Lemma 8.5.3.BAC60298D0F2 For $f_0>0$ and $L>0$, assume the set
\[ K=\{ A\in C_{n,m}:\operatorname {tr}A\leq L, \ \sigma _ m(A)\geq f_0^ m\} \]
is nonempty. It is a compact subset of the interior, and there are $\vartheta ,\Theta >0$, depending only on $n,m,f_0,L$, such that
\[ \vartheta I\leq DF_ A\leq \Theta I\qquad (A\in K). \]
The function on all Hermitian matrices defined by
\[ \widehat F(B)=\inf _{A\in K} \bigl (F(A)+\operatorname {tr}(DF_ A(B-A))\bigr ) \]
is concave, agrees with $F$ on $K$, and satisfies, for $P\geq 0$,
\[ \vartheta \operatorname {tr}P\leq \widehat F(B+P)-\widehat F(B) \leq \Theta \operatorname {tr}P. \]
Proof.
In the closed $2$-cone,
\[ \sum _ i\lambda _ i^2=(\operatorname {tr}A)^2-2S_2(A) \leq L^2. \]
Thus $K$ is bounded and closed. Maclaurin gives $\sigma _ j(A)\geq \sigma _ m(A)^{j/m}\geq f_0^ j$ for $j\leq m$, so every matrix in $K$ is in the interior. In a unitary eigenframe,
\[ F_ i=\frac{F^{1-m}S_{m-1}(\lambda |i)}{m\binom nm}, \qquad \frac{f_0}{m!L}\leq F_ i\leq \frac{L^{m-1}}{n f_0^{m-1}}. \]
The lower bound is 8.2.5; the upper bound uses $|\lambda _ i|\leq L$ in the polynomial numerator and its positivity. These bounds give possible values of $\vartheta $ and $\Theta $. Concavity places each tangent plane above $F$ on the cone; the plane at $B$ gives equality when $B\in K$. An infimum of affine functions is concave. Each plane increases under $P\geq 0$ by a number between the two displayed trace bounds; taking infima preserves these bounds. The same estimates give global Lipschitz continuity of $\widehat F$.
$\square$
Pome
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