# If $p(v)=0\Leftrightarrow v=0$, $p(\lambda v)=|\lambda|p(v)$, then $p$ is a norm if and only if the unit ball is convex

I want to show that:

If $$p(v)=0\Leftrightarrow v=0$$, $$p(\lambda v)=|\lambda|p(v)$$, then $$p$$ is a norm if and only if the unit ball is convex.

One direction is easy: If $$p$$ is a norm, then for any $$\lambda\in[0,1]$$ one has $$\begin{equation*} p(\lambda x+(1-\lambda)y)\leq p(\lambda x)+(1-\lambda)p(y)=\lambda p(x)+(1-\lambda)p(y)\leq\lambda+ (1-\lambda)=1\text{,} \end{equation*}$$so $$\left\{\lambda x+(1-\lambda)y|\lambda\in[0,1]\right\}$$ is contained in the unit ball.

But I have no clue how to do the other direction. Can anyone give me a hint?

## 1 Answer

Take $$x,y\neq0$$ and set $$x^*=\frac{x}{\rho(x)}$$ and so $$y^*=\frac{y}{\rho(y)}$$ $$\rho(\frac{x+y}{\rho(x)+\rho(y)})=\rho(\frac{\rho(x)}{\rho(x)+\rho(y)}x^*+\frac{\rho(y)}{\rho(x)+\rho(y)}y^*)=\star$$Now, notice that $$\frac{\rho(y)}{\rho(x)+\rho(y)}=1-\frac{\rho(x)}{\rho(x)+\rho(y)}$$so set $$\lambda=\frac{\rho(x)}{\rho(x)+\rho(y)}$$ (can you prove that $$\lambda\in[0,1]$$?), so you have $$\star=\rho(\lambda x^*+(1-\lambda)y^*)$$ and $$x^*,y^*$$ belong to the unit ball. Can you continue from here?

• aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaah as usual a severe case of "not seeing the wood for all the trees". Apr 16 at 15:07