Self consistent projector iterations and certified mixing

The vector $e_1=(1,0,\dots ,0)$ belongs to ${\mathcal{C}}_{s,C}$ because its entries sum to $1$ and its ${\mathrm{\ell }}^1$ norm is $1\le C$. Thus the coefficient set is nonempty. For $c,d\in {\mathbb{R}}^s$, each $\mathrm{\mid }{c}_i-d_i\mathrm{\mid }\le \mathrm{\parallel }c-d{\mathrm{\parallel }}_2$, and [F3] gives $\mid \mathrm{\mid }{c}_i\mathrm{\mid }-\mathrm{\mid }{d}_i\mathrm{\mid }\mid \le \mathrm{\mid }{c}_i-d_i\mathrm{\mid }$; hence both $c\mapsto {\sum }_i{c}_i$ and $c\mapsto {\sum }_i\mathrm{\mid }{c}_i\mathrm{\mid }$ are continuous. Their closed level sets show that ${\mathcal{C}}_{s,C}$ is closed. It is bounded because $\mathrm{\parallel }c{\mathrm{\parallel }}_1^2={\sum }_i{c}_i^2+2{\sum }_{i<j}\mathrm{\mid }{c}_i\mathrm{\mid }\mathrm{\mid }{c}_j\mathrm{\mid }\ge \mathrm{\parallel }c{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }c{\mathrm{\parallel }}_1\le C$. Therefore [F2] makes ${\mathcal{C}}_{s,C}$ compact.

Put $T(c)={\sum }_i{c}_i{R}_i$. By the triangle inequality and homogeneity of the Frobenius norm, $\mathrm{\parallel }T(c)-T(d){\mathrm{\parallel }}_F\le {\sum }_i\mathrm{\mid }{c}_i-d_i\mathrm{\mid }\mathrm{\parallel }{R}_i{\mathrm{\parallel }}_F\le \mathrm{\parallel }c-d{\mathrm{\parallel }}_2{\sum }_i\mathrm{\parallel }{R}_i{\mathrm{\parallel }}_F$, so $T$ is continuous; the reverse triangle inequality then makes $c\mapsto J(c)=\mathrm{\parallel }T(c){\mathrm{\parallel }}_F^2$ continuous. By [F2], $J$ attains a minimum $m$ on the nonempty compact set ${\mathcal{C}}_{s,C}$. The set $M={\mathcal{C}}_{s,C}\cap J^{-1}(\{m\})$ is closed and bounded in ${\mathbb{R}}^s$, hence compact by [F2], and it is nonempty by attainment.

For $c,d\in {\mathcal{C}}_{s,C}$ and $0\le t\le 1$, the coefficients $e=(1-t)c+td$ satisfy ${\sum }_i{e}_i=(1-t){\sum }_i{c}_i+t{\sum }_i{d}_i=1$. By [F3], ${\sum }_i\mathrm{\mid }{e}_i\mathrm{\mid }\le (1-t){\sum }_i\mathrm{\mid }{c}_i\mathrm{\mid }+t{\sum }_i\mathrm{\mid }{d}_i\mathrm{\mid }\le C$, so ${\mathcal{C}}_{s,C}$ is convex. Expanding each complex matrix entry gives $\mathrm{\parallel }(1-t)A+tB{\mathrm{\parallel }}_F^2=(1-t)\mathrm{\parallel }A{\mathrm{\parallel }}_F^2+t\mathrm{\parallel }B{\mathrm{\parallel }}_F^2-t(1-t)\mathrm{\parallel }A-B{\mathrm{\parallel }}_F^2$ for real $t$. Applying this identity to $A=T(c)$ and $B=T(d)$ shows $J((1-t)c+td)\le (1-t)J(c)+tJ(d)=m$ when $c,d\in M$; since the combination is feasible and $m$ is the minimum, it also lies in $M$. Thus $M$ is convex. The function $c\mapsto \mathrm{\parallel }c{\mathrm{\parallel }}_2^2={\sum }_i{c}_i^2$ is continuous, so compactness of $M$ gives a minimizer $c_{\ast }\in M$.

