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4
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'seminorm'
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| Title of object: |
seminorm |
| Canonical Name: |
Seminorm |
| Type: |
Definition |
| Created on: |
2002-02-20 22:51:15-05 |
| Modified on: |
2002-11-25 15:30:25.875085-05 |
| Synonyms: |
seminorm=semi-norm |
Preamble:
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcommand{\reals}{\mathbb{R}}
\newcommand{\natnums}{\mathbb{N}}
\newcommand{\cnums}{\mathbb{C}}
\newcommand{\znums}{\mathbb{Z}}
\newcommand{\lp}{\left(}
\newcommand{\rp}{\right)}
\newcommand{\lb}{\left[}
\newcommand{\rb}{\right]}
\newcommand{\supth}{^{\text{th}}}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}[proposition]{Definition}
\newtheorem{theorem}[proposition]{Theorem} |
Content:
Let $V$ be a real, or a complex vector space, with $K$ denoting the
corresponding field of scalars. A seminorm is a function
$$p:V\to\reals^+,$$
from $V$ to the set of positive real numbers, that satisfies the
following two properties.
\begin{align*}
p(k \bu) &= \vert k\vert p(\bu),\quad k\in K,\; \bu\in V & \text{\bf
Homogeneity}\\
p(\bu+\bv) &\leq p(\bu)+p(\bv),\quad \bu,\bv\in U, &\text{\bf Convexity}
\end{align*}
A seminorm differs from a norm in that it is permitted that $p(\bu)=0$
for certain non-zero $\bu\in V.$
Finally, it is possible to characterize the seminorms properties
geometrically. For $k>0$, let
$$B_k = \{ \bu \in V: p(\bu)\leq k\}$$ denote
the ball of radius $k$. The homogeneity property is equivalent to the
assertion that
$$B_k = k B_1,$$
in the sense that
$\bu \in B_k$ if and only if
$$\frac{1}{k} \bv\in B_1.$$
Thus, we see that a seminorm is fully determined by its unit ball.
Indeed, given $B\subset V$ we may define a function
$p_B:V\to \reals^+$
$$p_B(\bu) = \inf\{ \lambda\in\reals^+ : \lambda^{-1}\bu\in B\}.$$
The geometric nature of the unit ball is described by the following.
\begin{proposition}
The function $p_B$ satisfies the homegeneity property if and only if
for every $\bu\in
V$, there exists a $k\in\reals^+\cup \{\infty\}$ such that
$$\lambda\bu \in B\quad\text{if and only if}\quad -k\leq \lambda \leq k.$$
\end{proposition}
\begin{proposition}
Taking the homogeneity property as a given, the second property is
equivalent to the assertion that the unit ball
is a convex subset of $U$.
\end{proposition}
\emph{Proof.} Suppose that the seminorm obeys both of the properties
above. Let $\bu,\bv\in B_1$, and let $t$ be a real number between $0$
and $1$. We must show that the weighted average $t\bu+(1-t)\bv$ is in
$B_1$ as well. By assumption,
$$p(t\bu +(1-t)\bv) \leq tp(\bu) +(1-t)p(\bv).$$
The right side is a
weighted average of two numbers between $0$ and $1$, and is therefore
between $0$ and $1$ itself. Therefore
$$t\bu+(1-t)\bv \in B_1,$$
as desired.
Conversely, suppose that the seminorm function is homogeneous, and
that the unit ball is convex. Let $\bu,\bv\in
V$ be given, and let us show that
$$p(\bu+\bv)\leq p(\bu)+p(\bv).$$
The essential complication here is that we do not exclude the
possibility that $p(\bu)=0$ does not necessarily imply $\bu=0$.
First, let us consider the case where
$$p(\bu)=p(\bv)=0.$$
By homogeneity, for every $k>0$ we have
$$k\bu,\,k\bv\in B_1,$$
and hence
$$\frac{k}{2}\, \bu + \frac{k}{2}\,\bv\in B_1, $$
as well. By homogeneity, again,
$$p(\bu+\bv)\leq \frac{2}{k}.$$
Sending $k$ to zero, we infer that
$$p(\bu+\bv) = 0,$$
as desired.
Next suppose that $p(\bu)=0$, but that $p(\bv)\neq 0$. We will show
that in this case, necessarily,
$$p(\bu+\bv) = p(\bv).$$
Owing to the homogeneity assumption, we may without loss of generality
assume that
$$p(\bv)=1.$$
For every $k$ such that $0\leq k\le 1$ we have
$$ k\bu+k\bv = (1-k)\lp \frac{k}{1-k} \bu \rp + k\bv.$$
The right-side expression is an element of $B_1$ because
$$\frac{k}{1-k}\bu,\, \bv \in B_1.$$
Hence
$$kp(\bu+\bv) \leq 1,$$
and since this holds for $k$ arbitrarily close to $1$ we conclude that
$$p(\bu+\bv)\leq 1.$$
Finally, suppose that neither $p(\bu)$ nor $p(\bv)$ is zero. Hence,
$$\frac{\bu}{p(u)},\, \frac{\bv}{p(v)}$$
are both in $B_1$, and hence
$$\frac{p(u)}{p(u)+p(v)} \frac{\bu}{p(u)}+
\frac{p(v)}{p(u)+p(v)} \frac{\bv}{p(v)} = \frac{\bu+\bv}{p(u)+p(v)}
is in $B_1$ also. Using homogeneity, we conclude that
$$p(u+v)\leq p(u)+p(v),$$
as desired. |
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