PlanetMath: seminorm

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seminorm

(Definition)

Let be a real, or a complex vector space, with denoting the corresponding field of scalars. A seminorm is a function

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

from

to the set of non-negative real numbers, that satisfies the following two properties.

$\displaystyle \operatorname{p}(k\, \mathbf{u})$	$\displaystyle = \vert k\vert \operatorname{p}(\mathbf{u}),\quad k\in K,\; \mathbf{u}\in V$	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 properties geometrically. For , let

$\displaystyle B_k = \{ \mathbf{u}\in V: \operatorname{p}(\mathbf{u})\leq k\}$

denote the ball of radius

. The homogeneity property is equivalent to the assertion that

$\displaystyle B_k = k B_1,$

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

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

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 if for every $\mathbf{u}\in V$ , there exists a $k\in\mathbb{R}^+\cup \{\infty\}$ such that

$\displaystyle \lambda\,\mathbf{u}\in B$ if and only if $\displaystyle \quad \Vert \lambda\Vert \leq k.$

Proposition 2 Suppose that $\operatorname{p}$ is homogeneous. Then, it is sublinear if and only if its unit ball, , is a convex subset of .

Proof. First, let us suppose that the seminorm is both sublinear and homogeneous, and prove that

is necessarily convex. Let $\mathbf{u},\mathbf{v}\in B_1$ , and let

be a real number between 0 and

. We must show that the weighted average $k\mathbf{u}+(1-k)\mathbf{v}$ is in

as well. By assumption,

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

The right side is a weighted average of two numbers between 0 and

, and is therefore between 0 and

itself. Therefore

$\displaystyle k\,\mathbf{u}+(1-k)\mathbf{v}\in B_1,$

as desired.

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

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

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

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

By homogeneity, for every

we have

$\displaystyle k\mathbf{u},\,k\mathbf{v}\in B_1,$

and hence

$\displaystyle \frac{k}{2}\, \mathbf{u}+ \frac{k}{2}\,\mathbf{v}\in B_1,$

as well. By homogeneity, again,

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

Since the above is true for all positive

, we infer that

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

as desired.

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

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

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

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

For every

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

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

The right-side expression is an element of

because

$\displaystyle \frac{k\,\mathbf{u}}{1-k},\, \mathbf{v}\in B_1.$

Hence

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

and since this holds for

arbitrarily close to

we conclude that

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

The same argument also shows that

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

and hence

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

as desired.

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

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

are both in

, and hence

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