Find a norm so that its closed unit ball is the area between $y=x^2-1$ and $y=1-x^2$ As the title specifies, I need to find an explicit formula for a norm $|||\cdot|||$ so that:
$$B_{|||\cdot|||}=\{\mathbf{x} : ||| \mathbf{x}|||\le1  \}$$ where $\mathbf{x}=(x,y)\in\mathbb R^2$,  is the area between $y=x^2-1$ and $y=1-x^2$. 
I have already found many norms but none of them satisfies that $|||\alpha \mathbf{x}|||=\alpha |||\mathbf{x}|||$ for $ \alpha \in R$

Any help will be appreciated.
A necessary and sufficient condition for a subset $B$ of a vector space $V$ to the unit ball of a norm on $V$ is that $B$ is non-empty, convex and symmetric about the origin ($-B = B$). If $B = \{(x, y) : x^2-1 \le y \le 1-x^2\}\subseteq\mathbb{R}^2$, then $B$ is non-empty, convex and symmetric about the origin and so it is the unit ball of a norm on $\mathbb{R}^2$.

Given the unit ball $B$ of a normed space there is only one possible definition of the norm, namely $\|v\| = \inf\{x &gt; 0 : v/x \in B\}$. (Because, for $x &gt; 0$, $v/x \in B$ iff $\|v/x\| \le 1$ iff $\|v\| \le x$.)
Balls in general have to be convex. I dont think its possible.