Operator domains, deficiency spaces and boundary realizations

Density follows from [F3]. For $f,g\in C_c^{\mathrm{\infty }}(0,L)$, integration by parts gives $⟨-f^{\mathrm{\prime }\mathrm{\prime }},g⟩=⟨f,-g^{\mathrm{\prime }\mathrm{\prime }}⟩$, since both functions and their derivatives vanish near the endpoints. Thus $T_c$ is dense and symmetric.

Suppose $u\in L^2(0,L)$ has distributional second derivative $D^{2}u=h\in L^2(0,L)$. By [F6], $h\in L^1(0,L)$. Define $v(x)={\int }_0^{x}h(t)dt$. Then $\mathrm{\mid }v(x)\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_{L^1}\le \sqrt{L}\mathrm{\parallel }h{\mathrm{\parallel }}_2$, so $v\in L^2(0,L)$. For every $\phi \in C_c^{\mathrm{\infty }}(0,L)$, choose a measurable representative of $h$ and define $G(t,x):={\mathbf{1}}_{\{0<t<x<L\}}h(t){\phi }^{\mathrm{\prime }}(x),(t,x)\in (0,L{)}^2\mathrm{.}$ The function $q(t,x)=\mathrm{\mid }h(t)\mathrm{\mid }\mathrm{\mid }{\phi }^{\mathrm{\prime }}(x)\mathrm{\mid }$ is nonnegative and product-measurable. By [F6], $h\in L^1(0,L)$. The test derivative ${\phi }^{\mathrm{\prime }}$ is continuous with compact support in $(0,L)$, hence belongs to $L^2(0,L)$; [F6] then gives ${\phi }^{\mathrm{\prime }}\in L^1(0,L)$. Tonelli [F8], applied only to this nonnegative product, gives ${\int }_{(0,L{)}^2}qd(\lambda \times \lambda )=({\int }_0^L\mathrm{\mid }h(t)\mathrm{\mid }dt)({\int }_0^L\mathrm{\mid }{\phi }^{\mathrm{\prime }}(x)\mathrm{\mid }dx)<\mathrm{\infty }\mathrm{.}$ Since $\mathrm{\mid }G\mathrm{\mid }\le q$, this establishes $G\in L^1((0,L{)}^2)$ without using Fubini to prove its own hypothesis. Its sections are integrable for every parameter because $h,{\phi }^{\mathrm{\prime }}\in L^1(0,L)$. Fubini [F9] therefore gives ${\int }_0^{L}v(x){\phi }^{\mathrm{\prime }}(x)dx={\int }_0^L{\int }_0^{x}h(t){\phi }^{\mathrm{\prime }}(x)dtdx={\int }_0^{L}h(t)({\int }_t^L{\phi }^{\mathrm{\prime }}(x)dx)dt=-{\int }_0^{L}h(t)\phi (t)dt,$ where [F7] applied to real and imaginary parts gives ${\int }_t^L{\phi }^{\mathrm{\prime }}(x)dx=\phi (L)-\phi (t)=-\phi (t)$, since $\phi$ vanishes near $L$. Hence $Dv=h$ weakly and $v\in H^1$. The distribution $Du-v$ has derivative zero. For every test function $\phi$ of integral zero, its primitive $\psi (x)={\int }_0^x\phi (t)dt$ is compactly supported in $(0,L)$, and hence $(Du-v)(\phi )=(Du-v)(D\psi )=-(D(Du-v))(\psi )=0$. Obtain a test function $\eta$ with integral one as follows. The constant function $1$ lies in $L^2(0,L)$ with norm $\sqrt{L}$. By the density in [F3], choose $\chi \in C_c^{\mathrm{\infty }}(0,L)$ with $\mathrm{\parallel }\chi -1{\mathrm{\parallel }}_2<\sqrt{L}\mathrm{/}2$; then $\mathrm{\parallel }\chi {\mathrm{\parallel }}_2>\sqrt{L}\mathrm{/}2$. Set $\eta =\mathrm{\mid }\chi {\mathrm{\mid }}^2\mathrm{/}\mathrm{\parallel }\chi {\mathrm{\parallel }}_2^2$. It is a smooth compactly supported test function and ${\int }_0^L\eta =1$. For arbitrary $\phi$, $\phi -({\int }_0^L\phi )\eta$ has integral zero, so $(Du-v)(\phi )=(Du-v)(\eta ){\int }_0^L\phi$. Therefore $Du-v$ is a constant distribution, $Du=v+c\in L^2$, and $u\in H^2(0,L)$ by [F1].

