Radial symmetry This is a full theorem and proof copied from PDE Evans, 2nd edition, pages 558-559. My two questions about two parts of the proof are on the bottom of this post.

THEOREM 2 (Radial symmetery). Let $u \in C^2(\bar{U})$ satisfy $\text{(2)}$, $\text{(3)}$. Then $U$ is radial; that is, $$u(x)=v(r) \quad (r=|x|)$$ for some strictly decreasing function $v : [0,1] \to [0,\infty)$.

Note that $(2)$ and $(3)$ can be seen on pages 555-556 or here in my previous question.

Proof. 1. We consider for each $0 \le \lambda < 1$ the statement $$u(x) < u(x_\lambda) \text{ for each point } x \in E_\lambda. \tag{$15_\lambda$}$$
$\quad$ 2. According to Lemma 2, $(15_\lambda)$ is valid for each $\lambda < 1$, $\lambda$ sufficiently close to $1$. Set $$\lambda_0 := \inf\{0 \le \lambda < 1 \mid (15_\mu) \text{ holds for each }\lambda \le \mu < 1\}. \tag{16}$$ We will prove $$\lambda_0 = 0. \tag{17}$$ Assume instead $\lambda_0 > 0$. Write $w(x) := u(x_{\lambda_0})-u(x)$ $(x \in E_{\lambda_0})$. Then $$-\Delta w = f(x_{\lambda_0})-f(u(x))=-cw \quad \text{in }E_{\lambda_0},$$ for $c(x) := -\int_0^1 f'(su(x_{\lambda_0})+(1-s)u(x)) \, ds$. As $w \ge 0$ in $E_{\lambda_0}$, we deduce from Lemma 1 (applied to $V =E_{\lambda_0}$) that $w > 0$ in $E_{\lambda_0}$, $w_{x_n} > 0$ on $P_{\lambda_0} \cap U$. Thus $$u(x) < u(x_{\lambda_0}) \quad \text{in } E_{\lambda_0}, \tag{18}$$ and $$u_{x_n} < 0 \quad \text{on } P_{\lambda_0} \cap U. \tag{19}$$ Using $(18)$, $(19)$ and Lemma 2, we conclude $$u(x) < u(x_{\lambda_0-\epsilon}) \quad \text{in }E_{\lambda_0-\epsilon} \text{ for all }0 \le \epsilon \le \epsilon_0, \tag{20}$$ if $\epsilon_0$ is small enough. Assertion $(20)$ contradicts our choice $(16)$ of $\lambda_0$, if $\lambda_0 > 0$.
$\quad$ 3. Since $\lambda = 0$, we see $u(x_1,\ldots,x_{n-1},-x_n) \le u(x_1,\ldots,x_n)$ for all $x \in U \cap \{x_n > 0\}$. A similar argument in $U \cap \{x_n < 0\}$ shows $u(x_1,\ldots,x_{n-1},-x_n) \le u(x_1,\ldots,x_n)$ for all $x \in U \cap \{x_n < 0\}$. Thus $u$ is symmetric in the plane $P_0$ and $u_{x_n}=0$ on $P_0$.
$\quad$ This argument applies as well after any rotation of coordinate axes, and so the theorem follows. $\blacksquare$

My questions:
(There are two parts of this proof I do not understand.)

*

*Why is $w \ge 0$ in $E_{\lambda_0}$? Initially, I thought $u(x) < u(x_{\lambda_0})$, which is taken from $(15_\lambda)$ (replaced $\lambda \to \lambda_0)$ for this purpose. However, it turns out that $u(x) < u(x_{\lambda_0})$ is what we were trying to prove in $(18)$.


*After establishing $\lambda_0=0$, how can we conclude that $u(x_1,\ldots,x_{n-1}, -x_n) \ge u(x_1,\ldots,x_n)$ on $U \cap \{x_n > 0\}$?
 A: *

*Your initial thoughts are correct here. Let $\{\lambda_n\}$ be a minimizing sequence in $(16)$ converging to $\lambda_0$. Then $x_{\lambda_n} \to x_{\lambda_0}$, while $E_{\lambda_0} = \bigcup_{n = 1}^{\infty}E_{\lambda_n}$, with $E_{\lambda_n} \subset E_{\lambda_{n+1}}$. Moreover, $u(x) < u(x_{\lambda_n})$ for each $x \in E_{\lambda_n}$.
Now let $x \in E_{\lambda_0}$. Then there is $N$ for which $x \in E_{\lambda_n}$ for all $n \geq N$. This implies in particular that $u(x) < u(x_{\lambda_n})$ for all $n \geq N$. As $u$ is continuous we find upon taking the limit that $u(x) \leq u(x_{\lambda_0})$. Thus $w \geq 0$ in $E_{\lambda_0}$. Note equation $(18)$ is an improvement, where we eliminate the possibility of equality: $w > 0$.

*We already know that $u(x) \leq u(x_{\lambda_0})$ for all $x = (x_1,\ldots,x_n)$ belonging to $E_{\lambda_0} = \{x \in U\,:\, 0 < \lambda < 1\} = U \cap \{x_n > 0\}$, where $x_{\lambda_0} = (x_1,\ldots,x_{n-1}, 2\lambda_0 - x_n)$. 
(Recall the equality for $E_{\lambda_0}$ follows because in this section Evans denotes $U$ as the open ball of radius $1$.)
Therefore with $\lambda_0 = 0$, this implies that $u(x_1,\ldots, x_n) \leq u(x_1,\ldots,x_{n-1},-x_n)$ for all $x \in U \cap\{x_n > 0\}$.
