\[ \DeclareMathOperator{\supp}{supp} \]

Let \(\vartheta\) be as defined in the proof of (0x68c9184d) . Then

\[ |\nabla \vartheta(x,0)|^2 \leq \tfrac{C (1+\alpha_-^2)}{\delta^2R_2^2}, \]

where \(C>0\) is a numerical constant.

Proof

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

Since

\[ \vartheta_2(0) = 1, \quad \vartheta_2'(0) = 0, \]

we obtain

\[ |\nabla \vartheta(x,0)|^2 = \lvert \vartheta_1'(z(x,0))\rvert^2 \lvert \nabla z(x,0)\rvert^2. \]

We choose \( \vartheta_1 \) such that \( |\vartheta_1'(t)| \leq \tfrac{2}{R_2} \) (cf. (0x68edf29b) ). Combining this with the bound \(\lvert \nabla z(x,0)\rvert^2\lesssim \frac{1+\alpha_-^2}{\delta^2}\) given in (0x68ecae27) , we obtain

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

which completes the proof.

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