**Theorem 4.5.2.**  Let τ be a Nevanlinna family in G. Then there exist an exit Hilbert space H and a self-adjoint relation A in H ⊕ H such that A is an extension of S and the compressed resolvent of A is given by the Kre˘ın–Na˘ımark formula:

$$P\_{\mathfrak{H}}(\tilde{A} - \lambda)^{-1} \iota\_{\mathfrak{H}} = (A\_0 - \lambda)^{-1} - \gamma(\lambda) \left( M(\lambda) + \tau(\lambda) \right)^{-1} \gamma(\overline{\lambda})^\*, \quad \lambda \in \mathbb{C} \backslash \mathbb{R}. \tag{4.5.11}$$

Furthermore, the self-adjoint relation A satisfies the following minimality condition:

$$\mathfrak{H} \oplus \mathfrak{H}' = \overline{\text{span}} \left\{ \mathfrak{H}, \text{ran} \left( \tilde{A} - \mu \right)^{-1} \iota\_{\mathfrak{H}} : \mu \in \mathbb{C} \mid \mathbb{R} \right\}. \tag{4.5.12}$$

Proof. Define the function R as in Lemma 4.5.1. Then R is a **B**(H)-valued holomorphic function on C \ R which satisfies R(λ) = R(λ)∗ and so, by Lemma 4.5.1, R is a generalized resolvent. Hence, by Theorem 4.4.8, the function R is a compressed resolvent, that is, there exist a Hilbert space H and a self-adjoint relation A in H ⊕ Hsuch that

$$R(\lambda) = P\_{\mathfrak{H}}(\tilde{A} - \lambda)^{-1} \iota\_{\mathfrak{H}}, \quad \lambda \in \mathbb{C} \backslash \mathbb{R};$$

this implies (4.5.11) Moreover, it follows from Theorem 4.4.8 that A satisfies the minimality condition (4.5.12).

It remains to prove that S ⊂ A. Observe first that by (4.5.11) the Straus ˇ family corresponding to A satisfies

$$\begin{aligned} T(\lambda) &= \left\{ \left\{ R(\lambda)h, (I + \lambda R(\lambda))h \right\} : h \in \mathfrak{H} \right\} \\ &\subset \left\{ R\_0(\lambda)h, (I + \lambda R\_0(\lambda))h : h \in \mathfrak{H} \right\} + \left\{ \left\{ \gamma(\lambda)\varphi, \lambda\gamma(\lambda)\varphi \right\} : \varphi \in \mathfrak{G} \right\}. \end{aligned}$$

Since each relation on the right-hand side is contained in S∗, so is the relation T(λ). As T(λ)∗ = T(λ), it follows that S ⊂ T(λ). Now let {f,f- } ∈ S so that for each λ ∈ C \ R there exists h ∈ Hsuch that

$$\left\{ \begin{pmatrix} f \\ h \end{pmatrix}, \begin{pmatrix} f' \\ \lambda h \end{pmatrix} \right\} \in \tilde{A}.$$

The relation A is self-adjoint and, in particular, symmetric. Therefore, one sees that (f- , f) + λ(h, h) ∈ R, while by definition (f- , f) ∈ R. Since λ ∈ C \ R, it follows that h = 0, and thus

$$\left\{ \begin{pmatrix} f \\ 0 \end{pmatrix}, \begin{pmatrix} f' \\ 0 \end{pmatrix} \right\} \in \tilde{A}.$$

This shows that S ⊂ A. -

## **4.6 Orthogonal coupling of boundary triplets**

In this section a different look is taken at the Kre˘ın–Na˘ımark formula. By means of an abstract coupling method for direct orthogonal sums of symmetric relations and corresponding boundary triplets, a particular self-adjoint extension A of the direct sum is identified, and it is shown that the compressed resolvent of A is of the same form as in the Kre˘ın–Na˘ımark formula. When combined with Theorem 4.2.4, this coupling procedure provides a constructive approach to the exit space extension in Theorem 4.5.2 in the special case where the Nevanlinna family τ is a uniformly strict Nevanlinna function.

First a slightly more general, abstract point of view is adopted. In the following let S and T be closed symmetric relations in the Hilbert spaces H and H- , respectively, and assume that the defect numbers of S and T coincide:

$$n\_+(S) = n\_-(S) = n\_+(T) = n\_-(T) \le \infty.$$

Let {G, Γ0, Γ1} be a boundary triplet for S∗ with A0 = ker Γ0 and let {G, Γ- 0, Γ- 1} be a boundary triplet for T ∗ with B0 = ker Γ- 0. Then it is easy to see that the direct orthogonal sum S ⊕ T is a closed symmetric relation in H⊕H and {G⊕G, Γ0, Γ1}, where

$$
\widetilde{\Gamma}\_0 \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \begin{pmatrix} \Gamma\_0 \widehat{f} \\ \Gamma\_0' \widehat{g} \end{pmatrix} \quad \text{and} \quad \widetilde{\Gamma}\_1 \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \begin{pmatrix} \Gamma\_1 \widehat{f} \\ \Gamma\_1' \widehat{g} \end{pmatrix}, \quad \widehat{f} \in S^\*, \widehat{g} \in T^\*, \tag{4.6.1}
$$

is a boundary triplet for (S ⊕ T)∗ = S∗ ⊕ T ∗, and that

$$
\dot{A}\_0 := A\_0 \hat{\oplus} B\_0 = \ker \dot{\Gamma}\_0 \tag{4.6.2}
$$

is a self-adjoint extension of S ⊕ T in H ⊕ H- . Furthermore, if γ and γ denote the γ-fields corresponding to the boundary triplets {G, Γ0, Γ1} and {G, Γ- 0, Γ- 1}, and M and τ are the Weyl functions corresponding to {G, Γ0, Γ1} and {G, Γ- 0, Γ- 1}, respectively, then it is clear that for λ ∈ ρ(A0) = ρ(A0) ∩ ρ(B0) the γ-field γ and the Weyl function M 5 corresponding to the boundary triplet {G ⊕ G, Γ0, Γ1} have the forms

$$
\widetilde{\gamma}(\lambda) = \begin{pmatrix} \gamma(\lambda) & 0 \\ 0 & \gamma'(\lambda) \end{pmatrix} \quad \text{and} \quad \widetilde{M}(\lambda) = \begin{pmatrix} M(\lambda) & 0 \\ 0 & \tau(\lambda) \end{pmatrix} . \tag{4.6.3}
$$

Let A be a self-adjoint extension of S ⊕ T in H ⊕ H- . Then Kre˘ın's formula in Theorem 2.6.1 has the form

$$(\tilde{A} - \lambda)^{-1} = (\tilde{A}\_0 - \lambda)^{-1} + \tilde{\gamma}(\lambda) \left(\tilde{\Theta} - \widehat{M}(\lambda)\right)^{-1} \tilde{\gamma}(\tilde{\lambda})^\*$$

for all λ ∈ ρ(A) ∩ ρ(A0), where γ and M 5 denote the γ-field and the Weyl function corresponding to the boundary triplet {G ⊕ G, Γ0, Γ1} in (4.6.3). If Θ = {A, B} with A, B ∈ **B**(G ⊕ G), then

$$(\tilde{A} - \lambda)^{-1} = (\tilde{A}\_0 - \lambda)^{-1} - \tilde{\gamma}(\lambda)\mathcal{A}\left(\widetilde{M}(\lambda)\mathcal{A} - \mathcal{B}\right)^{-1}\tilde{\gamma}(\tilde{\lambda})^\*,$$

see Corollary 2.6.3. Writing A and B as block operators

$$\mathcal{A} = \begin{pmatrix} \mathcal{A}\_{11} & \mathcal{A}\_{12} \\ \mathcal{A}\_{21} & \mathcal{A}\_{22} \end{pmatrix} \quad \text{and} \quad \mathcal{B} = \begin{pmatrix} \mathcal{B}\_{11} & \mathcal{B}\_{12} \\ \mathcal{B}\_{21} & \mathcal{B}\_{22} \end{pmatrix},$$

where Aij , Bij ∈ **B**(G), it follows that

$$
\widetilde{M}(\lambda)\mathcal{A} - \mathcal{B} = \begin{pmatrix} M(\lambda)\mathcal{A}\_{11} - \mathcal{B}\_{11} & M(\lambda)\mathcal{A}\_{12} - \mathcal{B}\_{12} \\ \tau(\lambda)\mathcal{A}\_{21} - \mathcal{B}\_{21} & \tau(\lambda)\mathcal{A}\_{22} - \mathcal{B}\_{22} \end{pmatrix},
$$

so that

$$\mathcal{A}\left(\widetilde{M}(\lambda)\mathcal{A}-\mathcal{B}\right)^{-1} = \begin{pmatrix} \mathcal{A}\_{11} & \mathcal{A}\_{12} \\ \mathcal{A}\_{21} & \mathcal{A}\_{22} \end{pmatrix} \begin{pmatrix} M(\lambda)\mathcal{A}\_{11} - \mathcal{B}\_{11} & M(\lambda)\mathcal{A}\_{12} - \mathcal{B}\_{12} \\ \tau(\lambda)\mathcal{A}\_{21} - \mathcal{B}\_{21} & \tau(\lambda)\mathcal{A}\_{22} - \mathcal{B}\_{22} \end{pmatrix}^{-1}.$$

The following proposition exhibits a particular self-adjoint extension of S ⊕ T in H ⊕ H- .

**Proposition 4.6.1.** Let S and T be closed symmetric relations in the Hilbert spaces H and H with boundary triplets {G, Γ0, Γ1} and {G, Γ- 0, Γ- 1} as above, respectively. Then

$$\tilde{A} = \left\{ \begin{pmatrix} \hat{f} \\ \hat{g} \end{pmatrix} : \hat{f} \in S^\*, \hat{g} \in T^\*, \ \Gamma\_0 \hat{f} = \Gamma\_0' \hat{g}, \ \Gamma\_1 \hat{f} = -\Gamma\_1' \hat{g} \right\} \tag{4.6.4}$$

is a self-adjoint relation in H ⊕ H and for all λ ∈ C \ R the resolvent of A has the form

$$(\tilde{A} - \lambda)^{-1} = (\tilde{A}\_0 - \lambda)^{-1} - \tilde{\gamma}(\lambda) \begin{pmatrix} (M(\lambda) + \tau(\lambda))^{-1} & (M(\lambda) + \tau(\lambda))^{-1} \\ (M(\lambda) + \tau(\lambda))^{-1} & (M(\lambda) + \tau(\lambda))^{-1} \end{pmatrix} \tilde{\gamma}(\tilde{\lambda})^\*,$$

where A0 and γ are as in (4.6.2), and M and τ denote the Weyl functions corresponding to {G, Γ0, Γ1} and {G, Γ- 0, Γ- 1}, respectively.

Proof. Consider the boundary triplet {G ⊕ G, Γ0, Γ1} in (4.6.1) and observe that the relation

$$\tilde{\Theta} := \left\{ \left\{ \begin{pmatrix} \varphi \\ \varphi \end{pmatrix}, \begin{pmatrix} \psi \\ -\psi \end{pmatrix} \right\} : \varphi, \psi \in \mathfrak{G} \right\} \tag{4.6.5}$$

is self-adjoint in G ⊕ G. Hence, by Corollary 2.1.4 and (4.6.1),

$$\left\{ \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} \in S^\* \oplus T^\* : \widetilde{\Gamma} \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \left\{ \begin{pmatrix} \Gamma\_0 \widehat{f} \\ \Gamma\_0' \widehat{g} \end{pmatrix}, \begin{pmatrix} \Gamma\_1 \widehat{f} \\ \Gamma\_1' \widehat{g} \end{pmatrix} \right\} \in \widetilde{\Theta} \right\} \subset S^\* \oplus T^\* \tag{4.6.6}$$

is a self-adjoint relation H ⊕ H- . Now it follows from the particular form of Θ in (4.6.5) that the self-adjoint relation in (4.6.6) coincides with A in (4.6.4).

Next the resolvent of A will be computed. Recall first that Kre˘ın's formula in Theorem 2.6.1 implies

$$(\tilde{A} - \lambda)^{-1} = (\tilde{A}\_0 - \lambda)^{-1} - \tilde{\gamma}(\lambda) \left(\tilde{\Theta} - \hat{M}(\lambda)\right)^{-1} \tilde{\gamma}(\tilde{\lambda})^\* \tag{4.6.7}$$

for all λ ∈ ρ(A) ∩ ρ(A0), where γ and M 5 denote the γ-field and the Weyl function corresponding to {G ⊕ G, Γ0, Γ1} in (4.6.3). From (4.6.5) and (4.6.3) one obtains

$$\left(\widetilde{\Theta} - \widetilde{M}(\lambda)\right)^{-1} = \left\{ \left\{ \begin{pmatrix} \psi - M(\lambda)\varphi\\ -\psi - \tau(\lambda)\varphi \end{pmatrix}, \begin{pmatrix} \varphi\\ \varphi \end{pmatrix} \right\} : \varphi, \psi \in \mathcal{G} \right\} \dots$$

Setting φ := ψ −M(λ)ϕ and χ := −ψ − τ (λ)ϕ one has φ + χ = −(M(λ) + τ (λ))ϕ. For λ ∈ C \ R it follows from Lemma 1.11.5 (see also Proposition 1.12.6) that (M(λ) + τ (λ))−1 ∈ **B**(G), and hence

$$\varphi = -\left(M(\lambda) + \tau(\lambda)\right)^{-1}\phi - \left(M(\lambda) + \tau(\lambda)\right)^{-1}\chi, \qquad \lambda \in \mathbb{C} \backslash \mathbb{R}.$$

This yields

$$\begin{pmatrix} \widetilde{\Theta} - \widetilde{M}(\lambda) \end{pmatrix}^{-1} = - \begin{pmatrix} (M(\lambda) + \tau(\lambda))^{-1} & (M(\lambda) + \tau(\lambda))^{-1} \\ (M(\lambda) + \tau(\lambda))^{-1} & (M(\lambda) + \tau(\lambda))^{-1} \end{pmatrix},$$

and the statement about the resolvent of A follows from (4.6.7). -

The compressions of the resolvent of the self-adjoint relation A in (4.6.4) to H and H are of interest. Note that the resolvent of A0 in (4.6.2) is given by the direct orthogonal sum of the resolvents of A0 and B0, and hence for λ ∈ ρ(A0) the compressions to H and Hare

$$P\_{\mathfrak{H}}(\tilde{A}\_0 - \lambda)^{-1} \iota\_{\mathfrak{H}} = (A\_0 - \lambda)^{-1} \quad \text{and} \quad P\_{\mathfrak{H}'}(\tilde{A}\_0 - \lambda)^{-1} \iota\_{\mathfrak{H}'} = (B\_0 - \lambda)^{-1},$$

respectively. The next statement follows directly from Proposition 4.6.1 and (4.6.3).

**Corollary 4.6.2.** Let S and T be closed symmetric relations in the Hilbert spaces H and H with boundary triplets {G, Γ0, Γ1} and {G, Γ- 0, Γ- 1}, and corresponding γ-fields and Weyl functions γ, γ and M, τ , respectively. Then for all λ ∈ C \ R the following statements hold:

(i) The compression of the resolvent of the self-adjoint relation A in (4.6.4) to H is given by

$$P\_{\mathfrak{H}}(\widetilde{A} - \lambda)^{-1} \iota\_{\mathfrak{H}} = (A\_0 - \lambda)^{-1} - \gamma(\lambda) \left( M(\lambda) + \tau(\lambda) \right)^{-1} \gamma(\overline{\lambda})^\*.$$

(ii) The compression of the resolvent of the self-adjoint relation A in (4.6.4) to His given by

$$P\_{\mathfrak{H}'} (\tilde{A} - \lambda)^{-1} \iota\_{\mathfrak{H}'} = (B\_0 - \lambda)^{-1} - \gamma'(\lambda) \left( M(\lambda) + \tau(\lambda) \right)^{-1} \gamma'(\overline{\lambda})^\* .$$

Corollary 4.6.2 and Proposition 4.6.1 can also be viewed as an alternative approach to the Kre˘ın–Na˘ımark formula in the special case where the Nevanlinna family τ in Theorem 4.5.2 is a uniformly strict Nevanlinna function. In fact, according to Theorem 4.2.4 every uniformly strict **B**(G)-valued Nevanlinna function

$$\mathbb{D}$$

can be realized as a Weyl function, that is, there exist a (reproducing kernel) Hilbert space H- (= H(Nτ )), a closed simple symmetric operator T (= Sτ ) in H- , and a boundary triplet {G, Γ- 0, Γ- 1} for the adjoint T ∗ such that τ is the corresponding Weyl function. In this situation the relation A in (4.6.4) is self-adjoint in H ⊕ H- = H ⊕ H(Nτ ) and its compressed resolvent in Corollary 4.6.2 coincides with the one in the Kre˘ın–Na˘ımark formula in Theorem 4.5.2. Summing up, the following special case of Theorem 4.5.2 is a consequence of the coupling method in Proposition 4.6.1 and Corollary 4.6.2.