The Hermite cubic polynomials $h_{00}(s)=2s^3-3s^2+1$, $h_{01}(s)=-2s^3+3s^2$, $h_{10}(s)=s^3-2s^2+s$, and $h_{11}(s)=s^3-s^2$ satisfy the endpoint data $(h_{00}(0),h_{00}(1),h_{00}^{\mathrm{\prime }}(0),h_{00}^{\mathrm{\prime }}(1))=(1,0,0,0)$, $(h_{01}(0),h_{01}(1),h_{01}^{\mathrm{\prime }}(0),h_{01}^{\mathrm{\prime }}(1))=(0,1,0,0)$, $(h_{10}(0),h_{10}(1),h_{10}^{\mathrm{\prime }}(0),h_{10}^{\mathrm{\prime }}(1))=(0,0,1,0)$, and $(h_{11}(0),h_{11}(1),h_{11}^{\mathrm{\prime }}(0),h_{11}^{\mathrm{\prime }}(1))=(0,0,0,1)$. Consequently, for arbitrary $a_0,a_L,d_0,d_L\in \mathbb{C}$, the polynomial $p(x)=a_0{h}_{00}(x\mathrm{/}L)+a_L{h}_{01}(x\mathrm{/}L)+L{d}_0{h}_{10}(x\mathrm{/}L)+L{d}_L{h}_{11}(x\mathrm{/}L)$ has $p(0)=a_0$, $p(L)=a_L$, $p^{\mathrm{\prime }}(0)=d_0$, and $p^{\mathrm{\prime }}(L)=d_L$. It belongs to $H^2(0,L)$, proving that $\gamma$ is onto.

If $g\in H^2(0,L)$, [F2] applied with a compactly supported test function shows $⟨T_{c}f,g⟩=⟨f,-D^{2}g⟩$; thus $g\in D(T_c^{\ast })$ and $T_c^{\ast }g=-D^{2}g$. Conversely, if $g\in D(T_c^{\ast })$ and $h=T_c^{\ast }g$, testing the adjoint identity against every $f\in C_c^{\mathrm{\infty }}(0,L)$ gives distributionally $D^{2}g=-h\in L^2$. Step 1.2 then gives $g\in H^2(0,L)$. Hence $T_c^{\ast }=T_{\max }$ with exactly the claimed domain.

By [F5] and step 2.1, $T_{\min }=T_c^{\ast \ast }=T_{\max }^{}$. If $u\in D(T_{\max }^{})$ and $h=T_{\max }^{}u$, testing against $C_c^{\mathrm{\infty }}(0,L)\subset H^2(0,L)$ gives distributionally $D^{2}u=-h$. Step 1.2 implies $u\in H^2(0,L)$. The adjoint identity for all $g\in H^2(0,L)$ and [F2] now reads $⟨{\mathrm{\Gamma }}_{1}g,{\mathrm{\Gamma }}_{0}u⟩-⟨{\mathrm{\Gamma }}_{0}g,{\mathrm{\Gamma }}_{1}u⟩=0$. By step 1.3 the two traces of $g$ can be prescribed independently: first set ${\mathrm{\Gamma }}_{0}g=0$ and vary ${\mathrm{\Gamma }}_{1}g$ to obtain ${\mathrm{\Gamma }}_{0}u=0$; then set ${\mathrm{\Gamma }}_{1}g=0$ and vary ${\mathrm{\Gamma }}_{0}g$ to obtain ${\mathrm{\Gamma }}_{1}u=0$. Conversely, if $u\in H^2$ has both traces zero, [F2] gives $⟨T_{\max }g,u⟩=⟨g,-D^{2}u⟩$ for every $g\in H^2$, so $u\in D(T_{\max }^{})$. Therefore $D(T_{\min })=\{u\in H^2:{\mathrm{\Gamma }}_{0}u={\mathrm{\Gamma }}_{1}u=0\}$ and its action is $-D^{2}u$.

The preceding steps establish density, symmetry, the exact adjoint, the closure domain, and trace surjectivity as stated.

