0x68f5f313

On \(\{-3R_2\le z,\, \tfrac{R_1}{4a} < y<\tfrac{R_1}{2a}\} \), we deduce

\begin{align*} \phi_{\delta,+}(x,y) &= \frac{\alpha_+}{\delta} y + \frac{\beta}{2\delta^2} y^2 - \frac{|x|^2}{2\delta^2} \\[4pt] &\leq \frac{\alpha_+}{\delta} \biggl(\frac{R_1}{2a}\biggr) +\frac{\beta}{2\delta^2} \biggl( \frac{R_1}{2a} \biggr)^2 \\[4pt] &= \biggl( 1 + \frac{\beta R_1}{4\alpha_+^2} \biggr) \frac{R_1}{2} \\[4pt] &\leq R_1, \end{align*}

since \( R_1 \leq \tfrac{\alpha_-^2}{18\beta} \) and \(\alpha_- < \alpha_+\).

On \(\{-3R_2\le z\le -2R_2,\, y<\tfrac{R_1}{2a}\}\cap \mathbb{R}^d_+ \), it follows that

\begin{align*} \phi_{\delta,+}(x,y) &= z(x,y) + \frac{\alpha_+ - \alpha_-}{\delta} y + \frac{\beta}{\delta^2} y^2 \\[4pt] &\leq -2R_2 + \frac{\alpha_+ - \alpha_-}{\delta} \left( \frac{R_1}{2a} \right) +\frac{\beta}{\delta^2} \left( \frac{R_1}{2a} \right)^2 \\[4pt] &= -2R_2 + \left( \frac{1}{2} + \frac{\beta R_1}{4\alpha_+^2} \right) R_1 \\[4pt] &\leq -\tfrac{5}{4}R_2, \end{align*}

where in the last step we used \( R_1 < \tfrac{\alpha_-^2}{18\beta} \), \(\alpha_- < \alpha_+\), and \( R_2 < R_1 \).

Finally, on \(\{-3R_2\le z\le -2R_2\}\cap \mathbb{R}^d_- \), we obtain, by employing a lower bound on \(y\) (see (0x68ecae27) ),

\begin{align*} \phi_{\delta,-}(x,y) &= z(x,y) + \frac{\beta}{\delta^2} y^2 \\[4pt] &\leq -2R_2 + \frac{\beta}{\delta^2} \left( \frac{\alpha_-^2 \delta^2}{\beta^2} \right) \left( 1 - \sqrt{1 + \tfrac{6\beta}{\alpha_-^2 } R_2} \right)^2 \\[4pt] &= -2R_2 + \frac{\alpha_-^2}{\beta} \left( \frac{1 - \left( 1 + \tfrac{6\beta}{\alpha_-^2} R_2 \right)} {1 + \sqrt{1 + \tfrac{6\beta}{\alpha_-^2} R_2}} \right)^2 \\[4pt] &\leq -2R_2 + \tfrac{\alpha_-^2}{\beta} \tfrac{36 \beta^2}{4\alpha_-^2} R_2^2 \\[4pt] &\leq - \frac{3}{2} R_2\\ &\leq - \frac{5}{4} R_2, \end{align*}

since

\[ \Biggl(\frac{1}{1 + \sqrt{1 + \tfrac{6\beta}{\alpha_-^2 } R_2}}\Biggr)^2 \le \frac{1}{4}, \]

and \( R_2 \leq \tfrac{\alpha_-^2 \delta^2}{18\beta} \).