**Corollary 4.6.3.**  Let τ be a uniformly strict **B**(G)-valued Nevanlinna function. Then there exist an exit Hilbert space H and a self-adjoint relation A in H ⊕ H such that A is an extension of S and the compressed resolvent of A is given by the Kre˘ın–Na˘ımark formula:

$$P\_{\mathfrak{H}}(\tilde{A} - \lambda)^{-1} \iota\_{\mathfrak{H}} = (A\_0 - \lambda)^{-1} - \gamma(\lambda) \left( M(\lambda) + \tau(\lambda) \right)^{-1} \gamma(\overline{\lambda})^\*, \quad \lambda \in \mathbb{C} \backslash \mathbb{R}.$$

Furthermore, the self-adjoint relation A satisfies the minimality condition

$$\mathfrak{H} \oplus \mathfrak{H}' = \overline{\text{span}} \left\{ \mathfrak{H}, \text{ran} \left( \tilde{A} - \mu \right)^{-1} \iota\_{\mathfrak{H}} : \mu \in \mathbb{C} \backslash \mathbb{R} \right\}. \tag{4.6.8}$$

Proof. All statements except the minimality condition (4.6.8) follow from Proposition 4.6.1, Corollary 4.6.2, and Theorem 4.2.4 as explained above. For (4.6.8) recall first that the closed symmetric operator Sτ (= T) in Theorem 4.2.4 is simple, and hence

$$\mathfrak{H}' = \overline{\text{span}}\left\{ \text{ker}\left( T^\* - \mu \right) : \mu \in \mathbb{C} \mid \mathbb{R} \right\} = \overline{\text{span}}\left\{ \text{ran}\,\gamma'(\mu) : \mu \in \mathbb{C} \mid \mathbb{R} \right\}.\tag{4.6.9}$$

It follows from Proposition 4.6.1 that

$$P\_{\mathfrak{H}'} (\tilde{A} - \mu)^{-1} \iota\_{\mathfrak{H}} = -\gamma'(\mu) \left( M(\mu) + \tau(\mu) \right)^{-1} \gamma(\tilde{\mu})^\*,$$

and since ran γ(μ)∗ = G and dom (M(μ) + τ (μ)) = G, one sees that

$$\text{ran}\left(P\_{\mathfrak{H}'} (\tilde{A} - \mu)^{-1} \iota\_{\mathfrak{H}}\right) = \text{ran}\,\gamma'(\mu), \qquad \mu \in \mathbb{C} \backslash \mathbb{R}.$$

With (4.6.9) one then concludes that

$$\mathfrak{H}' = \overline{\text{span}} \left\{ \text{ran} \left( P\_{\mathfrak{H}'} (\tilde{A} - \mu)^{-1} \iota\_{\mathfrak{H}} \right) : \mu \in \mathbb{C} \backslash \mathbb{R} \right\},$$

which in turn yields (4.6.8). -

In the next proposition a particular boundary triplet {G ⊕ G, Γ0, Γ1} is specified such that the self-adjoint relation A in (4.6.4) coincides with the kernel of the boundary mapping Γ0. The corresponding Weyl function M is useful for the spectral analysis of A; cf. Chapter 6.

**Proposition 4.6.4.** Let S and T be closed symmetric relations in the Hilbert spaces H and H with boundary triplets {G, Γ0, Γ1} and {G, Γ- 0, Γ- 1} and corresponding Weyl functions M and τ , respectively, as in the beginning of this section. Then {G ⊕ G, Γ0, Γ1}, where

$$
\widehat{\Gamma}\_0 \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \begin{pmatrix} -\Gamma\_1 \widehat{f} - \Gamma\_1' \widehat{g} \\ \Gamma\_0 \widehat{f} - \Gamma\_0' \widehat{g} \end{pmatrix} \quad \text{and} \quad \widehat{\Gamma}\_1 \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \begin{pmatrix} \Gamma\_0 \widehat{f} \\ -\Gamma\_1' \widehat{g} \end{pmatrix}, \quad \widehat{f} \in S^\*, \widehat{g} \in T^\*,
$$

is a boundary triplet for S∗ ⊕ T ∗ such that the self-adjoint relation A in (4.6.4) corresponds to the boundary mapping Γ0, that is,

$$
\bar{A} = \ker \bar{\Gamma}\_0.
$$

The Weyl function of {G ⊕ G, Γ0, Γ1} is given by

$$\begin{aligned} \widehat{M}(\lambda) &= -\begin{pmatrix} M(\lambda) & -I \\ -I & -\tau(\lambda)^{-1} \end{pmatrix}^{-1} \\ &= \begin{pmatrix} -(M(\lambda) + \tau(\lambda))^{-1} & (M(\lambda) + \tau(\lambda))^{-1}\tau(\lambda) \\ \tau(\lambda)(M(\lambda) + \tau(\lambda))^{-1} & \tau(\lambda)(M(\lambda) + \tau(\lambda))^{-1}M(\lambda) \end{pmatrix} \end{aligned} \tag{4.6.10}$$

for λ ∈ C \ R.

Proof. Instead of a direct proof the assertions will be obtained as consequences of the results in Section 2.5. For this consider the boundary triplet {G ⊕ G, Γ0, Γ1} in (4.6.1) with the Weyl function M 5 given in (4.6.3), let

$$\mathcal{A} = \frac{1}{\sqrt{2}} \begin{pmatrix} I & 0 \\ I & 0 \end{pmatrix} \quad \text{and} \quad \mathcal{B} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & I \\ 0 & -I \end{pmatrix},$$

and observe that Θ = {A, B}, with Θ in ( 4.6.5). It is easy to see that A and B satisfy the conditions in Corollary 2.5.11. Therefore, {G2, Γ˘0, Γ˘1}, where

$$
\check{\Gamma}\_0 = \mathcal{B}^\* \widetilde{\Gamma}\_0 - \mathcal{A}^\* \widetilde{\Gamma}\_1 \quad \text{and} \quad \check{\Gamma}\_1 = \mathcal{A}^\* \widetilde{\Gamma}\_0 + \mathcal{B}^\* \widetilde{\Gamma}\_1,
$$

is a boundary triplet with corresponding Weyl function

$$\check{M}(\lambda) = \left(\mathcal{A}^\* + \mathcal{B}^\* \widehat{M}(\lambda)\right) \left(\mathcal{B}^\* - \mathcal{A}^\* \widehat{M}(\lambda)\right)^{-1}, \quad \lambda \in \mathbb{C} \backslash \mathbb{R}.$$

It follows that

$$
\overset{\circ}{\Gamma}\_0 \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} -\Gamma\_1 \widehat{f} - \Gamma\_1' \widehat{g} \\ \Gamma\_0 \widehat{f} - \Gamma\_0' \widehat{g} \end{pmatrix}, \quad \widehat{f} \in S^\*, \widehat{g} \in T^\*, \widehat{g}
$$

and

$$
\widehat{\Gamma}\_1 \begin{pmatrix} \widehat{f} \\ \widehat{g} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} \Gamma\_0 \widehat{f} + \Gamma\_0' \widehat{g} \\ \Gamma\_1 \widehat{f} - \Gamma\_1' \widehat{g} \end{pmatrix}, \qquad \widehat{f} \in S^\*, \widehat{g} \in T^\*.
$$

Furthermore, it is easily seen from the above that

$$\begin{aligned} \check{M}(\lambda) &= \begin{pmatrix} 1 & 1 \\ M(\lambda) & -\tau(\lambda) \end{pmatrix} \begin{pmatrix} -M(\lambda) & -\tau(\lambda) \\ 1 & -1 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & 1 \\ M(\lambda) & -\tau(\lambda) \end{pmatrix} \begin{pmatrix} -(M(\lambda) + \tau(\lambda))^{-1} & (M(\lambda) + \tau(\lambda))^{-1}\tau(\lambda) \\ -(M(\lambda) + \tau(\lambda))^{-1} & -(M(\lambda) + \tau(\lambda))^{-1}M(\lambda) \end{pmatrix}, \end{aligned}$$

where the last step used the identity

$$M(\lambda)(M(\lambda) + \tau(\lambda))^{-1}\tau(\lambda) = \tau(\lambda)(M(\lambda) + \tau(\lambda))^{-1}M(\lambda).$$

Thus, it is clear that

$$
\check{M}(\lambda) = \begin{pmatrix}
(\tau(\lambda) - M(\lambda))(M(\lambda) + \tau(\lambda))^{-1} & 2M(\lambda)(M(\lambda + \tau(\lambda))^{-1}\tau(\lambda))
\end{pmatrix}
$$

holds for all λ ∈ C \ R. Now let

$$D = \frac{1}{\sqrt{2}} \begin{pmatrix} I & 0 \\ 0 & I \end{pmatrix} = D^\* \quad \text{and} \quad P = \frac{1}{2} \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix},$$

and apply Corollary 2.5.5 to conclude that

$$
\widehat{\Gamma}\_0 = D^{-1} \check{\Gamma}\_0 \quad \text{and} \quad \widehat{\Gamma}\_1 = D^\* \check{\Gamma}\_1 + PD^{-1} \check{\Gamma}\_0.
$$

give a boundary triplet for S∗ ⊕ T ∗. According to Corollary 2.5.5, the corresponding Weyl function is given by

$$
\widehat{M}(\lambda) = D^\* \check{M}(\lambda) D + P, \qquad \lambda \in \mathbb{C} \backslash \mathbb{R},
$$

and one verifies that the first identity in (4.6.10) holds. It is straightforward to check the second identity in (4.6.10). -

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## **Chapter 5**

## **Boundary Triplets and Boundary Pairs for Semibounded Relations**

Semibounded relations in a Hilbert space automatically have equal defect numbers, so that there are always self-adjoint extensions. In this chapter the semibounded self-adjoint extensions of a semibounded relation will be investigated. Special attention will be paid to the Friedrichs extension, which is introduced with the help of closed semibounded forms. Section 5.1 provides a self-contained introduction to closed semibounded forms and their representations via semibounded self-adjoint relations. Closely related is the discussion of the ordering for closed semibounded forms and for semibounded self-adjoint relations in Section 5.2; this section also contains a general monotonicity principle about monotone sequences of semibounded relations. The Friedrichs extension of a semibounded relation is defined and its central properties are studied in Section 5.3. Particular attention is paid to semibounded self-adjoint extensions which are transversal to the Friedrichs extension. Section 5.4 is devoted to special semibounded extensions, namely the Kre˘ın type extensions. In the nonnegative case these extensions include the wellknown Kre˘ın–von Neumann extension. The Friedrichs extension and the Kre˘ın type extensions act as extremal elements to describe the semibounded self-adjoint extensions with a given lower bound. In Section 5.5 there is a return to boundary triplets and Weyl functions for symmetric relations which are semibounded. Of special interest is the case where the self-adjoint extensions determined by the boundary triplet are semibounded and one of them coincides with the Friedrichs extension. In particular, this leads to a useful abstract version of the first Green formula. The notion of a boundary pair for semibounded relations is developed in Section 5.6. In conjunction with the above first Green formula, this notion serves as a link between boundary triplet methods and form methods when semibounded self-adjoint extensions are described; in a wider sense it establishes the connection with the Birman–Kre˘ın–Vishik method.

## **5.1 Closed semibounded forms and their representations**

A sesquilinear form t[·, ·] in a Hilbert space H with inner product (·, ·) is a mapping from D × D to C, where D is a linear subspace of H, such that t[·, ·] is linear in the first entry and anti-linear in the second entry. The domain dom t is defined by dom t = D. The form is said to be symmetric if t[ϕ, ψ] = t[ψ, ϕ] for all ϕ, ψ ∈ dom t. The corresponding quadratic form t[·] is defined by t[ϕ] = t[ϕ, ϕ], ϕ ∈ dom t. The polarization formula

$$\mathbf{t}[\varphi,\psi] = \frac{1}{4}\{\mathbf{t}[\varphi+\psi] - \mathbf{t}[\varphi-\psi]\} + \frac{i}{4}\{\mathbf{t}[\varphi+i\psi] - \mathbf{t}[\varphi-i\psi]\}\tag{5.1.1}$$

for ϕ, ψ ∈ dom t is easily checked. In the following the term sesquilinear will be dropped; whenever a form t[·, ·] is mentioned it is assumed to be sesquilinear and it will be denoted by t. For instance, the inner product (·, ·) is a form defined on all of H.

**Definition 5.1.1.** Let t1 and t2 be forms in H. Then the inclusion t2 ⊂ t1 means that

$$\operatorname{dom} \mathfrak{t}\_2 \subset \operatorname{dom} \mathfrak{t}\_1, \quad \mathfrak{t}\_2[\varphi] = \mathfrak{t}\_1[\varphi], \quad \varphi \in \operatorname{dom} \mathfrak{t}\_2. \tag{5.1.2}$$

If t2 ⊂ t1, then t2 is said to be a restriction of t1 and t1 is said to be an extension of t2. The sum t1 + t2 is defined by

$$(\mathfrak{t}\_1 + \mathfrak{t}\_2)[\varphi, \psi] = \mathfrak{t}\_1[\varphi, \psi] + \mathfrak{t}\_2[\varphi, \psi], \quad \varphi, \psi \in \text{dom}\left(\mathfrak{t}\_1 + \mathfrak{t}\_2\right),$$

where dom (t1 + t2) = dom t1 ∩ dom t2.

If α ∈ C the sum t[·, ·] + α(·, ·) is given by

$$\mathfrak{t}[\varphi,\psi] + \alpha(\varphi,\psi), \quad \varphi,\psi \in \text{dom}\,\mathfrak{t}.$$

This sum will be denoted by t+α. It is symmetric when t is symmetric and α ∈ R.

**Definition 5.1.2.** A symmetric form t in H is bounded from below if there exists a constant c ∈ R such that

$$\mathfrak{t}[\varphi] \ge c \|\varphi\|^2, \quad \varphi \in \text{dom}\,\mathfrak{t}.$$

This inequality will be denoted by t ≥ c. The lower bound m(t) is the largest of such numbers c ∈ R:

$$m(\mathbf{t}) = \inf \left\{ \frac{\mathbf{t}[\varphi]}{||\varphi||^2} : \varphi \in \text{dom}\,\mathbf{t}, \ \varphi \neq 0 \right\}.$$

If m(t) ≥ 0, then t is called nonnegative.

#### 5.1. Closed semibounded forms and their representations 283

In the following the terminology semibounded form is used for a symmetric form which is bounded from below. Note that t is a semibounded form if and only if for some, and hence for all α ∈ R the form t + α is semibounded. For a semibounded form t the lower bound m(t) will often be denoted by γ. Note that the form t − γ, γ = m(t), is nonnegative with lower bound 0. Therefore, one has the Cauchy–Schwarz inequality

$$|(\mathbf{t} - \gamma)[\varphi, \psi]| \le (\mathbf{t} - \gamma)[\varphi]^{\frac{1}{2}}(\mathbf{t} - \gamma)[\psi]^{\frac{1}{2}}, \quad \varphi, \psi \in \text{dom } \mathbf{t}, \tag{5.1.3}$$

and, hence the triangle inequality

$$[(\mathfrak{t}-\gamma)[\varphi+\psi]^{\frac{1}{2}} \le (\mathfrak{t}-\gamma)[\varphi]^{\frac{1}{2}} + (\mathfrak{t}-\gamma)[\psi]^{\frac{1}{2}}, \quad \varphi, \psi \in \text{dom}\,\mathfrak{t}.\tag{5.1.4}$$

It follows from (5.1.4) that

$$\left| (\mathbf{t} - \gamma)[\varphi]^{\frac{1}{2}} - (\mathbf{t} - \gamma)[\psi]^{\frac{1}{2}} \right| \le (\mathbf{t} - \gamma)[\varphi - \psi]^{\frac{1}{2}}, \quad \varphi, \psi \in \text{dom } \mathbf{t}.\tag{5.1.5}$$

The following continuity property is a simple consequence of (5.1.5). For a sequence (ϕn) in dom t and ϕ ∈ dom t one has

$$(\mathbf{t} - \gamma)[\varphi - \varphi\_n] \to 0 \quad \Rightarrow \quad (\mathbf{t} - \gamma)[\varphi\_n] \to (\mathbf{t} - \gamma)[\varphi]. \tag{5.1.6}$$

Let t be a semibounded form in H with lower bound γ and let a<γ. Equip the space dom t ⊂ H with the form

$$(\varphi, \psi)\_{\mathbf{t}-a} = \mathbf{t}[\varphi, \psi] - a(\varphi, \psi), \quad \varphi, \psi \in \text{dom } \mathbf{t}.\tag{5.1.7}$$

By rewriting this definition as

$$(\varphi, \psi)\_{t-a} = (t-\gamma)[\varphi, \psi] + (\gamma - a)(\varphi, \psi), \quad \varphi, \psi \in \text{dom } \mathfrak{t},\tag{5.1.8}$$

one sees that (·, ·)t−a is the sum of the semidefinite inner product t − γ and the inner product (γ − a)(·, ·). Hence, (·, ·)t−a is an inner product on dom t and the corresponding norm · t−a satisfies the inequality

$$\|\varphi\|\_{\mathfrak{t}-a}^2 \ge (\gamma - a) \|\varphi\|^2, \quad \varphi \in \text{dom } \mathfrak{t}.\tag{5.1.9}$$

When dom t is equipped with the inner product (·, ·)t−a, the resulting inner product space will be denoted by Ht−a. Note that if γ > 0, then obviously a = 0 is a natural choice in the above and the following arguments.

**Lemma 5.1.3.** Let t be a semibounded form in H with lower bound γ and let a<γ. Let (ϕn) be a sequence in dom t. Then (ϕn) is a Cauchy sequence in Ht−a if and only if

$$\text{At}[\varphi\_n - \varphi\_m] \to 0 \quad \text{and} \quad \|\varphi\_n - \varphi\_m\| \to 0. \tag{5.1.10}$$

Proof. According to (5.1.8), (ϕn) is a Cauchy sequence in Ht−a if and only if

$$(\mathbf{t} - \gamma)[\varphi\_n - \varphi\_m] \to 0 \quad \text{and} \quad \left\| \varphi\_n - \varphi\_m \right\|^2 \to 0. \tag{5.1.11}$$

Now assume that (ϕn) is a Cauchy sequence in Ht−a. Then it follows from (5.1.11) that (ϕn) is a Cauchy sequence in H and that

$$\mathbf{t}[\varphi\_n - \varphi\_m] = (\mathbf{t} - \gamma)[\varphi\_n - \varphi\_m] + \gamma ||\varphi\_n - \varphi\_m||^2 \to 0,$$

which shows (5.1.10). Conversely, if the sequence (ϕn) satisfies (5.1.10), then it follows from (5.1.7) that (ϕn) is a Cauchy sequence in Ht−a. -

Let (ϕn) be a Cauchy sequence in Ht−a. Since H is a Hilbert space, it follows from Lemma 5.1.3 that there is an element ϕ ∈ H such that ϕn → ϕ in H.

**Definition 5.1.4.** Let t be a semibounded form in H. A sequence (ϕn) in dom t is said to be t-convergent to an element ϕ ∈ H, not necessarily belonging to dom t, if

ϕn → ϕ in H and t[ϕn − ϕm] → 0, n, m → ∞.

This type of convergence will be denoted by ϕn →t ϕ.

The following result is a direct consequence of Lemma 5.1.3 and the completeness of H.

