0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaah as usual a severe case of "not seeing the wood for all the trees". $\endgroup$ Apr 16 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.