Pome

Lemma 8.4.261C6DAF48B41

Lemma 8.4.2.61C6DAF48B41 Suppose $f\geq f_0>0$, $C_ H,C_ G\geq 0$, and $q$ satisfies

\[ q\geq 8,\qquad q\geq \frac{2nm!C_ H}{f_0},\qquad q^2\geq \frac{4m!C_ G}{f_0}. \]

For $0\leq r\leq 1/2$, $a,b\in \mathbb C^ n$ and $M\in C_{n,m}$, put $W_ M=\sum _ iF_ iM_{i\bar i}$ and

\begin{align*} E={}& q(W_ M-\mathcal F)-\frac{C_ H}{\lambda _1}|M_{1\bar1}| -\frac{C_ G}{S}\sum _ i|a_ i|^2\\ & +\sum _ iF_ i\bigl (12q^2|a_ i-b_ i|^2 -(9q^2+2q+rq^2)|a_ i|^2-q|a_ i|\bigr ). \end{align*}

Then

\[ E\geq \frac q2W_ M+\frac{q^2}{4}\sum _ iF_ i|a_ i|^2 -(q+1+84q^2\max _ i|b_ i|^2)\mathcal F. \]

Proof. 8.4.1 gives

\[ 12q^2|a_ i-b_ i|^2-(9q^2+2q)|a_ i|^2-q|a_ i| \geq q^2|a_ i|^2-1-84q^2|b_ i|^2. \]

Subtracting $rq^2|a_ i|^2$ leaves at least $q^2|a_ i|^2/2$. Since $S\leq n\lambda _1$, 8.2.5 and the hypothesis on $q$ imply

\[ \frac{C_ H}{\lambda _1}|M_{1\bar1}| \leq \frac{nC_ H}{S}\operatorname {tr}M\leq \frac q2 W_ M. \]

The lower bound for each $F_ i$ similarly gives

\[ \frac{C_ G}{S}\sum _ i|a_ i|^2 \leq \frac{q^2}{4}\sum _ iF_ i|a_ i|^2. \]

Combining these inequalities yields the assertion. The error depends on the fixed comparison vector $b$ and is not discarded. $\square$

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