Lemma 14.19.48D049243EB2FA

Proof. If $\lambda (P)\in \Gamma $, then $\mathcal A_ U+tP$ is admissible for $t\geq 0$. The function $t\mapsto F(\mathcal A_ U+tP)$ is positive and concave on $[0,\infty )$, so its derivative at $t=0$ cannot be negative; thus $\operatorname {tr}(BP)\geq 0$. Since $F$ is concave and $F(0)=0$, the tangent inequality at $\mathcal A_ U$, evaluated at the origin, gives $\operatorname {tr}(B\mathcal A_ U)\leq h$. Choose $a_*>0$ with $F(a_*\operatorname {Id})>H_0+1$. Concavity gives

Thus $T\geq a_*^{-1}$. Since $\mathcal L(U-V)=\operatorname {tr}(B\mathcal A_ U)-\operatorname {tr}(B\mathcal A_ V)$ and $\mathcal L(\underline U-U)=\operatorname {tr}(B\mathcal A_{\underline U})-\operatorname {tr}(B\mathcal A_ U)$, the remaining two bounds follow from the nonnegativity of $\operatorname {tr}(B\mathcal A_ V)$ and $\operatorname {tr}(B\mathcal A_{\underline U})$.

We derive the alternative from the strict gap. Fix $r,H,\delta $ from lemma 14.19.43, and put $\mu =\lambda (\mathcal A_{\underline U})$. Compactness and the gap give one $S>r$ such that

everywhere. We first prove a scalar alternative. At a point of $X$, let $\lambda \in \Gamma $ satisfy $f(\lambda )=\sigma $, where $\sigma =h$ at that point, and consider the set

It is contained in $\Gamma $ because $\mu -r\mathbf1\in \Gamma $, it is closed, and the matrix bound of lemma 14.19.43, applied to diagonal matrices, makes it bounded; thus $K$ is compact. When nonempty, it contains $\mu -r\mathbf1$ and is star-shaped about that point: moving along a segment toward it decreases each coordinate, hence decreases $f$. Thus $K$ is connected. Let $N=\nabla f(\lambda )$, whose entries are positive, and let $\Pi $ be the tangent hyperplane $N\cdot (v-\lambda )=0$.

If $f(\mu -r\mathbf1)>\sigma $, concavity gives $N\cdot (\mu -\lambda )>r\sum _ iN_ i$. Otherwise $\mu -r\mathbf1\in K$. The diagonal ray $\mu -r\mathbf1+t\mathbf1$, $t\geq 0$, meets $\{ f=\sigma \} $, since $f$ diverges along it; at the meeting point $v$, the tangent inequality gives $N\cdot (v-\lambda )\geq 0$. If $\Pi $ misses $K$, connectedness makes $N\cdot (v-\lambda )$ strictly positive throughout $K$, and in particular at $\mu -r\mathbf1$; this again gives $N\cdot (\mu -\lambda )>r\sum _ iN_ i$.

If $v\in \Pi \cap K$, put $z_ i=v-r\mathbf1+Se_ i$. Then $z_ i\geq \mu -2r\mathbf1+Se_ i$, so $f(z_ i)>\sigma $. The tangent inequality at $\lambda $ gives

Thus $N_ i>(r/S)\sum _ jN_ j$ for every $i$. We have proved that $N\cdot (\mu -\lambda )>r\sum _ iN_ i$ or $N_ i>(r/S)\sum _ jN_ j$ for all $i$, at every positive level $\sigma $, with constants independent of $\sigma $.

To pass to matrices, take $\lambda =\lambda (\mathcal A_ U)$, and order $\lambda $ and $\mu $ decreasingly. By concavity and symmetry, the numbers $f_ i(\lambda )$ are then ordered increasingly. In the eigenframe of $\mathcal A_ U$, the diagonal of $\mathcal A_{\underline U}$ is a doubly stochastic average of permutations of $\mu $, so the rearrangement inequality gives

Since $N=(f_ i(\lambda ))_ i=(B_ i)_ i$, taking $\kappa $ smaller than $\min \{ r,r/S\} $ proves the matrix alternative. $\square$