All self-adjoint interval Laplacian boundary conditions

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.

The preceding steps give a bijection from $U(2)$ to the entire set of self-adjoint extensions between $T_{\min }$ and $T_{\max }$. By [F4], that same extension set is uniquely parameterized by the abstract unitaries between the deficiency subspaces. Composing the two bijections identifies the boundary-matrix labels with the abstract deficiency labels and proves that the displayed boundary family is exhaustive.