**Corollary 5.1.5.** Let t be a semibounded form in H with lower bound γ and let a<γ. Then any Cauchy sequence in Ht−a is t-convergent. Conversely, any tconvergent sequence in dom t is a Cauchy sequence in Ht−a.

If the sequence (ϕn) in dom t is t-convergent, then by Definition 5.1.4

(t − γ)[ϕn − ϕm] → 0 and ϕn − ϕm → 0.

Thus, one has the following result.

**Corollary 5.1.6.** Let t be a semibounded form in H with lower bound γ and let the sequence (ϕn) in dom t be t-convergent. Then the sequences ((t − γ)[ϕn]), (t[ϕn]), and ( ϕn ) converge and, consequently, they are bounded.

Proof. Since γ is the lower bound of t, one has (t − γ)[ϕn − ϕm] → 0. Hence, (5.1.5) shows that ((t − γ)[ϕn]) is a Cauchy sequence. Then the same is true for the sequence (t[ϕn]) and it is also clear that ( ϕn ) is a Cauchy sequence. In particular, the sequences ((t − γ)[ϕn]), (t[ϕn]), and ( ϕn ) are bounded. -

The t-convergence is preserved when one takes a sum of sequences. To see this, let (ϕn) and (ψn) be sequences in dom t such that

$$
\varphi\_n \to\_\mathfrak{t} \varphi \quad \text{and} \quad \psi\_n \to\_\mathfrak{t} \psi
$$

for some ϕ, ψ ∈ H. Then clearly ϕn + ψn → ϕ + ψ in H and

$$\begin{aligned} \left[ (\mathfrak{t} - \gamma) [\varphi\_n + \psi\_n - (\varphi\_m + \psi\_m)] \right]^{\frac{1}{2}} \\ \leq (\mathfrak{t} - \gamma) [\varphi\_n - \varphi\_m]^{\frac{1}{2}} + (\mathfrak{t} - \gamma) [\psi\_n - \psi\_m]^{\frac{1}{2}}, \end{aligned}$$

by the triangle inequality in (5.1.4). Therefore,

$$
\varphi\_n \to\_\mathbf{t} \varphi \quad \text{and} \quad \psi\_n \to\_\mathbf{t} \psi \quad \Rightarrow \quad \varphi\_n + \psi\_n \to\_\mathbf{t} \varphi + \psi. \tag{5.1.12}
$$

As a consequence, one sees that

$$
\varphi\_n \to\_\mathfrak{t} \varphi \quad \text{and} \quad \psi\_n \to\_\mathfrak{t} \psi \quad \Rightarrow \quad \lim\_{n \to \infty} \mathfrak{t}[\varphi\_n, \psi\_n] \quad \text{exists.} \tag{5.1.13}
$$

This last implication follows easily from Corollary 5.1.6 and (5.1.12) by the polarization formula in (5.1.1).

Assume that ϕn ∈ dom t and that ϕn →t ϕ for some ϕ ∈ H. Now the question is when ϕ ∈ dom t and, if this is the case, when t[ϕn − ϕ] → 0 ? This question gives rise to the notions of closed form and closable form in Definition 5.1.7 and Definition 5.1.11.

**Definition 5.1.7.** A semibounded form t in H is said to be closed if

ϕn →t ϕ ⇒ ϕ ∈ dom t and t[ϕn − ϕ] → 0.

The statement in (5.1.13) can now be made more precise when the form is closed.

**Lemma 5.1.8.** Let t be a closed semibounded form in H. Then

$$
\varphi\_n \to\_\mathbf{t} \varphi \quad \Rightarrow \quad \varphi \in \text{dom}\,\mathbf{t} \quad \text{and} \quad \mathbf{t}[\varphi\_n] \to \mathbf{t}[\varphi], \tag{5.1.14}
$$

and, consequently,

$$
\varphi\_n \to\_\mathbf{t} \varphi, \quad \psi\_n \to\_\mathbf{t} \psi \quad \Rightarrow \quad \varphi, \psi \in \text{dom } \mathbf{t} \quad \text{and} \quad \mathbf{t}[\varphi\_n, \psi\_n] \to \mathbf{t}[\varphi, \psi]. \tag{5.1.15}
$$

Proof. Assume that t is a closed semibounded form and ϕn →t ϕ. Then ϕ ∈ dom t and t[ϕn − ϕ] → 0, and since ϕn → ϕ it follows that (t − γ)[ϕn − ϕ] → 0. Hence, (t−γ)[ϕn] → (t−γ)[ϕ] by (5.1.6), and therefore t[ϕn] → t[ϕ]. This shows (5.1.14). Now polarization and (5.1.12) yield the assertion (5.1.15). -

**Lemma 5.1.9.** Let t be a semibounded form in H with lower bound γ and let a<γ. Then the following statements are equivalent:

In particular, t is closed if and only if t − x is closed for some, and hence for all x ∈ R.

Proof. (i) ⇒ (ii) Assume that Ht−a is complete. To show that t is closed, assume that ϕn →t ϕ, so

$$
\varphi\_n \to \varphi \quad \text{and} \quad \mathfrak{t}[\varphi\_n - \varphi\_m] \to 0.
$$

In particular, this implies by Lemma 5.1.3 that ϕn − ϕm t−a → 0. Since Ht−a is complete there is an element ϕ0 ∈ Ht−a = dom t such that ϕn − ϕ0 t−a → 0. Hence, by (5.1.9),

$$\|\varphi\_n - \varphi\_0\| \to 0.$$

Thus ϕ = ϕ0 ∈ dom t. Therefore, ϕn − ϕ t−a → 0 and by (5.1.7) one sees that t[ϕn − ϕ] → 0. This proves that t is closed.

(ii) ⇐ (i) Assume that t is closed. To show that Ht−a is complete, let (ϕn) be a Cauchy sequence in Ht−a. This implies that ϕn →t ϕ for some ϕ ∈ H; cf. Corollary 5.1.5. The closedness of t gives that ϕ ∈ dom t = Ht−a and t[ϕn−ϕ] → 0. By (5.1.7) this leads to ϕn − ϕ t−a → 0, so that Ht−a is complete.

Since t − x is a semibounded form in H with lower bound γ − x, the last statement follows from Ht−a = Ht−x−(a−x) and the equivalence of (i) and (ii). -

Let t be a semibounded form in H with lower bound γ and let Ht−a be the corresponding inner product space with a<γ. In general t is not closed and hence Ht−a is not complete; cf. Lemma 5.1.9. If t1 is a semibounded form with lower bound γ1, which extends the semibounded form t with lower bound γ, then

γ1 ≤ γ.

Note that for a<γ1 one has that t1 is closed if and only if Ht1−a is a Hilbert space. The question is when such a closed extension t1 exists and, if so, to determine the smallest such extension of t. In order to construct an extension of t, note that Lemma 5.1.8 suggests the following definition.

**Definition 5.1.10.** Let t be a semibounded form in H. The linear subspace domt is the set of all ϕ ∈ H for which there exists a sequence (ϕn) in dom t such that ϕn →t ϕ.

It is clear that domt is an extension of dom t. To establish the linearity of domt, recall the property (5.1.12). According to (5.1.13), it would now be natural to define the form t on domt as an extension of t by

$$\tilde{\mathfrak{t}}[\varphi,\psi] = \lim\_{n \to \infty} \mathfrak{t}[\varphi\_n,\psi\_n] \quad \text{for any} \quad \varphi\_n \to\_{\mathfrak{t}} \varphi, \quad \psi\_n \to\_{\mathfrak{t}} \psi,\tag{5.1.16}$$

as the limit on the right-hand side exists. However, in general the limit on the right-hand side of (5.1.16) depends on the choice of the sequences (ϕn) and (ψn), so that t may not be well defined as a form.

**Definition 5.1.11.** A semibounded form t in H is said to be closable if for any sequence (ϕn) in dom t

$$
\varphi\_n \to\_\mathbf{t} 0 \quad \Rightarrow \quad \mathbf{t}[\varphi\_n] \to 0.
$$

It will be shown that the extension procedure in (5.1.16) defines a form extension of t if t is closable. In fact, in this case the resulting form t is unique, being the smallest closed extension, and will be called the closure of t.

**Theorem 5.1.12.** Let t be a semibounded form in H with lower bound γ and let a<γ. Then t has a closed extension if and only if t is closable. In fact, if t is closable, then

and the inner product space Ht−a is dense in the Hilbert space H˜t−a. Moreover, t is closable if and only t− x is closable for some, and hence for all x ∈ R, in which case

$$
\widetilde{\mathfrak{f}-x} = \widetilde{\mathfrak{f}} - x.\tag{5.1.17}
$$

Proof. (⇒) Let t1 be a closed extension of t. In order to show that t is closable, assume that ϕn →t 0. The form t1 is an extension of t and this implies ϕn →t1 0. Since t1 is closed, it follows that

$$\mathfrak{t}[\varphi\_n] = \mathfrak{t}\_1[\varphi\_n] \to 0.$$

Hence, t is closable.

(⇐) Assume that t is closable. It will be shown that t in (5.1.16) is a well-defined form on domt. It is clear from (5.1.13) that the limit on the right-hand side of (5.1.16) exists. To verify that this limit depends only on the elements ϕ, ψ and not on the particular sequences (ϕn), (ψn), let (ϕ- n), (ψ- n) be other sequences such that ϕ- n →t ϕ and ψ- n →t ψ. Then

$$
\varphi\_n' - \varphi\_n \to\_\mathbf{t} 0 \quad \text{and} \quad \psi\_n' - \psi\_n \to\_\mathbf{t} 0;
$$

cf. (5.1.12). In particular, this gives

$$
\varphi\_n' - \varphi\_n \to 0 \quad \text{and} \quad \psi\_n' - \psi\_n \to 0,
$$

while the closability of t implies that

$$\mathfrak{t}[\varphi\_n' - \varphi\_n] \to 0 \quad \text{and} \quad \mathfrak{t}[\psi\_n' - \psi\_n] \to 0.$$

To see that the sequences t[ϕ- n, ψ- n] and t[ϕn, ψn] have the same limit, consider the inequalities

$$\begin{split} &|(\mathbf{t}-\gamma)[\varphi\_{n}^{\prime},\psi\_{n}^{\prime}]-(\mathbf{t}-\gamma)[\varphi\_{n},\psi\_{n}]| \\ &=|(\mathbf{t}-\gamma)[\varphi\_{n}^{\prime}-\varphi\_{n},\psi\_{n}^{\prime}]+(\mathbf{t}-\gamma)[\varphi\_{n},\psi\_{n}^{\prime}-\psi\_{n}]| \\ &\leq|(\mathbf{t}-\gamma)[\varphi\_{n}^{\prime}-\varphi\_{n},\psi\_{n}^{\prime}]|+|(\mathbf{t}-\gamma)[\varphi\_{n},\psi\_{n}^{\prime}-\psi\_{n}]| \\ &\leq(\mathbf{t}-\gamma)[\varphi\_{n}^{\prime}-\varphi\_{n}]^{\frac{1}{2}}(\mathbf{t}-\gamma)[\psi\_{n}^{\prime}]^{\frac{1}{2}}+(\mathbf{t}-\gamma)[\varphi\_{n}]^{\frac{1}{2}}(\mathbf{t}-\gamma)[\psi\_{n}^{\prime}-\psi\_{n}]^{\frac{1}{2}}. \end{split}$$

Clearly, due to the closability assumption, the terms

$$(\mathbf{t} - \gamma)[\varphi\_n' - \varphi\_n] \quad \text{and} \quad (\mathbf{t} - \gamma)[\psi\_n' - \psi\_n].$$

converge to 0 as n → ∞, while the terms

$$(\mathbf{t} - \gamma)[\psi\_n'] \quad \text{and} \quad (\mathbf{t} - \gamma)[\varphi\_n].$$

are bounded since ψ- n →t ψ and ϕn →t ϕ, respectively; cf. Corollary 5.1.6. It follows that t[ϕ- n, ψ- n] − t[ϕn, ψn] → 0 and hence t in (5.1.16) is a well-defined form. Moreover, it is clear that t extends t: t ⊂ t.

The form t is semibounded. To see this, let ϕ ∈ domt. Then there exists a sequence (ϕn) in dom t such that ϕn →t ϕ. In particular, ϕn → ϕ and hence ϕn → ϕ . According to (5.1.16),

$$\mathfrak{t}[\varphi] = \lim\_{n \to \infty} \mathfrak{t}[\varphi\_n],$$

where t[ϕn] ≥ γ ϕn 2. Therefore,

$$\widetilde{\mathfrak{t}}[\varphi] \ge \gamma \|\varphi\|^2, \quad \varphi \in \text{dom}\, \widetilde{\mathfrak{t}},$$

so that t is semibounded. Moreover, this argument shows that the lower bound of the extension is at least γ. Hence, t and t have the same lower bound.

The argument to show that t is closed, is based on the observation that for the extension t:

$$
\varphi\_n \to\_\mathbf{t} \varphi \quad \Rightarrow \quad \mathsf{f}[\varphi - \varphi\_n] \to 0. \tag{5.1.18}
$$

To see this, let ϕn →t ϕ, that is ϕn → ϕ and limm,n→∞ t[ϕn − ϕm] = 0. Now fix n ∈ N, then ϕm →t ϕ implies that

$$
\varphi\_m - \varphi\_n \to\_\mathfrak{t} \varphi - \varphi\_n \quad \text{as} \quad m \to \infty,
$$

so that, by definition,

$$\widetilde{\mathfrak{t}}[\varphi - \varphi\_n] = \lim\_{m \to \infty} \mathfrak{t}[\varphi\_m - \varphi\_n].$$

Now taking n → ∞ gives (5.1.18).

The following three steps will establish that t is closed or, equivalently, that H˜t−a, a<γ, is complete.

Step 1. Ht−a is dense in H˜t−a. Indeed, let ϕ ∈ H˜t−a = domt. Then there is a sequence (ϕn) in Ht−a = dom t such that ϕn →t ϕ. It follows from this assumption and (5.1.18) that

$$
\varphi\_n \to \varphi \quad \text{and} \quad \mathfrak{k}[\varphi - \varphi\_n] \to 0,
$$

in other words,

$$\|\varphi - \varphi\_n\|\_{\mathfrak{k}-a}^2 = \widetilde{\mathfrak{t}}[\varphi - \varphi\_n] - a\|\varphi - \varphi\_n\|^2 \to 0.$$

This shows that Ht−a is dense in H˜t−a.

Step 2. Every Cauchy sequence in Ht−a is convergent in H˜t−a. To see this, let (ϕn) be a Cauchy sequence in Ht−a. Then clearly there exists an element ϕ ∈ H such that ϕn →t ϕ; cf. Corollary 5.1.5. Again by (5.1.18) it follows that

$$\|\varphi - \varphi\_n\|\_{\overline{\mathfrak{t}} - a} \to 0,$$

which now shows that the Cauchy sequence (ϕn) in Ht−a is convergent in H˜t−a to, in fact, ϕ ∈ domt = H˜t−a.

Step 3. H˜t−a is a Hilbert space. To see this, let (χn) be a Cauchy sequence in H˜t−a. By Step 1, there is an element ϕn ∈ Ht−a such that

$$\|\chi\_n - \varphi\_n\|\_{\mathfrak{k}-a} \le \frac{1}{n}.$$

Hence, the approximating sequence (ϕn) is a Cauchy sequence in Ht−a. By Step 2, (ϕn) converges in H˜t−a, which implies that the original sequence (χn) converges in H˜t−a.

Next it will be shown that t is the smallest closed extension of t. Assume that t1 is a closed extension of t: t ⊂ t1. Let ϕ ∈ domt; then there exists a sequence (ϕn) in dom t with ϕn →t ϕ. Then also ϕn →t1 ϕ and hence ϕ ∈ dom t1. Therefore, domt ⊂ dom t1. For every ϕ, ψ ∈ domt it follows via corresponding sequences (ϕn), (ψn) in dom t with ϕn →t ϕ and ψn →t ψ that

$$\mathfrak{t}[\varphi,\psi] = \lim\_{n \to \infty} \mathfrak{t}[\varphi\_n,\psi\_n] = \lim\_{n \to \infty} \mathfrak{t}\_1[\varphi\_n,\psi\_n] = \mathfrak{t}\_1[\varphi,\psi],$$

where the first equality follows from (5.1.16), the second equality is valid as t1 extends t, and the third equality follows from (5.1.15). Therefore, t ⊂ t1, and t is the smallest closed extension of t.

As to the last statement, observe that Definition 5.1.11 implies that t is closable if and only t − x is closable for some, and hence for all x ∈ R. Finally, (5.1.17) follows from (5.1.16). -

Thus, a closed semibounded form t1 which extends t contains the closure t. The next corollary is a simple but useful description of the gap between t1 and t.

**Corollary 5.1.13.** Let the semibounded form t with lower bound γ be closable and let the closed form t1 with lower bound γ1 be an extension of t, so that γ1 ≤ γ. Assume that a<γ1, then

$$\mathfrak{H}\_{\mathfrak{t}\_1-a} = \left\{ \varphi \in \mathfrak{H}\_{\mathfrak{t}\_1-a} : (\varphi, \psi)\_{\mathfrak{t}\_1-a} = 0, \psi \in \mathfrak{H}\_{\mathfrak{t}\_1-a} \right\} \oplus\_{\mathfrak{t}\_1-a} \mathfrak{H}\_{\mathfrak{t}-a}.$$

Let t be a closed semibounded form in H. Let D ⊂ dom t be a linear subspace and consider the restriction tD of t to D,

$$\mathfrak{t}\_{\mathfrak{D}}[\varphi,\psi] = \mathfrak{t}[\varphi,\psi], \quad \varphi,\psi \in \mathfrak{D}.$$

Since tD is a restriction of a closed form, it is closable, see Theorem 5.1.12. Let tD be the closure of tD. Then by definition domtD is the set of all ϕ ∈ H for which there exists a sequence (ϕn) in D with ϕn →tD ϕ, which means ϕn →t ϕ. Since t is closed, one sees in particular that domtD ⊂ dom t. Moreover, one has

$$\widetilde{\mathfrak{t}}\_{\mathfrak{D}}[\varphi,\psi] = \lim\_{n \to \infty} \mathfrak{t}\_{\mathfrak{D}}[\varphi\_n,\psi\_n] = \lim\_{n \to \infty} \mathfrak{t}[\varphi\_n,\psi\_n] = \mathfrak{t}[\varphi,\psi]$$

for ϕ, ψ ∈ domtD, where the first equality is by definition, and the third equality follows from Lemma 5.1.8. Hence, the closure tD of tD is the restriction of t to domtD. Since t is closed, it follows that ϕ ∈ domtD if and only if there is a sequence (ϕn) in D such that

$$
\varphi\_n \to \varphi \quad \text{and} \quad \mathfrak{t}[\varphi\_n - \varphi] \to 0.
$$

**Definition 5.1.14.** Let t be a closed semibounded form in H. A linear subspace D of dom t is said to be a core of t if the closure tD of the restriction tD of t to D coincides with t.

