PomeFDAFFF47B8A3Proposition 8.4.3.FDAFFF47B8A3 Fix $a>0$, $b,d,e\geq 0$ and $0<\delta <1$. Let $K\geq 1$ and $0<f_0\leq f\leq f_1$. Suppose the following inequality holds according to the sign of $\lambda _ n+\delta \lambda _1$:
If $\lambda _ n<-\delta \lambda _1$, then $0\geq aQ/K-bK\mathcal F-d\lambda _1\mathcal F-e$.
If $\lambda _ n\geq -\delta \lambda _1$, then $0\geq aQ/K-bKF_1-e$.
Then $S\leq CK$ and $\lambda _1\leq CK$, with $C$ depending only on the fixed displayed data and $n,m$.
Proof. In the first case, symmetry and concavity of $F$ imply $F_ n\geq F_ i$ for every $i$, so $Q\geq \delta ^2\lambda _1^2\mathcal F/n$. To see the derivative ordering, apply monotonicity of the gradient of a concave function to $\lambda $ and the vector obtained by exchanging its $i$th and $n$th coordinates. Divide the assumed inequality by $K\mathcal F$, use $\mathcal F\geq 1$, and set $t=\lambda _1/K$. It yields $a\delta ^2t^2/n\leq b+dt+e$, which bounds $t$. Since $S\leq n\lambda _1$, the trace bound follows. In the second case, either $\lambda _1^2\leq 2bK^2/a$, which gives the same trace bound directly, or $aQ/(2K)\geq bKF_1$. The assumed inequality then gives $Q\leq 2eK/a$. 8.2.6 bounds $S$ by a multiple of $K$. Finally the $\Gamma _2$ identity gives $\lambda _1\leq (\sum _ i\lambda _ i^2)^{1/2}<S$. $\square$
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