Seminorm 2002-02-20 22:51:15 2008-03-02 11:06:20 Definition normed vector space homogeneous \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} \newcommand{\nl}[1]{\PMlinkescapetext{{#1}}} \newtheorem{theorem}[proposition]{Theorem} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\pnorm}{\operatorname{p}} \newcommand{\snorm}[1]{\pnorm(#1)} Let $V$ be a real, or a complex vector space, with $K$ denoting the corresponding field of scalars. A \emph{seminorm} is a function $$\operatorname{p}:V\to\reals^+,$$ from $V$ to the set of non-negative real numbers, that satisfies the following two properties. \begin{align*} \snorm{k\, \bu} &= \vert k\vert \snorm{\bu},\quad k\in K,\; \bu\in V & \text{\bf Homogeneity}\\ \snorm{\bu+\bv} &\leq \snorm{\bu}+\snorm{\bv},\quad \bu,\bv\in V, &\text{\bf Sublinearity} \end{align*} A seminorm differs from a norm in that it is permitted that $\snorm{\bu}=0$ for some non-zero $\bu\in V.$ It is possible to characterize the seminorms properties geometrically. For $k>0$, let $$B_k = \{ \bu \in V: \snorm{\bu}\leq k\}$$ denote the ball of radius $k$. The homogeneity property is equivalent to the assertion that $$B_k = k B_1,$$ in the sense that $\bu \in B_1$ if and only if $k\bu\in B_k.$ Thus, we see that a seminorm is fully determined by its unit ball. Indeed, given $B\subset V$ we may define a function $\pnorm_B:V\to \reals^+$ by $$\pnorm_B(\bu) = \inf\{ \lambda\in\reals^+ : \lambda^{-1}\bu\in B\}.$$ The geometric nature of the unit ball is described by the following. \begin{proposition} The function $\pnorm_B$ satisfies the homegeneity property if and only if for every $\bu\in V$, there exists a $k\in\reals^+\cup \{\infty\}$ such that $$\lambda\,\bu \in B\quad\text{if and only if}\quad \Vert \lambda\Vert \leq k.$$ \end{proposition} \begin{proposition} Suppose that $\pnorm$ is homogeneous. Then, it is sublinear if and only if its unit ball, $B_1$, is a convex subset of $V$. \end{proposition} \emph{Proof.} First, let us suppose that the seminorm is both sublinear and homogeneous, and prove that $B_1$ is necessarily convex. Let $\bu,\bv\in B_1$, and let $k$ be a real number between $0$ and $1$. We must show that the weighted average $k\bu+(1-k)\bv$ is in $B_1$ as well. By assumption, $$\snorm{k\,\bu +(1-k)\bv} \leq k\snorm{\bu} +(1-k)\snorm{\bv}.$$ The right side is a weighted average of two numbers between $0$ and $1$, and is therefore between $0$ and $1$ itself. Therefore $$k\,\bu+(1-k)\bv \in B_1,$$ 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 $$\snorm{\bu+\bv}\leq \snorm{\bu}+\snorm{\bv}.$$ The essential complication here is that we do not exclude the possibility that $\snorm{\bu}=0$, but that $\bu\neq 0$. First, let us consider the case where $$\snorm{\bu}=\snorm{\bv}=0.$$ 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, $$\snorm{\bu+\bv}\leq \frac{2}{k}.$$ Since the above is true for all positive $k$, we infer that $$\snorm{\bu+\bv} = 0,$$ as desired. Next suppose that $\snorm{\bu}=0$, but that $\snorm{\bv}\neq 0$. We will show that in this case, necessarily, $$\snorm{\bu+\bv} = \snorm{\bv}.$$ Owing to the homogeneity assumption, we may without loss of generality assume that $$\snorm{\bv}=1.$$ For every $k$ such that $0\leq k<1$ we have $$ k\,\bu+k\,\bv = (1-k) \frac{k\,\bu}{1-k} + k\,\bv.$$ The right-side expression is an element of $B_1$ because $$\frac{k\,\bu}{1-k},\, \bv \in B_1.$$ Hence $$k\snorm{\bu+\bv} \leq 1,$$ and since this holds for $k$ arbitrarily close to $1$ we conclude that $$\snorm{\bu+\bv}\leq \snorm{\bv}.$$ The same argument also shows that $$\snorm{\bv} = \snorm{-\bu+(\bu+\bv)} \leq \snorm{\bu+\bv},$$ and hence $$\snorm{\bu+\bv}=\snorm{\bv},$$ as desired. Finally, suppose that neither $\snorm{\bu}$ nor $\snorm{\bv}$ is zero. Hence, $$\frac{\bu}{\snorm{u}},\, \frac{\bv}{\snorm{v}}$$ are both in $B_1$, and hence $$\frac{\snorm{u}}{\snorm{u}+\snorm{v}} \frac{\bu}{\snorm{u}}+ \frac{\snorm{v}}{\snorm{u}+\snorm{v}} \frac{\bv}{\snorm{v}} = \frac{\bu+\bv}{\snorm{u}+\snorm{v}} $$ is in $B_1$ also. Using homogeneity, we conclude that $$\snorm{u+v}\leq \snorm{u}+\snorm{v},$$ as desired.