The graph of $A$ is the image of the graph of $S$ under the bounded invertible map $(f,g)\mapsto (f,E_{\ast }^{-1}g)$ on $H\oplus H$, so it is closed; real scaling preserves symmetry. If $g\in D(S^{\ast })$, then for every $f\in D(S)$, $⟨Af,g⟩=E_{\ast }^{-1}⟨Sf,g⟩=⟨f,E_{\ast }^{-1}{S}^{\ast }g⟩$, so $g\in D(A^{\ast })$ and $A^{\ast }g=E_{\ast }^{-1}{S}^{\ast }g$. Conversely, if $g\in D(A^{\ast })$, then $⟨Sf,g⟩=E_{\ast }⟨Af,g⟩=⟨f,E_{\ast }{A}^{\ast }g⟩$, so $g\in D(S^{\ast })$ and $S^{\ast }g=E_{\ast }{A}^{\ast }g$. Therefore $D(A^{\ast })=D(S^{\ast })$ and $A^{\ast }=E_{\ast }^{-1}{S}^{\ast }$. Since $A^{\ast }\mp iI=E_{\ast }^{-1}(S^{\ast }\mp i{E}_{\ast }I)$, the ordinary deficiency spaces of $A$ at $\pm i$ are exactly $N_{\pm }$.

Suppose $n_{+}=n_{-}$. Choose orthonormal bases $(e_j{)}_{j\in J_{+}}$ of $N_{+}$ and $(f_j{)}_{j\in J_{-}}$ of $N_{-}$. By [F6], choose a bijection $\pi :J_{+}\to J_{-}$. The map $V_0{e}_j=f_{\pi (j)}$ preserves norms on finite linear combinations by orthonormality, so it extends continuously to an isometry $V:N_{+}\to N_{-}$. Its range contains the finite span of the $f_j$, which is dense; the range of an isometry from the complete space $N_{+}$ is closed, so $V$ is onto and unitary.

For $f\in D(A)$, symmetry makes $⟨Af,f⟩$ real; expanding the two norms gives $\mathrm{\parallel }(A\pm iI)f{\mathrm{\parallel }}^2=\mathrm{\parallel }Af{\mathrm{\parallel }}^2+\mathrm{\parallel }f{\mathrm{\parallel }}^2$. Thus $A\pm iI$ are injective. If $(A\pm iI)f_k$ converges, the norm identity applied to differences makes $(f_k)$ Cauchy and then $(A{f}_k)$ Cauchy; closedness of $A$ shows the limit is in the range. Hence $M_{\pm }:=\mathrm{ran}(A\pm iI)$ are closed. By [F2] and [F3], $H=M_{+}\oplus N_{+}=M_{-}\oplus N_{-}$, with orthogonal sums.

Equip $D(A^{\ast })$ with the graph inner product $⟨f,g{⟩}_{A^{\ast }}=⟨f,g⟩+⟨A^{\ast }f,A^{\ast }g⟩$. It is complete by [F2, F4], and $D(A)$ is a closed subspace because $A$ is closed and $A^{\ast }f=Af$ there. If $z\in D(A^{\ast })$ is graph-orthogonal to $D(A)$, then for every $f\in D(A)$, $⟨Af,A^{\ast }z⟩=-⟨f,z⟩$. By the adjoint definition, this means $A^{\ast }z\in D(A^{\ast })$ and $A^{\ast 2}z=-z$. Therefore $z_{+}=(z-i{A}^{\ast }z)\mathrm{/}2$ and $z_{-}=(z+i{A}^{\ast }z)\mathrm{/}2$ satisfy $A^{\ast }{z}_{+}=i{z}_{+}$ and $A^{\ast }{z}_{-}=-i{z}_{-}$, and $z=z_{+}+z_{-}$. Conversely, $D(A)$ is graph-orthogonal to both kernels: for $u\in N_{+}$, $⟨f,u⟩+⟨Af,A^{\ast }u⟩=⟨f,u⟩+⟨Af,iu⟩=0$ by the adjoint identity and conjugate-linearity in the second argument, and the same computation with $A^{\ast }u=-iu$ holds for $u\in N_{-}$. If $u\in N_{+}$ and $v\in N_{-}$, then $⟨u,v⟩+⟨A^{\ast }u,A^{\ast }v⟩=⟨u,v⟩+⟨iu,-iv⟩=0$. Thus the two kernels are graph-orthogonal to each other. Applying [F3] in the graph Hilbert space proves the orthogonal decomposition $D(A^{\ast })=D(A)\oplus N_{+}\oplus N_{-}$.

