  they  were proved  before  the notion of  boundary  triplets appeared.

The index of  families of  self-adjoint  operators.   The  theory of  boundary  triplets  turns out  be a very efficient  tool  to study  the index of  families of  self-adjoint  operators.   It  was already used  by  the author  in  [$I_{\hskip 1.04996pt2}$]  to explain why  the index  theorem of  [$I_{\hskip 1.04996pt2}$]  applies only  to bundle-like boundary conditions.   See  [$I_{\hskip 1.04996pt2}$],   Section  13.   The key result  for  the applications  to  the index  theory  is  a corollary of  one of  the main  results of  the  theory of  boundary  triplets,   namely,   of  the  Krein–Naimark  resolvent  formula.   See  the identities  (5)  and  (6)  below.   Originally  the  Krein–Naimark  resolvent  formula and  this corollary are proved  for  the boundary  triplets constructed  abstractly  in  terms of  the operator  theory.   But  the boundary  triplets are unique in a very strong sense,   and  this allows  to apply  this corollary  to families of  differential  operators  after  the  Lagrange  identity  is  adjusted.

The  Krein–Naimark  resolvent  formula  is  concerned  with extensions of  symmetric operators,   but  in all  sources known  to  the author  it  is  proved without  referring  to  von  Neumann  theory of  extensions.   For  the sake of  the readers who may be,   like  the author,   not  quite comfortable about  such state of  affairs,   we provided  in  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  a direct  proof  of  the identities  (5)  and  (6)  based on  von  Neumann  theory and  bypassing  the Krein–Naimark  resolvent  formula.   For a proof  based on  the  Krein–Naimark  resolvent  formula see  [$I_{\hskip 1.04996pt2}$],   Section  12.

An abstract  construction of  boundary  triplets.   In  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  we develop an abstract  axiomatic version of  the adjustment  procedure of  Vishik  and  Grubb.   The starting  point  is  an axiomatic version of  the  “naïve”  Lagrange  identity,   closely  related  to  the abstract  boundary  problems  framework of  [$I_{\hskip 1.04996pt2}$],   Section  5.   Another  key  ingredient  is  a  reference operator,   a self-adjoint  and  invertible extension $A$ of  $D$,   which  is  assumed  to be naïvely defined  by  the boundary  condition  $\gamma_{0}\hskip 3.99994pt=\hskip 3.99994pt0$.   Under appropriate assumptions we construct  a boundary  triplet  generalizing  the boundary  triplet  implicitly  present  in  the results of  Vishik  and  Grubb.   An  important  role in  the construction of  Grubb  [$G_{\hskip 0.70004pt1}$]  is  played  by  two results of  J.L.  Lions  and  E.  Magenes  [$LM_{\hskip 0.35002pt2}$],  [$LM_{\hskip 0.35002pt3}$].   We prove an abstract  version of  them.   See  Theorems  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  and  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems.   These  two  theorems are proved  by an adaptation of  arguments of  Lions  and  Magenes.   Most  of  the other  proofs in  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  are adapted  from  arguments of  Grubb  [$G_{\hskip 0.70004pt1}$].

Our axiomatic approach  involves  the notion of  Gelfand  triples  in an essential  manner.   We need  only  the relevant  definitions and a couple of  the most  basic properties,   and we included  in  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  a self-contained exposition of  what  is  needed.

Families of  abstract  boundary  problems.   In  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  we add  parameters  to  the  theory developed  in  Sections  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  –  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  and apply  the parameterized  theory  to  the index.   Let  $W$  be a reasonable  topological  space.   Suppose  that  everything  in  sight  depends on  the parameter  $w\hskip 1.99997pt\in\hskip 1.99997ptW$.   We indicate  this dependence by a subscript.   The index of  the family  $A_{\hskip 0.70004ptw}\hskip 1.00006pt,\hskip 3.00003ptw\hskip 1.99997pt\in\hskip 1.99997ptW$  of  reference operators  is  equal  to $0$  because  these operators are invertible.   By  our assumptions  these operators are defined  by  the boundary conditions $\gamma_{w\hskip 1.04996pt0}\hskip 3.99994pt=\hskip 3.99994pt0$.   Let  $\mathcal{B}_{\hskip 0.35002ptw}\hskip 1.00006pt,\hskip 3.00003ptw\hskip 1.99997pt\in\hskip 1.99997ptW$  be a family of  naïve  boundary  conditions.   Replacing  the boundary conditions  $\gamma_{w\hskip 1.04996pt0}\hskip 3.99994pt=\hskip 3.99994pt0$  by  the boundary conditions  $\mathcal{B}_{\hskip 0.35002ptw}$ results in a family of  self-adjoint  operators with  the index equal  to  the index of  $\mathcal{B}_{\hskip 0.35002ptw}\hskip 1.99997pt-\hskip 1.99997ptM_{\hskip 0.70004ptw}\hskip 1.00006pt,\hskip 3.00003ptw\hskip 1.99997pt\in\hskip 1.99997ptW$,   where  $M_{\hskip 0.70004ptw}$ are  the adjusting operators,   analogues of  the  Dirichlet–to–Neumann  ones.   See  the  last  subsection of  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  for  the precise statement.   The appearance of  the summand  $-\hskip 1.99997ptM_{\hskip 0.70004ptw}$  is  fairly surprising,   but  the operators  $\mathcal{B}_{\hskip 0.35002ptw}$ are usually  invertible and  then  the index of  the uncorrected  family  $\mathcal{B}_{\hskip 0.35002ptw}\hskip 1.00006pt,\hskip 3.00003ptw\hskip 1.99997pt\in\hskip 1.99997ptW$  is  $0$.

Differential  boundary  problems of  order one.   Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  is  devoted  to  the application of  the abstract  theory  to differential  operators of  order one.   The key  role  is  played  by  the results related  to  the  Calderón  projector.   They  are used  to prove  that  the adjusting operators $M_{\hskip 0.70004ptw}$ are pseudo-differential  operators of  order zero continuously depending on $w\hskip 1.99997pt\in\hskip 1.99997ptW$.   We identify  the symbols of  the adjusting operators $M_{\hskip 0.70004ptw}$ and  prove  that  the graph of  $M_{\hskip 0.70004ptw}$  is  the space of  the  Cauchy  data of  solutions of  the equation  $D^{\hskip 0.35002pt*}\hskip 1.00006ptu\hskip 3.99994pt=\hskip 3.99994pt0$.

In  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  we use  the results of  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  to give an analytic and  fairly  elementary  proof  of  the index  theorem  for  Dirac-like boundary problems from  [$I_{\hskip 1.04996pt2}$],   Section  15.   See  Theorem  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems.   In  [$I_{\hskip 1.04996pt2}$]  this  theorem  was deduced  from  its analogue for  the  topological  index and  the general  index  theorem  for operators of  order one.   In a sense  the proof  from  [$I_{\hskip 1.04996pt2}$]  provides  topological  reasons for  the need  to adjust  the boundary operators and  predicts  the symbols of  adjusting  operators.   The desire  to find  a direct  analytic proof  of  this  theorem was  the starting  point  of  the present  paper.

In  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  we modify  the arguments of  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  in order  to prove an  Agranovich–Dynin  type  theorem computing  the difference of  indices of  two families of  self-adjoint  problems differing only  by  the boundary conditions.   The formula for  the difference  is  more complicated  than  the classical  one and  involves an appropriate form of  the adjusting  operators $M_{\hskip 0.70004ptw}$.

