Seminorm

Remarks

The two conditions imply $p(0) = 0$ and $p(x) \ge 0$ for all $x$. Indeed $p(0) = p(0 \cdot 0) = 0 \cdot p(0) = 0$, and then $0 = p(x - x) \le p(x) + p(-x) = 2 p(x)$. A seminorm is a norm exactly when $p(x) = 0$ implies $x = 0$. Example: on the space $C([0, 1])$ of continuous real functions, $p(f) = \sup_{0 \le t \le 1/2} \abs{f(t)}$ is a seminorm; both conditions follow from the same inequalities for the absolute value at each point. It is not a norm, since it vanishes on a nonzero function supported in $(1/2, 1]$. Non-example: on $\R$, $p(x) = x^2$ is not a seminorm, since $p(2 \cdot 1) = 4 \ne 2 = \abs{2} \, p(1)$. The kernel $\set{x : p(x) = 0}$ is a linear subspace, and $p$ induces a norm on the quotient of $V$ by it.