seminorm (original) (raw)

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

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

| p⁡(k⁢𝐮) | =|k|⁢p⁡(𝐮),k∈K,𝐮∈V | Homogeneity | | --------- | ----------------------- | ------------ | | p⁡(𝐮+𝐯) | ≤p⁡(𝐮)+p⁡(𝐯),𝐮,𝐯∈V, | Sublinearity |

A seminorm differs from a norm in that it is permitted that p⁡(𝐮)=0for some non-zero 𝐮∈V.

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

denote the ball of radius k. The homogeneity property is equivalent to the assertion that

in the sense that𝐮∈B1 if and only ifk⁢𝐮∈Bk.Thus, we see that a seminorm is fully determined by its unit ball. Indeed, given B⊂V we may define a functionpB:V→ℝ+by

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

Proposition 1

The function pB satisfies the homegeneity property if and only if for every u∈V, there exists a k∈R+∪{∞} such that

Proof. First, let us suppose that the seminorm is both sublinear and homogeneous, and prove that B1 is necessarily convex. Let𝐮,𝐯∈B1, and let k be a real number between 0 and 1. We must show that the weighted average k⁢𝐮+(1-k)⁢𝐯 is in B1 as well. By assumption,

The right side is a weighted average of two numbers between 0 and 1, and is therefore between 0 and 1 itself. Therefore

as desired.

Conversely, suppose that the seminorm function is homogeneous, and that the unit ball is convex. Let 𝐮,𝐯∈V be given, and let us show that

The essential complication here is that we do not exclude the possibility that p⁡(𝐮)=0, but that 𝐮≠0. First, let us consider the case where

By homogeneity, for every k>0 we have

and hence

as well. By homogeneity, again,

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

as desired.

Next suppose that p⁡(𝐮)=0, but that p⁡(𝐯)≠0. We will show that in this case, necessarily,

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

For every k such that 0≤k<1 we have

The right-side expression is an element of B1 because

Hence

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

The same argument also shows that

and hence

as desired.

Finally, suppose that neither p⁡(𝐮) nor p⁡(𝐯) is zero. Hence,

are both in B1, and hence

is in B1 also. Using homogeneity, we conclude that

as desired.