Support of an integrated finite trajectory family

Each $E_a$ is closed and bounded by its defining inequalities. Its strict inequalities have a solution, for example $s_i={\alpha }_a+i({\beta }_a-{\alpha }_a)\mathrm{/}(n_a+1)$, so it is nonempty and [F4] makes it compact. Its closedness also makes it Lebesgue measurable by [F3], so [F9] supplies the restricted measure ${\mu }_a$. For a continuous real scalar function on $E_a$, continuity in [F8] makes the inverse image of an open set relatively open: it is the union of relatively open neighbourhoods mapped into that set. Thus it has the form $E_a\cap O$ for some ambient open set $O$. Since $E_a$ is closed, these inverse images are Borel, hence Lebesgue measurable by [F3]. Thus its extension by zero off $E_a$ is measurable (an open set containing zero also includes the measurable complement of $E_a$ in its inverse image). By [F4] it is bounded. Apply this argument to the real and imaginary parts of each entry of $z(K_a(s))z(K_a(s){)}^{†}$ and, for fixed $X$, of $K_a(s)X{K}_a(s{)}^{†}$, and to their moduli. All are continuous on $E_a$. A bounded modulus is at most some finite constant $M$ there, so [F5] and [F9] bound its integral by $M{\mu }_a({\mathbb{R}}^{n_a})=M{\lambda }_{n_a}(E_a)<\mathrm{\infty }$, using [F3]. This proves every asserted entrywise integral exists.

Entrywise complex conjugation commutes with integration by the real-and-imaginary-part definition in [F5], so $Q=Q^{†}$. For $v\in H\otimes K$, finite-sum linearity gives $v^{†}Qv={\sum }_b\mathrm{\mid }{v}^{†}z(D_b){\mathrm{\mid }}^2+{\sum }_a{\int }_{E_a}\mathrm{\mid }{v}^{†}z(K_a(s)){\mathrm{\mid }}^{2}d{\lambda }_{n_a}(s)$. These terms are finite by step 1.1 and nonnegative by [F5]. Hence $Q$ is positive semidefinite.

If the continuous scalar $g(s)=v^{†}z(K_a(s))$ is nonzero at an interior point $s_0$, [F8] applied to the open disc of radius $\mathrm{\mid }g(s_0)\mathrm{\mid }\mathrm{/}2$ about $g(s_0)$ gives a relatively open neighbourhood of $s_0$ on which $\mathrm{\mid }g(s)\mathrm{\mid }\ge \mathrm{\mid }g(s_0)\mathrm{\mid }\mathrm{/}2>0$. Interiority lets us choose an ambient ball $B(s_0,r)\subseteq E_a$ inside that neighbourhood. On that ball [F9] gives ${\mu }_a(B(s_0,r))={\lambda }_{n_a}(B(s_0,r))$. By [F3] and [F5], its contribution to the integral in step 2.1 is at least $\mathrm{\mid }g(s_0){\mathrm{\mid }}^2{\lambda }_{n_a}(B(s_0,r))\mathrm{/}4>0$. Thus a zero quadratic form forces orthogonality to every isolated vector and every interior trajectory vector. Conversely, orthogonality to these vectors extends to every boundary trajectory vector by continuity: fix a strict interior point $c$ from step 1.1 and use $(1-\epsilon )s+\epsilon c$ for $0<\epsilon <1$, which is interior and tends to any given $s\in E_a$ as $\epsilon \downarrow 0$. Indeed its distance to $s$ is $\epsilon \mathrm{\parallel }c-s\mathrm{\parallel }$, so the metric continuity condition in [F8] makes $g((1-\epsilon )s+\epsilon c)$ tend to $g(s)$. All integrands and isolated terms then vanish.

Let $W$ be the complex span in $H\otimes K$ of the isolated vectors and interior trajectory vectors. Orthogonality to its generators is equivalent to membership in $W^{\perp }$, since the vectors orthogonal to any fixed $v$ form a linear subspace. The boundary-limit argument in step 3.1 shows every boundary vector is orthogonal to $W^{\perp }$, hence belongs to $W^{\perp \perp }=W$ by [F2]. Thus $W$ is also the span of all the vectorized generators. In a spectral decomposition of $Q$ from [F6], the quadratic form is a sum of nonnegative eigenvalues times squared coordinates, so it vanishes exactly on $\ker Q$. Steps 2.1 and 3.1 therefore give $\ker Q=W^{\perp }$, and $\mathrm{supp}Q=(\ker Q{)}^{\perp }=W$. The positive eigenvectors also span the image of $Q$, so its rank is $\dim W$. Finally $z(U_{ji})=e_i\otimes f_j$, so [F1] makes $z$ a linear bijection from the matrix space to $H\otimes K$; consequently $W=z(S)$ and $\dim W={\dim }_{\mathbb{C}}S$. These arguments include the zero subspace.

The span of a family supported in a coordinate set remains supported in that set, because these support conditions define a linear subspace. If the sets for distinct families are disjoint and a sum of one vector from each span is zero, restriction to the coordinates of any one family forces its vector to be zero. The sum of the spans is therefore direct. Each is finite-dimensional as a subspace of the common finite matrix space, and [F2] gives the asserted sum of dimensions, including empty families. In vector coordinates these subspaces are also orthogonal by [F1].

The entrywise integrability in step 1.1 and linearity in [F5] show that $\mathrm{\Phi }$ is a complex-linear map. For an arbitrary $A:H\to K$, expand [F7]: $J(X\mapsto AX{A}^{†})={\sum }_{i,l}\mathrm{\mid }{e}_i⟩⟨e_l\mathrm{\mid }\otimes \mathrm{\mid }A{e}_i⟩⟨A{e}_l\mathrm{\mid }=z(A)z(A{)}^{†}$. This calculation uses only the common bases and finite tensor coordinates in [F1]. Apply it to each isolated matrix and to $K_a(s)$ at each $s$. The finite Choi sum passes through each entrywise integral by [F5], giving $J(\mathrm{\Phi })={\sum }_{b}z(D_b)z(D_b{)}^{†}+{\sum }_a{\int }_{E_a}z(K_a(s))z(K_a(s){)}^{†}d{\lambda }_{n_a}(s)=Q$. Thus the Choi assertion holds for the stated linear map without a trace-preservation assumption.