Define $C_A:M_{+}\to M_{-}$ by $C_A((A+iI)f)=(A-iI)f$. Injectivity of $A+iI$ makes this well defined, the norm identity in step 2.1 makes it an isometry, and it is onto $M_{-}$. Define $W(m+u)=C_{A}m-Vu$ for $m\in M_{+}$ and $u\in N_{+}$. The decompositions from step 2.1 make $W$ well defined; it is unitary because its two orthogonal summands are mapped isometrically onto $M_{-}$ and $N_{-}$. For $f\in D(A)$, $(I-W)(A+iI)f=2if$, so $\mathrm{ran}(I-W)$ is dense. If $Wz=z$, then $z$ is orthogonal to $\mathrm{ran}(I-W)$ by unitarity, hence to the dense domain $D(A)$, so $z=0$. The Cayley correspondence [F5] therefore gives a self-adjoint operator $A_V$ with Cayley transform $W$.

For $m=(A+iI)f$ and $u\in N_{+}$, direct calculation gives $(I-W)(m+u)=2if+u+Vu$ and $i(I+W)(m+u)=2iAf+iu-iVu$. The Cayley inverse is $A_V=i(I+W)(I-W{)}^{-1}$ on $\mathrm{ran}(I-W)$, so $D(A_V)=\mathrm{ran}(I-W)=D(A)+\{u+Vu:u\in N_{+}\}$ and $A_V(2if+u+Vu)=2iAf+iu-iVu$. Equivalently, after renaming $2if$ as $f$, $A_V(f+u+Vu)=Af+iu-iVu$. The sum is direct: if $f=u+Vu\in D(A)$, then $(I-W)(A+iI)f=2if=(I-W)(2iu)$; injectivity of $I-W$ gives $(A+iI)f=2iu$. Its left side lies in $M_{+}$ and its right side in $N_{+}$, whose intersection is zero, so $u=f=0$. Multiplication by $E_{\ast }$ gives the displayed domain and action for $S_V=E_{\ast }{A}_V$. The two-way scalar-adjoint calculation in step 1.1 applies to $A_V$ as well, so $S_V^{\ast }=E_{\ast }{A}_V^{\ast }=S_V$. Thus $S_V$ is a self-adjoint extension of $S$.

Conversely, let $T$ be any self-adjoint extension of $S$ and put $R=E_{\ast }^{-1}T$. The two-way scalar-adjoint calculation in step 1.1 gives $R^{\ast }=E_{\ast }^{-1}{T}^{\ast }=R$, so $R$ is a self-adjoint extension of $A$. Its Cayley transform $C_R$ is unitary by [F5] and agrees with $C_A$ on $M_{+}$, since $R$ equals $A$ on $D(A)$. A unitary carrying $M_{+}$ onto $M_{-}$ carries their orthogonal complements $N_{+}$ onto $N_{-}$ by step 2.1 and [F2]; hence $C_R(N_{+})=N_{-}$. Put $V=-C_R{\mathrm{\mid }}_{N_{+}}$; this is unitary. The unitary $W$ constructed from this $V$ agrees with $C_R$ on both orthogonal summands of $H$, so uniqueness in [F5] gives $R=A_V$. This determines $V$ uniquely from $T$ and proves that every self-adjoint extension of $S$ is uniquely $S_V$.

If $n_{+}=n_{-}$, step 1.2 supplies a unitary and step 4.1 supplies a self-adjoint extension. If an extension exists, step 4.2 supplies a unitary $N_{+}\to N_{-}$, so its orthonormal bases have bijective index sets and $n_{+}=n_{-}$. This proves the existence equivalence and the complete parameterization.

Let $T$ be any linear operator with $T_{\min }\subseteq T\subseteq T_{\max }$. For this proof write ${\gamma }_{b}f=({\mathrm{\Gamma }}_{0}f,{\mathrm{\Gamma }}_{1}f)\in {\mathbb{C}}^2\times {\mathbb{C}}^2$. This is the four-coordinate trace map of [F1] with the last coordinate sign reversed, so it is onto and has kernel $D(T_{\min })$. Its image $M:={\gamma }_b(D(T))$ is a linear subspace. Since $\ker {\gamma }_b=D(T_{\min })\subseteq D(T)$, every $f\in H^2$ whose trace lies in $M$ differs from a member of $D(T)$ by an element of $D(T_{\min })$; thus $D(T)={\gamma }_b^{-1}(M)$. Conversely, every linear subspace $M$ gives such a domain because ${\gamma }_b$ is onto.

