If $p$, $r$ satisfy $|p|,\, |r| \le 1$ and $1=\lambda p + (1-\lambda)r$ for some $\lambda \in (0,1)$ then $p=r=1$.

Question

If $p$, $r$ are real numbers and $|p|,\, |r| \le 1$ and there exists $\lambda \in (0,1)$ such that $1=\lambda p + (1-\lambda)r$ then $p=r=1$.

Not sure why this is true? Can anybody help?

Context

It was in relation to extreme points. I'm trying to prove this and then use it to show that $(1,0)$ is an extreme point of $S=\{x \in R^2 : \|x\|_{L^1} \le 1\}$. Causing me some issues.

Answer 1

Suppose without loss of generality that $p \le r$. Conclude that $1 \le r$ and hence $r=1$. Then rearrange to obtain $0 = (p-1)\epsilon$.