Let \((x, y) \in E\). Then,

\begin{equation}\label{eq:z} \frac{|x|^2}{2\delta} + \frac{\beta y^2}{2\delta^2} - \frac{\alpha_- y}{\delta} \leq 3 R_2. \end{equation}

With square correction, it follows that

\begin{equation}\label{eq:ellipsoid} -z(x,y) + \frac{\alpha_-^2}{2\beta} = \frac{|x|^2}{2\delta} + \frac{\beta}{2\delta^2}\Bigl(y - \tfrac{\alpha_- \delta}{\beta}\Bigr)^2 \leq 3 R_2 + \frac{\alpha_-^2}{2 \beta}. \end{equation}

This also proves \(z(x,y)\le \tfrac{\alpha_-^2}{2}\), since

\[ -z(x,y) + \frac{\alpha^2_-}{2 \beta} \ge 0. \]

This proves that \(E\) is an ellipsoid with center in \((0, \tfrac{\alpha \delta}{\beta})\). Furthermore, from relation \eqref{eq:z}, we obtain that

\[ |x| \leq \sqrt{6 \delta R_2}, \]

and relation \eqref{eq:ellipsoid} implies

\[ y \in \left[ \frac{\delta \alpha_-}{\beta} \biggl(1 - \sqrt{1 + \tfrac{6\beta R_2}{\alpha_-^2}}\biggr),\; \frac{\delta \alpha_-}{\beta} \biggl(1 + \sqrt{1 + \tfrac{6\beta R_2}{\alpha_-^2}}\biggr) \right]. \]

The gradient of \(z\) is given by

\[ \nabla z(x, y) = \left( -\frac{x}{\delta}, \frac{\alpha_-}{\delta} - \frac{\beta y}{\delta^2} \right). \]

Then,

\begin{align*} |\nabla z(x, y)|^2 & = \frac{x^2}{\delta^2} + \left( \frac{\alpha_-}{\delta} - \frac{\beta y}{\delta^2} \right)^2 \\ & \le \frac{x^2}{\delta^2} + \frac{2\alpha_-^2}{\delta^2} + \frac{2 \beta^2 y^2}{\delta^4}. \end{align*}

Combining \eqref{eq:x}, \eqref{eq:y}, \(R_2\le \frac{\alpha_-^2}{18\beta}\) and the bounds \(R_2< 1\), \(\delta< 1\) yields

\begin{equation*} \begin{aligned} |\nabla z(x, y)|^2 &\le \frac{6}{\delta} + \frac{2 \alpha_-^2}{\delta^2} + \frac{2 \alpha_-^2}{\delta^2} \Biggl(1 + \sqrt{1+\tfrac{6\beta}{\alpha_-^2} R_2}\Biggr)^2, \\ &\lesssim \frac{1}{\delta} + \frac{\alpha_-^2}{\delta^2}\\ &\le \frac{1+\alpha_-^2}{\delta^2}. \end{aligned} \end{equation*}

On the other hand, on \(E\cap \mathbb{R}^{d-1}\times \{0\}\), it follows that

\begin{align*} |\nabla z(x, 0)|^2 &= \frac{x^2}{\delta^2} + \frac{\alpha^2_-}{\delta^2}\\ &\le \frac{6}{\delta} + \frac{\alpha_-^2}{\delta^2}\\ &\lesssim \frac{1+\alpha_-^2}{\delta^2}. \end{align*}

Finally, since \(\delta< 1\) and \(\beta>1\),

\[ \lvert \Delta z(x, y)\rvert^2 = \Biggl\lvert -\frac{1}{\delta} - \frac{\beta}{\delta^2}\Biggr\rvert^2 \lesssim \frac{\beta^2}{\delta^4}. \]