Norms are non-negative

More specifically, $\forall x\in X$ we have that $p(x)>0$.

Proof

We need the absolute homogeneity of $p$ and the triangle inequality:

$$\begin{array}{rll}0& =p(0)& \text{By the triangle inequality}\\ & \le p(x)+p(-x)& \text{By absolute homegeneity}\\ & =p(x)+|1|p(x)& \\ & =2p(x)& \\ & \Downarrow & \\ p(x)& \ge 0& \forall x\in X\end{array}$$