Let $V$ be a real, or a complex vector space, with $K$ denoting the corresponding field of scalars. A seminorm is a function $$\operatorname{p}:V\to\reals^+,$$ from $V$ to the set of non-negative real numbers, that satisfies the following two properties.

		Homogeneity
		Sublinearity

It is possible to characterize the seminorms properties geometrically. For $k>0$ , let $$B_k = \{ \bu \in V: \snorm{\bu}\leq k\}$$ denote the ball of radius $k$ . The homogeneity property is equivalent to the assertion that $$B_k = k B_1,$$ in the sense that $\bu \in B_1$ if and only if $k\bu\in B_k.$ Thus, we see that a seminorm is fully determined by its unit ball. Indeed, given $B\subset V$ we may define a function $\pnorm_B:V\to \reals^+$ by $$\pnorm_B(\bu) = \inf\{ \lambda\in\reals^+ : \lambda^{-1}\bu\in B\}.$$ The geometric nature of the unit ball is described by the following.

Conversely, suppose that the seminorm function is homogeneous, and that the unit ball is convex. Let $\bu,\bv\in V$ be given, and let us show that $$\snorm{\bu+\bv}\leq \snorm{\bu}+\snorm{\bv}.$$ The essential complication here is that we do not exclude the possibility that $\snorm{\bu}=0$ , but that $\bu\neq 0$ . First, let us consider the case where $$\snorm{\bu}=\snorm{\bv}=0.$$ By homogeneity, for every $k>0$ we have $$k\bu,\,k\bv\in B_1,$$ and hence $$\frac{k}{2}\, \bu + \frac{k}{2}\,\bv\in B_1, $$ as well. By homogeneity, again, $$\snorm{\bu+\bv}\leq \frac{2}{k}.$$ Since the above is true for all positive $k$ , we infer that $$\snorm{\bu+\bv} = 0,$$ as desired.

Next suppose that $\snorm{\bu}=0$ , but that $\snorm{\bv}\neq 0$ . We will show that in this case, necessarily, $$\snorm{\bu+\bv} = \snorm{\bv}.$$ Owing to the homogeneity assumption, we may without loss of generality assume that $$\snorm{\bv}=1.$$ For every $k$ such that $0\leq k<1$ we have $$ k\,\bu+k\,\bv = (1-k) \frac{k\,\bu}{1-k} + k\,\bv.$$ The right-side expression is an element of $B_1$ because $$\frac{k\,\bu}{1-k},\, \bv \in B_1.$$ Hence $$k\snorm{\bu+\bv} \leq 1,$$ and since this holds for $k$ arbitrarily close to $1$ we conclude that $$\snorm{\bu+\bv}\leq \snorm{\bv}.$$ The same argument also shows that $$\snorm{\bv} = \snorm{-\bu+(\bu+\bv)} \leq \snorm{\bu+\bv},$$ and hence $$\snorm{\bu+\bv}=\snorm{\bv},$$ as desired.

Finally, suppose that neither $\snorm{\bu}$ nor $\snorm{\bv}$ is zero. Hence, $$\frac{\bu}{\snorm{u}},\, \frac{\bv}{\snorm{v}}$$ are both in $B_1$ , and hence $$\frac{\snorm{u}}{\snorm{u}+\snorm{v}} \frac{\bu}{\snorm{u}}+ \frac{\snorm{v}}{\snorm{u}+\snorm{v}} \frac{\bv}{\snorm{v}} = \frac{\bu+\bv}{\snorm{u}+\snorm{v}} $$ is in $B_1$ also. Using homogeneity, we conclude that $$\snorm{u+v}\leq \snorm{u}+\snorm{v},$$ as desired.