maximum on ${\mathcal}S^2_\pm$ shows that $\psi_{ss}{\big|_{{\mathcal}S^2_\pm}}=(\psi_\pm)_{ss}{\big|_{{\mathcal}S^2_\pm}}\leq0$. The result follows.

First derivative estimates {#FirstOrder}
--------------------------

Based on the one-sided Kähler stability established in Lemma \[NearCalabi\], we now derive estimates on the first derivatives of $f$ and $g$, and consequently on the curvatures which depend on these first derivatives. We first state an immediate corollary of Lemma \[NearCalabi\], which controls $|g_s|$.

\[cor-gs\] Solutions originating from initial data satisfying our Closeness Assumptions have $|g_s|\leq1$ for as long as they exist.

Because, as noted above, the ordering $f\leq g$ is preserved by the flow, it follows from Lemma \[NearCalabi\] that $f^2g_s^2\leq g^2g_s^2\leq f^2$. The result follows.

Next, we obtain a bound for $|f_s|$.

\[lem-fs\] If $f\leq g$ initially, then for as long as the flow exists, $$|f_s|\leq\max\left\{\frac{2}{\sqrt3},\,\max|f_s(\cdot,0)|\right\}.$$

Using equation  (21) of [@IKS14] together with the fact that $$\Delta\Omega=\Omega_{ss}+(f_s/f+2g_s/g)\Omega_s$$ holds for any smooth function $\Omega(s,t)$, we see that $f_s$ evolves by $$\label{f_s-evolution}
(f_s)_t=(f_s)_{ss}+\left\{2\frac{g_s}{g}-\frac{f_s}{f}\right\}(f_s)_s
    -\left\{6\frac{f^2}{g^4}+2\frac{g_s^2}{g^2}\right\}f_s
    +8\frac{f^3}{g^5}\,g_s.$$ Because $f_s{\big|_{{\mathcal}S^2_\pm}}=\pm1$, we do not need to worry about a maximum of $|f_s|$ on ${\mathcal}S^2_\pm$.

We apply the weighted Cauchy–Schwarz inequality $|ab|\leq\epsilon a^2+(1/4\epsilon)b^2$ to the term $8f^3g_s/g^5$ above, with $a=g_s/g$ and $b=f^3/g^4$. Thus if $(f_s)_{\max}=C>0$ at some time, we obtain $$\begin{aligned}
\frac{d}{dt} (f_s)_{\max}
&\leq -(f_s)_{\max}\, \left(6 \frac{f^2}{g^4} + 2 \frac{g_s^2}{g^2}\right) 