Therefore, D ⊂ dom t is a core of t if and only if for every ϕ ∈ dom t there is a sequence (ϕn) in D such that

$$
\varphi\_n \to \varphi \quad \text{and} \quad \mathfrak{t}[\varphi\_n - \varphi] \to 0. \tag{5.1.19}
$$

This leads to the following corollary.

**Corollary 5.1.15.** Let t be a closed semibounded form in H with lower bound γ, let a<γ, and let D ⊂ dom t be a linear subspace. Then D is a core of t if and only if D is dense in the Hilbert space Ht−a.

Note that in the situation of Theorem 5.1.12 the original domain dom t is a core of the closuret of t (recall that the formt is closed). The following fact is useful: If t and s are closed semibounded forms in H which coincide on D ⊂ dom t∩dom s and D is a core of both t and s, then t = s.

Recall the definition of the sum of two forms in Definition 5.1.1 and observe that a sum of semibounded forms is also semibounded. The following result is concerned with additive perturbations of forms: it provides a sufficient condition so that the sum of a closed semibounded form and a symmetric form remains closed and semibounded. Sometimes this result is referred to as KLMN theorem, named after Kato, Lions, Lax, Milgram, and Nelson. For a typical application to Sturm–Liouville operators, see, e.g., Lemma 6.8.3.

**Theorem 5.1.16.** Assume that t is a closed semibounded form in H and let s be a symmetric form in H such that dom t ⊂ dom s and

$$|\mathfrak{s}[\varphi]| \le a \|\varphi\|^2 + b\mathfrak{t}[\varphi], \qquad \varphi \in \text{dom } \mathfrak{t},\tag{5.1.20}$$

holds for some a ≥ 0 and b ∈ [0, 1). Then the symmetric form

$$\mathfrak{t} + \mathfrak{s}, \qquad \text{dom}\,(\mathfrak{t} + \mathfrak{s}) = \text{dom}\,\mathfrak{t},$$

is closed and semibounded in H. Furthermore, if D is a core of t, then D is also a core of t + s.

Proof. Let γ be the lower bound of t. Fix some a- < γ and assume a- < 0. For all ϕ ∈ dom t, ϕ = 0, one obtains from (5.1.20) that

$$\mathfrak{a}[\varphi] \ge -a \|\varphi\|^2 - b\mathfrak{t}[\varphi],\tag{5.1.21}$$

and hence

$$(\mathfrak{t} + \mathfrak{s})[\varphi] \ge (1 - b)\mathfrak{t}[\varphi] - a\|\varphi\|^2 \\ > \left((1 - b)a' - a\right)\|\varphi\|^2 = c'\|\varphi\|^2,$$

where c- = (1 − b)a- − a < 0. This shows that t + s is semibounded from below. Furthermore, the estimate (5.1.21) also shows that

$$\begin{aligned} (1 - b) \|\varphi\|\_{\mathfrak{t} - a'}^2 &= (1 - b)\mathfrak{t}[\varphi] - (1 - b)a' \|\varphi\|^2 \\ &= \mathfrak{t}[\varphi] - b\mathfrak{t}[\varphi] - a \|\varphi\|^2 - \left( (1 - b)a' - a \right) \|\varphi\|^2 \\ &\le (\mathfrak{t} + \mathfrak{s})[\varphi] - \left( (1 - b)a' - a \right) \|\varphi\|^2 \\ &= \|\varphi\|\_{\mathfrak{t} + \mathfrak{s} - \mathfrak{e}'}^2 \end{aligned}$$

Using (5.1.20) one obtains

$$\begin{aligned} \|\varphi\|\_{\mathfrak{t}+\mathfrak{a}-c'}^2 &= \mathfrak{t}[\varphi] + \mathfrak{s}[\varphi] - c' \|\varphi\|^2 \\ &\le (1+b)\mathfrak{t}[\varphi] - (c'-a) \|\varphi\|^2 \\ &\le b' \|\varphi\|\_{\mathfrak{t}-a'}^2, \end{aligned}$$

where b- = max {(1 + b),(c- − a)/a- }. Therefore, the above estimates imply that the norms · 2 t−a and · 2 t+s−c are equivalent on dom t = dom (s + t). Since t is closed, Ht−a is a Hilbert space and hence Ht+s−c is a Hilbert space, that is, the form t + s is closed; cf. Lemma 5.1.9. The assertion about the core D is clear from Corollary 5.1.15. -

Semibounded relations in a Hilbert space generate closable semibounded forms as will be shown in the following lemma. Note that if a relation is semibounded, then so is its closure, with the same lower bound; this follows directly from Definition 1.4.5. Furthermore, the closure will generate the same form. The particular situation of semibounded self-adjoint relations will be considered in detail in Theorem 5.1.18 and Proposition 5.1.19.

**Lemma 5.1.17.** Let S be a semibounded relation in H with lower bound m(S). Then the form tS given by

$$\mathfrak{t}\_S[\varphi,\psi] = (\varphi',\psi), \quad \{\varphi,\varphi'\}, \{\psi,\psi'\} \in S,\tag{5.1.22}$$

with dom tS = dom S, is well defined, semibounded with the lower bound m(S), and closable. The closure tS of tS is a semibounded closed form whose lower bound is equal to m(S), and

$$\text{dom}\,\widetilde{\mathfrak{t}}\_S \subset \overline{\text{dom}}\,S.\tag{5.1.23}$$

Moreover, dom S = dom tS is a core of tS. Furthermore, with the closure S of S one has

$$
\widetilde{\mathbf{t}}\_S = \widetilde{\mathbf{t}}\_{\widetilde{\mathbf{S}}}.\tag{5.1.24}
$$

Proof. As a semibounded relation S is automatically symmetric, it follows that mul S ⊂ mul S∗ = (dom S)⊥, and hence

$$(\varphi', \psi) = (\varphi'', \psi), \quad \{\varphi, \varphi'\}, \{\varphi, \varphi''\}, \{\psi, \psi'\} \in S.$$

Thus, the form in (5.1.22) is well defined with dom tS = dom S. By definition tS is semibounded and its lower bound is clearly equal to γ = m(S).

In order to show that tS is closable, let ϕn →tS 0. Then, equivalently,

ϕn → 0 and (tS − γ)[ϕn − ϕm] → 0.

It suffices to verify that (tS − γ)[ϕn] → 0. Note that there exists ϕ- n ∈ H such that {ϕn, ϕ- n} ∈ S. Then

$$(\mathbf{t}\_S - \gamma)[\varphi\_n] = (\mathbf{t}\_S - \gamma)[\varphi\_n, \varphi\_n] = (\mathbf{t}\_S - \gamma)[\varphi\_n, \varphi\_n - \varphi\_m] + (\mathbf{t}\_S - \gamma)[\varphi\_n, \varphi\_m],$$

and it follows with the help of the Cauchy–Schwarz inequality (5.1.3) for the nonnegative form (tS − γ) and (5.1.22) that

$$\begin{split} |(\mathsf{f}\_{S} - \gamma)[\varphi\_{n}]| &\leq |(\mathsf{f}\_{S} - \gamma)[\varphi\_{n}, \varphi\_{n} - \varphi\_{m}]| + |(\mathsf{f}\_{S} - \gamma)[\varphi\_{n}, \varphi\_{m}]| \\ &\leq (\mathsf{t}s - \gamma)[\varphi\_{n}]^{\frac{1}{2}}(\mathsf{t}s - \gamma)[\varphi\_{n} - \varphi\_{m}]^{\frac{1}{2}} + |(\varphi\_{n}' - \gamma\varphi\_{n}, \varphi\_{m})|. \end{split}$$

By Corollary 5.1.6, the sequence ((tS −γ)[ϕn]) is bounded by M2 for some M > 0. Moreover, for every ε > 0 there exists N ∈ N such that (tS − γ)[ϕn − ϕm] ≤ ε2 for n, m ≥ N. Therefore,

$$|(\mathfrak{t}\_S - \gamma)[\varphi\_n]| \le M\varepsilon + |(\varphi\_n' - \gamma \varphi\_n, \varphi\_m)|, \qquad n, m \ge N.$$

Fix n ≥ N and let m → ∞. From |(ϕ- n − γϕn, ϕm)|≤ ϕ- n − γϕn ϕm and ϕm → 0 it follows that |(tS − γ)[ϕn]| ≤ Mε for n ≥ N. This shows that (tS − γ)[ϕn] → 0 as n → ∞, and hence tS is closable.

By Theorem 5.1.12, it is clear that the closuretS of tS is a semibounded closed form whose lower bound is equal to m(S). It also follows from the definition of t that the inclusion (5.1.23) holds. Furthermore, dom S = dom tS is a core of tS.

It remains to show (5.1.24). The inclusion tS ⊂ tS is clear. For the opposite inclusion, let ϕ ∈ dom tS = dom S and ϕ- ∈ H such that {ϕ, ϕ- } ∈ S, in which case

$$\mathfrak{t}\_{\overline{\mathcal{S}}}[\varphi,\varphi] = (\varphi',\varphi).$$

Then there exists a sequence ({ϕn, ϕ- n}) in S with ϕn → ϕ and ϕ- n → ϕ- , and hence

$$\text{tr}\_{\mathbf{S}}[\varphi\_n - \varphi\_m] = (\varphi\_n' - \varphi\_m', \varphi\_n - \varphi\_m) \to 0.$$

Therefore, ϕn →tS ϕ, so that ϕ ∈ domtS. Moreover,

$$\mathfrak{tr}\_{\mathfrak{S}}[\varphi,\varphi] = (\varphi',\varphi) = \lim\_{n \to \infty} (\varphi'\_n,\varphi\_n) = \lim\_{n \to \infty} \mathfrak{t}\_{\mathbb{S}}[\varphi\_n,\varphi\_n] = \dot{\mathfrak{t}}\_{\mathbb{S}}[\varphi,\varphi],$$

where in the last equality the definition of the closure in Theorem 5.1.12 was used. This implies tS ⊂ tS and hence tS ⊂ tS. Therefore, tS = tS. -

In the next theorem it is shown that every closed semibounded form can be represented by a semibounded self-adjoint relation.

**Theorem 5.1.18** (First representation theorem)**.** Assume that t is a closed semibounded form in H. Then there exists a semibounded self-adjoint relation H in H such that the following statements hold:

(i) dom H ⊂ dom t and

$$\mathbf{t}[\varphi, \psi] = (\varphi', \psi) \tag{5.1.25}$$

for every {ϕ, ϕ- } ∈ H and ψ ∈ dom t;

$$\mathbf{t}[\varphi, \psi] = (\varphi', \psi) \tag{5.1.26}$$

for every ψ in a core of t, then {ϕ, ϕ- } ∈ H;

(iv) mul H = (dom t)⊥ and t[ϕ, ψ]=(Hop ϕ, ψ) (5.1.27)

for every ϕ ∈ dom H and ψ ∈ dom t.

The semibounded self-adjoint relation H is uniquely determined by (i). The closed form t and the corresponding semibounded self-adjoint relation H have the same lower bound: m(t) = m(H). Moreover, for each x ∈ R the closed semibounded form t − x corresponds to the semibounded self-adjoint relation H − x.

Proof. (i) Let m(t) = γ and choose a<γ. Then the assumption that t is closed is equivalent to the inner product space Ht−a being complete, where Ht−a = dom t is equipped with the inner product of (·, ·)t−a as in (5.1.7)–(5.1.8); cf. Lemma 5.1.9. For any fixed ω ∈ H consider the linear functional

$$\psi \mapsto (\psi, \omega)$$

defined for all ψ ∈ Ht−a = dom t ⊂ H. It follows from (5.1.9) that

$$|\langle \psi, \omega \rangle| \le \|\psi\| \|\omega\| \le \left(\frac{1}{\sqrt{\gamma - a}} \|\omega\|\right) \|\psi\|\_{\mathfrak{t}-a}, \quad \psi \in \mathfrak{H}\_{\mathfrak{t}-a} \omega$$

Hence, the mapping ψ → (ψ, ω) from Ht−a to C is bounded with bound at most ω / √γ − a. Therefore, by the Riesz representation theorem, there exists an element ω in Ht−a such that for all ψ ∈ Ht−a:

$$(\psi, \omega) = (\psi, \widehat{\omega})\_{\mathfrak{t}-a}, \quad \|\widehat{\omega}\|\_{\mathfrak{t}-a} \le \frac{1}{\sqrt{\gamma - a}} \|\omega\|.$$

Taking conjugates for convenience, it follows from the definition (5.1.7) of (·, ·)t−a that

$$(\omega, \psi) = (\widehat{\omega}, \psi)\_{\mathfrak{t}-a} = \mathfrak{t}[\widehat{\omega}, \psi] - a(\widehat{\omega}, \psi), \tag{5.1.28}$$

or, in other words,

$$\mathfrak{A}[\widehat{\omega}, \psi] = (\omega + a\widehat{\omega}, \psi), \quad \psi \in \mathfrak{H}\_{\mathfrak{t}-a}.\tag{5.1.29}$$

Note that the linear mapping A from H to Ht−a defined by Aω = ω satisfies

$$\left|\sqrt{\gamma - a}\right| \|A\omega\| \le \|A\omega\|\_{\mathfrak{t}-a} \le \frac{1}{\sqrt{\gamma - a}} \|\omega\|;$$

where in the first inequality (5.1.9) was used. In other words, if A is interpreted as a mapping from H to H, then

$$\|A\omega\| \le \frac{1}{\gamma - a} \|\omega\|.$$

By means of A define the linear relation H in H by

$$H = \left\{ \left\{ A\omega, \omega + aA\omega \right\} \colon \omega \in \mathfrak{H} \right\},$$

so that

$$A = (H - a)^{-1}.$$

One sees that dom H = ran A ⊂ dom t and mul H = ker A. Moreover, every element {ϕ, ϕ- } ∈ H can be written as {ϕ, ϕ- } = {ω, ω + aω} for some ω, so that by the identity (5.1.29) one obtains

$$\mathfrak{st}[\varphi,\psi] = (\varphi',\psi), \quad \{\varphi,\varphi'\} \in H, \qquad \psi \in \mathfrak{H}\_{\mathfrak{t}-a} = \text{dom } \mathfrak{t}.\tag{5.1.30}$$

It follows from (5.1.30) with ψ = ϕ that H is a semibounded relation with lower bound

$$m(H) \ge m(\mathbf{t}) = \gamma. \tag{5.1.31}$$

It is clear that H is symmetric. According to the definition of H one sees that ran (H − a) = H, which, since a<γ, implies that H is self-adjoint; cf. Proposition 1.5.6. Thus, (i) has been proved.

(ii) The statement that dom H is a core of t is equivalent to the statement that dom H is dense in the Hilbert space Ht−a. To verify denseness, assume that the element ψ ∈ Ht−a is orthogonal to dom H = ran A, i.e.,

$$0 = (A\omega, \psi)\_{\mathfrak{t}-a} = (\widehat{\omega}, \psi)\_{\mathfrak{t}-a} = (\omega, \psi),$$

for all ω ∈ H; cf. (5.1.28). This leads to ψ = 0. Hence, dom H = ran A is dense in the Hilbert space Ht−a, and the assertion follows from Corollary 5.1.15.

(iii) Let ϕ ∈ dom t and ϕ- ∈ H satisfy (5.1.26) for every ψ in a core D of the form t. Then (5.1.26) holds for all ψ ∈ dom t. To see this, let ψ ∈ dom t. Then there exists a sequence (ψn) in D such that ψn →t ψ, which implies that t[ψn − ψ] → 0. Since ψn ∈ D, the assumption yields

$$\mathfrak{t}[\varphi,\psi] = \lim\_{n \to \infty} \mathfrak{t}[\varphi,\psi\_n] = \lim\_{n \to \infty} (\varphi',\psi\_n) = (\varphi',\psi), \quad \psi \in \text{dom } \mathfrak{t},$$

so that (5.1.26) holds for all ψ ∈ dom t. Due to the symmetry of t this result may also be written as

$$\mathbf{t}[\psi,\varphi] = (\psi,\varphi'), \qquad \psi \in \text{dom } \mathbf{t}. \tag{5.1.32}$$

Now let {ψ, ψ- } ∈ H. Then ψ ∈ dom H ⊂ dom t and, by (i),

$$\mathfrak{t}[\psi,\varphi] = (\psi',\varphi),\tag{5.1.33}$$

because ϕ ∈ dom t. Comparing (5.1.32) and (5.1.33) gives

$$(\psi, \varphi') = (\psi', \varphi) \quad \text{for all} \quad \{\psi, \psi'\} \in H,$$

which leads to {ϕ, ϕ- } ∈ H∗ = H. This proves (iii).

(iv) It follows from (i) that if {0, ϕ- } ∈ H, then (ϕ- , ψ) = 0 for all ψ ∈ dom t, and hence mul H ⊂ (dom t)⊥. Conversely, as dom H ⊂ dom t by (i) and H is self-adjoint, (dom t)⊥ ⊂ (dom H)⊥ = mul H. This shows that mul H = (dom t)⊥.

To see (5.1.27), let {ϕ, ϕ- } ∈ H. Then ϕ- = Hop ϕ + χ, where χ ∈ mul H. Hence, from (5.1.25) one obtains

$$\mathfrak{t}[\varphi,\psi] = (\varphi',\psi) = (H\_{\mathrm{op}}\,\varphi+\chi,\psi) = (H\_{\mathrm{op}}\,\varphi,\psi),$$

which gives (5.1.27). This completes the proof of (iv).