If distinct $c,d\in M$ both had minimum Euclidean norm, their midpoint would lie in $M$ and satisfy ${\parallel \frac{c+d}{2}\parallel }_2^2=\frac{\mathrm{\parallel }c{\mathrm{\parallel }}_2^2+\mathrm{\parallel }d{\mathrm{\parallel }}_2^2}{2}-\frac{\mathrm{\parallel }c-d{\mathrm{\parallel }}_2^2}{4}<\frac{\mathrm{\parallel }c{\mathrm{\parallel }}_2^2+\mathrm{\parallel }d{\mathrm{\parallel }}_2^2}{2},$ contradicting minimality. Thus $c_{\ast }$ is the unique minimum-norm least-squares minimizer. No nonsingularity of the Gram matrix is needed for existence or tie-breaking.

Since ${\lambda }_{r+1}(H_0)={\lambda }_r(H_0)+g$, we have $R={\lambda }_r(H_0)-{\lambda }_1(H_0)+g\mathrm{/}2\ge g\mathrm{/}2>0$. Reversing the descending-index convention in [F1] gives the same perturbation bound for the ascending eigenvalues used here. Thus if $\mathrm{\parallel }H-H_0{\mathrm{\parallel }}_{op}\le g\mathrm{/}4$, then ${\lambda }_1(H)\ge c-g\mathrm{/}4$ and ${\lambda }_r(H)\le c+R-g\mathrm{/}4$. Since $R-g\mathrm{/}4\ge g\mathrm{/}4$, every first-$r$ eigenvalue therefore satisfies $\mathrm{\mid }{\lambda }_j(H)-c\mathrm{\mid }\le R-g\mathrm{/}4$, so it lies at least $g\mathrm{/}4$ inside the circle centered at $c$.

The same eigenvalue estimate gives ${\lambda }_{r+1}(H)\ge c+R+g\mathrm{/}4$, so all remaining eigenvalues satisfy $\mathrm{\mid }{\lambda }_j(H)-c\mathrm{\mid }\ge R+g\mathrm{/}4$ and lie at least $g\mathrm{/}4$ outside the circle. Also ${\lambda }_{r+1}(H)-{\lambda }_r(H)\ge g-2(g\mathrm{/}4)=g\mathrm{/}2$. Thus the contour separates exactly the occupied cluster and the stated gap persists.

By [F2], the spectrum of $H$ is real and $H$ diagonalizes unitarily. For every $z\in \mathrm{\Gamma }$, reverse triangle gives $\mathrm{\mid }z-{\lambda }_j(H)\mathrm{\mid }\ge \mathrm{\mid }\mathrm{\mid }z-c\mathrm{\mid }-\mathrm{\mid }{\lambda }_j(H)-c\mathrm{\mid }\mathrm{\mid }\ge g\mathrm{/}4$ by steps 1.1 and 2.1. In the unitary eigenbasis, $(zI-H{)}^{-1}$ is diagonal with entries of modulus at most $4\mathrm{/}g$. Applying [F3] to a unit vector in this basis gives $\mathrm{\parallel }(zI-H{)}^{-1}{\mathrm{\parallel }}_{op}\le 4\mathrm{/}g$.

By [F1], choose an orthonormal eigenbasis $(u_1,\dots ,u_n)$ of $F(X)$ with the eigenvalues in the given order. The strict gap prevents an eigenspace for one eigenvalue from straddling positions $r$ and $r+1$, so the sum of the first $r$ eigenspaces is the span of exactly the first $r$ basis vectors.