We  limited ourselves  by  the differential  operators of  order one in  order  to avoid  technical  complications and  present  the main  ideas in  the most  transparent  form.   Most  of  the results can  be extended  to pseudo-differential  operators satisfying  the  transmission condition.

Rellich  example.   In  Section  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  we illustrate  our abstract  theory by an example,   due  to  F.  Rellich,   of  a family of  self-adjoint  boundary problems parameterized  by  the circle for  a fixed operator  of  order $2$.   The operator  is  $-\hskip 1.99997ptd^{\hskip 0.70004pt2}/\hskip 1.00006ptdx^{\hskip 0.70004pt2}$ on  the interval  $[\hskip 1.49994pt0\hskip 0.50003pt,\hskip 1.99997pt1\hskip 1.00006pt]$.   The index of  this family can be computed  by  brute force and  is  equal  to $1$.   We prove  this result  as an application of  our abstract  theory.   In  dimension one  there  is  no need  to adjust  the  Lagrange  identity,   but  one can still  do  this.   Each way  leads  to a proof  that  the index  is  equal  to $1$.

2. Boundary  triplets  and  self-adjoint  extensions

The main abstract  example.   Let  us consider  the following abstract  situation  taken  from  [Schm],   Example  14.5.   Let $H$ be a separable  Hilbert  space,  $T$ be a densely defined closed symmetric operator in $H$,   and $A$ be a fixed self-adjoint  extension of  $T$.   By  $\dotplus$ we will  denote  the direct,   but  not  necessarily orthogonal,   sum of  subspaces of  $H$.   Let  us fix a number  $\mu\hskip 1.99997pt\in\hskip 1.99997pt\mathbf{C}\hskip 1.99997pt\smallsetminus\hskip 1.99997pt\mathbf{R}$.   Then  the numbers $\mu\hskip 1.00006pt,\hskip 3.99994pt\overline{\mu}$  belong  to  the resolvent  set  of  $A$.   In particular,   the operators  $(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}$  and  $(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}$  are well  defined.   Let

$$\quad\mathcal{K}_{\hskip 0.70004pt+}\hskip 3.99994pt=\hskip 3.99994pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.99997pt-\hskip 1.99997pt\hskip 0.24994pt\mu\hskip 1.99997pt)\hskip 3.99994pt=\hskip 3.99994pt\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt\perp}\quad\mbox{and}\quad$$

$$\quad\mathcal{K}_{\hskip 0.70004pt-}\hskip 3.99994pt=\hskip 3.99994pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 3.99994pt=\hskip 3.99994pt\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.99997pt)^{\hskip 0.70004pt\perp}\hskip 3.00003pt.$$

Then  the domain  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$  of  $T^{\hskip 0.70004pt*}$  is  equal  to

(2)

$$\quad\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)\hskip 1.99997pt\dotplus\hskip 1.99997ptA\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 1.99997pt\dotplus\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 3.99994pt.$$

See  [Schm],   Proposition  14.11.   Hence every  $z\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$  can be uniquely  written as

(3)

$$\quad z\hskip 3.99994pt=\hskip 3.99994ptz_{\hskip 1.04996ptT}\hskip 1.99997pt+\hskip 1.99997ptA\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt0}\hskip 1.99997pt+\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt1}\hskip 3.99994pt$$

with $z_{\hskip 1.04996ptT}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)$ and  $z_{\hskip 0.70004pt0}\hskip 1.00006pt,\hskip 1.99997ptz_{\hskip 0.70004pt1}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}$.   Then $K\hskip 3.99994pt=\hskip 3.99994pt\mathcal{K}_{\hskip 0.70004pt-}$ and  the maps  $\Gamma_{i}\hskip 1.00006pt\colon\hskip 1.00006ptz\hskip 3.99994pt\longmapsto\hskip 3.99994ptz_{\hskip 0.70004pti}$,  $i\hskip 3.99994pt=\hskip 3.99994pt0\hskip 0.50003pt,\hskip 1.99997pt1$  form a boundary  triplet  for $T^{\hskip 0.70004pt*}$.   This  is  the boundary  triplet  from  [Schm],   Example  14.5.

The main example from  the point  of  view of  von  Neumann  theory.   By  von  Neuman  theory

$$\quad\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)\hskip 1.99997pt\dotplus\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt+}\hskip 1.99997pt\dotplus\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 3.00003pt$$

and  $A$ defines an  isometry $V\hskip 1.00006pt\colon\mathcal{K}_{\hskip 0.70004pt+}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}$ such  that  $A$  is  the restriction of  $T^{\hskip 0.70004pt*}$  to

$$\quad\bigl{\{}\hskip 3.00003ptx\hskip 1.99997pt+\hskip 1.99997pty\hskip 1.99997pt-\hskip 1.99997ptVy\hskip 3.00003pt\bigl{|}\hskip 3.00003ptx\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)\hskip 0.50003pt,\hskip 3.99994pty\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt+}\hskip 3.00003pt\bigr{\}}\hskip 3.99994pt.$$

2.1. Lemma.   $\displaystyle Vy\hskip 3.99994pt=\hskip 3.99994pt\frac{A\hskip 1.99997pt-\hskip 1.99997pt\mu}{\hskip 1.00006ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.00006pt}\hskip 1.99997pty$   for every  $y\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt+}$.   

Proof.   If  $y\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt+}$,   then

$$\quad(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 1.99997pt(\hskip 1.49994pty\hskip 1.99997pt-\hskip 1.99997ptVy\hskip 1.99997pt)\hskip 3.99994pt=\hskip 3.99994pt\mu\hskip 1.99997pty\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 3.00003ptVy\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pty\hskip 1.99997pt+\hskip 1.99997pt\overline{\mu}\hskip 3.00003ptVy\hskip 3.99994pt=\hskip 3.99994pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 1.99997pty\hskip 3.00003pt,$$

$$\quad y\hskip 1.99997pt-\hskip 1.99997ptVy\hskip 3.99994pt=\hskip 3.99994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 1.99997pty\hskip 3.00003pt,$$

and  hence  $\displaystyle Vy\hskip 3.99994pt=\hskip 3.99994pty\hskip 1.99997pt-\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 1.99997pty\hskip 3.99994pt=\hskip 3.99994pt\frac{A\hskip 1.99997pt-\hskip 1.99997pt\mu}{\hskip 1.00006ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.00006pt}\hskip 1.99997pty$.    $\blacksquare$

2.2. Lemma.   Suppose  that  $z\hskip 3.99994pt=\hskip 3.99994ptz_{\hskip 1.04996ptT}\hskip 1.99997pt+\hskip 1.99997ptz_{\hskip 0.70004pt+}\hskip 1.99997pt+\hskip 1.99997ptz_{\hskip 0.70004pt-}$  with  $z_{\hskip 1.04996ptT}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)$,  $z_{\hskip 0.70004pt+}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt+}$,   and  $z_{\hskip 0.70004pt-}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}$.   Let  $z_{\hskip 1.04996pt0}\hskip 3.99994pt=\hskip 3.99994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}$ and  $z_{\hskip 0.70004pt1}\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}$.   Then  the equality  (3)  holds.   

