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The admissible Hermitian density domain is closed, convex, and complete

Statement

For $n \geq 2$ and $1 \leq r < n$ , the set $D_{r}$ of The fixed-particle admissible Hermitian density domain is nonempty, closed, and convex in $H_{n}$ , and complete in the Frobenius metric.

Facts & Assumptions

Given: Integers $n \geq 2$ , $1 \leq r < n$ , the real vector space $H_{n}$ with Frobenius norm, and $D_{r} = {D = D^{*} : 0 ⪯ D ⪯ I_{n}, tr D = r}$ .

[F1]

The domain and its order conditions are as in The fixed-particle admissible Hermitian density domain.

[F2]

Every Euclidean Cauchy sequence in a finite-dimensional real coordinate space converges ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ).

Proof

1.1givenF1construct

The diagonal matrix $D_{0} = diag (1, \dots, 1, 0, \dots, 0)$ with $r$ ones is Hermitian, satisfies $0 ⪯ D_{0} ⪯ I_{n}$ , and has trace $r$ . Thus $D_{0} \in D_{r}$ , so the domain is nonempty.

1.2givenF2constructalgebra

Identify $H_{n}$ with $R^{n^{2}}$ by listing the real diagonal entries and, for each $i < j$ , the coordinates $\sqrt{2} Re D_{i j}$ and $\sqrt{2} Im D_{i j}$ . This real-linear map is bijective, since the lower-triangular entries are recovered by conjugation, and its squared Euclidean norm is $\sum_{i} D_{i i}^{2} + 2 \sum_{i < j} ∣ D_{i j} ∣^{2} = ∥ D ∥_{F}^{2}$ . Thus it is an isometry. If $(D_{k})$ is Frobenius-Cauchy, its image is Cauchy in $R^{n^{2}}$ , so [F2] gives a limit whose inverse image $D \in H_{n}$ satisfies $D_{k} \to D$ in Frobenius norm.

2.1step 1.2F1algebra

For each fixed $x \in C^{n}$ , $∣ x^{*} (D_{k} - D) x ∣ \leq (\sum_{i} ∣ x_{i} ∣)^{2} ∥ D_{k} - D ∥_{F}$ , since each matrix entry is bounded by the Frobenius norm. The same estimate applies to $x^{*} ((I_{n} - D_{k}) - (I_{n} - D)) x$ . Both quadratic forms therefore converge; their terms are nonnegative, so their limits are nonnegative and $0 ⪯ D ⪯ I_{n}$ . Also $∣ tr (D_{k} - D) ∣ \leq n ∥ D_{k} - D ∥_{F}$ , so $tr D = \lim_{k} tr D_{k} = r$ . Hence $D \in D_{r}$ .

3.1step 1.2step 2.1F1algebra

Steps 1.2 and 2.1 show that every Frobenius-Cauchy sequence in $D_{r}$ converges to a member of $D_{r}$ , so the domain is complete. To prove closedness directly, for each fixed $x \in C^{n}$ the real-valued maps $D \mapsto x^{*} D x$ and $D \mapsto x^{*} (I_{n} - D) x$ are continuous, since their differences at $D, E$ have modulus at most $(\sum_{i} ∣ x_{i} ∣)^{2} ∥ D - E ∥_{F}$ . The trace is continuous as $∣ tr (D - E) ∣ \leq n ∥ D - E ∥_{F}$ . Thus $D_{r}$ is the intersection of the closed preimages of $[0, \infty)$ under these quadratic-form maps and of ${r}$ under the trace map. An arbitrary intersection of closed sets is closed.

4.1F1algebra∎

If $D_{1}, D_{2} \in D_{r}$ 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 $t tr D_{1} + (1 - t) tr D_{2} = t r + (1 - t) r = r$ . Hence it lies in $D_{r}$ , proving convexity.

Depends on

Used by

A uniform derivative bound certifies contraction of relaxed density mixing Theorem

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Eric Cancès, Gaspard Kemlin, and Antoine Levitt, Convergence analysis of direct minimization and self-consistent iterations, §4.4 and Appendix Lemma 4.1 (standard reference, not scraped)