To show uniqueness, assume that H is a semibounded self-adjoint relation in H such that dom H-⊂ dom t and

$$\mathfrak{t}[\varphi,\psi] = (\varphi',\psi)$$

for every {ϕ, ϕ- } ∈ H and ψ ∈ dom t. Then, in particular, one concludes that ϕ ∈ dom H- ⊂ dom t and ϕ- ∈ H, so that by (iii) it follows that {ϕ, ϕ- } ∈ H. Hence, H- ⊂ H and one obtains equality as Hand H are both self-adjoint.

Recall that it has been shown in the proof of (i) that m(H) ≥ m(t); cf. (5.1.31). The equality follows from the fact that dom H is a core of t; see (ii). In fact, if ϕ ∈ dom t, then there exists a sequence (ϕn) in dom H such that ϕn →t ϕ. Therefore, if ϕ = 0, then

$$\frac{\mathfrak{t}[\varphi]}{||\varphi||^2} = \lim\_{n \to \infty} \frac{\mathfrak{t}[\varphi\_n]}{||\varphi\_n||^2} = \lim\_{n \to \infty} \frac{(H\_{\text{op }}\varphi\_n, \varphi\_n)}{||\varphi\_n||^2} \ge m(H).$$

Since this inequality holds for every nontrivial ϕ ∈ dom t one concludes that

$$m(\mathbf{t}) = \inf \left\{ \frac{\mathbf{t}[\varphi]}{||\varphi||^2} : \varphi \in \text{dom } \mathbf{t}, \,\varphi \neq 0 \right\} \ge m(H),$$

and so m(t) = m(H).

Finally, note that for x ∈ R the form t − x is semibounded and closed, and the relation H − x is semibounded and self-adjoint. For {ϕ, ϕ- } ∈ H,

$$(\mathbf{t} - x)[\varphi, \psi] = (\varphi', \psi) - x(\varphi, \psi) = (\varphi' - x\varphi, \psi) \tag{5.1.34}$$

for all ψ ∈ dom t = dom t−x. Observe from (iii) that {ϕ, ϕ- −xϕ} ∈ H −x belongs to the semibounded self-adjoint relation corresponding to t − x. As H − x is selfadjoint and contained in the semibounded self-adjoint relation corresponding to t−x both coincide, i.e., H−x corresponds to the closed semibounded form t−x. -

The representation result in Theorem 5.1.18 gives assertions concerning the semibounded self-adjoint relation associated with a given semibounded form. In fact, every semibounded self-adjoint relation appears in such a context, as is shown in the following proposition; cf. Lemma 5.1.17.

**Proposition 5.1.19.** Let A be a semibounded self-adjoint relation in H. Then the semibounded, closable form defined by

$$\mathfrak{t}\_A[\varphi,\psi] = (\varphi',\psi), \quad \{\varphi,\varphi'\}, \{\psi,\psi'\} \in A,$$

has a closure whose corresponding semibounded self-adjoint relation is given by A.

Proof. Since A is semibounded and self-adjoint, Lemma 5.1.17 shows that the form tA is well defined, semibounded, and closable. Moreover, dom A is a core of its closure t. Let H be the semibounded self-adjoint relation corresponding to t. Since t is an extension of t, one has

$$\mathfrak{t}[\varphi,\psi] = \mathfrak{t}[\varphi,\psi] = (\varphi',\psi), \quad \{\varphi,\varphi'\}, \{\psi,\psi'\} \in A.$$

Therefore, Theorem 5.1.18 (iii) implies that {ϕ, ϕ- } ∈ H, since dom A is a core of t. Consequently, A ⊂ H and since A and H are both self-adjoint, one concludes A = H. -

The following observation, based on Theorem 5.1.18 and Proposition 5.1.19, is included for completeness.

**Corollary 5.1.20.** There is a one-to-one correspondence between all closed semibounded forms and all closed semibounded self-adjoint relations via the identity (5.1.25) or, equivalently, via the identity (5.1.27) in the first representation theorem.

The correspondence between closed semibounded forms and semibounded self-adjoint relations in Theorem 5.1.18 can be illuminated further in the context of nonnegative forms and nonnegative self-adjoint relations. As a preparation, observe that a typical way to define forms is via linear operators.

**Lemma 5.1.21.** Let T be a linear operator from a Hilbert space H to a Hilbert space K and define a nonnegative form t in H by

$$\mathfrak{tt}[\varphi, \psi] = (T\varphi, T\psi), \quad \varphi, \psi \in \text{dom } \mathfrak{t} = \text{dom}\, T.$$

Then

t is a closable form ⇔ T is a closable operator,

and in this case the closure of t is given by

$$\overline{\mathfrak{t}}[\varphi,\psi] = (\overline{T}\varphi,\overline{T}\psi), \quad \varphi,\psi \in \text{dom}\,\overline{\mathfrak{t}} = \text{dom}\,\overline{T}.\tag{5.1.35}$$

Proof. (⇒) Assume that t is closable. Let (ϕn) be a sequence in dom T such that ϕn → 0 in H and T ϕn → ψ in K. Then

$$\mathfrak{t}[\varphi\_n - \varphi\_m] = \|T(\varphi\_n - \varphi\_m)\|^2 \to 0,$$

which implies that ϕn →t 0. Since t is closable, one obtains

$$\|T\varphi\_n\|^2 = \mathfrak{t}[\varphi\_n] \to 0,$$

so that T ϕn → 0. It follows that T is closable.

(⇐) Assume that T is closable. Let (ϕn) in dom t with ϕn →t 0. Then ϕn → 0 in H and (T ϕn) is a Cauchy sequence in K. Hence, T ϕn → ψ for some ψ ∈ K and since T is closable one sees that ψ = 0. Therefore, t[ϕn] = T ϕn 2 → 0. It follows that t is closable.

Finally, assume that t or, equivalently, T is closable. Then one has

$$\text{dom}\,\widetilde{\mathfrak{t}} = \text{dom}\,\overline{T}.\tag{5.1.36}$$

Indeed, for the inclusion (⊂) in (5.1.36) consider ϕ ∈ domt. Then there exists a sequence (ϕn) in dom t with ϕn →t ϕ; cf. Theorem 5.1.12. Hence, ϕn → ϕ in H and (T ϕn) is a Cauchy sequence in K. Thus, there exists ϕ- ∈ K such that T ϕn → ϕ- . Since T is closable it follows that ϕ ∈ dom T and ϕ- = T ϕ. Moreover, by Theorem 5.1.12 and (5.1.16) it follows that

$$\widetilde{\mathfrak{t}}[\varphi,\varphi] = \lim\_{n \to \infty} \mathfrak{t}[\varphi\_n,\varphi\_n] = \lim\_{n \to \infty} (T\varphi\_n, T\varphi\_n) = (\overline{T}\varphi, \overline{T}\varphi),$$

and polarization leads to the identity in (5.1.35). For the inclusion (⊃) in (5.1.36) let ϕ ∈ dom T. Then T ϕ = ϕ for some ϕ- ∈ K, and there exists a sequence (ϕn) in dom T for which ϕn → ϕ while T ϕn → ϕ- . In particular, it follows that ϕn →t ϕ. Therefore, ϕ ∈ domt; this proves (5.1.36). -

The following result specializes the first representation theorem to closed nonnegative forms as in Lemma 5.1.21. For a class of closed nonnegative forms it identifies the associated self-adjoint relations. Recall that for a closed operator R a linear subspace D ⊂ dom R is a core if the closure of the restriction R D of R to D coincides with R; cf. Lemma 1.5.10.

**Proposition 5.1.22.** Let T be a closed relation from a Hilbert space H to a Hilbert space K and let Top = P T be the closed orthogonal operator part of T, where P is the orthogonal projection in K onto (mul T)⊥; cf. Theorem 1.3.15. Then the rule

$$\mathbf{t}[\varphi, \psi] = (T\_{\text{op}}\varphi, T\_{\text{op}}\psi), \quad \varphi, \psi \in \text{dom}\,\mathbf{t} = \text{dom}\,T\_{\text{op}} = \text{dom}\,T,\tag{5.1.37}$$

defines a closed nonnegative form t in H. The nonnegative self-adjoint relation corresponding to the form t is given by T ∗T. Moreover, a subset of dom t = dom T is a core of the form t if and only if it is a core of the operator Top .

Proof. Since the operator Top is closed, the nonnegative form t in (5.1.37) is closed, with dom t = dom Top = dom T; cf. Lemma 5.1.21. Recall that T ∗T is a nonnegative self-adjoint relation in H; cf. Lemma 1.5.8. Assume that ϕ ∈ dom T ∗T and ψ ∈ dom T. Let ϕ- ∈ H be any element such that {ϕ, ϕ- } ∈ T ∗T. This implies that {ϕ, η} ∈ T and {η, ϕ- } ∈ T ∗ for some η ∈ K. Clearly, η = Top ϕ + ω for some ω ∈ mul T. Since {Top ϕ + ω,ϕ- } ∈ T ∗ and {ψ, Top ψ} ∈ T, one sees that

$$0 = (\varphi', \psi) - (T\_{\text{op }}\varphi + \omega, T\_{\text{op }}\psi) = (\varphi', \psi) - (T\_{\text{op }}\varphi, T\_{\text{op }}\psi), \quad \psi \in \text{dom } T,$$

i.e.,

$$\mathfrak{t}[\varphi,\psi] = (\varphi',\psi), \quad \{\varphi,\varphi'\} \in T^\*T, \quad \psi \in \text{dom}\,T.$$

Let H be the nonnegative self-adjoint relation associated with t via Theorem 5.1.18. According to (iii) of Theorem 5.1.18, the nonnegative self-adjoint relation T ∗T satisfies T ∗T ⊂ H, which gives T ∗T = H.

Now let D ⊂ dom t = dom T be a linear subset. Then D is a core of t if and only if for every ϕ ∈ dom t = dom T there is a sequence (ϕn) in D such that

$$
\varphi\_n \to \varphi \quad \text{and} \quad \mathfrak{t}[\varphi\_n - \varphi] \to 0;
$$

cf. (5.1.19). In view of the definition of t, this condition reads as

$$
\varphi\_n \to \varphi \quad \text{and} \quad T\_{\text{op}}\varphi\_n \to T\_{\text{op}}\varphi,
$$

in other words D is a core of Top . -

The so-called second representation theorem may be seen as a corollary of Theorem 5.1.18 and Proposition 5.1.22.

**Theorem 5.1.23** (Second representation theorem)**.** Assume that the closed semibounded form t and the semibounded self-adjoint relation H are connected as in Theorem 5.1.18, so that m(H) = m(t) = γ, and let x ≤ γ. Then

$$\text{dom}\,\mathbf{t} = \text{dom}\,(H - x)^{\frac{1}{2}}$$

and the form t is represented by

$$\mathfrak{a}[\varphi,\psi] = \left( (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \varphi, (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \psi \right) + x(\varphi, \psi), \quad \varphi, \psi \in \mathrm{dom}\, \mathfrak{t}.$$

Moreover, a subset of dom t = dom (H − x) 1 2 is a core of the form t if and only if it is a core of the operator (Hop − x) 1 2 .

Proof. For x ≤ γ define the form sx by

$$\mathfrak{s}\_x[\varphi,\psi] = \left( (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \varphi, (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \psi \right), \quad \varphi, \psi \in \mathrm{dom}\,\mathfrak{s}\_x,$$

on the domain dom sx = dom (Hop − x)1/2 = dom (H − x)1/2. By Proposition 5.1.22, the form sx is closed and nonnegative. The corresponding nonnegative self-adjoint relation is given by

$$\left((H-x)^{\frac{1}{2}}\right)^{\*}(H-x)^{\frac{1}{2}}=H-x,$$

and hence sx[ϕ, ψ]=(ϕ- , ψ) holds for all {ϕ, ϕ- } ∈ H−x and ψ ∈ dom sx. It follows as in the proof of Theorem 5.1.18 (see (5.1.34)) that the closed semibounded form

$$(\mathfrak{a}\_x + x)[\varphi, \psi] = \left( (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \varphi, (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \psi \right) + x(\varphi, \psi), \quad \varphi, \psi \in \text{dom}\, \mathfrak{a}\_x,$$

is represented by the semibounded self-adjoint relation H. Furthermore,

$$(\mathfrak{a}\_x + x)[\varphi, \psi] = ((H\_{\text{op}} - x)\varphi, \psi) + x(\varphi, \psi) = (H\_{\text{op}}\varphi, \psi)$$

for all ϕ, ψ ∈ dom H, and hence the restrictions of the form sx +x and of the form t to dom Hop coincide; cf. Theorem 5.1.18 (iv). According to Proposition 5.1.22 and Lemma 1.5.10, dom Hop is a core of sx and hence also of sx +x. On the other hand, by Theorem 5.1.18 (ii), dom Hop = dom H is also a core of t. Hence, the forms sx + x and t coincide on the common core dom Hop . This implies that the forms sx + x and t coincide. Therefore,

$$\text{dom}\,\mathfrak{t} = \text{dom}\,(\mathfrak{s}\_x + x) = \text{dom}\,(H - x)^{\frac{1}{2}}, \qquad x \le \gamma,$$

and

$$\mathfrak{st}[\varphi,\psi] = \left( (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \varphi, (H\_{\mathrm{op}} - x)^{\frac{1}{2}} \psi \right) + x(\varphi, \psi), \quad \varphi, \psi \in \mathrm{dom}\, \mathfrak{t}.$$

Finally, Proposition 5.1.22 shows that a subset of dom t is a core of t if and only if it is a core of the operator (Hop − x) 1 2 . This completes the proof. -

## **5.2 Ordering and monotonicity**

In this section an ordering will be introduced for semibounded closed forms t1 and t2, and for semibounded self-adjoint relations H1 and H2 in a Hilbert space H. It will be shown that these orderings are compatible if t1 and H1, and t2 and H2 are related via the first representation theorem (Theorem 5.1.18), respectively. An alternative formulation of the ordering of semibounded self-adjoint relations will be given in terms of their resolvent operators. The last part of the section is devoted to a general monotonicity principle in the context of semibounded selfadjoint relations or, equivalently, of closed semibounded forms.

First an ordering will be defined for semibounded forms that are not necessarily closed.

**Definition 5.2.1.** Let t1 and t2 be semibounded forms in H that are not necessarily closed. Then one writes t1 ≤ t2, if

$$\text{dom}\,\mathfrak{t}\_2 \subset \text{dom}\,\mathfrak{t}\_1, \quad \mathfrak{t}\_1[\varphi] \le \mathfrak{t}\_2[\varphi], \quad \varphi \in \text{dom}\,\mathfrak{t}\_2. \tag{5.2.1}$$

Note that if t1 ≤ t2, then t2-convergence implies t1-convergence. Indeed, let ϕn →t2 ϕ. By Definition 5.1.4, this means that

ϕn ∈ dom t2, ϕn → ϕ, and t2[ϕn − ϕm] → 0.

Since t1 ≤ t2, this implies that

ϕn ∈ dom t1 and t1[ϕn − ϕm] → 0,

which shows that ϕn →t1 ϕ. Definition 5.2.1 generates a number of simple but useful observations.

**Lemma 5.2.2.** Let t1, t2, and t3 be semibounded forms in H that are not necessarily closed. Then the following statements hold:

Proof. (i) This follows from the definition of t2 ⊂ t1; cf. (5.1.2).

(ii) It follows from (5.2.1) that

$$\begin{aligned} \inf \left\{ \frac{\mathfrak{t}\_1[\varphi]}{||\varphi||^2} : \varphi \in \text{dom}\, \mathfrak{t}\_1, \,\varphi \neq 0 \right\} &\leq \inf \left\{ \frac{\mathfrak{t}\_1[\varphi]}{||\varphi||^2} : \varphi \in \text{dom}\, \mathfrak{t}\_2, \,\varphi \neq 0 \right\} \\ &\leq \inf \left\{ \frac{\mathfrak{t}\_2[\varphi]}{||\varphi||^2} : \varphi \in \text{dom}\, \mathfrak{t}\_2, \,\varphi \neq 0 \right\}. \end{aligned}$$

Hence, Definition 5.1.2 implies that m(t1) ≤ m(t2).

(iii) This is an immediate consequence of Definition 5.2.1.

(iv) If t1 ≤ t2 and t2 ≤ t1, then it follows from (5.2.1) that dom t1 = dom t2 and that t1[ϕ] = t2[ϕ] for all ϕ ∈ dom t1 = dom t2. The conclusion now follows by polarization; cf. (5.1.1).

(v) Assume that t1 and t2 are closable forms. Let ϕ ∈ domt2; then, by Definition 5.1.10, there exists a sequence (ϕn) in dom t2 such that ϕn →t2 ϕ. Recall that t2-convergence implies t1-convergence and thus ϕ ∈ domt1. This shows domt2 ⊂ domt1. Therefore, Theorem 5.1.12 implies that for ϕ ∈ domt2 one has

$$\widetilde{\mathfrak{t}}\_1[\varphi] = \lim\_{n \to \infty} \mathfrak{t}\_1[\varphi\_n] \le \lim\_{n \to \infty} \mathfrak{t}\_2[\varphi\_n] = \widetilde{\mathfrak{t}}\_2[\varphi],$$

which shows (v). -

Next an ordering will be defined for semibounded self-adjoint relations. It will be shown in Proposition 5.2.6 below that this ordering is in agreement with the notation γ ≤ H for a semibounded self-adjoint relation H with lower bound γ; cf. Definition 1.4.5. Note that the following definition relies on Lemma 1.5.10.

**Definition 5.2.3.** Let H1 and H2 be semibounded self-adjoint relations in H, with lower bounds m(H1) and m(H2), respectively. Then the relations H1 and H2 are said to be ordered, and one writes H1 ≤ H2, if

$$\begin{aligned} \text{dom}\,(H\_2 - x)^{\frac{1}{2}} &\subset \text{dom}\,(H\_1 - x)^{\frac{1}{2}},\\ \|(H\_{1, \text{op}} - x)^{\frac{1}{2}}\varphi\| &\le \|(H\_{2, \text{op}} - x)^{\frac{1}{2}}\varphi\|, \quad \varphi \in \text{dom}\,(H\_2 - x)^{\frac{1}{2}},\end{aligned} \tag{5.2.2}$$

is satisfied for some, and hence for all x ≤ min {m(H1), m(H2)}.

