below. We define $$A^2:=2\int_{s_-}^{s_+} f(s)\,{\mathrm}d s,$$ noting that we are free to let the difference $s_+-s_-$, and hence $A^2$, be as large as we wish. We then choose $\alpha$ and $\delta$ to be any positive parameters satisfying $$\label{2d}
\alpha^2+\delta^2\leq\frac{A^2}{2}.$$

To define $g$, and hence a metric $(f,g)\in{\mathcal}G_{\alpha,\delta,{\varepsilon}}$, we choose ${\varphi}$ to be any smooth function satisfying $1-{\varepsilon}\leq{\varphi}\leq1$, requiring that it be nonconstant unless ${\varepsilon}=0$. Clearly ${\varepsilon}$ controls how much ${\varphi}$ can stray from being constant. We then set $$g^2(s):=\alpha^2+2\int_{s_-}^s {\varphi}(\bar s)f(\bar s)\,{\mathrm}d\bar s.$$ We readily verify that $g$ defined in this way satisfies closing conditions  and .

To verify that part (a) of our Closeness Assumptions is satisfied, we observe that the gradient restriction $|f_s|\leq1$ implies that $$\begin{aligned}
f^2(s)&=2 \int_{s_-}^s f(\bar s)f_{\bar s}(\bar s)\,{\mathrm}d\bar s\\
    &\leq2 \int_{s_-}^s \big\{{\varphi}+(1-{\varphi})\big\}f(\bar s)\,{\mathrm}d\bar s\\
    &\leq g^2-\alpha^2+{\varepsilon}A^2\\
    &\leq g^2,\end{aligned}$$ provided that $${\varepsilon}\leq\frac{\alpha^2}{A^2}.$$ It immediately follows that part (b) holds for all ${\varepsilon}\in[0,1)$, with equality — which Lemma \[CalabiCondition\] (below) shows is equivalent to the metric being Kähler — if and only if ${\varepsilon}=0$. Part (c) holds as a consequence of the gradient restriction $|f_s|\leq1$.

To verify that part (d) holds, we observe that one has $$\begin{aligned}
g^2(s_+)-3g^2(s_-)&\geq\big\{\alpha^2+(1-{\varepsilon})A^2\big\}-3\alpha^2\\
&=(A^2-2\alpha^2)-{\varepsilon}A^2\\
&\geq2\delta^2-{\varepsilon}A^2,\end{aligned}$$ with the last inequality following from 