Proof.   Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  implies  that

$$\quad z_{\hskip 1.04996pt0}\hskip 3.99994pt=\hskip 3.99994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 3.99994pt=\hskip 3.99994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 1.99997pt\frac{A\hskip 1.99997pt-\hskip 1.99997pt\mu}{\hskip 1.00006ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.00006pt}\hskip 3.99994ptz_{\hskip 0.70004pt+}\quad\mbox{and}$$

$$\quad z_{\hskip 0.70004pt1}\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.99994pt\frac{A\hskip 1.99997pt-\hskip 1.99997pt\mu}{\hskip 1.00006ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.00006pt}\hskip 3.99994ptz_{\hskip 0.70004pt+}\hskip 3.00003pt.$$

It  follows  that

$$\quad A\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt0}\hskip 3.99994pt=\hskip 3.99994ptA\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 1.99997ptA\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt+}\hskip 3.00003pt,$$

$$\quad(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt1}\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt+}\hskip 3.00003pt,$$

and  hence

$$\quad A\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt0}\hskip 1.99997pt+\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt1}$$

$$\quad=\hskip 3.99994ptA\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 3.99994pt+\hskip 3.99994ptA\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt+}\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006ptz_{\hskip 0.70004pt+}$$

$$\quad=\hskip 3.99994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 1.99997ptz_{\hskip 0.70004pt+}\hskip 3.00003pt.$$

The equality  (3)  follows.    $\blacksquare$

The  Lagrange  identity.   In  the notations of  Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems,   let

$$\quad\Gamma_{0}\hskip 1.00006ptz\hskip 3.99994pt=\hskip 3.99994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\quad\mbox{and}\quad\Gamma_{1}\hskip 1.00006ptz\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 3.00003pt.$$

We claim  that  then  for every  $x\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$

(4)

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}x\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.00003pt-\hskip 3.00003pt\langle\hskip 1.49994pt\hskip 1.00006ptx\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}y\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Gamma_{1}\hskip 1.00006ptx\hskip 1.00006pt,\hskip 1.99997pt\Gamma_{0}\hskip 1.00006pty\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Gamma_{0}\hskip 1.00006ptx\hskip 1.00006pt,\hskip 1.99997pt\Gamma_{1}\hskip 1.00006pty\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt,$$

where all  scalar products are  taken  in  $H$.   In order  to prove  this,   let  us write $x$  and  $y$  in  the form of  Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems,  $x\hskip 3.99994pt=\hskip 3.99994ptx_{\hskip 1.04996ptT}\hskip 1.99997pt+\hskip 1.99997ptx_{\hskip 0.70004pt+}\hskip 1.99997pt+\hskip 1.99997ptx_{\hskip 0.70004pt-}$  and  $y\hskip 3.99994pt=\hskip 3.99994pty_{\hskip 1.04996ptT}\hskip 1.99997pt+\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.99997pt+\hskip 1.99997pty_{\hskip 0.70004pt-}$.   Since  $T$  is  symmetric,  $x_{\hskip 1.04996ptT}$ and  $y_{\hskip 1.04996ptT}$ do not  affect  the validity of  the  Lagrange  identity and  hence we can assume  that  $x_{\hskip 1.04996ptT}\hskip 3.99994pt=\hskip 3.99994pty_{\hskip 1.04996ptT}\hskip 3.99994pt=\hskip 3.99994pt0$.   Then  $T^{\hskip 0.70004pt*}\hskip 1.00006ptx\hskip 3.99994pt=\hskip 3.99994pt\mu\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 1.99997pt+\hskip 3.00003pt\overline{\mu}\hskip 1.99997ptx_{\hskip 0.70004pt-}$ and  $T^{\hskip 0.70004pt*}\hskip 1.00006pty\hskip 3.99994pt=\hskip 3.99994pt\mu\hskip 1.00006pty_{\hskip 0.70004pt+}\hskip 1.99997pt+\hskip 3.00003pt\overline{\mu}\hskip 1.99997pty_{\hskip 0.70004pt-}$.   Therefore  the  left  hand side of  (4)  is  equal  to

$$\quad\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle$$

$$\quad-\hskip 3.99994pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle$$

$$\quad=\hskip 3.99994pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.00003pt.$$

The right  hand side of  (4)  is  equal  to

$$\quad\left\langle\hskip 1.49994pt\hskip 1.49994pt-\hskip 1.99997pt\mu\hskip 1.49994ptx_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.00003ptV\hskip 0.50003ptx_{\hskip 0.70004pt+}\hskip 1.00006pt,\hskip 3.99994pty_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 3.00003ptV\hskip 0.50003pty_{\hskip 0.70004pt+}\hskip 1.49994pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.49994ptx_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 3.00003ptV\hskip 0.50003ptx_{\hskip 0.70004pt+}\hskip 1.00006pt,\hskip 3.99994pt-\hskip 1.99997pt\mu\hskip 1.49994pty_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.00003ptV\hskip 0.50003pty_{\hskip 0.70004pt+}\hskip 1.49994pt\hskip 1.49994pt\right\rangle$$

$$\quad=\hskip 3.99994pt-\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 1.99997ptVy_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptVx_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptVx_{\hskip 0.70004pt+}\hskip 1.00006pt,\hskip 1.99997ptVy_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle$$

$$\quad\phantom{=\hskip 3.99994pt}+\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 1.99997ptVy_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptVx_{\hskip 0.70004pt+}\hskip 1.00006pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt\mu\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptVx_{\hskip 0.70004pt+}\hskip 1.00006pt,\hskip 1.99997ptVy_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle$$

$$\quad=\hskip 3.99994pt-\hskip 1.99997pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptVx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997ptVy_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle$$

$$\quad=\hskip 3.99994pt-\hskip 1.99997pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt-}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt-}\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 1.99997pt+\hskip 1.99997pt(\hskip 1.49994pt\mu\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.00006pt\langle\hskip 1.49994pt\hskip 1.00006ptx_{\hskip 0.70004pt+}\hskip 0.50003pt,\hskip 1.99997pty_{\hskip 0.70004pt+}\hskip 1.00006pt\hskip 1.49994pt\rangle$$

because  $V$  is  an  isometry.   The  identity  (4)  follows.

The boundary  triplet.   Clearly,   the map  $z\hskip 3.99994pt\longmapsto\hskip 3.99994pt(\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 1.99997ptz_{\hskip 0.70004pt+}\hskip 1.49994pt)$  from $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$  to  $\mathcal{K}_{\hskip 0.70004pt-}\hskip 1.00006pt\oplus\hskip 1.00006pt\mathcal{K}_{\hskip 0.70004pt+}$  is  surjective.   Since  $\mu\hskip 1.99997pt\not\in\hskip 1.99997pt\mathbf{R}$,   this implies  that  the map

$$\quad\Gamma_{0}\hskip 1.00006pt\oplus\hskip 1.00006pt\Gamma_{1}\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 1.00006pt\oplus\hskip 1.00006pt\mathcal{K}_{\hskip 0.70004pt-}$$

is  also surjective.   In view of  (4)  this implies  that  $(\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 1.00006pt,\hskip 3.99994pt\Gamma_{0}\hskip 1.00006pt,\hskip 3.99994pt\Gamma_{1}\hskip 1.49994pt)$  is  a boundary  triplet  for  $T^{\hskip 0.70004pt*}$.   Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  implies  that  this  is  the same boundary  triplet  as  in  [Schm],   Example  14.5.

