Let us consider the following event: for a fixed $x \in \mathbb{R}^n$ with $\|x\|=1$ and $\textbf{z} \sim N(0,\textbf{I}_n)$, $$ A_x(\textbf{z}) = \{\exists w \in \mathbb{R}^n \text{ such that } \|w - \textbf{z}\| \le R, \text{ and } \text{sign}(\langle w, x\rangle) \ne \text{sign}(\langle \textbf{z}, x\rangle) \}. $$ I am trying to show that $$ A_x(\textbf{z}) \text{ happens } \iff |\langle \textbf{z}, x\rangle| < R. $$
($\implies$) This direction is readily done as follow. Suppose $A_x(\textbf{z})$ happens. Then there exists $w \in \mathbb{R}^n$ such that $\|w - \textbf{z}\| \le R$ and $\text{sign}(\langle w, x\rangle) \ne \text{sign}(\langle \textbf{z}, x\rangle)$. Therefore, $$ R \ge \|w - \textbf{z}\| \ge |\langle (w - \textbf{z}), x\rangle | \ge |\langle \textbf{z}, x\rangle | $$ where the third inequality uses $\text{sign}(\langle w, x\rangle) \ne \text{sign}(\langle \textbf{z}, x\rangle)$.
However, I am not sure why the reverse direction is true. Here is my attempt.
($\impliedby$) Suppose $R\ge |\langle \textbf{z}, x\rangle |$. Then our goal is to find $w \in \mathbb{R}^n$ satisfying (i) $\|w - \textbf{z}\| \le R$ and (ii) $\text{sign}(\langle w, x\rangle) \ne \text{sign}(\langle \textbf{z}, x\rangle)$. I was trying to construct $w$ by setting $w = -a\textbf{z}$ for some positive constant $a$. By doing so, the condition (ii) is automatically satisfied. And I was hoping that (i) can be satisfied by carefully choosing $a$. However, this only works under the condition of $\|\textbf{z}\| \le R$, not $|\langle \textbf{z}, x\rangle | \le R$. This is becuase, by setting $w = -a\textbf{z}$, we have $$ \|w - \textbf{z}\| = (1+a)\|\textbf{z}\|. $$ In order for this to be less or equal to $R$, $$ a + 1 \le \frac{R}{\|\textbf{z}\|} \iff a \le \frac{R-\|\textbf{z}\|}{\|\textbf{z}\|}. $$ Since $a$ is positive, we need $\|\textbf{z}\| < R$.
Any suggestions/answers/comments will be very appreciated.
You can simply let $w=z-R\cdot \text{sign}(\langle z,x\rangle)x$. Then we have that $\|w-z\|=R\|x\|=R$ and $$ \langle w,x\rangle=\langle z,x\rangle-R\cdot\text{sign}(\langle z,x\rangle). $$ We can see that $$ \langle w,x\rangle\langle z,x\rangle=|\langle z,x\rangle|^2-R\cdot |\langle z,x\rangle|\le 0 $$ so $\langle w,x\rangle$ and $\langle z,x\rangle$ cannot be both positive or negative.