Define $B((a,b),(a^{\mathrm{\prime }},b^{\mathrm{\prime }}))=⟨b,a^{\mathrm{\prime }}⟩-⟨a,b^{\mathrm{\prime }}⟩$ and $M^{\omega }=\{z:B(m,z)=0\text{ for every }m\in M\}$. By [F2], integration by parts gives $T^{\ast }=T_{{\gamma }_b^{-1}(M^{\omega })}$: indeed, since $T_{\min }\subseteq T$, adjoint inclusion reversal puts $D(T^{\ast })$ inside $D(T_{\min }^{})=H^2$, and for $g\in H^2$ the adjoint identity for every $f\in D(T)$ is exactly $B({\gamma }_{b}f,{\gamma }_{b}g)=0$. Hence $T$ is self-adjoint if and only if $M=M^{\omega }$.

Put ${\gamma }_{\mathrm{\ell }}f=(a,b):=({\mathrm{\Gamma }}_{0}f,\mathrm{\ell }{\mathrm{\Gamma }}_{1}f)$ and let $\widehat{M}={\gamma }_{\mathrm{\ell }}(D(T))$. Equivalently, $\widehat{M}=\{(a,b):(a,b\mathrm{/}\mathrm{\ell })\in M\}$, where $M$ is the unscaled trace image from step 1.1. Define $\widehat{B}((a,b),(a^{\mathrm{\prime }},b^{\mathrm{\prime }}))=⟨b,a^{\mathrm{\prime }}⟩-⟨a,b^{\mathrm{\prime }}⟩$ and ${\widehat{M}}^{\widehat{\omega }}:=\{z:\widehat{B}(m,z)=0\text{ for all }m\in \widehat{M}\}$. Since $\widehat{B}=\mathrm{\ell }B$ after undoing the derivative scaling and $\mathrm{\ell }>0$, step 1.2 gives $\widehat{M}={\widehat{M}}^{\widehat{\omega }}$ exactly when $T$ is self-adjoint. The map ${\gamma }_{\mathrm{\ell }}$ is onto because the trace map in [F1] is onto and multiplying its derivative coordinates by $\mathrm{\ell }>0$ is invertible. For $x=a+ib$ and $y=a-ib$, direct expansion using the first-variable-linear inner product gives $\widehat{B}((a,b),(a^{\mathrm{\prime }},b^{\mathrm{\prime }}))=\frac{⟨x,x^{\mathrm{\prime }}⟩-⟨y,y^{\mathrm{\prime }}⟩}{2i}\mathrm{.}$ Thus $\widehat{M}={\widehat{M}}^{\widehat{\omega }}$ implies that $(a,b)\mapsto (x,y)$ sends $\widehat{M}$ to the graph of an inner-product-preserving linear map $R:X\to {\mathbb{C}}^2$, where $X=\{a+ib:(a,b)\in \widehat{M}\}$. It is well defined: if a member of $\widehat{M}$ has $x=0$, isotropy gives $\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2=0$, so $y=0$. Maximality forces $X={\mathbb{C}}^2$. Indeed, define $X^{\perp }=\{z:⟨x,z⟩=0\text{ for every }x\in X\}$ with $⟨x,z⟩=x_1\overline{z_1}+x_2\overline{z_2}$. Every proper complex linear subspace $X$ has a nonzero vector in $X^{\perp }$, by the following coordinate argument. If $X=\{0\}$, take $x_0=(1,0)$. Otherwise take $0\mathrm{\ne }v=(v_1,v_2)\in X$. Every $w\in X$ is a multiple of $v$: if $v_1{w}_2-v_2{w}_1\mathrm{\ne }0$, the inverse of the two-column matrix $(v,w)$ expresses both standard coordinate vectors as linear combinations of $v,w$, implying $X={\mathbb{C}}^2$; if this determinant is zero, solving in a nonzero coordinate of $v$ gives $w=cv$. Thus for proper $X$ one has $X=\mathrm{span}\{v\}$, and $x_0=(-\overline{v_2},\overline{v_1})\mathrm{\ne }0$ satisfies $⟨cv,x_0⟩=c(-v_1{v}_2+v_2{v}_1)=0$. No closed-subspace decomposition or additional choice principle is used. If $X$ were proper, take this $x_0$ and set $a=x_0\mathrm{/}2$, $b=x_0\mathrm{/}(2i)$, so $(x,y)=(x_0,0)$. The displayed identity makes this boundary pair orthogonal for $\widehat{B}$ to all of $\widehat{M}$, hence it lies in ${\widehat{M}}^{\widehat{\omega }}=\widehat{M}$; then $x_0\in X\cap X^{\perp }$, a contradiction. Therefore $R$ is a linear isometry from ${\mathbb{C}}^2$ into ${\mathbb{C}}^2$. Let $e_1,e_2$ be the standard basis and write $R{e}_1=({\alpha }_1,{\beta }_1)$ and $R{e}_2=({\alpha }_2,{\beta }_2)$. Isometry gives $\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\beta }_1{\mathrm{\mid }}^2=\mathrm{\mid }{\alpha }_2{\mathrm{\mid }}^2+\mathrm{\mid }{\beta }_2{\mathrm{\mid }}^2=1,{\alpha }_1\overline{{\alpha }_2}+{\beta }_1\overline{{\beta }_2}=0.$ Direct expansion gives $\mathrm{\mid }{\alpha }_1{\beta }_2-{\alpha }_2{\beta }_1{\mathrm{\mid }}^2=(\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\beta }_1{\mathrm{\mid }}^2)(\mathrm{\mid }{\alpha }_2{\mathrm{\mid }}^2+\mathrm{\mid }{\beta }_2{\mathrm{\mid }}^2)-\mathrm{\mid }{\alpha }_1\overline{{\alpha }_2}+{\beta }_1\overline{{\beta }_2}{\mathrm{\mid }}^2=1.$ Thus the matrix $U$ with columns $R{e}_1,R{e}_2$ is invertible, so these vectors span ${\mathbb{C}}^2$ and $R$ is onto. Their orthonormality gives $U^{\ast }U=I_2$; since $U$ is invertible, this implies $U^{\ast }=U^{-1}$ and hence $U{U}^{\ast }=I_2$. Thus $U\in U(2)$, and its graph condition is $U(a+ib)=a-ib$, equivalently $(U-I_2)a+i(U+I_2)b=0$. This $U$ is unique because its value is fixed for every $x\in {\mathbb{C}}^2$.