In the next theorem it is shown that the ordering for semibounded forms in Definition 5.2.1 and the ordering for semibounded self-adjoint relations in Definition 5.2.3 are compatible. Here the second representation theorem (Theorem 5.1.23) plays an essential role.

**Theorem 5.2.4.** Let t1 and t2 be closed semibounded forms in H and let H1 and H2 be the corresponding semibounded self-adjoint relations. Then

$$\mathfrak{t}\_1 \le \mathfrak{t}\_2 \quad \Leftrightarrow \quad H\_1 \le H\_2.$$

Proof. Assume first that t1 ≤ t2. Then, by Definition 5.2.1,

$$\operatorname{dom} \mathbf{t}\_2 \subset \operatorname{dom} \mathbf{t}\_1, \quad \mathbf{t}\_1[\varphi] \le \mathbf{t}\_2[\varphi], \quad \varphi \in \operatorname{dom} \mathbf{t}\_2,$$

and for all x ≤ min {m(t1), m(t2)} it follows from Theorem 5.1.23 that (5.2.2) holds. Hence, H1 ≤ H2 by Definition 5.2.3.

Conversely, assume that H1 ≤ H2. Then, by Definition 5.2.3, (5.2.2) holds for all x ≤ min {m(H1), m(H2)} and hence Theorem 5.1.23 implies t1 ≤ t2. -

**Lemma 5.2.5.** Let H1, H2, and H3 be semibounded self-adjoint relations in H. Then the following statements hold:

Proof. Let ti be the closed semibounded form corresponding to Hi, i = 1, 2, 3. For the proof of (i) it is sufficient to observe that

$$
\overline{\operatorname{dom}}\,H\_2 = \overline{\operatorname{dom}}\,\mathbf{t}\_2 \subset \overline{\operatorname{dom}}\,\mathbf{t}\_1 = \overline{\operatorname{dom}}\,H\_1,
$$

where Theorem 5.2.4 and Theorem 5.1.18 (iv) were used. Taking orthogonal complements then gives mul H1 ⊂ mul H2. For (ii) recall that

$$m(H\_1) = m(\mathfrak{t}\_1) \le m(\mathfrak{t}\_2) = m(H\_2),$$

as follows from Lemma 5.2.2 and Theorem 5.1.18. Statements (iii) and (iv) are translations of similar statements in Lemma 5.2.2. The statement (v) is clear from Theorem 5.2.4. -

Assume that in Definition 5.2.3 the self-adjoint relation H1 has a closed domain dom H1. Then the operator part H1,op of H1 is a bounded operator which implies that dom H1 = dom (H1 − x) 1 2 . Thus, in this case H1 ≤ H2 if and only if

$$\begin{aligned} \text{dom}\,(H\_2 - x)^{\frac{1}{2}} &\subset \text{dom}\,H\_1, \\\|(H\_{1, \text{op}} - x)\varphi, \varphi\rangle &\le \|(H\_{2, \text{op}} - x)^{\frac{1}{2}}\varphi\|^2, \quad \varphi \in \text{dom}\,(H\_2 - x)^{\frac{1}{2}}.\end{aligned} \tag{5.2.3}$$

The following proposition gives an alternative version of this statement.

**Proposition 5.2.6.** Let H1 and H2 be semibounded self-adjoint relations in H and assume that dom H1 is closed. Then the following statements are equivalent:

Moreover, if H1 ∈ **B**(H), then these statements are equivalent to

(iii) (H1ϕ, ϕ) ≤ (H2,op ϕ, ϕ), ϕ ∈ dom H2;

(iv) (H1ϕ, ϕ) ≤ (ϕ- , ϕ), {ϕ, ϕ- } ∈ H2;

and in the particular case that H1 = γ1IH to

Proof. (i) ⇒ (ii) Let (i) be satisfied. Then dom H2 ⊂ dom (H2 − x) 1 2 ⊂ dom H1 by (5.2.3), and for all ϕ ∈ dom H2 the inequality in (5.2.3) takes the form

$$((H\_{1, \text{op}} - x)\varphi, \varphi) \le ((H\_{2, \text{op}} - x)\varphi, \varphi),$$

which implies (ii).

(ii) ⇒ (i) Let (ii) be satisfied and let ϕ ∈ dom (H2 − x) 1 2 . Then there exists a sequence (ϕn) in dom H2 such that

$$
\varphi\_n \to \varphi \quad \text{and} \quad (H\_{2, \text{op}} - x)^{\frac{1}{2}} \varphi\_n \to (H\_{2, \text{op}} - x)^{\frac{1}{2}} \varphi, \quad n \to \infty,
$$

since dom H2 is a core of (H2 − x) 1 2 ; see Lemma 1.5.10. Due to the assumption one has ϕn ∈ dom H1 and

$$\left( (H\_{1, \text{op}} - x)\varphi\_n, \varphi\_n \right) \le \left( (H\_{2, \text{op}} - x)\varphi\_n, \varphi\_n \right) = \|(H\_{2, \text{op}} - x)^{\frac{1}{2}}\varphi\_n\|^2.$$

Since dom H1 is closed it follows by taking the limit that

$$\|\left( (H\_{1, \text{op}} - x)\varphi, \varphi \right) \le \|(H\_{2, \text{op}} - x)^{\frac{1}{2}}\varphi\|^2, \quad \varphi \in \text{dom}\left( H\_2 - x \right)^{\frac{1}{2}}.$$

Hence, (5.2.3) is satisfied or, equivalently, H1 ≤ H2.

If H1 ∈ **B**(H), then dom H1 = H and hence the rest of the statements is clear. -

In particular, the inequality in (v)–(vi) of Proposition 5.2.6 shows that the ordering γIH ≤ H is equivalent to H being semibounded with lower bound γ as defined in Definition 1.4.5. Furthermore, if both H1 and H2 are self-adjoint operators in **B**(H), then they are semibounded and Proposition 5.2.6 (iii) shows that H1 ≤ H2 in the sense of Definition 5.2.3 agrees with the usual definition (H1ϕ, ϕ) ≤ (H2ϕ, ϕ) for all ϕ ∈ H.

The ordering for semibounded relations H1 and H2 can also be expressed in terms of their resolvent operators. The next proposition is an immediate consequence of Proposition 1.5.11 (for the special case ρ = 1).

**Proposition 5.2.7.** Let H1 and H2 be semibounded self-adjoint relations in H. Then the following statements are equivalent:

$$\left(H\_2 - x\right)^{-1} \le \left(H\_1 - x\right)^{-1}.$$

The next corollary slightly extends Proposition 5.2.7 and gives a further interpretation of the inequality H1 ≤ H2 when x ≤ min {m(H1), m(H2)}. The equivalence in (5.2.4) below is an example of the antitonicity property.

**Corollary 5.2.8.** Let H1 and H2 be semibounded self-adjoint relations in H. Then

$$H\_1 \le H\_2$$

if and only if for γ ≤ min {m(H1), m(H2)} one has

$$(H\_2 - \gamma)^{-1} \le (H\_1 - \gamma)^{-1}.$$

In particular, if H1 and H2 are nonnegative self-adjoint relations, then

$$H\_1 \le H\_2 \quad \Leftrightarrow \quad H\_2^{-1} \le H\_1^{-1}. \tag{5.2.4}$$

Proof. Let H be a semibounded self-adjoint relation with γ ≤ m(H). Then H − γ is nonnegative and hence also (H − γ)−1 is a nonnegative self-adjoint relation. Now write for x<γ

$$H - x = H - \gamma - (x - \gamma),$$

and apply Corollary 1.1.12 (with H replaced by (H − γ)−1 and λ replaced by (x − γ)−1), obtaining

$$\left(\left(H-x\right)^{-1} - \frac{1}{x-\gamma} - \frac{1}{(x-\gamma)^2} \left( (H-\gamma)^{-1} - \frac{1}{x-\gamma} \right)^{-1} \right)$$

Hence, for the pair of semibounded self-adjoint relations H1 and H2 and with γ ≤ min {m(H1), m(H2)} one obtains for each x<γ:

$$\begin{aligned} &\left( (H\_1 - x)^{-1} - (H\_2 - x)^{-1} \right) \\ &= \frac{1}{(x - \gamma)^2} \left[ \left( (H\_2 - \gamma)^{-1} - \frac{1}{x - \gamma} \right)^{-1} - \left( (H\_1 - \gamma)^{-1} - \frac{1}{x - \gamma} \right)^{-1} \right]. \end{aligned}$$

Since x−γ < 0, a repeated application of Proposition 5.2.7 shows the equivalence. In fact, H1 ≤ H2 if and only if (H2 − x)−1 ≤ (H1 − x)−1 by Proposition 5.2.7, which by the above formula is equivalent to

$$\left( (H\_1 - \gamma)^{-1} - \frac{1}{x - \gamma} \right)^{-1} \le \left( (H\_2 - \gamma)^{-1} - \frac{1}{x - \gamma} \right)^{-1}.\tag{5.2.5}$$

Another application of Proposition 5.2.7 shows that the inequality (5.2.5) is equivalent to the inequality (H2 − γ)−1 ≤ (H1 − γ)−1. -

As a corollary to Proposition 5.2.7 it will be shown that in the case H1 ≤ H2 the difference (H1 − x)−1 − (H2 − x)−1, x < min {m(H1), m(H2)}, can be used to describe the gap between the corresponding form domains

$$\operatorname{dom}\left(H\_2 - x\right)^{\frac{1}{2}} \subset \operatorname{dom}\left(H\_1 - x\right)^{\frac{1}{2}}.$$

**Corollary 5.2.9.** Let H1 and H2 be semibounded self-adjoint relations in H and assume that

$$H\_1 \le H\_2.$$

Then for all x < min {m(H1), m(H2)} the operator (H1−x)−1−(H2−x)−1 ∈ **B**(H) is nonnegative and

$$\operatorname{dom}\left(H\_1 - x\right)^{\frac{1}{2}} = \operatorname{ran}\left(\left(H\_1 - x\right)^{-1} - \left(H\_2 - x\right)^{-1}\right)^{\frac{1}{2}} + \operatorname{dom}\left(H\_2 - x\right)^{\frac{1}{2}}.$$

Proof. Since H1 ≤ H2, the operator R(x) ∈ **B**(H), defined by

$$R(x) = \left(H\_1 - x\right)^{-1} - \left(H\_2 - x\right)^{-1},$$

is nonnegative for x < min {m(H1), m(H2)}; cf. Proposition 5.2.7. Hence, one can write

$$\begin{aligned} \left( (H\_1 - x)^{-1} = R(x) + (H\_2 - x)^{-1} \\ = \left( R(x)^{\frac{1}{2}} \quad (H\_2 - x)^{-\frac{1}{2}} \right) \begin{pmatrix} R(x)^{\frac{1}{2}} \\ (H\_2 - x)^{-\frac{1}{2}} \end{pmatrix}. \end{aligned} \tag{5.2.6}$$

Now recall that if T = (A B) is a row operator with A, B ∈ **B**(H), then it follows from ran (T T ∗) 1 2 = ran |T ∗| = ran T, cf. Corollary D.6, that

$$\text{ran}\,(AA^\* + BB^\*)^{\frac{1}{2}} = \text{ran}\,(A\,\,B) = \text{ran}\,A + \text{ran}\,B.\tag{5.2.7}$$

Hence, taking square roots in the identity (5.2.6) and applying (5.2.7) shows that

$$\text{ran}\,(H\_1 - x)^{-\frac{1}{2}} = \text{ran}\,R(x)^{\frac{1}{2}} + \text{ran}\,(H\_2 - x)^{-\frac{1}{2}},$$

which yields the desired decomposition

$$\operatorname{dom}\left(H\_1 - x\right)^{\frac{1}{2}} = \operatorname{ran} R(x)^{\frac{1}{2}} + \operatorname{dom}\left(H\_2 - x\right)^{\frac{1}{2}}$$

for x < min {m(H1), m(H2)}. -

Now the ordering for semibounded self-adjoint relations and for semibounded closed forms will be used to reinterpret and extend the monotonicity result in Proposition 1.9.9

For the proof of the following theorem it is useful to have available an auxiliary result concerning the interchange of limits. Let (fn) be a nondecreasing sequence of real nondecreasing functions defined on an open interval (a, b). Thus, for all x ∈ (a, b) one has

$$f\_m(x) \le f\_n(x), \quad m \le n,\tag{5.2.8}$$

and for all n ∈ N

$$f\_n(x) \le f\_n(y), \quad a < x \le y < b. \tag{5.2.9}$$

In view of (5.2.8) the pointwise limit

$$f\_{\infty}(x) = \lim\_{n \to \infty} f\_n(x), \quad x \in (a, b), \tag{5.2.10}$$

gives a function f∞ : (a, b) → R ∪ {∞} that is nondecreasing, thanks to (5.2.9). This is clear when all f∞(x) are finite, in which case limx→b f∞(x) is proper or improper. However, if f∞(x0) = ∞ for some x0 ∈ (a, b), then (5.2.9) shows that f∞(x) = ∞ for all x0 ≤ x<b. In this case the function f∞ is also called nondecreasing (in the sense of R ∪ {∞}) and one defines limx→b f∞(x) = ∞. In view of (5.2.9) the limit

$$f\_n(b) = \lim\_{x \to b^{-}} f\_n(x), \quad n \in \mathbb{N}, \tag{5.2.11}$$

gives a sequence with values in R ∪ {∞} that is nondecreasing, thanks to (5.2.8). This is again clear when all limits fn(b) are finite in which case limn→∞ fn(b) is proper or improper. However, if there exists some m ∈ N for which fm(b) = ∞, then for all n ≥ m one has fn(b) = ∞. In this case one defines limn→∞ fn(b) = ∞.

**Lemma 5.2.10.** *Let* (fn) *be a nondecreasing sequence of nondecreasing functions defined on some open interval* (a, b)*. Let* f∞ *be the nondecreasing limit function in* (5.2.10) *and let* (fn(b)) *be the nondecreasing sequence of limits in* (5.2.11)*. Then*

$$\lim\_{x \to b} f\_{\infty}(x) = \lim\_{n \to \infty} f\_n(b). \tag{5.2.12}$$

*In particular, both limits in* (5.2.12) *are finite or infinite simultaneously.*

*Proof.* Consider the case that all values of f∞ are real. Since fn(x) ≤ f∞(x) for all x ∈ (a, b), it follows that for any n ∈ N

$$f\_n(b) = \lim\_{x \to b} f\_n(x) \le \lim\_{x \to b} f\_{\infty}(x).$$

This implies

$$\lim\_{n \to \infty} f\_n(b) \le \lim\_{x \to b} f\_{\infty}(x),\tag{5.2.13}$$

where the limits may be infinite. Assume that there is strict inequality in (5.2.13). First consider the case limx→b f∞(x) < ∞. Then clearly there exists some δ > 0 for which

$$\delta + \lim\_{n \to \infty} f\_n(b) < \lim\_{x \to b} f\_{\infty}(x). \tag{5.2.14}$$

Next consider the case limx→b f∞(x) = ∞. Then limn→∞ fn(b) < ∞ (otherwise there would be equality in (5.2.13)) and (5.2.14) holds for any δ > 0. In each case, there exists some x ∈ (a, b) such that

$$
\delta + \lim\_{n \to \infty} f\_n(b) < f\_{\infty}(x).
$$

From this one concludes

$$
\delta + f\_{\infty}(x) = \delta + \lim\_{n \to \infty} f\_n(x) \le \delta + \lim\_{n \to \infty} f\_n(b) < f\_{\infty}(x);
$$

a contradiction. Hence, there is equality in (5.2.13). It remains to consider the situation where f∞(x0) = ∞ for some a<x0 < b. In this case f∞(x) = ∞ for all x0 <x<b and limx→b f∞(x) = ∞. Assume that

$$L = \lim\_{n \to \infty} f\_n(b) < \infty.$$

For any x0 ≤ x<b one has

$$f\_n(x) \le f\_n(b) \le L,$$

which implies that limn→∞ fn(x) ≤ L; a contradiction. Again, there is equality in (5.2.13). - **Theorem 5.2.11** (Monotonicity principle)**.** Let (Hn) be a nondecreasing sequence of semibounded self-adjoint relations in H and let γ ≤ m(H1). Then there exists a semibounded self-adjoint relation H∞ with γ ≤ m(H∞) and Hn ≤ H∞ such that Hn → H∞ in the strong resolvent sense, i.e.,

$$(H\_n - \lambda)^{-1}\varphi \to \left(H\_\infty - \lambda\right)^{-1}\varphi, \quad \varphi \in \mathfrak{H}, \quad \lambda \in \mathbb{C} \backslash [\gamma, \infty). \tag{5.2.15}$$

Furthermore, H∞ satisfies

$$\begin{aligned} &\text{dom}\,(H\_{\infty}-\gamma)^{\frac{1}{2}}\\ &= \left\{ \varphi \in \bigcap\_{n=1}^{\infty} \text{dom}\,(H\_n-\gamma)^{\frac{1}{2}} : \lim\_{n \to \infty} \|(H\_{n,\text{op}}-\gamma)^{\frac{1}{2}}\varphi\| < \infty \right\} \end{aligned} \tag{5.2.16}$$

and for all ϕ ∈ dom (H∞ − γ) 1 2 it holds that

$$\|(H\_{\infty, \text{op}} - \gamma)^{\frac{1}{2}} \varphi\| = \lim\_{n \to \infty} \|(H\_{n, \text{op}} - \gamma)^{\frac{1}{2}} \varphi\|. \tag{5.2.17}$$

Proof. The assumption Hn ≤ Hm for n ≤ m and Proposition 5.2.7 lead to

$$0 \le (H\_m - x)^{-1} \le (H\_n - x)^{-1}, \quad x < \gamma,$$

where γ ≤ m(H1). Hence, by Proposition 1.9.14, there exists a semibounded selfadjoint relation H∞ with γ ≤ m(H∞) such that

$$0 \le (H\_{\infty} - x)^{-1} \le (H\_n - x)^{-1}, \quad x < \gamma,\tag{5.2.18}$$

and Hn converges to H∞ in the strong resolvent sense on C \ [γ, ∞), that is, (5.2.15) holds.

