Recall the definition of a dual norm is
$$ {\|x \|}_* = {\max}_{\|z \|\leq 1} z^Tx $$
Positivity: if $x=0$, then it's obvious that $ {\|x\|}_* $ is also $0$. $$ {\|x\|}_* = \max_{\|z \|\leq 1} \sum z_ix_i \leq \max_{\|z \|\leq 1} \sum |z_ix_i|$$
I'm stuck here and I don't how to show this greater zero.
Also have no idea on how to prove scaling, triangle inequality.
Please help! Thanks
It is clear that $\|0\|_* = 0$.
If $x \neq 0$ then $\|x\| \neq 0$ and taking $z={x \over \|x\|}$, we have $\|x\|_* \ge {\|x\|_2^2 \over \|x\|} >0$. In particular, if $\|x\|_* = 0$ we must have $x=0$.
Note that $\|x\|_* = \max_{\|z\| \le 1} |z^T x|$ and so $\|tx\|_* = \max_{\|z\| \le 1} |z^T tx|= \max_{\|z\| \le 1} |t||z^T x|= |t| \max_{\|z\| \le 1} |z^T x| = |t| \|x\|_*$
Note that the functions $x \mapsto |z^T x|$ are convex, hence $x \mapsto \|x\|_*$ is convex.
By noting that $x+y = 2( {x +y \over 2})$, we can combine the last two facts to show the triangle inequality.