8.4 A trace estimate with a prescribed potential54309E0621C2

Proof. Young's inequality gives $|a-b|^2\geq (7/8)|a|^2-7|b|^2$. Since $\chi ''=12q^2$, the coefficient of $|a|^2$ after subtracting $9q^2$ is $3q^2/2$. Moreover, $2q\leq q^2/4$ and $q|a|\leq q^2|a|^2/4+1$. The remaining coefficient is at least $q^2$, and the error involving $b$ is $84q^2|b|^2$. Finally $q\geq 1/(36\epsilon )$, which proves the last assertion. The weighted version follows by summing the scalar inequality with positive coefficients. $\square$

Proof. 8.4.1 gives

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

The lower bound for each $F_ i$ similarly gives

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

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 $\lambda _1\leq (\sum _ i\lambda _ i^2)^{1/2}<S$. $\square$

Proof. Choose a smooth unitary frame parallel for the Chern connection at the point and diagonalizing $H$ there. The positive semidefinite tensor $B$, diagonal in this frame with entries $0,\beta ,\ldots ,\beta $, satisfies $\nabla B=0$ at the point and $|\nabla ^2B|\leq C\beta $. For $0<\beta <1$, the largest eigenvalue of $W-B$ is simple, lies below that of $W$, and agrees with it at the point. This gives the required smooth support. Only its largest eigenvalue is perturbed; the equation still uses $F(W)$. Differentiating the normalized eigenvector equation gives

The denominators are at most $(n+1)\lambda _1$. Let $U=\partial \bar\partial u$, and take $T^ p_{ki}=\Gamma ^ p_{ki}-\Gamma ^ p_{ik}$. The first commutations are

Apply the first identity, commute one mixed pair of covariant derivatives, and apply the conjugate identity. The result has the form

Indeed the mixed commutator contracts curvature with $U$, differentiating the first torsion coefficient contracts $\nabla T$ with $U$, and the second use of the first identities produces contractions of two torsion coefficients with $U$. Thus all the omitted terms satisfy the displayed bound. Since $g$ is parallel, the same derivative relations apply to $H$ apart from the explicit $g$ terms in the first commutation. At the diagonal point $|U|\leq C\lambda _1$. For $p>1$, rewrite $U_{k\bar p;1}=U_{1\bar p;k}+T^ j_{k1}U_{j\bar p}$. Young's inequality absorbs its third derivative into the positive eigenvalue square, at the cost of $C\lambda _1$. Consequently

Differentiate $F(W)=f$ twice and sum this inequality with weights $F_ k$. Write $D_ k=H_{1\bar1;k}$ and $t_ k=T^1_{k1}$. The first commutation, using $U_{1\bar1}=\lambda _1-1$, gives

The negative logarithmic square and the remaining torsion term therefore satisfy

For example, expanding the left side leaves $-|Z_ k|^2-2\lambda _1^{-1}\operatorname {Re}(Z_ k\bar t_ k)-(1-\lambda _1^{-1})^2|t_ k|^2$, and $\lambda _1|Z_ k|\leq |D_ k|+C\lambda _1$. This proves the claim, uniformly for small $\beta $. $\square$

Proof. Differentiate the squared norm using metric compatibility. Its quadratic part is the displayed sum. Commuting the remaining third derivatives gives

The first differentiated equation is $\sum _ iF_ i u_{i\bar i;k}=f_ k$, because the background $g$ is parallel. At the diagonal point the torsion errors are bounded by $C\sqrt K\sum _ iF_ i(|\lambda _ i|+1)$, and the curvature errors by $CK\mathcal F$. Use

with a sufficiently small $\eta '$ for the finite sum. This gives the asserted bound. $\square$

Proof. For $\varepsilon \leq 1$ the density has fixed bounds $0<f_0\leq f\leq f_1$. Its first derivative is

The nonnegative $u_ k$ contribution can be dropped from the gradient lemma. At a point with $\lambda _1>1$, differentiating once more gives

Here $u_{1\bar1}=\lambda _1-1>0$, the square from differentiating the exponential is nonnegative, and all derivatives of $h$ are bounded on the fixed interval.

Set

Then $(4K)^{-1}\leq \Phi '\leq (2K)^{-1}$, $\Phi ''=2(\Phi ')^2$, $-3A\leq \Psi '\leq -A$, $\Psi ''=A\tau $, and $q=\chi '$ lies between $1/(36\varepsilon )$ and $1/(12\varepsilon )$, with $\chi ''=12q^2$. Take a maximum of $G=\log \lambda _1+\Phi (s)+\Psi (u)+\chi (v-\varphi )$ and use the smooth lower support from the eigenvalue lemma. At this point put

The first derivative condition and the torsion identity give

Thus $|e_ i|\leq E_0$, with $E_0$ independent of $A$ and of derivatives of $v$.

