# seminorm

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

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

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

$\mathrm{p}(k\mathbf{u})$ | $=|k|\mathrm{p}(\mathbf{u}),k\in K,\mathbf{u}\in V$ | Homogeneity | ||

$\mathrm{p}(\mathbf{u}+\mathbf{v})$ | $\le \mathrm{p}(\mathbf{u})+\mathrm{p}(\mathbf{v}),\mathbf{u},\mathbf{v}\in V,$ | Sublinearity |

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

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

$${B}_{k}=\{\mathbf{u}\in V:\mathrm{p}(\mathbf{u})\le k\}$$ |

denote
the ball of radius $k$. The homogeneity property is equivalent^{} to the
assertion that

$${B}_{k}=k{B}_{1},$$ |

in the sense that $\mathbf{u}\in {B}_{1}$ 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 ${\mathrm{p}}_{B}:V\to {\mathbb{R}}^{+}$ by

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

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

###### Proposition 1

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

$$\lambda \mathbf{u}\in B\mathit{\hspace{1em}}\mathit{\text{if and only if}}\mathit{\hspace{1em}}\parallel \lambda \parallel \le k.$$ |

###### Proposition 2

Suppose that $\mathrm{p}$ is homogeneous^{}. Then, it is sublinear if and
only if its unit ball, ${B}_{\mathrm{1}}$, 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^{},

$$\mathrm{p}(k\mathbf{u}+(1-k)\mathbf{v})\le k\mathrm{p}(\mathbf{u})+(1-k)\mathrm{p}(\mathbf{v}).$$ |

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

$$\mathrm{p}(\mathbf{u}+\mathbf{v})\le \mathrm{p}(\mathbf{u})+\mathrm{p}(\mathbf{v}).$$ |

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

$$\mathrm{p}(\mathbf{u})=\mathrm{p}(\mathbf{v})=0.$$ |

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,

$$\mathrm{p}(\mathbf{u}+\mathbf{v})\le \frac{2}{k}.$$ |

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

$$\mathrm{p}(\mathbf{u}+\mathbf{v})=0,$$ |

as desired.

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

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

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

$$\mathrm{p}(\mathbf{v})=1.$$ |

For every $k$ such that $$ 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

$$k\mathrm{p}(\mathbf{u}+\mathbf{v})\le 1,$$ |

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

$$\mathrm{p}(\mathbf{u}+\mathbf{v})\le \mathrm{p}(\mathbf{v}).$$ |

The same argument^{} also shows that

$$\mathrm{p}(\mathbf{v})=\mathrm{p}(-\mathbf{u}+(\mathbf{u}+\mathbf{v}))\le \mathrm{p}(\mathbf{u}+\mathbf{v}),$$ |

and hence

$$\mathrm{p}(\mathbf{u}+\mathbf{v})=\mathrm{p}(\mathbf{v}),$$ |

as desired.

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

$$\frac{\mathbf{u}}{\mathrm{p}(u)},\frac{\mathbf{v}}{\mathrm{p}(v)}$$ |

are both in ${B}_{1}$, and hence

$$\frac{\mathrm{p}(u)}{\mathrm{p}(u)+\mathrm{p}(v)}\frac{\mathbf{u}}{\mathrm{p}(u)}+\frac{\mathrm{p}(v)}{\mathrm{p}(u)+\mathrm{p}(v)}\frac{\mathbf{v}}{\mathrm{p}(v)}=\frac{\mathbf{u}+\mathbf{v}}{\mathrm{p}(u)+\mathrm{p}(v)}$$ |

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

$$\mathrm{p}(u+v)\le \mathrm{p}(u)+\mathrm{p}(v),$$ |

as desired.

Title | seminorm |
---|---|

Canonical name | Seminorm |

Date of creation | 2013-03-22 12:24:57 |

Last modified on | 2013-03-22 12:24:57 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 20 |

Author | rmilson (146) |

Entry type | Definition |

Classification | msc 46B20 |

Synonym | semi-norm |

Defines | homogeneous |