Pome

Lemma 8.2.57710D01DF6F5

Lemma 8.2.5.7710D01DF6F5 If $M$ belongs to the closed Hermitian $m$-cone, then

\[ W:=\sum _ iF_ iM_{i\bar i}\geq \frac{f}{m!S}\operatorname {tr}M, \qquad F_ i\geq \frac{f}{m!S},\qquad |M_{1\bar1}|\leq \operatorname {tr}M. \]

No simultaneous diagonalization of $M$ and $\Lambda $ is required.

Proof. In the wedge notation of the mixed-trace estimate, differentiation gives $DF_\Lambda (M)=fM_ m/\sigma _ m$. The estimate $M_ m\geq \sigma _ m M_1/(m!\sigma _1)$, with $M_1=\operatorname {tr}M/n$ and $\sigma _1=S/n$, proves the first inequality. The positive semidefinite rank-one coordinate projections belong to the closed $m$-cone; taking $M$ to be each projection proves the second inequality. The closed $m$-cone is contained in the closed $2$-cone. Its eigenvalues $\mu $ satisfy

\[ \operatorname {tr}M\geq 0,\qquad \sum _ i\mu _ i^2=(\operatorname {tr}M)^2-2S_2(\mu ) \leq (\operatorname {tr}M)^2. \]

Thus the operator norm of $M$ is at most its trace, which proves the last inequality. The wedge pairing computes the derivative in the diagonal frame of $\Lambda $, regardless of the eigenvectors of $M$. $\square$

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