Changing the orthonormal basis within a repeated eigenspace below the gap leaves that entire eigenspace in the sum; changing one within a repeated eigenspace above the gap leaves that entire eigenspace out. Thus the selected subspace does not depend on the eigenbasis chosen in step 1.1. Its orthogonal projector is $P={\sum }_{j=1}^r{u}_j{u}_j^{\ast }$, which satisfies $P=P^{\ast }$ and $P^2=P$ by orthonormality. By [F3], $\mathrm{tr}P={\sum }_{a=1}^n{P}_{aa}={\sum }_{a=1}^n{\sum }_{j=1}^r(u_j{)}_a\overline{(u_j{)}_a}={\sum }_{j=1}^r{\sum }_{a=1}^n\mathrm{\mid }(u_j{)}_a{\mathrm{\mid }}^2={\sum }_{j=1}^{r}1=r$, since each $u_j$ is a unit vector.

The operator in step 2.1 is independent of all allowed basis choices, lies in ${\mathcal{P}}_r$ by [L1], and is exactly the map specified by [F2]. Hence it is the unique rank-$r$ occupied spectral projector.

By [F2], choose a unitary eigenbasis of $P$ with real eigenvalues ${\mu }_j$. From $P^2=P$, each satisfies ${\mu }_j^2={\mu }_j$, hence ${\mu }_j\in \{0,1\}$. Therefore every eigenvalue lies between $0$ and $1$, so for each $x$ the quadratic forms $x^{\ast }Px$ and $x^{\ast }(I-P)x$ are nonnegative.

The matrix $P$ is Hermitian and has trace $r$ by [F1]; step 1.1 gives $0⪯P⪯I$. These are exactly the conditions in [F3], so $P\in {\mathcal{D}}_r$.

The diagonal matrix $D_0=\mathrm{diag}(1,\dots ,1,0,\dots ,0)$ with $r$ ones is Hermitian, satisfies $0⪯D_0⪯I_n$, and has trace $r$. Thus $D_0\in {\mathcal{D}}_r$, so the domain is nonempty.

Identify ${\mathbb{H}}_n$ with ${\mathbb{R}}^{n^2}$ by listing the real diagonal entries and, for each $i<j$, the coordinates $\sqrt{2}\mathrm{Re}{D}_{ij}$ and $\sqrt{2}\mathrm{Im}{D}_{ij}$. This real-linear map is bijective, since the lower-triangular entries are recovered by conjugation, and its squared Euclidean norm is ${\sum }_i{D}_{ii}^2+2{\sum }_{i<j}\mathrm{\mid }{D}_{ij}{\mathrm{\mid }}^2=\mathrm{\parallel }D{\mathrm{\parallel }}_F^2$. Thus it is an isometry. If $(D_k)$ is Frobenius-Cauchy, its image is Cauchy in ${\mathbb{R}}^{n^2}$, so [F2] gives a limit whose inverse image $D\in {\mathbb{H}}_n$ satisfies $D_k\to D$ in Frobenius norm.

For each fixed $x\in {\mathbb{C}}^n$, $\mathrm{\mid }{x}^{\ast }(D_k-D)x\mathrm{\mid }\le ({\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }{)}^2\mathrm{\parallel }{D}_k-D{\mathrm{\parallel }}_F$, since each matrix entry is bounded by the Frobenius norm. The same estimate applies to $x^{\ast }((I_n-D_k)-(I_n-D))x$. Both quadratic forms therefore converge; their terms are nonnegative, so their limits are nonnegative and $0⪯D⪯I_n$. Also $\mathrm{\mid }\mathrm{tr}(D_k-D)\mathrm{\mid }\le n\mathrm{\parallel }{D}_k-D{\mathrm{\parallel }}_F$, so $\mathrm{tr}D={\lim }_k\mathrm{tr}{D}_k=r$. Hence $D\in {\mathcal{D}}_r$.

