that $$\begin{aligned}
 \left(D_{\tau} + \Delta\right){\hat{{\operatorname{R}}}}&=  -({\operatorname{R}}^2 + {\operatorname{R}}^{\#})\circ {\hat{{\operatorname{P}}}}+ 2\nabla_i {\operatorname{R}}\circ \nabla_i {\hat{{\operatorname{P}}}}+ {\operatorname{R}}\circ \Delta {\hat{{\operatorname{P}}}},\end{aligned}$$ which immediately yields $$\begin{aligned}
 \left|\left(D_{\tau} + \Delta\right){\hat{{\operatorname{R}}}}\right|&\leq C|{\operatorname{R}}|(|{\hat{{\operatorname{R}}}}| + |{\operatorname{B}}|) + C|\nabla {\operatorname{R}}| |{\operatorname{A}}| + |{\operatorname{R}}^{\#}\circ {\hat{{\operatorname{P}}}}|.\end{aligned}$$ So, to obtain , it remains only to estimate ${\operatorname{R}}^{\#}\circ {\hat{{\operatorname{P}}}}$.

Using $*$ to denote the adjoint of an element in ${\operatorname{End}}({\wedge^2 T^*M})$, we may write $${\operatorname{R}}= ({\hat{{\operatorname{P}}}}+ {\tilde{{\operatorname{P}}}}) \circ {\operatorname{R}}=  {\hat{{\operatorname{P}}}}\circ {\operatorname{R}}+ {\tilde{{\operatorname{P}}}}\circ {\operatorname{R}}= {\hat{{\operatorname{R}}}}^* + {\tilde{{\operatorname{R}}}}^*.$$ By the symmetry and bilinearity of the $\#$ pairing we then see that $$\label{eq:sharpex}
 {\operatorname{R}}^{\#} \circ {\hat{{\operatorname{P}}}}= ({\operatorname{R}}\#{\operatorname{R}}) \circ {\hat{{\operatorname{P}}}}= ({\hat{{\operatorname{R}}}}^*\#{\hat{{\operatorname{R}}}}^* + 2{\tilde{{\operatorname{R}}}}^*\#{\hat{{\operatorname{R}}}}^*)\circ {\hat{{\operatorname{P}}}}+ ({\tilde{{\operatorname{R}}}}^* \#{\tilde{{\operatorname{R}}}}^*) \circ {\hat{{\operatorname{P}}}}.$$ But $({\tilde{{\operatorname{R}}}}^*\#{\tilde{{\operatorname{R}}}}^*)\circ {\hat{{\operatorname{P}}}}= 0$ since, by , $\left\langle [{\tilde{{\operatorname{P}}}}{\operatorname{R}}\varphi_i, {\tilde{{\operatorname{P}}}}{\operatorname{R}}\varphi_j], {\hat{{\operatorname{P}}}}\varphi_k\right\rangle = 0$ for all $i$, $j$, $k$. So $$|({\operatorname{R}}\#{\operatorname{R}}) \circ {\hat{{\operatorname{P}}}}| \leq C(|{\hat{{\operatorname{R}}}}^*| + |{\tilde{{\operatorname{R}}}}^*|)|{\hat{{\operatorname{R}}}}^*| \leq C|{\operatorname{R}}||{\hat{{\operatorname{R}}}}|$$ by , and follows.

Preservation of Kählerity from the vanishing of ${\operatorname{A}}$ and ${\hat{{\operatorname{R}}}}$
-----------------------------------------------------------------------------------------------------

We have seen that if the initial time-slice $(M, g(0))$ is Kähler, then ${\operatorname{A}}$, ${\operatorname{B}}$, ${\hat{{\operatorname{R}}}}$, and ${\hat{{\operatorname{S}}}}$ vanish on this slice. In the next section, we will show that the backward uniqueness theorem proven in [@KotschwarWangConical] implies that ${\operatorname{A}}$, ${\operatorname{B}}$, ${\hat{{\operatorname{R}}}}$, and ${\hat{{\operatorname{S}}}}$ must vanish identically on $M\times[0, T]$ for the specific class of solutions $(M, g(\tau))$ 