seminorm

Let $V$ be a real, or a complex vector space, with $K$ denoting thecorresponding field of scalars. A seminorm is a function

\\operatorname{p}:V\\to\\mathbb{R}^{+},

from $V$ to the set of non-negative real numbers, that satisfies thefollowing two properties.

	$\\displaystyle\\operatorname{p}(k\\,\\mathbf{u})$	$\\displaystyle=\|k\|\\operatorname{p}(\\mathbf{u}),\\quad k\\in K,\\;\\mathbf{u}\\in V$	Homogeneity
	$\\displaystyle\\operatorname{p}(\\mathbf{u}+\\mathbf{v})$	$\\displaystyle\\leq\\operatorname{p}(\\mathbf{u})+\\operatorname{p}(\\mathbf{v}),%\\quad\\mathbf{u},\\mathbf{v}\\in V,$	Sublinearity

A seminorm differs from a norm in that it is permitted that $\\operatorname{p}(\\mathbf{u})=0$ for some non-zero $\\mathbf{u}\\in V.$

It is possible to characterize the seminorms propertiesgeometrically. For $k>0$ , let

B_{k}=\\{\\mathbf{u}\\in V:\\operatorname{p}(\\mathbf{u})\\leq k\\}

denotethe ball of radius $k$ . The homogeneity property is equivalent to theassertion that

B_{k}=kB_{1},

in the sense that $\\mathbf{u}\\in B_{1}$ if and only if $k\\mathbf{u}\\in B_{k}.$ Thus, we see that a seminorm is fully determined by its unit ball.Indeed, given $B\\subset V$ we may define a function $\\operatorname{p}_{B}:V\\to\\mathbb{R}^{+}$ by

\\operatorname{p}_{B}(\\mathbf{u})=\\inf\\{\\lambda\\in\\mathbb{R}^{+}:\\lambda^{-1}%\\mathbf{u}\\in B\\}.

The geometric nature of the unit ball is described by the following.

Proposition 1

The function $\\operatorname{p}_{B}$ satisfies the homegeneity property if and only iffor every $\\mathbf{u}\\in V$ , there exists a $k\\in\\mathbb{R}^{+}\\cup\\{\\infty\\}$ such that

\\lambda\\,\\mathbf{u}\\in B\\quad\\text{if and only if}\\quad\\|\\lambda\\|\\leq k.

Proposition 2

Suppose that $\\operatorname{p}$ is homogeneous. Then, it is sublinear if andonly if its unit ball, $B_{1}$ , is a convex subset of $V$ .

Proof. First, let us suppose that the seminorm is both sublinearand homogeneous, and prove that $B_{1}$ is necessarily convex. Let $\\mathbf{u},\\mathbf{v}\\in B_{1}$ , and let $k$ be a real number between $0$ and $1$ .We must show that the weighted average $k\\mathbf{u}+(1-k)\\mathbf{v}$ is in $B_{1}$ aswell. By assumption,

\\operatorname{p}(k\\,\\mathbf{u}+(1-k)\\mathbf{v})\\leq k\\operatorname{p}(\\mathbf{%u})+(1-k)\\operatorname{p}(\\mathbf{v}).

The right side is aweighted average of two numbers between $0$ and $1$ , and is thereforebetween $0$ and $1$ itself. Therefore

k\\,\\mathbf{u}+(1-k)\\mathbf{v}\\in B_{1},

as desired.

Conversely, suppose that the seminorm function is homogeneous, andthat the unit ball is convex. Let $\\mathbf{u},\\mathbf{v}\\in V$ be given, and let us show that

\\operatorname{p}(\\mathbf{u}+\\mathbf{v})\\leq\\operatorname{p}(\\mathbf{u})+%\\operatorname{p}(\\mathbf{v}).

The essential complication here is that we do not exclude thepossibility that $\\operatorname{p}(\\mathbf{u})=0$ , but that $\\mathbf{u}\eq 0$ .First, let us consider the case where

\\operatorname{p}(\\mathbf{u})=\\operatorname{p}(\\mathbf{v})=0.

By homogeneity, for every $k>0$ we have

k\\mathbf{u},\\,k\\mathbf{v}\\in B_{1},

and hence

\\frac{k}{2}\\,\\mathbf{u}+\\frac{k}{2}\\,\\mathbf{v}\\in B_{1},

as well. By homogeneity, again,

\\operatorname{p}(\\mathbf{u}+\\mathbf{v})\\leq\\frac{2}{k}.

Since the above is true for all positive $k$ , we infer that

\\operatorname{p}(\\mathbf{u}+\\mathbf{v})=0,

as desired.

Next suppose that $\\operatorname{p}(\\mathbf{u})=0$ , but that $\\operatorname{p}(\\mathbf{v})\eq 0$ . We will showthat in this case, necessarily,

\\operatorname{p}(\\mathbf{u}+\\mathbf{v})=\\operatorname{p}(\\mathbf{v}).

Owing to the homogeneity assumption, we may without loss of generalityassume that

\\operatorname{p}(\\mathbf{v})=1.

For every $k$ such that $0\\leq k<1$ we have

k\\,\\mathbf{u}+k\\,\\mathbf{v}=(1-k)\\frac{k\\,\\mathbf{u}}{1-k}+k\\,\\mathbf{v}.

The right-side expression is an element of $B_{1}$ because

\\frac{k\\,\\mathbf{u}}{1-k},\\,\\mathbf{v}\\in B_{1}.

Hence

k\\operatorname{p}(\\mathbf{u}+\\mathbf{v})\\leq 1,

and since this holds for $k$ arbitrarily close to $1$ we conclude that

\\operatorname{p}(\\mathbf{u}+\\mathbf{v})\\leq\\operatorname{p}(\\mathbf{v}).

The same argument also shows that

\\operatorname{p}(\\mathbf{v})=\\operatorname{p}(-\\mathbf{u}+(\\mathbf{u}+\\mathbf{%v}))\\leq\\operatorname{p}(\\mathbf{u}+\\mathbf{v}),

and hence

\\operatorname{p}(\\mathbf{u}+\\mathbf{v})=\\operatorname{p}(\\mathbf{v}),

as desired.

Finally, suppose that neither $\\operatorname{p}(\\mathbf{u})$ nor $\\operatorname{p}(\\mathbf{v})$ is zero. Hence,

\\frac{\\mathbf{u}}{\\operatorname{p}(u)},\\,\\frac{\\mathbf{v}}{\\operatorname{p}(v)}

are both in $B_{1}$ , and hence

\\frac{\\operatorname{p}(u)}{\\operatorname{p}(u)+\\operatorname{p}(v)}\\frac{%\\mathbf{u}}{\\operatorname{p}(u)}+\\frac{\\operatorname{p}(v)}{\\operatorname{p}(u%)+\\operatorname{p}(v)}\\frac{\\mathbf{v}}{\\operatorname{p}(v)}=\\frac{\\mathbf{u}+%\\mathbf{v}}{\\operatorname{p}(u)+\\operatorname{p}(v)}

is in $B_{1}$ also. Using homogeneity, we conclude that

\\operatorname{p}(u+v)\\leq\\operatorname{p}(u)+\\operatorname{p}(v),

as desired.