Steps 1.2 and 2.1 show that every Frobenius-Cauchy sequence in ${\mathcal{D}}_r$ converges to a member of ${\mathcal{D}}_r$, so the domain is complete. To prove closedness directly, for each fixed $x\in {\mathbb{C}}^n$ the real-valued maps $D\mapsto x^{\ast }Dx$ and $D\mapsto x^{\ast }(I_n-D)x$ are continuous, since their differences at $D,E$ have modulus at most $({\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }{)}^2\mathrm{\parallel }D-E{\mathrm{\parallel }}_F$. The trace is continuous as $\mathrm{\mid }\mathrm{tr}(D-E)\mathrm{\mid }\le n\mathrm{\parallel }D-E{\mathrm{\parallel }}_F$. Thus ${\mathcal{D}}_r$ is the intersection of the closed preimages of $[0,\mathrm{\infty })$ under these quadratic-form maps and of $\{r\}$ under the trace map. An arbitrary intersection of closed sets is closed.

If $D_1,D_2\in {\mathcal{D}}_r$ and $t\in [0,1]$, then $t{D}_1+(1-t)D_2$ is Hermitian; its quadratic form and that of $I_n-t{D}_1-(1-t)D_2$ are convex combinations of nonnegative quadratic forms, and its trace is $t\mathrm{tr}{D}_1+(1-t)\mathrm{tr}{D}_2=tr+(1-t)r=r$. Hence it lies in ${\mathcal{D}}_r$, proving convexity.

Since $\mathrm{\parallel }A{\mathrm{\parallel }}_{op}\le \mathrm{\parallel }A{\mathrm{\parallel }}_F$ by [F6], every point of the operator-norm ball has a Frobenius neighborhood contained in it; hence the stated domain is open for the real Frechet definition. Diagonalize $H=U\mathrm{diag}({\mu }_1,\dots ,{\mu }_n)U^{\ast }$ with ${\mu }_1\le \cdots \le {\mu }_n$ by [F2]. Since [F1] places exactly ${\mu }_1,\dots ,{\mu }_r$ inside $\mathrm{\Gamma }$, the resolvent is $U\mathrm{diag}((z-{\mu }_j{)}^{-1})U^{\ast }$ on the contour.

For a sufficiently small Hermitian perturbation $E$ with $H+E$ still in the ball, the inverse identity gives $R_{H+E}=R_H+R_{H}E{R}_{H+E}$. Subtracting $R_H+R_{H}E{R}_H$ yields $R_{H+E}-R_H-R_{H}E{R}_H=R_{H}E{R}_{H}E{R}_{H+E}\mathrm{.}$ The identity follows by multiplying the two inverse equations for $zI-H$ and $zI-H-E$; all inverses exist by [F1].

Interpret a matrix-valued contour integral entrywise; each entry of the resolvent and derivative integrands is continuous on $\mathrm{\Gamma }$, so its scalar integral exists by [F4]. Finite matrix sums and constant left and right factors commute with this integral, and the Frobenius norm of an integral is at most the contour length times the supremum Frobenius norm of its integrand, by the triangle inequality for finite-dimensional Riemann sums. Integrating each diagonal entry and using [F3], linearity in [F4], and the finite spectral expansion gives $\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{R}_H(z)dz=U\mathrm{diag}(1,\dots ,1,0,\dots ,0)U^{\ast }={\mathrm{\Pi }}_r(H)\mathrm{.}$ Thus the contour expression represents the full occupied cluster projector, including repeated eigenvalues within that cluster.

For any matrix $A$ and matrix $X$ with columns $x_j$, the operator-norm definition gives $\mathrm{\parallel }AX{\mathrm{\parallel }}_F^2={\sum }_j\mathrm{\parallel }A{x}_j{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }A{\mathrm{\parallel }}_{op}^2\mathrm{\parallel }X{\mathrm{\parallel }}_F^2$. Applying this also to adjoints gives the right-multiplication bound $\mathrm{\parallel }XB{\mathrm{\parallel }}_F\le \mathrm{\parallel }X{\mathrm{\parallel }}_F\mathrm{\parallel }B{\mathrm{\parallel }}_{op}$. Iterating these inequalities, then using [F1] and [F6], bounds the Frobenius norm of the right side by $(4\mathrm{/}g{)}^3\mathrm{\parallel }E{\mathrm{\parallel }}_F^2$, uniformly for $z\in \mathrm{\Gamma }$. The matrix-valued integral estimate in step 2.1 bounds the integrated remainder by $\mathrm{\ell }(\mathrm{\Gamma })(4\mathrm{/}g{)}^3\mathrm{\parallel }E{\mathrm{\parallel }}_F^2\mathrm{/}(2\pi )$. Dividing by $\mathrm{\parallel }E{\mathrm{\parallel }}_F$ makes this tend to zero, so the contour formula in step 2.1 is Frechet differentiable in the sense of [F5], with the stated derivative; the integral is real-linear in $E$ by entrywise linearity. Each directional derivative is a limit of Hermitian difference quotients, so it is Hermitian-valued because ${\mathbb{H}}_n$ is closed.

