Let \(\vartheta\) be as defined in the proof of (0x68c9184d) , and let \((x,y) \in \supp \vartheta\). Using \(\alpha_+ > \alpha_-\) and \(a = \tfrac{\alpha_+}{\delta}\), yields
\[ \frac{R_1}{2a} < \frac{\delta}{2\alpha_-} R_1 < \frac{\delta}{2\alpha_-} \alpha_- r < \delta r \leq \delta r_0. \]Thus, \( y \in [-\delta r_0, \delta r_0]\). Furthermore, \( z(x,y) > -3R_2 \), i.e.
\begin{align*} \frac{|x|^2}{2\delta} &< 3R_2 + \frac{\alpha_-}{\delta} y - \frac{\beta}{2\delta^2} y^2\\ &\leq \frac{3\alpha_- r}{16} + \frac{\alpha_-}{\delta} \delta r\\ &\leq \frac{19\alpha_-}{16} r\\ &\leq \frac{\delta}{8}, \end{align*}where we used \(y \in [-\delta r_0, \delta r_0]\) and the bound of \(r\le \frac{2\delta}{19\alpha_-}\). Thus, \( |x| < \frac{\delta}{2} \), and therefore
\[ \supp \vartheta \subset B_{\delta/2}(0) \times [-\delta r_0, \delta r_0]. \]