# seminorm

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

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

from $V$ to the set of non-negative real numbers, that satisfies the following 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 properties geometrically. For $k>0$, let

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

denote the ball of radius $k$. The homogeneity property is equivalent to the assertion 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 if for 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 and only if its unit ball, $B_{1}$, is a convex subset of $V$.

Proof. First, let us suppose that the seminorm is both sublinear and 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 $B_{1}$ as well. 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 a weighted average of two numbers between $0$ and $1$, and is therefore between $0$ and $1$ itself. Therefore

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

as desired.

Conversely, suppose that the seminorm function is homogeneous, and that 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 the possibility that $\operatorname{p}(\mathbf{u})=0$, but that $\mathbf{u}\neq 0$. 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 $\operatorname{p}(\mathbf{v})\neq 0$. We will show that in this case, necessarily,

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

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

Title seminorm Seminorm 2013-03-22 12:24:57 2013-03-22 12:24:57 rmilson (146) rmilson (146) 20 rmilson (146) Definition msc 46B20 semi-norm homogeneous