The same left/right multiplication inequalities give $\mathrm{\parallel }{R}_H(z)E{R}_H(z){\mathrm{\parallel }}_F\le (4\mathrm{/}g{)}^2\mathrm{\parallel }E{\mathrm{\parallel }}_F$ by [F1] and [F6]. Integrating over length $2\pi R$ yields $\mathrm{\parallel }D{\mathrm{\Pi }}_r(H)[E]{\mathrm{\parallel }}_F\le R(4\mathrm{/}g{)}^2\mathrm{\parallel }E{\mathrm{\parallel }}_F=16R\mathrm{\parallel }E{\mathrm{\parallel }}_F\mathrm{/}{g}^2$. For nearby $H^{\mathrm{\prime }}$ in the ball, the resolvent identity and [F1] give uniformly on $\mathrm{\Gamma }$ $\mathrm{\parallel }{R}_{H^{\mathrm{\prime }}}-R_H{\mathrm{\parallel }}_{op}\le (4\mathrm{/}g{)}^2\mathrm{\parallel }{H}^{\mathrm{\prime }}-H{\mathrm{\parallel }}_{op}$. Subtracting the two derivative integrands gives $R_{H^{\mathrm{\prime }}}E(R_{H^{\mathrm{\prime }}}-R_H)+(R_{H^{\mathrm{\prime }}}-R_H)E{R}_H$. Each summand has Frobenius norm at most $(4\mathrm{/}g{)}^3\mathrm{\parallel }{H}^{\mathrm{\prime }}-H{\mathrm{\parallel }}_F\mathrm{\parallel }E{\mathrm{\parallel }}_F$ by the left/right multiplication bounds and [F6], so their sum is at most twice this quantity. The matrix-valued integral estimate from step 2.1 then bounds the difference of derivative values by $2R(4\mathrm{/}g{)}^3\mathrm{\parallel }{H}^{\mathrm{\prime }}-H{\mathrm{\parallel }}_F\mathrm{\parallel }E{\mathrm{\parallel }}_F$. Taking the supremum over $\mathrm{\parallel }E{\mathrm{\parallel }}_F=1$ shows that the derivative varies continuously in operator norm, which by [F7] proves the claimed $C^1$ regularity; the bound was proved above.

By [F1], there is a unique $X_{\ast }$ with $\mathrm{\Phi }(X_{\ast })=X_{\ast }$. In the projector application, [L1] supplies the stated map wherever the strict gap hypothesis holds.

For exact iterates, [F3] gives $d(X_{k+1},X_{\ast })=d(\mathrm{\Phi }(X_k),\mathrm{\Phi }(X_{\ast }))\le qd(X_k,X_{\ast })$. Induction from $k=0$ yields $d(X_k,X_{\ast })\le q^{k}d(X_0,X_{\ast })$.

For any $X\in K$, [F2] identifies the SCF metric residual with $d(X,\mathrm{\Phi }(X))$. The triangle inequality and contraction property give $d(X,X_{\ast })\le d(X,\mathrm{\Phi }(X))+d(\mathrm{\Phi }(X),\mathrm{\Phi }(X_{\ast }))\le d(X,\mathrm{\Phi }(X))+qd(X,X_{\ast })\mathrm{.}$ Since $1-q>0$, rearrangement proves the residual error bound.