Isometries  between subspaces.   Let  $W\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{K}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathcal{K}\hskip 0.50003pt^{\prime}$  be an  isometry  between  two closed subspaces $\mathcal{K},\hskip 3.99994pt\mathcal{K}\hskip 0.50003pt^{\prime}\hskip 1.99997pt\subset\hskip 1.99997ptH$.   We will  denote by  $W_{\hskip 0.70004pt0}$  the corresponding  partial  isometry  of  $H$,   i.e.  the operator equal  to $W$ on  $\mathcal{K}$ and  to $0$ on  $\mathcal{K}^{\hskip 0.70004pt\perp}$.   When  $\mathcal{K}\hskip 3.00003pt=\hskip 3.99994pt\mathcal{K}\hskip 0.50003pt^{\prime}$,   we will  denote by  $W_{\hskip 0.70004ptH}$  the operator  $H\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$ equal  to $W$ on  $\mathcal{K}$ and  to  the identity on  $\mathcal{K}^{\hskip 0.70004pt\perp}$.

The  Cayley  transforms.   As above,   let  $\mu\hskip 1.99997pt\in\hskip 1.99997pt\mathbf{C}\hskip 1.99997pt\smallsetminus\hskip 1.99997pt\mathbf{R}$ and  $T$  be a symmetric operator in  $H$.   The  $\mu$-Cayley  transform  $U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)$ of  $T$  is  the partial  isometry equal  to

$$\quad\frac{T\hskip 1.99997pt-\hskip 1.99997pt\mu}{\hskip 1.00006ptT\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.00006pt}$$

on  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)$ and as $0$ on  the orthogonal  complement  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)^{\hskip 0.70004pt\perp}\hskip 3.99994pt=\hskip 3.99994pt\mathcal{K}_{\hskip 0.70004pt+}$.   It  induces an  isometry  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\mu\hskip 1.49994pt)$.   If  $A$  is  a self-adjoint  extension of  $T$ as above,   then  $U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.49994pt)$ agrees with  $U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)$ on  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)$.   In  this case  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptA\hskip 1.99997pt-\hskip 1.99997pt\overline{\mu}\hskip 1.99997pt)\hskip 3.99994pt=\hskip 3.99994ptH$ and  $U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.49994pt)$  is  an  isometry  $H\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$.   Let  $V$ be related  to $A$ as above.

2.3. Lemma.   $U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994ptU_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptT\hskip 1.49994pt)\hskip 1.99997pt+\hskip 1.99997ptV_{\hskip 0.35002pt0}$.

Proof.   This immediately  follows  from  Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems.    $\blacksquare$

Changing  the extension $A$.   Let  $B\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}$  be a self-adjoint  operator.   Together with our boundary  triplet  $B$ defines another self-adjoint  extension $A^{\prime}$ of  $T$,   namely  the restriction of  $T^{\hskip 0.70004pt*}$  to  $\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994pt\Gamma_{1}\hskip 1.99997pt-\hskip 1.99997ptB\hskip 1.49994pt\Gamma_{0}\hskip 1.49994pt)$.   The domain  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)$  is  described  by  the equation

$$\quad-\hskip 1.99997pt\mu\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt-\hskip 3.00003pt\overline{\mu}\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 3.99994pt=\hskip 3.99994ptB\hskip 1.49994pt(\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 1.49994pt)\hskip 3.00003pt,$$

or,   equivalently,   by either of  the equations

$$\quad(\hskip 1.49994ptB\hskip 1.99997pt+\hskip 1.99997pt\mu\hskip 1.49994pt)\hskip 1.49994ptz_{\hskip 0.70004pt-}\hskip 1.99997pt+\hskip 1.99997pt(\hskip 1.49994ptB\hskip 1.99997pt+\hskip 3.00003pt\overline{\mu}\hskip 1.99997pt)\hskip 1.49994ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 3.99994pt=\hskip 3.99994pt0\hskip 3.00003pt,\quad z_{\hskip 0.70004pt-}\hskip 3.00003pt+\hskip 3.00003pt\frac{B\hskip 1.99997pt+\hskip 3.00003pt\overline{\mu}\hskip 0.50003pt}{B\hskip 1.99997pt+\hskip 1.99997pt\mu}\hskip 3.00003ptV\hskip 0.50003ptz_{\hskip 0.70004pt+}\hskip 3.99994pt=\hskip 3.99994pt0\hskip 3.00003pt.$$

Let  $V\hskip 0.50003pt^{\prime}$ be  related  to $A^{\prime}$  in  the same way as $V$  to $A$.   The  last  formula shows  that

$$\quad V\hskip 0.50003pt^{\prime}\hskip 3.99994pt=\hskip 3.99994pt\frac{B\hskip 1.99997pt+\hskip 3.00003pt\overline{\mu}\hskip 0.50003pt}{B\hskip 1.99997pt+\hskip 1.99997pt\mu}\hskip 3.00003ptV$$

Together  with  Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  this  implies  that

$$\quad U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)\hskip 3.99994pt\hskip 0.50003pt=\hskip 3.99994pt\hskip 1.00006pt\left(\hskip 1.99997pt\frac{B\hskip 1.99997pt+\hskip 3.00003pt\overline{\mu}\hskip 0.50003pt}{B\hskip 1.99997pt+\hskip 1.99997pt\mu}\hskip 1.99997pt\right)_{\hskip 0.70004ptH}U_{\hskip 0.70004pt\mu}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.49994pt)\hskip 3.00003pt.$$

Now,   let  us  take  $\mu\hskip 3.99994pt=\hskip 3.99994pti$  and observe  that  $U_{\hskip 0.35002pti}\hskip 1.49994pt(\hskip 1.49994pt\bullet\hskip 1.49994pt)$  is  the usual  Cayley  transform,   which we will  denote by  $U\hskip 1.49994pt(\hskip 1.49994pt\bullet\hskip 1.49994pt)$.   Hence  the  last  formula  implies  that

(5)

$$\quad U\hskip 1.49994pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994ptU\hskip 1.49994pt(\hskip 1.49994ptB\hskip 1.49994pt)_{\hskip 0.70004ptH}\hskip 1.49994ptU\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.49994pt)\hskip 3.00003pt.$$

Since  $U_{\hskip 0.35002pt-\hskip 0.70004pti}\hskip 1.49994pt(\hskip 1.49994pt\bullet\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994ptU\hskip 1.49994pt(\hskip 1.49994pt\bullet\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}$,   for  $\mu\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pti$  we get  $U\hskip 1.49994pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 3.99994pt=\hskip 3.99994ptU\hskip 1.49994pt(\hskip 1.49994ptB\hskip 1.49994pt)_{\hskip 0.70004ptH}^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.49994ptU\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.49994pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}$  and  hence

(6)

$$\quad U\hskip 1.49994pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994ptU\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.49994pt)\hskip 1.49994ptU\hskip 1.49994pt(\hskip 1.49994ptB\hskip 1.49994pt)_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

Here  $U\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.49994pt)\hskip 0.50003pt,\hskip 3.00003ptU\hskip 1.49994pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)$ have  the same meaning as before,   but  $B$  is  a operator  in  $\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.99997pt-\hskip 1.99997pti\hskip 1.99997pt)$ and  not  in  $\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.99997pt+\hskip 1.99997pti\hskip 1.99997pt)$ as before.   The identity  (6)  is  a minor  generalization of  the  last  identity  in  [Schm],   Theorem  14.20.   The same arguments work  for self-adjoint  relations  $\mathcal{B}\hskip 1.99997pt\subset\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}\hskip 1.99997pt\oplus\hskip 1.99997pt\mathcal{K}_{\hskip 0.70004pt-}$.   In  this case we get  $U\hskip 1.49994pt(\hskip 1.49994ptA^{\prime}\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994ptU\hskip 1.49994pt(\hskip 1.49994pt\mathcal{B}\hskip 1.49994pt)_{\hskip 0.70004ptH}\hskip 1.49994ptU\hskip 1.49994pt(\hskip 1.49994ptA\hskip 1.49994pt)$  for  $\mu\hskip 3.99994pt=\hskip 3.99994pti$.

