# Proof of equivalence in strictly convex spaces

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:

1. $(X, \lVert\cdot\rVert)$ is strictly convex.
2. $\forall x_1,x_2\in B(0,1)\colon \lVert\frac{x_1 +x_2}{2}\rVert=1\Rightarrow x_1=x_2$
3. $\forall x_0,x_1,x_2\in X \forall r>0\colon x_1,x_2\in B(0,1)\wedge \lVert x_1-x_2\rVert=2r\Rightarrow \frac{x_1+x_2}{2}=x_0$

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 $\frac{x_1-x_0}{r}= -\frac{x_2-x_0}{r}\Rightarrow\frac{x_1+x_2}{2}=x_0$

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Did you want to write $B(x_0,r)$ instead of $B(0,1)$ in the 3rd condition? In the way it is given now, no conditions on $x_0$ are given. – Martin Sleziak Jul 22 '11 at 13:54
I double-checked it the script in front of me. This writes $B(0,1)$. However if there is a $B(x_0,r)$ the proof would obviously work. I'll test the other statements. – qbi Jul 22 '11 at 14:07
@qbi: I think your comment is to the point. If your transcription is faithful this must be a typo for what you write. – t.b. Jul 22 '11 at 14:34
It was a typo as far as I see it. All equivalences work with $B(x_0,r)$ and, what is even better, also a proof of another proposition works now. Thanks for your help. I'll close this question. – qbi Jul 22 '11 at 15:24