# A problem about using the moving plane method to prove radial symmetry of the $C^{2}$ global solution of a elliptic PDE in $R^{2}$

Recently I'm learning the use of moving plane method to prove radial symmetry of $$C^{2}$$ global solution of a PDE in $$R^{2}$$, and I'm reading a paper where this method is applied: precisely I'm reading "Monotonicity and symmetry of solutions of fully nonlinear elliptic equations on unbounded domains" written by Congming Li.

Since the Laplacian operator is invariant under rotations, we only have to prove the symmetry about a line across origin: thus I try to prove $$u(x,y)=u(-x,y)$$.

Using maximum principle I have proved that $$w \leqslant 0 \ in \ \Sigma(\lambda) \ for \ \lambda \leqslant 0$$ and $$w>0 \ in \ \Sigma(\lambda) \ for \ \lambda>0.$$ By using Hopf lemma we have $$w_{1}(0,0)>0 \ for \ y \ \in \mathbb{R},$$ where

• $$\Sigma(\lambda)=\left\{(x, y) \in \mathbb{R}^{2} \mid x<\lambda\right\}$$ and
• $$v=u(2 \lambda-x, y)$$, $$w(x,\lambda)=v-u$$.

What I'm confused about is that in his paper Dr.Li says we should use the maximality of 0: trying to follow his work, I set a sequence such that $$\lambda^{k} \searrow 0, x^{k} \in \sum\left(\lambda^{k}\right), w\left(x^{k}, \lambda^{k}\right)>0$$ $$w\left(x^{k}, \lambda^{k}\right)=\max _{x \in \Sigma\left(\lambda^{k}\right) \atop 0\leq\lambda \leq \lambda^{k}} w(x, \lambda)>0$$ $$\nabla_{x} w\left(x^{k}, \lambda^{k}\right)=0,\left\{w_{i j}\left(x^{k}, \lambda^{k}\right)\right\} \leq 0$$ The paper said that we can 'assume' $$x^{k} \longrightarrow 0 \in \overline{\sum(0)}$$, and taking the limit we have $$\nabla_{x} w(0,0)=0$$ this contradicts to the results I get using Hopf lemma that $$w_{1}(0,0)>0 \ \text{ for } y \ \in \mathbb{R}$$ Actually I'm so confused about the construction of this sequence so my question is: how can I prove that $$x^{k} \longrightarrow 0 \in \overline{\sum(0)}\;?$$