Conversely, let $U\in U(2)$ and set ${\widehat{M}}_U=\{(a,b):U(a+ib)=a-ib\}$ in the scaled boundary coordinates. Every $x\in {\mathbb{C}}^2$ occurs, by taking $a=(x+Ux)\mathrm{/}2$ and $b=(x-Ux)\mathrm{/}(2i)$. Unitarity gives $⟨x,x^{\mathrm{\prime }}⟩=⟨Ux,U{x}^{\mathrm{\prime }}⟩$ for every pair, so the displayed formula in step 2.1 makes ${\widehat{M}}_U$ isotropic. If $(a,b)\in {\widehat{M}}_U^{\widehat{\omega }}$, then for every $x^{\mathrm{\prime }}\in {\mathbb{C}}^2$ that formula gives $⟨x^{\mathrm{\prime }},x⟩=⟨U{x}^{\mathrm{\prime }},y⟩=⟨x^{\mathrm{\prime }},U^{\ast }y⟩$, hence $x=U^{\ast }y$ and $y=Ux$; therefore $(a,b)\in {\widehat{M}}_U$. Thus ${\widehat{M}}_U={\widehat{M}}_U^{\widehat{\omega }}$. Since ${\gamma }_{\mathrm{\ell }}$ is onto, the restriction to ${\gamma }_{\mathrm{\ell }}^{-1}({\widehat{M}}_U)$ is self-adjoint by the criterion in step 1.2 and the scaling equivalence in step 2.1. Its boundary equation is exactly the Statement's equation, and its domain lies between $T_{\min }$ and $T_{\max }$ because it is a restriction of $T_{\max }$ whose domain contains the zero-trace domain.