It remains to prove (5.2.16) and (5.2.17). It follows from Corollary 1.1.12 with H replaced by Hn − γ and H∞ − γ, respectively, that for x < 0 one has

$$\begin{aligned} &\left(\left\{(H\_n-\gamma)^{-1}-x\right\}^{-1}\varphi,\varphi\right)-\left(\left\{(H\_\infty-\gamma)^{-1}-x\right\}^{-1}\varphi,\varphi\right) \\ &=\frac{1}{x^2}\left[\left(\left((H\_\infty-\gamma)-\frac{1}{x}\right)^{-1}\varphi,\varphi\right)-\left(\left((H\_n-\gamma)-\frac{1}{x}\right)^{-1}\varphi,\varphi\right)\right].\end{aligned}$$

Since γ + 1/x < γ, the right-hand side tends to zero monotonically from below for n → ∞, as follows from (5.2.15) and (5.2.18); but then also the left-hand side tends to zero monotonically from below.

To complete the proof, consider the functions, defined for ϕ ∈ H and x < 0 by

$$f\_n(x) = \left(\left((H\_n - \gamma)^{-1} - x\right)^{-1}\varphi, \varphi\right)$$

and

$$f\_{\infty}(x) = \left(\left((H\_{\infty} - \gamma)^{-1} - x\right)^{-1}\varphi, \varphi\right).$$

The above argument shows that the sequence fn is nondecreasing with f∞ as pointwise limit. It follows from Lemma 1.5.12 (with H replaced by Hn − γ and H∞ − γ, respectively), that both functions fn and f∞ are nondecreasing on the interval (−∞, 0) and that

$$\begin{split} f\_n(0) &= \lim\_{x \uparrow 0} \left( ((H\_n - \gamma)^{-1} - x)^{-1} \varphi, \varphi \right) \\ &= \begin{cases} \| (H\_{n, \text{op}} - \gamma)^{\frac{1}{2}} \varphi \|^{2}, & \varphi \in \text{dom} \left( H\_n - \gamma \right)^{\frac{1}{2}}, \\ \infty, & \text{otherwise}, \end{cases} \end{split} \tag{5.2.19}$$

while

$$\begin{split} f\_{\infty}(0) &= \lim\_{x \uparrow 0} \left\{ ((H\_{\infty} - \gamma)^{-1} - x)^{-1} \varphi, \varphi \right\} \\ &= \begin{cases} \| (H\_{\infty, \text{op}} - \gamma)^{\frac{1}{2}} \varphi \| ^2, & \varphi \in \text{dom} \left( H\_{\infty} - \gamma \right)^{\frac{1}{2}}, \\ \infty, & \text{otherwise}. \end{cases} \end{split} \tag{5.2.20}$$

Hence, by Lemma 5.2.10,

$$\lim\_{n \to \infty} f\_n(0) = f\_{\infty}(0),\tag{5.2.21}$$

where the limits in (5.2.21) are finite or infinite simultaneously.

Assume that ϕ ∈ dom (H∞ − γ) 1 2 . Then, by (5.2.20), f∞(0) < ∞, which, in view of (5.2.21), implies that all fn(0) < ∞. Hence, ϕ ∈ 8∞ n=1 dom (Hn − γ) 1 2 by (5.2.19), and (5.2.21) reads

$$\lim\_{n \to \infty} \|(H\_{n, \text{op}} - \gamma)^{\frac{1}{2}} \varphi\|^2 = \|(H\_{\infty, \text{op}} - \gamma)^{\frac{1}{2}} \varphi\|^2. \tag{5.2.22}$$

Thus, ϕ belongs to the right-hand side of (5.2.16). This shows the inclusion (⊂) in (5.2.16), and (5.2.22) gives (5.2.17).

Conversely, assume that ϕ belongs to the right-hand side of (5.2.16), that is, ϕ ∈ 8∞ n=1 dom (Hn − γ) 1 2 and

$$\lim\_{n \to \infty} \|(H\_{n, \text{op}} - \gamma)^{\frac{1}{2}} \varphi\| < \infty.$$

By (5.2.19) one sees that fn(0) < ∞ and that limn→∞ fn(0) < ∞. It follows from (5.2.21) that f∞(0) < ∞. Now apply (5.2.20) to conclude that ϕ ∈ dom (H∞−γ) 1 2 . This shows the inclusion (⊃) in (5.2.16). -

**Corollary 5.2.12.** Let (Hn) be a nondecreasing sequence of semibounded self-adjoint relations and let H∞ be the strong resolvent limit as in Theorem 5.2.11. Then the following statements hold:

Proof. (i) Assume that Hn ≤ K. Then for all x<γ ≤ m(H1)

$$0 \le ((K - x)^{-1} \varphi, \varphi) \le ((H\_n - x)^{-1} \varphi, \varphi), \quad \varphi \in \mathfrak{H}.$$

By (5.2.15), (Hn − x)−1ϕ → (H∞ − x)−1ϕ for ϕ ∈ H and one concludes that

$$0 \le ((K - x)^{-1} \varphi, \varphi) \le ((H\_{\infty} - x)^{-1} \varphi, \varphi), \quad \varphi \in \mathfrak{H}.$$

Hence, by Proposition 5.2.7 it follows that H∞ ≤ K.

(ii) Assume that {ϕ, ϕ- } ∈ S. Then {ϕ, ϕ- } ∈ Hn by assumption and hence for all n ∈ N one has

$$(H\_n - \lambda)^{-1}(\varphi' - \lambda \varphi) = \varphi, \quad \lambda \in \mathbb{C} \backslash [\gamma, \infty).$$

By (5.2.15), (Hn − x)−1ψ → (H∞ − x)−1ψ for ψ ∈ H and one concludes that

$$(H\_{\infty} - \lambda)^{-1}(\varphi' - \lambda \varphi) = \lim\_{n \to \infty} (H\_n - \lambda)^{-1}(\varphi' - \lambda \varphi) = \varphi,$$

which gives {ϕ, ϕ- } ∈ H∞. Hence, S ⊂ H∞. -

Now consider the special case of a nondecreasing sequence of self-adjoint operators (Hn) in **B**(H). Then it is clear that H∞ is a semibounded self-adjoint relation which is an operator in **B**(H) if and only if the sequence (Hn) is uniformly bounded; cf. Corollary 1.9.10 and the beginning of Section 1.9. The following corollary shows that the domain of the square root of H∞ − γ, γ ≤ m(H1), is given by those ϕ ∈ H for which (Hnϕ, ϕ) has a finite limit as n → ∞.

**Corollary 5.2.13.** Let (Hn) be a nondecreasing sequence of self-adjoint operators in **B**(H) with γ ≤ m(H1) and define

$$\mathfrak{E} = \left\{ \varphi \in \mathfrak{H} : \lim\_{n \to \infty} (H\_n \varphi, \varphi) < \infty \right\}.$$

Let H∞ be the semibounded self-adjoint limit of the sequence Hn. Then

$$\mathfrak{E} = \text{dom}\,(H\_{\infty} - \gamma)^{\frac{1}{2}}.$$

In particular, one has

A useful variant of Corollary 5.2.13 is concerned with a nondecreasing function M : (a, b) → **B**(H), whose values are self-adjoint operators; cf. Corollary 2.3.8. Then there exists a self-adjoint limit at the right endpoint b, which can be retrieved via sequences converging to b. For the existence of the self-adjoint limit at the left endpoint a consider the function x → −M(x), which is nondecreasing when x ∈ (a, b) tends to a.

**Corollary 5.2.14.** Let M : (a, b) → **B**(H) be a nondecreasing function, whose values are self-adjoint operators. Then there exist self-adjoint relations M(a) and M(b) in H such that M(x) → M(b) in the strong resolvent sense when x → b and M(x) → M(a) in the strong resolvent sense when x → a. Furthermore,

$$M(x) \le M(b) \quad \text{and} \quad -M(x) \le -M(a), \qquad x \in (a, b).$$

Define

$$\mathfrak{E}\_b = \left\{ \varphi \in \mathfrak{H} : \lim\_{x \uparrow b} (M(x)\varphi, \varphi) < \infty \right\}.$$

and

$$\mathfrak{E}\_a = \left\{ \varphi \in \mathfrak{H} : \lim\_{x \downarrow a} (M(x)\varphi, \varphi) > -\infty \right\}.$$

Then for c = a or c = b one has

(i) Ec = H ⇔ M(c) ∈ **B**(H);

Let (tn) be a nondecreasing sequence of closed semibounded forms in H which satisfy γ ≤ m(t1). By Theorem 5.1.18, there exist unique semibounded self-adjoint relations Hn, bounded from below by γ, which correspond to tn. According to Theorem 5.2.4, the sequence (Hn) is nondecreasing. By the monotonicity principle in Theorem 5.2.11 the strong resolvent limit of the sequence (Hn) exists as a semibounded self-adjoint relation H∞ with lower bound γ such that Hn ≤ H∞. Let t∞ be the form corresponding to H∞ by Proposition 5.1.19. Then t∞ is bounded below by γ and tn ≤ t∞ by Theorem 5.2.4. Therefore, the following theorem concerning a nondecreasing sequence of forms may be seen as a direct consequence of Theorem 5.2.11.

**Theorem 5.2.15** (Monotonicity principle)**.** Let (tn) be a nondecreasing sequence of closed semibounded forms in H and let γ ≤ m(t1). Then there exists a closed semibounded form t∞ with γ ≤ m(t∞) such that tn ≤ t∞ and

$$\text{dom}\,\mathbf{t}\_{\infty} = \left\{ \varphi \in \bigcap\_{n=1}^{\infty} \text{dom}\,\mathbf{t}\_{n} : \lim\_{n \to \infty} \mathbf{t}\_{n}[\varphi] < \infty \right\} \tag{5.2.23}$$

and

$$\mathfrak{t}\_{\infty}[\varphi] = \lim\_{n \to \infty} \mathfrak{t}\_{n}[\varphi], \quad \varphi \in \text{dom } \mathfrak{t}\_{\infty}. \tag{5.2.24}$$

Moreover, the relations Hn corresponding to the forms tn converge in the strong resolvent sense to the relation H∞ corresponding to the form t∞.

Proof. It is clear from Theorem 5.1.23 and the formulas (5.2.16) and (5.2.17) in Theorem 5.2.11 that the limit form t∞ satisfies (5.2.23) and (5.2.24). -

## **5.3 Friedrichs extensions of semibounded relations**

A semibounded, not necessarily closed, relation S in a Hilbert space H has equal defect numbers, and hence admits self-adjoint extensions in H. It will be shown that such a relation S has a distinguished semibounded self-adjoint extension SF, the so-called Friedrichs extension of S, with m(SF) = m(S). The construction of this extension involves a closed semibounded form associated with S. The characteristic properties of this extension will be investigated in detail.

Let S be a semibounded relation in H. Recall from Lemma 5.1.17 that the form tS given by

$$\text{tr}\_S[f,g] = (f',g), \quad \{f,f'\}, \{g,g'\} \in S,\tag{5.3.1}$$

is well defined and that it is semibounded with the lower bound m(S). Moreover, it has been shown that tS is closable and that the closure tS of tS is a semibounded closed form whose lower bound is equal to m(S). Also, dom tS = dom S is a core of tS.

**Lemma 5.3.1.** Let S be a semibounded relation in H with lower bound m(S). Let tS be the closure of the form tS in (5.3.1). Then the unique relation SF corresponding to tS via Theorem 5.1.18 is a semibounded self-adjoint extension of S with the lower bound m(SF) = m(S). In fact, SF is a self-adjoint extension of the semibounded relation S + N ∞(S∗), so that

$$S \subset S \stackrel{\frown}{+} \dot{\mathfrak{N}}\_{\infty}(S^\*) \subset S\_{\sf F}, \quad where \quad \dot{\mathfrak{N}}\_{\infty}(S^\*) = \{0\} \times \operatorname{mult} S^\*. \tag{5.3.2}$$

Moreover,

$$S\_{\mathcal{F}} = (\overline{S})\_{\mathcal{F}}.\tag{5.3.3}$$

Proof. By Theorem 5.1.18, the closed form tS induces a unique semibounded selfadjoint relation SF in H such that

> tS[f,g]=(f- , g), {f,f- } ∈ SF, g ∈ domtS.

To show that SF is an extension of S, let {f,f- } ∈ S. As f ∈ dom tS, it follows that for all g ∈ dom tS

$$\mathfrak{t}\_S[f,g] = \mathfrak{t}\_S[f,g] = (f',g).$$

Since dom tS = dom S is a core of tS, one obtains {f,f- } ∈ SF from Theorem 5.1.18 (iii). Hence, SF is a self-adjoint extension of S with lower bound m(SF) = m(S).

In order to verify (5.3.2) it now suffices to see that {0} × mul S∗ ⊂ SF. Let ϕ ∈ mul S∗ and let g ∈ dom S, then clearly tS[0, g] = 0 and (ϕ, g) = 0. Therefore,

$$\text{ts}[0, g] = (\varphi, g) \quad \text{for all} \quad g \in \text{dom}\, S.$$

Since dom S is a core of tS, it follows that {0, ϕ} ∈ SF.

To see that (5.3.3) holds, it suffices to recall from Lemma 5.1.17 that tS = tS holds for the closures of tS and tS. - **Definition 5.3.2.** Let S be a semibounded relation in H. The semibounded selfadjoint relation SF associated with the closure of the form tS in (5.3.1) is called the Friedrichs extension of S. The closure tS of the form tS will be denoted by tSF , so that tSF = tS.

Let S be a semibounded relation in H and let a<m(S). Then S − a is a nonnegative relation and it is a consequence of (5.3.1) that tS−a = tS − a. The translation invariance of the closures, cf. (5.1.17), leads to

$$\mathbf{t}\_{(S-a)\mathbb{F}} = \widehat{\mathbf{t}s\_{-a}} = \widehat{\mathbf{t}s - a} = \mathbf{\tilde{t}s} - a = \mathbf{t}\_{\mathbb{F}} - a.$$

The nonnegative self-adjoint relation (S − a)F corresponding to the form on the left-hand side is equal to the nonnegative self-adjoint relation corresponding to the form on the right-hand side. Thus, one obtains

$$(S - a)\_{\mathcal{F}} = S\_{\mathcal{F}} - a, \quad a < m(S). \tag{5.3.4}$$

In other words, the Friedrichs extension is translation invariant.

By Lemma 5.3.1, the Friedrichs extension SF is a semibounded self-adjoint extension of S. As a restriction of S∗ the Friedrichs extensions can be characterized as follows.

**Theorem 5.3.3.** Let S be a semibounded relation in H. The Friedrichs extension SF of S admits the representation

$$S\_{\mathcal{F}} = \left\{ \{f, f'\} \in S^\* \, : \, f \in \text{dom}\, \mathbf{t}\_{S\_{\mathcal{F}}} \right\} \tag{5.3.5}$$

with mul SF = mul S∗. Furthermore, if H is a self-adjoint extension of S, that is not necessarily semibounded, then

$$\text{dom}\,H \subset \text{dom}\,\mathfrak{t}\_{\text{SF}} \quad \Rightarrow \quad H = S\_{\text{F}}.$$

Proof. In order to show that SF is contained in the right-hand side of (5.3.5), let {f,f- } ∈ SF. Clearly, S ⊂ SF and since SF is self-adjoint this implies SF ⊂ S∗, so that {f,f- } ∈ S∗. Note also that f ∈ dom SF ⊂ dom tSF . Hence, SF is contained in the right-hand side of (5.3.5).

To show the opposite inclusion, let {f,f- } ∈ S∗ be such that f ∈ dom tSF . Then there exists a sequence (fn) in dom tS = dom S with

$$f\_n \to\_{\mathfrak{t}\_{\mathcal{S}\_{\mathbb{P}}}} f.$$

Let {fn, f- n} be corresponding elements in S and let {g, g- } ∈ S be arbitrary. Then tS ⊂ tSF and S ⊂ S∗ imply that

$$\mathfrak{t}\_{\mathbb{S}^{\mathbb{P}}}[f\_n, g] = \mathfrak{t}\_{\mathbb{S}}[f\_n, g] = (f'\_n, g) = (f\_n, g').$$

Since tSF is a closed form, it follows that

$$\mathfrak{t}\_{S^\mathbb{P}}[f,g] = (f,g').$$

As {f,f- } ∈ S∗ and {g, g- } ∈ S, one obtains (f,g- )=(f- , g), so that

$$\mathfrak{t}\_{\mathrm{Sp}}[f,g] = (f',g).$$

This identity holds for an arbitrary element g ∈ dom S. Since dom S = dom tS is a core of the form tSF it follows from Theorem 5.1.18 (iii) that {f,f- } ∈ SF.

Now let H be any self-adjoint extension of S with dom H ⊂ dom tSF . Hence, if {f,f- } ∈ H, then {f,f- } ∈ S∗ and f ∈ dom H ⊂ dom tSF . By (5.3.5), one has {f,f- } ∈ SF. This shows H ⊂ SF, and since both H and SF are self-adjoint, it follows that H = SF. -

According to Theorem 5.3.3, the inclusion dom H ⊂ dom tSF for any selfadjoint extension H of S implies H = SF. Note that in general a self-adjoint extension of a semibounded relation is not necessarily semibounded. The situation is different when S has finite defect numbers; cf. Proposition 5.5.8.

The construction of SF in (5.3.5) results in a description of SF by means of approximating elements from the graph of S.

**Corollary 5.3.4.** Let S be a semibounded relation in H. Then SF is the set of all elements {f,f- } ∈ S∗ for which there exists a sequence ({fn, f- n}) in S such that

> fn → f and (f- n, fn) → (f- , f).

