Norm

A mapping $x\rightarrow\lVert x\rVert$ from a vector space $X$ over the field of real or complex numbers into the real numbers, subject to the conditions:

1. $\lVert x\rVert\geq 0$, and $\lVert x\rVert=0$ for $x=0$ only;
2. $\lVert\lambda x\rVert=\lvert\lambda\rvert\cdot\lVert x\rVert$ for every scalar $\lambda$;
3. $\lVert x+y\rVert\leq\lVert x\rVert+\lVert y\rVert$ for all $x,y\in X$ (the triangle axiom).

The number $\lVert x\rVert$ is called the norm of the element $x$.

A vector space $X$ with a distinguished norm is called a normed space. A norm induces on $X$ a metric by the formula $dist(x,y)=\lVert x-y\rVert$, hence also a topology compatible with this metric. And so a normed space is endowed with the natural structure of a topological vector space. A normed space that is complete in this metric is called a Banach space. Every normed space has a Banach completion.

A topological vector space is said to be normable if its topology is compatible with some norm. Normability is equivalent to the existence of a convex bounded neighborhood of zero (a theorem of Kolmogorov, 1934).

The norm in a normed vector space $X$ is generated by an inner product (that is, $X$ is isometrically isomorphic to a pre-Hilbert space) if and only if for all $x,y\in X$, \begin{equation} \lVert x+y\rVert^2 + \lVert x-y\rVert^2 = 2(\lVert x\rVert^2 + \lVert y\rVert^2). \end{equation}

Two norms $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ on one and the same vector space $X$ are called equivalent if they induce the same topology. This comes to the same thing as the existence of two constants $C_1$ and $C_2$ such that \begin{equation} \lVert\cdot\rVert_1 \leq C_1\lVert\cdot\rVert_2 \leq C_2\lVert\cdot\rVert_1\quad \text{for all}\; x\in X. \end{equation}

If $X$ is complete in both norms, then their equivalence is a consequence of compatibility. Here compatibility means that the limit relations \begin{equation} \lVert x_n-a\rVert_1\rightarrow 0,\quad\lVert x_n-b\rVert_2\rightarrow 0. \end{equation} imply that $a=b$.

Not every topological vector space, even if it is assumed to be locally convex, has a continuous norm. For example, there is no continuous norm on an infinite product of straight lines with the topology of coordinate-wise convergence. The absence of a continuous norm can be an obvious obstacle to the continuous imbedding of one topological vector space in another.

If $Y$ is a closed subspace of a normed space $X$, then the quotient space $X/Y$ of cosets by $Y$ can be endowed with the norm \begin{equation} \lVert\tilde{x}\rVert=\inf\{\lVert x\rVert\colon x\in\tilde{x}\}, \end{equation} under which it becomes a normed space. The norm of the image of an element $x$ under the quotient mapping $X\rightarrow X/Y$ is called the quotient norm of $x$ with respect to $Y$.

The totality $X^*$ of continuous linear functionals $\psi$ on a normed space $X$ forms a Banach space relative to the norm \begin{equation} \lVert\psi\rVert=\sup\{\lvert\psi(x)\rvert\colon \lVert x\rVert\leq 1\}. \end{equation} The norms of all functionals are attained at suitable points of the unit ball of the original space if and only if the space is reflexive.

The totality $L(X,Y)$ of continuous (bounded) linear operators $A$ from a normed space $X$ into a normed space $Y$ is made into a normed space by introducing the operator norm: \begin{equation} \lVert A\rVert=\sup\{\lVert Ax\rVert\colon \lVert x\rVert\leq 1\}. \end{equation} Under this norm $L(X,Y)$ is complete if $Y$ is. When $X=Y$ is complete, the space $L(X)=L(X,X)$ with multiplication (composition) of operators becomes a Banach algebra, since for the operator norm \begin{equation} \lVert AB\rVert \leq \lVert A\rVert\cdot\lVert B\rVert,\quad\lVert I\rVert=1, \end{equation} where $I$ is the identity operator (the unit element of the algebra). Other equivalent norms on $L(x)$ subject to the same condition are also interesting. Such norms are sometimes called algebraic or ringed. Algebraic norms can be obtained by renorming $X$ equivalently and taking the corresponding operator norms; however, even for $\dim X=2$ not all algebraic norms on $L(x)$ can be obtained in this manner.

A pre-norm, or semi-norm, on a vector space $X$ is defined as a mapping $p$ with the properties of a norm except non-degeneracy: $p(x)=0$ does not preclude that $x\neq 0$. If $\dim X<\infty$, a non-zero pre-norm $p$ on $L(x)$ subject to the condition $p(AB)\leq p(A)p(B)$ actually turns out to be a norm (since in this case $L(x)$ has no non-trivial two-sided ideals). But for infinite-dimensional normed spaces this is not so. If $X$ is a Banach algebra over $C$, then the spectral radius \begin{equation} \lvert x\rvert=\lim_{n\rightarrow\infty}\lVert x^n\rVert^{1/n} \end{equation} is a semi-norm if and only if it is uniformly continuous on $X$, and this condition is equivalent to the fact that the quotient algebra by the radical is commutative.

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The theorem that the norms of all functionals are attained at points of the unit ball of the original space $X$ if and only if $X$ is reflexive is called James' theorem.

For norms in algebra see Norm on a field or ring (see also Valuation).

The norm of a group is the collection of group elements that commute with all subgroups, that is, the intersection of the normalizers of all subgroups (cf. Normalizer of a subset). The norm contains the centre of a group and is contained in the second hypercentre $Z_2$. For groups with a trivial centre the norm is the trivial subgroup $E$.

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