3. Gelfand  triples

Gelfand  triples.   Let  $H$  be a separable  Hilbert  space and  $K\hskip 1.99997pt\subset\hskip 1.99997ptH$  be a dense vector subspace which  is  also a  Hilbert  space in  its own  right.   Let  $\iota\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$  be  the inclusion.   It  is  assumed  that  $\iota$  is  bounded.   Let  $K\hskip 0.50003pt^{\prime}$  be  the space of  anti-linear  maps  $K\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathbf{C}$.   Then  the dual  of  $\iota$  is  the map  $\iota\hskip 0.50003pt^{\prime}\hskip 1.00006pt\colon\hskip 1.00006ptH\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$.   Since $\iota$  is  injective,  $\iota\hskip 0.50003pt^{\prime}\hskip 1.00006pt(\hskip 1.49994ptH\hskip 1.49994pt)$  is  dense in  $K\hskip 0.50003pt^{\prime}$.   Since $\iota\hskip 1.49994pt(\hskip 1.49994ptK\hskip 1.49994pt)$  is  dense in  $H$,   the map  $\iota\hskip 0.50003pt^{\prime}$  is  injective.   We will  identify  $H$  with  $H\hskip 0.50003pt^{\prime}$ in  the usual  manner,   but  not  $K$  with  $K\hskip 0.50003pt^{\prime}$,   despite  the fact  that  $K$  is  a  Hilbert  space.   In  fact,   it  is  impossible  to satisfactory  identify  both  $H$  with  $H\hskip 0.50003pt^{\prime}$ and  $K$  with  $K\hskip 0.50003pt^{\prime}$  simultaneously.   Still,   since $K$  is  a  Hilbert  space,  $K\hskip 0.50003pt^{\prime}$  is  also a  Hilbert  space.   Usually  we  will  treat  the maps  $\iota$  and  $\iota\hskip 0.50003pt^{\prime}$ as inclusions  (of  sets).   Then we  get  a  triple  $K\hskip 1.99997pt\subset\hskip 1.99997ptH\hskip 1.99997pt\subset\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$  of  Hilbert  spaces,   called  the  Gelfand  triple  associated  with  the pair  $K\hskip 0.50003pt,\hskip 1.99997ptH$.   We will  denote  the  Hilbert  scalar  product  in  $K$  by  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK}$ and use similar notations for other  Hilbert  spaces.   Let

$$\quad\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 1.99997pt\colon\hskip 1.99997ptK\hskip 0.50003pt^{\prime}\hskip 1.00006pt\times\hskip 1.00006ptK\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathbf{C}$$

be  the canonical  pairing $\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.99994pt=\hskip 3.99994pty\hskip 1.49994pt(\hskip 1.49994ptx\hskip 1.49994pt)$,   If  $(\hskip 1.49994pty\hskip 0.50003pt,\hskip 1.99997ptx\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997ptH\hskip 1.00006pt\times\hskip 1.00006ptK$,   then

$$\quad\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.99994pt=\hskip 3.99994pt\iota\hskip 0.50003pt^{\prime}\hskip 0.50003pty\hskip 1.49994pt(\hskip 1.49994ptx\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 3.00003pt\iota\hskip 1.00006ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

In other words,  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}$  is  equal  to  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptH}$  on  $H\hskip 1.00006pt\times\hskip 1.00006ptK$.   Let

$$\quad\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}\hskip 1.99997pt\colon\hskip 1.99997ptK\hskip 1.00006pt\times\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathbf{C}$$

be  the pairing  $\langle\hskip 1.49994pt\hskip 1.00006ptx\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}\hskip 3.99994pt=\hskip 3.99994pt\overline{y\hskip 1.49994pt(\hskip 1.49994ptx\hskip 1.49994pt)}$.   If  $(\hskip 1.49994ptx\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997ptK\hskip 1.00006pt\times\hskip 1.00006ptH$,   then

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptx\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}\hskip 3.99994pt=\hskip 3.99994pt\overline{\iota\hskip 0.50003pt^{\prime}\hskip 0.50003pty\hskip 1.49994pt(\hskip 1.49994ptx\hskip 1.49994pt)}\hskip 3.99994pt=\hskip 3.99994pt\overline{\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle}_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\langle\hskip 1.49994pt\hskip 1.00006ptx\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

In other words,  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}$  is  equal  to  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptH}$  on  $K\hskip 1.00006pt\times\hskip 1.00006ptH$.   The  Hilbert  space $K\hskip 0.50003pt^{\prime}$ can be also constructed as  the completion of  $H$  with  respect  to  a new scalar  product.   Let  $\iota^{\hskip 0.70004pt*}\hskip 1.00006pt\colon\hskip 1.00006ptH\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$  be  the operator adjoint  to $\iota$,   i.e.  such  that

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pt\iota\hskip 1.49994ptx\hskip 0.50003pt,\hskip 3.00003pty\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006ptx\hskip 0.50003pt,\hskip 3.00003pt\iota^{\hskip 0.70004pt*}y\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK}$$

for every  $x\hskip 1.99997pt\in\hskip 1.99997ptK$,  $y\hskip 1.99997pt\in\hskip 1.99997ptH$.   For $y\hskip 1.99997pt\in\hskip 1.99997ptH$ the  linear  functional  $\iota\hskip 0.50003pt^{\prime}\hskip 0.50003pty\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathbf{C}$  is  equal  to

$$\quad\iota\hskip 0.50003pt^{\prime}\hskip 0.50003pty\hskip 1.00006pt\colon\hskip 1.00006ptx\hskip 3.99994pt\longmapsto\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 3.00003pt\iota\hskip 1.00006ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\iota^{\hskip 0.70004pt*}y\hskip 0.50003pt,\hskip 3.00003ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK}\hskip 3.00003pt.$$

It  follows  that  for every  $u\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.99997pt\in\hskip 1.99997ptH$  the scalar  product  in  $K\hskip 0.50003pt^{\prime}$  is

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pt\iota\hskip 0.50003pt^{\prime}\hskip 0.50003ptu\hskip 0.50003pt,\hskip 3.00003pt\iota\hskip 0.50003pt^{\prime}\hskip 0.50003ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\iota^{\hskip 0.70004pt*}u\hskip 0.50003pt,\hskip 3.00003pt\iota^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK}\hskip 3.00003pt.$$

Therefore  $K\hskip 0.50003pt^{\prime}$  can  be identified with  the completion of  $H$  with respect  to  the scalar  product

(7)

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006ptu\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle^{\prime}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\iota^{\hskip 0.70004pt*}u\hskip 0.50003pt,\hskip 3.00003pt\iota^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK}\hskip 3.00003pt.$$

In  particular,  $\left\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\right\rangle^{\prime}$.   The equality  (7)  implies  that  the operator  $\iota^{\hskip 0.70004pt*}$  defines an  isometry  from  the subspace  $H$  of  $K\hskip 0.50003pt^{\prime}$  into  $K$.   Extending  $\iota^{\hskip 0.70004pt*}$  by continuity,   we get  an  isometric operator  $\mathbf{I}\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$.   In  fact,  $\mathbf{I}$  is  surjective.   Indeed,   the image  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt\mathbf{I}$  is  closed,   and  if  $x\hskip 1.99997pt\in\hskip 1.99997ptK$  is  orthogonal  to  this image,   then $x$  is  orthogonal  to  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt\iota^{\hskip 0.70004pt*}$ and  hence $x\hskip 3.99994pt=\hskip 3.99994pt0$.

