Lemma 14.19.49.BD78CDB91D24 Put $Q=1+|\partial U|_\omega ^2$. For any smooth real $\xi $, one has
\begin{align*} \mathrm e^{-\xi }\mathcal L(\mathrm e^\xi Q) & \geq \sum _{i,k}B_ i|U_{ki}+U_ k\xi _ i|^2 +\frac14\sum _ iB_ i\lambda _ i^2+Q\mathcal L\xi +|\partial \xi |_ B^2\\ & \quad +2\operatorname{Re}\sum _ iB_ i\lambda _ iU_ i\xi _{\bar i} -2\operatorname{Re}\sum _{i,k}B_ i\chi _{i\bar k}U_ k\xi _{\bar i} -C_ gTQ-2H_1|\partial U|_\omega , \end{align*}
where $C_ g$ depends only on $\omega $ and $\chi $.
Proof.
Write $\mathscr T^ p_{ik}$ and $R^ p_{i\bar jk}$ for the torsion and curvature of the Chern connection. The commutation formula and the differentiated equation read
\begin{gather*} \nabla _{\bar j}\nabla _ iU_ k =\nabla _ kU_{i\bar j}+\sum _ p\bigl (R^ p_{i\bar jk}U_ p-\mathscr T^ p_{ik}U_{p\bar j}\bigr ),\\ \sum _ iB_ i\nabla _ kU_{i\bar i}=h_ k-\sum _ iB_ i\nabla _ k\chi _{i\bar i}. \end{gather*}
The product rule for $|\partial U|^2$ gives both squares $\sum B_ i|U_{ki}|^2$ and $\sum B_ i|U_{k\bar i}|^2$, the term $2\operatorname{Re}\sum _ kh_ kU_{\bar k}$, a curvature error bounded by $CT|\partial U|^2$, and the errors from $\nabla \chi $ and torsion. The latter are bounded by
\[ CT|\partial U|+CT|\partial U|^2 +C|\partial U|T^{1/2}\Bigl (\sum _{i,k}B_ i|U_{k\bar i}|^2\Bigr )^{1/2}. \]
Absorb the last term into half of the mixed square. Since $U_{i\bar j}=\lambda _ i\delta _{ij}-\chi _{i\bar j}$, the remaining half is at least $\frac14\sum _ iB_ i\lambda _ i^2-CT$. Hence
\[ \mathcal L|\partial U|^2\geq \sum _{i,k}B_ i|U_{ki}|^2 +\frac14\sum _ iB_ i\lambda _ i^2-C_ gTQ-2H_1|\partial U|. \]
Finally,
\[ \partial _ i|\partial U|^2 =\sum _ kU_{ki}U_{\bar k}+\lambda _ iU_ i-\sum _ k\chi _{i\bar k}U_ k. \]
In the product rule for $\mathrm e^\xi Q$, combine the cross term coming from the first sum with $|\partial U|^2|\partial \xi |_ B^2$ and with the square $\sum B_ i|U_{ki}|^2$. This gives $\sum B_ i|U_{ki}+U_ k\xi _ i|^2$, while the remaining weight term is $|\partial \xi |_ B^2$.
$\square$
Pome
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