For an inexact update, the triangle inequality gives $e_{k+1}=d(X_{k+1},X_{\ast })\le d(X_{k+1},\mathrm{\Phi }(X_k))+d(\mathrm{\Phi }(X_k),\mathrm{\Phi }(X_{\ast }))\le {\eta }_k+q{e}_k\mathrm{.}$ This is the one-step recurrence.

Repeated substitution in the recurrence yields $e_k\le q^k{e}_0+{\sum }_{j<k}{q}^{k-1-j}{\eta }_j$: it holds at $k=0$ with an empty sum, and substituting its $k$th bound into $e_{k+1}\le q{e}_k+{\eta }_k$ gives the formula at $k+1$.

Assume $PF=FP$. Then both $\mathrm{ran}P$ and $\mathrm{ran}(I-P)$ are invariant under $F$: if $x=Px$, then $P(Fx)=F(Px)=Fx$, and the analogous identity holds when $Px=0$. Relative to their orthogonal direct sum, $F$ is block diagonal and $F_s$ leaves the occupied block unchanged while adding $sI$ on the virtual block.

By [F2], the eigenvalues of $F_s$ are exactly those of $F_{occ}$ together with those of $F_{vir}$ shifted by $s$. If $\max \sigma (F_{occ})<\min \sigma (F_{vir})+s$, the lowest $r$ eigenvalues are precisely the occupied block and the gap is strict; [L1] gives ${\mathrm{\Pi }}_r(F_s)=P$.

Conversely, if ${\mathrm{\Pi }}_r(F_s)=P$, its defining strict gap makes $P$ the sum of the eigenspace projectors of $F_s$ for the lowest $r$ eigenvalues, so $P$ commutes with $F_s$. Since $[s(I-P),P]=0$, this gives $[F,P]=0$, and the two restrictions are the occupied and virtual blocks of $F$. Their eigenvalues in $F_s$ are respectively $\sigma (F_{occ})$ and $\sigma (F_{vir})+s$. Because $P$ is exactly the lowest-$r$ spectral projector with a strict gap, the largest occupied eigenvalue is strictly smaller than the smallest shifted virtual eigenvalue; hence the displayed ordering holds.

Since $D$ and $\mathrm{\Phi }(D)$ are Hermitian, $D^{+}$ is Hermitian. For every $x$, $x^{\ast }{D}^{+}x=(1-\alpha )x^{\ast }Dx+\alpha x^{\ast }\mathrm{\Phi }(D)x\ge 0$. Likewise $x^{\ast }(I-D^{+})x=(1-\alpha )x^{\ast }(I-D)x+\alpha x^{\ast }(I-\mathrm{\Phi }(D))x\ge 0$, since $D\in {\mathcal{D}}_r$ and $\mathrm{\Phi }(D)\in {\mathcal{D}}_r$ by [F1], [F2], [F4].

The trace is linear, so $\mathrm{tr}{D}^{+}=(1-\alpha )r+\alpha r=r$. The two positive-semidefinite inequalities and this trace identity give $D^{+}\in {\mathcal{D}}_r$.

To verify that idempotence need not be preserved, take $n=2$, $r=1$, the constant Hermitian response $F(X)=\mathrm{diag}(0,1)$ on $U={\mathbb{H}}_2$, $D=\mathrm{diag}(0,1)$, and $\alpha =1\mathrm{/}2$. Directly, $D$ is Hermitian, lies between $0$ and $I_2$, and has trace $1$, so $D\in {\mathcal{D}}_1$. The response matrix has gap $1$, so [F2] gives $\mathrm{\Phi }(D)=\mathrm{diag}(1,0)$. Then $D^{+}=\frac{1}{2}{I}_2$ and $(D^{+}{)}^2=\frac{1}{4}{I}_2\mathrm{\ne }{D}^{+}$. Thus the update belongs to ${\mathcal{D}}_1$ by step 2.1 but is not an orthogonal projector.