It  turns out  that  $\mathbf{I}$  admits a canonical  presentation as  the composition of  two isometries  $K\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$.   In order  to see  this,   let  us consider  the operator  $j\hskip 3.99994pt=\hskip 3.99994pt\iota\hskip 1.00006pt\circ\hskip 1.00006pt\iota^{\hskip 0.70004pt*}\hskip 1.00006pt\colon\hskip 1.00006ptH\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$.   Clearly,  $j$  is  self-adjoint  and  $\langle\hskip 1.49994pt\hskip 1.00006ptj\hskip 1.00006ptu\hskip 0.50003pt,\hskip 1.99997ptu\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\langle\hskip 1.49994pt\hskip 1.00006pt\iota^{\hskip 0.70004pt*}u\hskip 0.50003pt,\hskip 1.99997pt\iota^{\hskip 0.70004pt*}u\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK}\hskip 1.99997pt\geqslant\hskip 1.99997pt0$  for every  $u\hskip 1.99997pt\in\hskip 1.99997ptH$,   i.e  $j$  is  a non-negative operator.   Hence  the square root  $\Lambda\hskip 3.99994pt=\hskip 3.99994pt\sqrt{j}$  is  well  defined.   Clearly,

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda\hskip 1.00006ptu\hskip 0.50003pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{2}\hskip 1.00006ptu\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006ptj\hskip 0.50003ptu\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\iota^{\hskip 0.70004pt*}u\hskip 0.50003pt,\hskip 1.99997pt\iota^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006ptu\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}}$$

for every  $u\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.99997pt\in\hskip 1.99997ptH$.   Hence  $\Lambda$ defines an  isometry  from  the subspace  $H\hskip 1.99997pt\subset\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$  into  $H$.   Extending $\Lambda$  by continuity  to  $K\hskip 0.50003pt^{\prime}$  we get  an  isometric operator  $\Lambda^{\prime}\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$.   We claim  that  it  is  surjective.   Indeed,   the image  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt\Lambda^{\prime}$  is  closed,   and  if  $u\hskip 1.99997pt\in\hskip 1.99997ptH$  is  orthogonal  to  this image,   then $u$  is  orthogonal  to  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt\Lambda\hskip 3.99994pt=\hskip 3.99994pt\Lambda\hskip 1.00006pt(\hskip 1.49994ptH\hskip 1.49994pt)$.   Since $\Lambda$  is  self-adjoint,   in  this case  $\Lambda\hskip 1.00006ptu\hskip 3.99994pt=\hskip 3.99994pt0$ and  hence  $\iota^{\hskip 0.70004pt*}u\hskip 3.99994pt=\hskip 3.99994pt0$.   In  turn,   this implies  that  $u\hskip 3.99994pt=\hskip 3.99994pt0$.   It  follows  that  $\Lambda^{\prime}$  is  surjective,   and  hence  is  an  isomorphism  $K\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$.   Next,   we claim  that

$$\quad\Lambda\hskip 1.00006pt\circ\hskip 1.00006pt\Lambda^{\prime}\hskip 3.99994pt=\hskip 3.99994pt\iota\hskip 1.00006pt\circ\hskip 1.00006pt\mathbf{I}\hskip 3.00003pt.$$

Indeed,   on  the subspace $H\hskip 1.99997pt\subset\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$  both compositions are equal  to  $j\hskip 3.99994pt=\hskip 3.99994pt\iota\hskip 1.00006pt\circ\hskip 1.00006pt\iota^{\hskip 0.70004pt*}$.   By continuity  this equality extends  to  the whole space $K\hskip 0.50003pt^{\prime}$.   This equality  implies  that  the image  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt\Lambda$  is  contained  in  $\operatorname{Im}\hskip 1.49994pt\hskip 0.50003pt\iota\hskip 3.99994pt=\hskip 3.99994ptK$.   Hence we may consider  $\Lambda$ as an operator  $H\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$.   Then  the above equality  turns into  $\Lambda\hskip 1.00006pt\circ\hskip 1.00006pt\Lambda^{\prime}\hskip 3.99994pt=\hskip 3.99994pt\mathbf{I}$,   where $\Lambda$  is  a map $H\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$ and  $\Lambda^{\prime}$  is  a map  $K\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$.   Since  $\Lambda^{\prime}$ and  $\mathbf{I}$ are isometric  isomorphisms,   this implies  that  $\Lambda$  is  also an  isometric  isomorphism.   The equality  $\mathbf{I}\hskip 3.99994pt=\hskip 3.99994pt\Lambda\hskip 1.00006pt\circ\hskip 1.00006pt\Lambda^{\prime}$  is  the promised  presentation of  $\mathbf{I}$ as  the composition of  two isometries  $K\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$.

3.1. Lemma.   For every  $x\hskip 1.99997pt\in\hskip 1.99997ptK$,  $y\hskip 1.99997pt\in\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$.   

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{\prime}\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997pt\Lambda^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.00006ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

Proof.   Let  $u\hskip 3.99994pt=\hskip 3.99994pt\Lambda^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.00006ptx$.   If  $y\hskip 1.99997pt\in\hskip 1.99997ptH$,   then

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{\prime}\hskip 1.00006pty\hskip 1.00006pt,\hskip 1.99997ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

This proves  the  lemma for $y\hskip 1.99997pt\in\hskip 1.99997ptH$.   The  general  case follows by continuity.    $\blacksquare$

3.2. Lemma.   The operators  $\Lambda\hskip 1.00006pt\colon\hskip 1.00006ptH\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$  and  $\Lambda^{\prime}\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$  are adjoint  to each other.   

