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?


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?

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

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