I'm trying to understand a proof and from my current understanding there is one thing not clear/wrong. Maybe you can help:
Let $(X, \lVert\cdot\rVert)$ a normed space and $B(0,1)$ the closed unit ball. Then the following is equivalent:
- $(X, \lVert\cdot\rVert)$ is strictly convex.
- $\forall x_1,x_2\in B(0,1)\colon \lVert\frac{x_1 +x_2}{2}\rVert=1\Rightarrow x_1=x_2$
- $\forall x_0,x_1,x_2\in X \forall r>0\colon x_1,x_2\in B(0,1)\wedge \lVert x_1-x_2\rVert=2r\Rightarrow \frac{x_1+x_2}{2}=x_0$
The part $2\Rightarrow3$ states: Let $x_0,x_1,x_2\in X,r>0, x_1,x_2\in B(0,1)$ and $\lVert x_1-x_2\rVert=2r$. Then $y_1 \colon=\frac{x_1-x_0}{r}, y_2\colon= -\frac{x_2-x_0}{r}\in B(0,1)$. But from my understanding this is not true. For instance $1,-1\in B(0,1)$ and $y_1=\frac{1-(-1)}{1}=2\notin B(0,1)$. What do I miss here? The proof continues: We have $\lVert\frac{y_1+y_2}{2}\rVert= \lVert\frac{x_1-x_2}{2r}\rVert=1\Rightarrow y_1=y_2$ and $\frac{x_1-x_0}{r}= -\frac{x_2-x_0}{r}\Rightarrow\frac{x_1+x_2}{2}=x_0$