Proof.   We need  to check  that  $\Lambda^{\prime}\hskip 1.00006pty\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pty\hskip 1.49994pt(\hskip 1.49994pt\Lambda\hskip 1.00006ptu\hskip 1.49994pt)$  for every  $y\hskip 1.99997pt\in\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$,  $u\hskip 1.99997pt\in\hskip 1.99997ptH$.   Recall  that  we identify $H$ with $H\hskip 0.50003pt^{\prime}$ in  the standard way.   In view of  this identification

$$\quad\Lambda^{\prime}\hskip 1.00006pty\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{\prime}\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

Also,  $y\hskip 1.49994pt(\hskip 1.49994pt\Lambda\hskip 1.00006ptu\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}$  by  the definition.   If  $y\hskip 1.99997pt\in\hskip 1.99997ptH$,   then  $\Lambda^{\prime}\hskip 1.00006pty\hskip 3.99994pt=\hskip 3.99994pt\Lambda\hskip 1.00006pty$ and

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.00003pt.$$

The operator $\Lambda$,   considered as an operator  $H\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$,   is  the square root  of  a self-adjoint  positive operator and  hence  is  self-adjoint.   Therefore,   if  $y\hskip 1.99997pt\in\hskip 1.99997ptH$,   then

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{\prime}\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pty\hskip 0.50003pt,\hskip 1.99997pt\Lambda\hskip 1.00006ptu\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.00003pt.$$

It  follows  that  $\Lambda^{\prime}\hskip 1.00006pty\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pty\hskip 1.49994pt(\hskip 1.49994pt\Lambda\hskip 1.00006ptu\hskip 1.49994pt)$  for every  $y\hskip 0.50003pt,\hskip 1.99997ptu\hskip 1.99997pt\in\hskip 1.99997ptH$.   By continuity  this equality  holds for every  $y\hskip 1.99997pt\in\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$,  $u\hskip 1.99997pt\in\hskip 1.99997ptH$.   The  lemma  follows.    $\blacksquare$

Adjoints  in  the context  of  Gelfand  triples.   Let  us define  the  adjoint  of  a closed densely defined operator  $B\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$ as  the operator $B^{\hskip 0.35002pt*}\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK\hskip 0.50003pt^{\prime\prime}\hskip 3.99994pt=\hskip 3.99994ptK$  having as its domain of  definition  $\mathcal{D}\hskip 1.49994pt(\hskip 1.49994ptB^{\hskip 0.35002pt*}\hskip 1.49994pt)$  the subspace of  $x\hskip 1.99997pt\in\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$ such  that  the  linear  functional  $a\hskip 1.99997pt\longmapsto\hskip 1.99997pt\langle\hskip 1.49994pt\hskip 1.00006ptB\hskip 1.00006pta\hskip 1.00006pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}$  on  $\mathcal{D}\hskip 1.49994pt(\hskip 1.49994ptB\hskip 1.49994pt)$  extends  to a continuous functional  on $K\hskip 0.50003pt^{\prime}$ and such  that

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptB\hskip 1.00006pta\hskip 1.00006pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}\hskip 3.99994pt=\hskip 3.99994pt\langle\hskip 1.49994pt\hskip 1.00006pta\hskip 1.00006pt,\hskip 1.99997ptB^{\hskip 0.35002pt*}x\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}\hskip 3.00003pt$$

for every  $a\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.49994pt(\hskip 1.49994ptB\hskip 1.49994pt)\hskip 0.50003pt,\hskip 1.99997ptx\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.49994pt(\hskip 1.49994ptB^{\hskip 0.35002pt*}\hskip 1.49994pt)$.   Since $\mathcal{D}\hskip 1.49994pt(\hskip 1.49994ptB\hskip 1.49994pt)$ is  dense,   this uniquely determines $B^{\hskip 0.35002pt*}x$.   This notion  is  different  from  the usual  notion of  the adjoint  of  $B$ as an operator between  two  Hilbert  spaces,   the  latter  being an operator  $K\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$.   An operator $B\hskip 1.00006pt\colon\hskip 1.00006ptK\hskip 0.50003pt^{\prime}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK$  is  said  to be  self-adjoint  if  it  is  equal  to  its adjoint.   Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  implies  that  $B$  is  self-adjoint  if  and  only  if  the operator  $\Lambda^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.00006pt\circ\hskip 1.00006ptB\hskip 1.00006pt\circ\hskip 1.00006pt(\hskip 1.49994pt\Lambda^{\prime}\hskip 1.99997pt)^{\hskip 0.70004pt-\hskip 0.70004pt1}\hskip 1.00006pt\colon\hskip 1.00006ptH\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH$  is  self-adjoint  in  the usual  sense.

These definitions naturally extend  to relations  $\mathcal{B}\hskip 1.99997pt\subset\hskip 1.99997ptK\hskip 0.50003pt^{\prime}\hskip 1.00006pt\oplus\hskip 1.00006ptK$.   Namely,   the  adjoint  relation  $\mathcal{B}^{\hskip 0.35002pt*}$  is  defined as  the space of  pairs  $(\hskip 1.49994ptx\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997ptK\hskip 0.50003pt^{\prime}\hskip 1.00006pt\oplus\hskip 1.00006ptK$ such  that

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptb\hskip 1.00006pt,\hskip 1.99997ptx\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt,\hskip 1.39998ptK\hskip 0.35002pt^{\prime}}\hskip 3.99994pt=\hskip 3.99994pt\langle\hskip 1.49994pt\hskip 1.00006pta\hskip 1.00006pt,\hskip 1.99997pty\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptK\hskip 0.35002pt^{\prime}\hskip 0.35002pt,\hskip 1.39998ptK}$$

for every  $(\hskip 1.49994pta\hskip 0.50003pt,\hskip 1.99997ptb\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{B}$.   Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems  implies  that  $(\hskip 1.49994ptx\hskip 0.50003pt,\hskip 1.99997pty\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{B}^{\hskip 0.35002pt*}$  if  and only  if

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.00006ptb\hskip 1.00006pt,\hskip 1.99997pt\Lambda^{\prime}\hskip 1.00006ptx\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006pt\Lambda^{\prime}\hskip 1.00006pta\hskip 1.00006pt,\hskip 1.99997pt\Lambda^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.00006pty\hskip 1.00006pt\hskip 1.49994pt\right\rangle_{\hskip 0.70004ptH}$$

for every    $(\hskip 1.49994pta\hskip 0.50003pt,\hskip 1.99997ptb\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{B}$.   Then $\mathcal{B}$  is  a  self-adjoint  relation,   i.e.  is  equal  to its adjoint,   if  and  only  if  its  image  $\Lambda\hskip 0.50003pt^{\prime}\hskip 1.00006pt\oplus\hskip 1.00006pt\Lambda^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.49994pt(\hskip 1.49994pt\mathcal{B}\hskip 1.49994pt)$  in  $H\hskip 1.00006pt\oplus\hskip 1.00006ptH$  is  a self-adjoint  relation.

4. Abstract  boundary  problems

The operator $A$.   Let  $H_{\hskip 0.70004pt0}$  be a separable  Hilbert  space  and  let  $T$  be a closed densely defined symmetric operator in  $H_{\hskip 0.70004pt0}$.   We need  to  fix a closed self-adjoint  extension $A$ of  $T$,   which we will  call  the  reference operator.   The operator $A$  is contained  in  $T^{\hskip 0.70004pt*}$.   We will  say  that  $A$  is  invertible  if  $A$ has a bounded everywhere defined  inverse  $A^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.00006pt\colon\hskip 1.00006ptH_{\hskip 0.70004pt0}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH_{\hskip 0.70004pt0}$.

4.1. Lemma.   If  the operator $A$  is  invertible,   then  there  is  a  topological  direct  sum decomposition  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.00006pt)\hskip 1.99997pt\dotplus\hskip 1.99997pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003ptT^{\hskip 0.70004pt*}$,   where  the domains  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)$ and  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.00006pt)$ are equipped  with  graph  topologies.   The associated  projection  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.00006pt)$  is  equal  to  $A^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.49994ptT^{\hskip 0.70004pt*}$.   

