$\kappa=\frac12$, then, in view of the fact that $|\xi-\eta|\leq \kappa\leq 1$, the above inequality implies $$\begin{gathered}
K \geq C^{-1} c_0^2|\xi|^{2p}-Cc_0 -C\|\sigma\|^2_{L^\infty({\mathbb{R}^d})} C_0 (1+|\xi|^{p+1})  -C\|\sigma\|^2_{L^\infty({\mathbb{R}^d})}\|D_x\sigma\|^2_{L^\infty({\mathbb{R}^d})}|\xi|^2 \\
-C \|\sigma\|^2_{L^\infty({\mathbb{R}^d})} C_0  (1+|\xi|^{p}),\end{gathered}$$ where $C$ depends only on $p$. Since $p>1$, this yields that $|\xi|\leq R_K$, as desired.

Consider the case that $$\sigma(\xi,x) = \sigma(\xi) = \sqrt{2}\left( I_d- \frac{\xi}{\left| \xi \right|}\otimes \frac{\xi}{\left| \xi \right|} \right) \quad \mbox{and} \quad   H(\xi,x)= a(x)\left|\xi \right|,$$ where $a$ is a random field satisfying $c_0 \leq a(x) \leq C_0$. Then $$A(\xi) = \frac12\sigma(\xi) \sigma(\xi)^T =  I_d- \frac{\xi}{\left| \xi \right|}\otimes \frac{\xi}{\left| \xi \right|}$$ and so equation  is the forced mean curvature equation .

We claim that, as pointed out in [@LS0], the (LS) condition  is satisfied provided that $$\label{e.CondLSMCM}
\inf_{x\in {\mathbb{R}}^d} \left(a(x))^2 -(d-1) \left|D_xa(x)\right|\right) > 0.$$ To check this, we suppose $\mathcal C_{\sigma,H}(\xi,x) \leq K$. Then there exists $\eta\in B_\kappa(\xi)$ such that $$K \geq \theta(1-2\theta) a(x)^2|\eta|^2 -\theta (1+\kappa)^2(d-1) \left|D_xa(x)\right||\eta||\xi| 
-\theta \kappa (1+\kappa)^2 (d-1)^2 c_0 |\xi|.
$$ Using that $|\eta-\xi|\leq \kappa\leq 1$, we get $$K \geq \theta(1-2\theta) \left(  a(x)^2 -\frac{(1+\kappa)^2}{(1-2\theta)}(d-1) \left|D_xa(x) \right|\right)|\xi|^2 -C |\xi|.$$ In view of assumption , we may choose $\kappa$, $\delta$ and $\theta$ so small that $$a^2 -\frac{(1+\kappa)^2}{(1-2\theta)}(d-1) \left|D_xa \right|\geq \delta>0\quad \mbox{in} \ {\mathbb{R}}^d.$$ This implies that $|\xi|\leq R_K$ 