Proof. By Theorem 5.3.3, SF is the set of all elements {f,f- } ∈ S∗ for which f ∈ dom tSF . Hence, there exists a sequence ({fn, f- n}) in S such that

$$f\_n \to\_{\mathfrak{t}\_{\mathcal{S}\_\mathbb{P}}} f.$$

In particular, fn → f in H and, moreover,

$$\mathfrak{t}\left(f',f\right) = \mathfrak{t}\_{\mathrm{Sp}}[f,f] = \lim\_{n \to \infty} \mathfrak{t}\_{\mathrm{Sp}}[f\_n,f\_n] = \lim\_{n \to \infty} \mathfrak{t}\_S[f\_n,f\_n] = \lim\_{n \to \infty} \mathfrak{t}\left(f'\_n,f\_n\right).$$

Hence, SF is contained in the relation

$$\left\{ \{ f, f' \} \in S^\* \,:\, f\_n \to f \text{ and } (f'\_n, f\_n) \to (f', f) \text{ for some } \{ f\_n, f'\_n \} \in S \right\}.$$

Observe that this relation is symmetric since (f- n, fn) ∈ R implies (f- , f) ∈ R. Thus, the self-adjoint relation SF is contained in the symmetric relation above, and therefore they coincide. -

$$\text{Since } S \subset S\_{\mathcal{F}} \subset S^\* \text{ and } \text{dom } S\_{\mathcal{F}} \subset \text{dom } S \text{ by Corollary 5.3.4 one has that}$$

$$\text{dom } S \subset \text{dom } S\_{\mathcal{F}} \subset \left( \overline{\text{dom }} S \cap \text{dom } S^\* \right).$$

The next corollary shows when these inclusions are identities.

**Corollary 5.3.5.** Let S be a semibounded relation in H. Then

$$S \stackrel{\frown}{+} \dot{\mathfrak{N}}\_{\infty}(S^\*) = S\_{\text{F}}, \quad \dot{\mathfrak{N}}\_{\infty}(S^\*) = \{0\} \times \text{mult} \, S^\*,\tag{5.3.6}$$

if and only if

$$\operatorname{dom} S = \overline{\operatorname{dom}} S \cap \operatorname{dom} S^\*.$$

In particular, if dom S is closed, then SF has the form (5.3.6).

Proof. Recall from (5.3.2) that S + N ∞(S∗) ⊂ SF. Note that there is equality if and only if S + N ∞(S∗) is self-adjoint. Hence, the assertion follows from Lemma 1.5.7. -

The construction of the Friedrichs extension SF of a semibounded relation S via the form tS in (5.3.1) leads to an important characteristic property. First of all, recall that

$$\mathfrak{t}\_{\mathbb{S}^{\mathbb{F}}}[f,g] = (f',g), \quad \{f,f'\} \in S\_{\mathbb{F}}, \quad g \in \text{dom } \mathfrak{t}\_{\mathbb{S}^{\mathbb{F}}},$$

with lower bound m(SF) = m(S). Now assume that H is another semibounded self-adjoint extension of S. Then clearly m(H) ≤ m(S), and according to Proposition 5.1.19, the relation H generates a closed semibounded form tH on H with

$$\mathfrak{t}\_H[f,g] = (f',g), \quad \{f,f'\} \in H, \quad g \in \text{dom } \mathfrak{t}\_H.$$

By specializing {f,f- } ∈ S and g ∈ dom S, it follows that

$$\mathfrak{t}\_H[f,g] = (f',g) = \mathfrak{t}\_S[f,g]$$

and hence tS ⊂ tH. By construction, tS ⊂ t = tSF and hence tSF is the smallest closed form extension of tS; cf. Theorem 5.1.12. Therefore,

$$\mathbf{t}\_S \subset \mathbf{t}\_{S\circ} \subset \mathbf{t}\_H. \tag{5.3.7}$$

This leads to the extremality property of SF stated in the next result.

**Proposition 5.3.6.** Let S be a semibounded relation in H and let H be a semibounded self-adjoint extension of S. Then m(tH) = m(H) ≤ m(SF) = m(S) and

$$\mathfrak{t}\_H \le \mathfrak{t}\_{\mathbb{S}^\mathbb{P}} \quad or \quad H \le S\_{\mathbb{F}};$$

or, equivalently, for some, and hence for all a<m(H),

$$(S\_{\mathcal{F}} - a)^{-1} \le (H - a)^{-1}.$$

Proof. It is a consequence of (5.3.7) and Lemma 5.2.2 (i) that tH ≤ tSF . The rest of the statements follow from Theorem 5.2.4 and Proposition 5.2.7. -

According to Proposition 5.3.6, the Friedrichs extension SF is the largest semibounded self-adjoint extension of S in the sense of the ordering for forms or relations, and so it has the smallest form domain. Recall that for any semibounded self-adjoint extension H of S one has for a<m(H) ≤ m(SF) that

$$\operatorname{dom}\left(H - a\right)^{\frac{1}{2}} = \operatorname{ran} R(a)^{\frac{1}{2}} + \operatorname{dom}\left(S\_{\mathcal{F}} - a\right)^{\frac{1}{2}},\tag{5.3.8}$$

where the nonnegative operator R(a) is defined by

$$R(a) = (H - a)^{-1} - (S\_\mathcal{F} - a)^{-1} \in \mathbf{B}(\mathfrak{H});\tag{5.3.9}$$

cf. Corollary 5.2.9. The identity (5.3.8) will now be put in a geometric context. Recall that for a<m(H) the closed nonnegative form tH − a on H defines the following inner product on dom tH

$$\mathbf{t}\_{\mathbf{f}}(f,g)\_{\mathfrak{t}\_H-a} = \mathbf{t}\_H[f,g] - a(f,g), \quad f,g \in \text{dom } \mathfrak{t}\_H = \text{dom}\,(H-a)^{\frac{1}{2}},\tag{5.3.10}$$

which makes the space HtH−a = domtH = dom(H −a) 1 2 complete; cf. Lemma 5.1.8. Similarly, the closed nonnegative form tSF − a on H defines the following inner product on dom tSF :

$$(f,g)\_{\mathfrak{t}\_{\mathcal{F}\_{\mathcal{F}}}-a} = \mathfrak{t}\_{\mathcal{F}}[f,g] - a(f,g), \quad f,g \in \text{dom}\,\mathfrak{t}\_{\mathcal{F}} = \text{dom}\,(S\_{\mathcal{F}} - a)^{\frac{1}{2}},\tag{5.3.11}$$

which makes the space HtSF−a = dom tSF = dom (SF − a) 1 2 complete. Thus, in terms of inner product spaces one obtains

$$
\mathfrak{H}\_{\mathbf{t}\_{\mathrm{Sp}} - a} \subset \mathfrak{H}\_{\mathbf{t}\_H - a},
$$

and by (5.3.7) the restriction of the inner product in (5.3.10) to HtSF−a coincides with the inner product in (5.3.11). Therefore,

$$\mathfrak{H}\_{\mathfrak{t}\_H - a} = \left( \mathfrak{H}\_{\mathfrak{t}\_H - a} \ominus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_{\mathfrak{S}\_F} - a} \right) \oplus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_{\mathfrak{S}\_F} - a},\tag{5.3.12}$$

see Corollary 5.1.13. In terms of the spaces HtH−a and HtSF−a the sum decomposition in (5.3.8) may be rewritten as

$$\mathfrak{H}\_{\mathfrak{H}-a} = \text{ran}\, R(a)^{\frac{1}{2}} + \mathfrak{H}\_{\mathfrak{H}\_{\mathbb{S}^\*}-a}.\tag{5.3.13}$$

The connection between the decompositions in (5.3.12) and (5.3.13) is discussed in the next proposition.

**Proposition 5.3.7.** Let S be a semibounded relation in H, let SF be its Friedrichs extension, and let H be a semibounded self-adjoint extension of S with lower bound m(H). Furthermore, let a<m(H) and let R(a) be the nonnegative operator in (5.3.9). Then

$$\mathfrak{H}\_{\mathfrak{t}\_H - a} \ominus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_{\mathfrak{F}\_\mathbf{F}} - a} = \ker \left( S^\* - a \right) \cap \mathfrak{H}\_{\mathfrak{t}\_H - a} = \text{ran} \, R(a)^{\frac{1}{2}},\tag{5.3.14}$$

and, consequently, the Hilbert space HtH−a has the orthogonal decomposition

$$\begin{split} \mathfrak{H}\_{\mathsf{t}\_{H}-a} &= \left( \ker \left( S^\* - a \right) \cap \mathfrak{H}\_{\mathsf{t}\_{H}-a} \right) \oplus\_{\mathsf{t}\_{H}-a} \mathfrak{H}\_{\mathsf{t}\_{\mathsf{S}\_{\mathsf{F}}}-a} \\ &= \text{ran} \, R(a)^{\frac{1}{2}} \oplus\_{\mathsf{t}\_{H}-a} \mathfrak{H}\_{\mathsf{t}\_{\mathsf{S}\_{\mathsf{F}}}-a} . \end{split} \tag{5.3.15}$$

In particular, the sum decomposition in (5.3.8) is direct for every a<m(H).

Proof. First the identity

$$\mathfrak{H}\_{\mathfrak{t}\_H - a} \ominus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_{\mathfrak{S}\_H} - a} = \ker \left( S^\* - a \right) \cap \mathfrak{H}\_{\mathfrak{t}\_H - a} \tag{5.3.16}$$

in (5.3.14) will be shown. Recall from Lemma 5.1.17 and Theorem 5.1.12 that HtS−a is a dense subspace of HtSF−a, and hence it suffices to verify that

$$\mathfrak{H}\_{\mathfrak{t}\_H - a} \ominus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_S - a} = \ker \left( S^\* - a \right) \cap \mathfrak{H}\_{\mathfrak{t}\_H - a}. \tag{5.3.17}$$

Assume first that g belongs to the left-hand side of (5.3.17). Then (f,g)tH−a = 0 for all f ∈ dom S. Hence,

$$0 = (f,g)\_{\mathfrak{t}\_H - a} = \mathfrak{t}\_H[f,g] - a(f,g) = (f',g) - a(f,g) = (f' - af, g)\_{\mathfrak{t}\_H}$$

for all {f,f- } ∈ S ⊂ H, where in the third equality the first representation theorem was used. This implies g ∈ (ran (S − a))⊥ ∩ HtH−a = ker (S∗ − a) ∩ HtH−a. Conversely, assume that g ∈ ker (S∗ − a) ∩ HtH−a. Then the same reasoning as above shows that g belongs to the left-hand side of (5.3.17).

In order to prove the second equality in (5.3.14), one first shows that

$$\operatorname{ran} R(a)^{\frac{1}{2}} \subset \ker \left( S^\* - a \right). \tag{5.3.18}$$

To see this, let {f,f- } ∈ S. Then {f,f- } ∈ H∩SF and hence (H−a)−1(f- −af) = f and (SF − a)−1(f-− af) = f, which implies that for h ∈ H

$$(f' - af, R(a)h) = \left( (H - a)^{-1}(f' - af) - (S\_\mathcal{F} - a)^{-1}(f' - af), h \right) = 0.$$

Therefore,

$$\text{ran}\,R(a) \subset (\text{ran}\,(S - a))^\perp = \text{ker}\,(S^\* - a)^\perp$$

and hence also ran R(a) ⊂ ker (S∗ − a). Moreover, since R(a) is a nonnegative self-adjoint operator it follows from Corollary D.7 that

$$\operatorname{ran} R(a) \subset \operatorname{ran} R(a)^{\frac{1}{2}} \subset \overline{\operatorname{ran}} R(a)^{\frac{1}{2}} = \overline{\operatorname{ran}} R(a) \subset \ker \left( S^\* - a \right), \tag{5.3.19}$$

which shows (5.3.18). By (5.3.8), one has ran R(a) 1 2 ⊂ dom (H − a) 1 2 = HtH−a. Together with (5.3.19) one concludes that ran R(a) 1 2 ⊂ ker (S∗ −a)∩HtH−a. From (5.3.16) it is clear that

$$\operatorname{ran} R(a)^{\natural} \subset \mathfrak{H}\_{\mathfrak{t}\_H - a} \ominus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_S - a} \dots$$

Comparing (5.3.8) with (5.3.12), one then concludes that

$$\text{ran}\,R(a)^{\frac{1}{2}} = \mathfrak{H}\_{\mathfrak{t}\_H - a} \ominus\_{\mathfrak{t}\_H - a} \mathfrak{H}\_{\mathfrak{t}\_S - a} \dots$$

Together with (5.3.16) the identities in (5.3.14) follow. Furthermore, the above reasoning shows that the sum decomposition in (5.3.8) is direct. -

The semibounded self-adjoint extensions H of S for which the subspace ker (S∗ − a) ∩ HtH−a is not a proper subset of ker (S∗ − a) are of special interest. In fact, they coincide with the semibounded self-adjoint extensions H for which H and SF are transversal.

**Theorem 5.3.8.** Let S be a semibounded relation in H and let H be a semibounded self-adjoint extension of S. Then the following statements are equivalent:

Proof. (i) ⇔ (ii) In general, the self-adjoint extensions H and SF are transversal if and only if for some, and hence for all a<m(H)

$$\text{ran}\,R(a) = \ker\left(S^\*-a\right),$$

where R(a)=(H − a)−1 − (SF − a)−1, which follows from Theorem 1.7.8. Since R(a) is a nonnegative self-adjoint operator and since ker (S∗ −a) is closed, the last statement is equivalent to

$$\text{ran}\,R(a)^{\frac{1}{2}} = \text{ker}\,(S^\*-a);$$

cf. Corollary D.7. It follows from Proposition 5.3.7 that this condition is the same as

$$\ker\left(S^\*-a\right) \cap \text{dom}\left(H-a\right)^{\frac{1}{2}} = \ker\left(S^\*-a\right).$$

(ii) ⇒ (iii) This follows immediately from the direct sum decomposition

$$\operatorname{dom}\left(H - a\right)^{\frac{1}{2}} = \left(\ker\left(S^\* - a\right) \cap \operatorname{dom}\left(H - a\right)^{\frac{1}{2}}\right) + \operatorname{dom}\left(S\_\mathcal{F} - a\right)^{\frac{1}{2}};$$

cf. (5.3.8) and Proposition 5.3.7.

(iii) ⇒ (ii) This implication is trivial.

(i) ⇒ (iv) The identity S∗ = SF + H shows that

$$
\text{dom}\,S^\* = \text{dom}\,S\_\mathcal{F} + \text{dom}\,H,
$$

and note that dom H and dom SF are subsets of dom (H − a) 1 2 .

(iv) ⇒ (ii) This is clear. -

The following result is a consequence of Theorem 5.3.8; it describes the behavior of any semibounded self-adjoint extension H of S in the presence of a semibounded self-adjoint extension H such that H and SF are transversal.

**Corollary 5.3.9.** Let S be a semibounded relation in H and let H be a semibounded self-adjoint extension of S such that H and SF are transversal. Then every semibounded self-adjoint extension Hof S satisfies

$$\dim\left(H'-a\right)^{\frac{1}{2}} \subset \text{dom}\left(H-a\right)^{\frac{1}{2}}, \quad a < \min\left\{m(H), m(H')\right\},\tag{5.3.20}$$

in which case there exists C > 0 such that

$$\|(H\_{\rm op} - a)^{\frac{1}{2}}\varphi\| \le C \|((H')\_{\rm op} - a)^{\frac{1}{2}}\varphi\|\tag{5.3.21}$$

for all ϕ ∈ dom (H- − a) 1 2 . Moreover, there is equality in (5.3.20) if and only if Hand SF are transversal, in which case there exist c > 0 and C > 0 such that

$$\|c\|((H')\_{\rm op} - a)^{\frac{1}{2}}\varphi\| \le \|(H\_{\rm op} - a)^{\frac{1}{2}}\varphi\| \le C\|((H')\_{\rm op} - a)^{\frac{1}{2}}\varphi\|\tag{5.3.22}$$

for all ϕ ∈ dom (H- − a) 1 2 = dom (H − a) 1 2 .

Proof. By Theorem 5.3.8, the semibounded self-adjoint extension H of S satisfies all of the equivalent conditions (i)–(iv) in Theorem 5.3.8. Let H be another semibounded self-adjoint extension of S. Applying Proposition 5.3.7 to H one sees that for a<m(H- )

$$\operatorname{dom}\left(H'-a\right)^{\frac{1}{2}} = \left(\ker\left(S^\*-a\right)\cap\mathfrak{H}\_{\operatorname{t}\_{H'}-a}\right) + \operatorname{dom}\left(S\_{\mathcal{F}}-a\right)^{\frac{1}{2}}.$$

Choosing a < min {m(H), m(H- )} it follows from Theorem 5.3.8 (iii) for H that the inclusion (5.3.20) holds. The inequality (5.3.21) is a direct consequence of Proposition 1.5.11.

Assume that there is equality in (5.3.20). Then Theorem 5.3.8 (iii) implies that H and SF are transversal. Conversely, if H and SF are transversal, then it follows from Theorem 5.3.8 (iii) that there is equality in (5.3.20). If there is equality in (5.3.20), then (5.3.21) also holds for some c > 0 when H and H are interchanged. Thus, the inequalities in (5.3.22) hold. -

The extreme case of equality of H and SF is described in the following immediate corollary of Proposition 5.3.7 and (5.3.8).

**Corollary 5.3.10.** Let S be a semibounded relation in H and let H be a semibounded self-adjoint extension of S. Then H = SF if and only if

$$\ker\left(S^\*-a\right) \cap \text{dom}\left(H-a\right)^{\frac{1}{2}} = \{0\}$$

for some, and hence for all a<m(H).

In the next corollary the form corresponding to a semibounded self-adjoint extension H which is transversal to SF is specified.

**Corollary 5.3.11.** Let S be a semibounded relation in H, let H be a semibounded self-adjoint extension of S such that H and SF are transversal, and let tSF and tH be the corresponding closed semibounded forms in H. Then

$$\operatorname{dom} \mathfrak{t}\_H = \ker \left( S^\* - a \right) \oplus\_{\mathfrak{t}\_H - a} \operatorname{dom} \mathfrak{t}\_{\mathbb{S}^p}, \qquad a < m(H), \tag{5.3.23}$$

and the restriction of tH to Na(S∗) = ker (S∗ − a) is a closed form in Na(S∗) which is bounded from below by m(H) and represented by a bounded self-adjoint operator La ∈ **B**(Na(S∗)). Furthermore, one has

$$\mathbf{t}\_H[f,g] - a(f,g) = \left( (L\_a - a)f\_a, g\_a \right) + \mathbf{t}\_{\mathrm{Sp}}[f\_\mathrm{F}, g\_\mathrm{F}] - a(f\_\mathrm{F}, g\_\mathrm{F}) \tag{5.3.24}$$

for all f = fa + fF, g = ga + gF ∈ dom tH, where fa, ga ∈ ker (S∗ − a) and fF, gF ∈ dom tSF .