Proof.   See  Grubb  [$G_{\hskip 0.70004pt1}$],   Lemma  II.1.1.   Let  us reproduce  the key part  of  the proof.   Clearly,   the right  hand side  is  contained  in  the  left  hand side.   If  $u\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)$ and $u\hskip 3.99994pt=\hskip 3.99994pta\hskip 1.99997pt+\hskip 1.99997ptz$  with $a\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.00006pt)$ and  $z\hskip 1.99997pt\in\hskip 1.99997pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003ptT^{\hskip 0.70004pt*}$,   then  $T^{\hskip 0.70004pt*}u\hskip 3.99994pt=\hskip 3.99994ptA\hskip 1.00006pta$ and  hence $a\hskip 3.99994pt=\hskip 3.99994ptA^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.49994ptT^{\hskip 0.70004pt*}u$.   It  follows  that  the presentation  $u\hskip 3.99994pt=\hskip 3.99994pta\hskip 1.99997pt+\hskip 1.99997ptz$,   if  exists,   is  unique.   If  $a\hskip 3.99994pt=\hskip 3.99994ptA^{\hskip 0.35002pt-\hskip 0.70004pt1}\hskip 1.49994ptT^{\hskip 0.70004pt*}u$ and  $z\hskip 3.99994pt=\hskip 3.99994ptu\hskip 1.99997pt-\hskip 1.99997pta$,   then

$$\quad T^{\hskip 0.70004pt*}z\hskip 3.99994pt=\hskip 3.99994ptT^{\hskip 0.70004pt*}u\hskip 1.99997pt-\hskip 1.99997ptA\hskip 1.00006pta\hskip 3.99994pt=\hskip 3.99994ptT^{\hskip 0.70004pt*}u\hskip 1.99997pt-\hskip 1.99997ptT^{\hskip 0.70004pt*}a\hskip 3.99994pt=\hskip 3.99994pt0$$

and  hence  $z\hskip 1.99997pt\in\hskip 1.99997pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003ptT^{\hskip 0.70004pt*}$.   This proves  the above decomposition on  the algebraic  level.   Passing  to  the  topological  decomposition  is  fairly  routine.    $\blacksquare$

Notations.   When  the operator  $A$  is  invertible,   we will  denote by  $p\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.00006pt)$  and  $k\hskip 1.00006pt\colon\hskip 1.00006pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997pt\operatorname{Ker}\hskip 1.49994pt\hskip 0.50003ptT^{\hskip 0.70004pt*}$  the projections associated with  the decomposition of  Lemma  Boundary  triplets  and  the  index  of  families  of  self-adjoint  elliptic  boundary  problems.   We will  denote  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004ptH_{\hskip 0.50003pt0}}$  simply  by  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 0.50003pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle$.

4.2. Lemma.   Suppose  that  $A$  is  a reference operator.   If  $u\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.00006pt)$,   then

$$\quad\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 0.50003pt,\hskip 1.99997ptv\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.00003pt-\hskip 3.00003pt\langle\hskip 1.49994pt\hskip 1.00006ptu\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\rangle\hskip 3.99994pt=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 1.00006pt,\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006ptk\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt.$$

Proof.   Since  $u\hskip 3.99994pt=\hskip 3.99994ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.99997pt+\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)$ and  $v\hskip 3.99994pt=\hskip 3.99994ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.99997pt+\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)$,   the  left  hand side  is  equal  to

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 1.00006pt,\hskip 1.99997ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.99997pt+\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.99997pt+\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle$$

$$\quad=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 1.00006pt,\hskip 1.99997ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt+\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 1.00006pt,\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt-\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptk\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt.$$

Since  $T^{\hskip 0.70004pt*}k\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994ptT^{\hskip 0.70004pt*}k\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 3.99994pt=\hskip 3.99994pt0$,   the  last  expression  is  equal  to

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}p\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt+\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 1.00006pt,\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt-\hskip 3.00003pt\left\langle\hskip 1.49994pt\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}p\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt-\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptk\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt$$

$$\quad=\hskip 3.99994pt\left\langle\hskip 1.49994pt\hskip 1.00006ptA\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt-\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptA\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt+\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptT^{\hskip 0.70004pt*}u\hskip 1.00006pt,\hskip 1.99997ptk\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt-\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptk\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptT^{\hskip 0.70004pt*}v\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.00003pt.$$

Since $A$  is  a self-adjoint  operator and  $p\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 0.50003pt,\hskip 3.00003ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptA\hskip 1.00006pt)$,

$$\quad\left\langle\hskip 1.49994pt\hskip 1.00006ptA\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 1.99997pt-\hskip 1.99997pt\left\langle\hskip 1.49994pt\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptu\hskip 1.49994pt)\hskip 1.00006pt,\hskip 1.99997ptA\hskip 1.00006ptp\hskip 1.49994pt(\hskip 1.49994ptv\hskip 1.49994pt)\hskip 1.00006pt\hskip 1.49994pt\right\rangle\hskip 3.99994pt=\hskip 3.99994pt0\hskip 3.00003pt.$$

The  lemma follows.    $\blacksquare$

Boundary  operators.   Let  $H_{\hskip 0.35002pt1}$  be a dense subspace of  $H_{\hskip 0.70004pt0}$,   which  is  a  Hilbert  space in  its own  right.   Let  $K^{\hskip 0.70004pt\partial}$  be another separable  Hilbert  space and  and  $K$  be a dense subspace of  $K^{\hskip 0.70004pt\partial}$,   which  is  a  Hilbert  space in  its own  right.   Suppose  that  the inclusion maps  $H_{\hskip 0.35002pt1}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH_{\hskip 0.70004pt0}$  and  $K\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK^{\hskip 0.70004pt\partial}$  are bounded operators with  respect  to  these  Hilbert  space structures.   Let  $K\hskip 1.99997pt\subset\hskip 1.99997ptK^{\hskip 0.70004pt\partial}\hskip 1.99997pt\subset\hskip 1.99997ptK\hskip 0.50003pt^{\prime}$  be  the  Gelfand  triple associated  with  the pair  $K\hskip 0.50003pt,\hskip 3.00003ptK^{\hskip 0.70004pt\partial}$.

We will  denote by  $\langle\hskip 1.49994pt\hskip 1.00006pt\bullet\hskip 1.00006pt,\hskip 1.99997pt\bullet\hskip 1.00006pt\hskip 1.49994pt\rangle_{\hskip 0.70004pt\partial}$  the scalar product  in  $K^{\hskip 0.70004pt\partial}$.   Suppose  that  $H_{\hskip 0.70004pt1}\hskip 1.99997pt\subset\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$,  $H_{\hskip 0.70004pt1}$  is  dense in  $\mathcal{D}\hskip 1.00006pt(\hskip 1.49994ptT^{\hskip 0.70004pt*}\hskip 1.49994pt)$,   the operator  $H_{\hskip 0.70004pt1}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptH_{\hskip 0.70004pt0}$  induced  by  $T^{\hskip 0.70004pt*}$ is  bounded,   and

$$\quad\gamma_{0}\hskip 1.00006pt,\hskip 3.00003pt\gamma_{1}\hskip 1.00006pt\colon\hskip 1.00006ptH_{\hskip 0.70004pt1}\hskip 1.99997pt\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\hskip 1.99997ptK\hskip 3.99994pt\subset\hskip 3.99994ptK^{\hskip 0.70004pt\partial}$$