We write \(\lesssim\) if \(\le\) holds up to a positive numerical constant.

Using the mean value inequality and the bound \(\lvert x\rvert\lesssim \sqrt{\delta R_2}\) on \(\supp \vartheta\), it follows that

\begin{align*} |\vartheta_1'(z(x,0)) - \vartheta_2'(z(y,0))| &\leq \sup_{t \in [2R_2, 3R_2]} |\vartheta_1''(t)| \, |z(x,0) - z(y,0)|\\ &\lesssim \frac{1}{R_2^2} \, \frac{1}{2\delta} |\lvert x\rvert^2 - \lvert y\rvert^2|\\ &\lesssim \frac{1}{\delta R_2^2} \lvert x+y\rvert \lvert x-y\rvert\\ &\lesssim \frac{1}{\delta R_2^2} \sqrt{\delta R_2}\, |x-y|\\ &= \frac{1}{\sqrt{\delta} R_2^{3/2}} \, |x-y|. \end{align*}

Combining this estimate with

$$ |\vartheta_1'(t)|^2 \lesssim \frac{1}{R_2^2}, \qquad \lvert \nabla z(x,0)\rvert^2 \lesssim \frac{1+\alpha_-^2}{\delta^2} \qquad |\nabla z(x,0) - \nabla z(y,0)|^2 = \frac{1}{\delta^2} |x-y|^2, $$

we obtain

\begin{align*} &\quad|\nabla \vartheta(x,0) - \nabla \vartheta(y,0)|^2\\ &= \big| \vartheta_1'(z(x,0)) \nabla z(x,0) - \vartheta_1'(z(y,0)) \nabla z(y,0) \big|^2\\ &= \big| \vartheta_1'(z(x,0)) \nabla z(x,0) - \vartheta_1'(z(x,0))\nabla z(y,0) + \vartheta_1'(z(x,0))\nabla z(y,0) - \vartheta_1'(z(y,0)) \nabla z(y,0) \big|^2\\ &\lesssim \tfrac{1}{R_2^2} |\nabla z(x,0) - \nabla z(y,0)|^2 + \tfrac{1+\alpha_-^2}{\delta^2}|\vartheta_1'(z(x,0)) - \vartheta_1'(z(y,0))|^2\\ &\lesssim \Bigl(\tfrac{1}{\delta^2 R_2^2} + \tfrac{1+\alpha^2_-}{\delta^3 R_2^3}\Bigr) \lvert x-y\rvert^2. \end{align*}

Using \(R_2< 1\), and \(\delta < 1\), we conclude

\[ \lvert \nabla \vartheta(x,0)-\nabla \vartheta(y,0)\rvert^2 \lesssim \frac{1+\alpha_-^2}{\delta^3R_2^3} \lvert x-y\rvert^2. \]