We make the initial maximum inequality explicit. The gradient lemma and the identity $u_{i\bar j}=(\lambda _ i-1)\delta _{ij}$ give, after multiplication by $\Phi '$,

To justify the constants, retain half of the positive quadratic term, use $(\lambda _ i-1)^2\geq \lambda _ i^2/2-1$, and choose the torsion parameter to cost at most $Q/(128K)$. The remaining density-gradient error is $C|a|/\sqrt K$. Young's inequality bounds it by $C_ G'|a|^2/S+cS/K$; choose $C_ G'$ large enough that the quadratic coercivity lemma absorbs $cS/K$ by $Q/(128K)$, with a fixed constant error. This leaves $3Q/(64K)$. The linear $H_{1\bar1;i}$ error in the eigenvalue lemma is estimated using the first derivative condition. Its $x_ i$ term costs at most half of the retained quadratic sum and $C\mathcal F$. Its $y_ i$ term costs at most $\Psi ''\sum _ iF_ i|u_ i|^2/4+\mathcal F$ whenever $\lambda _1\geq L_0(A)$, where $L_0(A)$ is fixed independently of $K,v$. Its $qa_ i$ term costs at most $q\sum _ iF_ i|a_ i|$ when $\lambda _1$ exceeds a fixed metric constant; its $qb_ i$ term costs $C\mathcal F$. If either eigenvalue threshold fails, the required bound follows directly. Otherwise, with fixed $C_ H,C_ G,C_0$ independent of $A$, the second derivative condition gives

The fixed $qL\varphi $ term is included in $C_0$. Choose $\varepsilon $ so the lower bound for $q$ meets all thresholds of the derivative-absorption lemma. These thresholds depend on $M,B,h$ and the metric, and not on derivatives of $\varphi $. This choice precedes that of $A$.

First suppose $\lambda _ n<-\delta \lambda _1$. Concavity gives $N\geq 0$, and

The derivative-absorption lemma applies after decreasing the right side further by $(3q^2+2q)|a_ i|^2$. Drop its nonnegative reserves and $\Psi ''$ term. Since $\Psi 'f\geq -3Af_1$, this gives

where $C_*$ is fixed independently of $A,\delta $.

Now suppose $\lambda _ n\geq -\delta \lambda _1$, and let $I=\{ i:F_ i>\delta ^{-1}F_1\} $. For $i\in I$, $\lambda _ i<\lambda _1$ and

The spectral formula for $D^2F$, including its nonpositive diagonal part, therefore gives $N\geq (1-2\delta )\sum _{i\in I}F_ i|Z_ i|^2$. This uses the original eigenvalues; repeated eigenvalues do not occur in $I$. Using $x_ i=-Z_ i-y_ i-qa_ i+e_ i$ and $|z+w|^2\geq \delta |z|^2-\delta |w|^2/(1-\delta )$, one obtains

Choose $\delta \leq \min \{ 1/17,\tau /(72A+\tau )\} $. Then the first coefficient is at most $\Psi ''/2$, and $r\leq 1/2$. On the complementary indices the preceding elementary bound gives

Apply the derivative-absorption lemma, with penalty $r$ on $I$ and zero outside it; adding the unused negative $(9q^2+2q)|a_ i|^2$ on $I$, and $(3q^2+2q)|a_ i|^2$ off $I$, only decreases its input. The constants in its fixed-comparison error and the high-index $rE_0^2$ error are bounded independently of $A,\delta $. Enlarge $C_*$ to include them. We obtain

Take $A\geq \max \{ 1,C_*+1\} $, then fix the displayed $\delta $ and the threshold $L_0(A)$. Both case inequalities now satisfy the hypotheses of the conditional eigenvalue proposition, with fixed data. Hence $\lambda _1\leq CK$ at the maximum point. Finally $\Phi ,\Psi ,\chi $ have oscillations bounded independently of $K$ and of derivatives of $v$. Comparing $G$ at any other point with its maximum gives the same bound for $\lambda _1$ everywhere. Since $S\leq n\lambda _1$, this proves the trace estimate. $\square$

Proof. The Hessian density is $F(W)^ m\leq f_1^ m$. Apply the trace-to-gradient blowup proposition to the fixed background $\theta =\omega $ and the preceding trace estimate. It gives the uniform gradient bound. The trace is then uniformly bounded, and the $\Gamma _2$ identity bounds the absolute value of every eigenvalue of $W$ by its trace. Subtracting the identity matrix gives the complex-Hessian bound. $\square$