1. Ball in metric space

Let $\mathbb{B}((x,y),r)$ on $\mathbb{R}^2$ be a ball in euclidean norm. We define a norm $||x||$ that ball in this norm satisfes:
$B=\{ x \in \mathbb{R}^2: ||x||<1 \} = ( (-1,1) \times (-1,1) \cup \mathbb{B}((1,0),1) \cup \mathbb{B}((-1,0),1)$.
Is this norm determined by a dot product?

I think that this problem need to use parallelogram law ... but how?
Also we know that ball in eclidean norm is determined by the dot product. Maybe any draw should solve everything?

2. Re: Ball in metric space

Originally Posted by Camille91
Let $\mathbb{B}((x,y),r)$ on $\mathbb{R}^2$ be a ball in euclidean norm. We define a norm $||x||$ that ball in this norm satisfes:
$B=\{ x \in \mathbb{R}^2: ||x||<1 \} = ( (-1,1) \times (-1,1) \cup \mathbb{B}((1,0),1) \cup \mathbb{B}((-1,0),1)$.
Is this norm determined by a dot product?

I think that this problem need to use parallelogram law ... but how?
Also we know that ball in eclidean norm is determined by the dot product. Maybe any draw should solve everything?
Right, ok, so go on. What exactly is your idea?

3. Re: Ball in metric space

Take two vectors $x, y \in B$ and show that $||x-y||^2+||x+y||^2=2(||x||^2+||y||^2)$? I don't get it.

4. Re: Ball in metric space

Now I see it seems to be simple. This norm doesn't satisfe parallelogram law. Let's take $e_1=(1,0)$ and $e_2=(0,1)$. With so defined ball we know that: $e_2, e_1-e_2, e_1+e_2 \in S$ so $||e_2||=1, ||e_1+e_2||=1, ||e_1-e_2||=1$ but $e_1 \notin S, ||e_1||=a, a \not= 1$. $S$ is the sphere of this ball.