# Norm

(Redirected from Operator 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 $x,y\in X$, $$\lVert x+y\rVert^2 + \lVert x-y\rVert^2 = 2(\lVert x\rVert^2 + \lVert y\rVert^2).$$

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 $$\lVert\cdot\rVert_1 \leq C_1\lVert\cdot\rVert_2 \leq C_2\lVert\cdot\rVert_1\quad \text{for all}\; x\in X.$$

If $X$ is complete in both norms, then their equivalence is a consequence of compatibility. Here compatibility means that the limit relations $$\lVert x_n-a\rVert_1\rightarrow 0,\quad\lVert x_n-b\rVert_2\rightarrow 0.$$ 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 $$\lVert\tilde{x}\rVert=\inf\{\lVert x\rVert\colon x\in\tilde{x}\},$$ 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 $$\lVert\psi\rVert=\sup\{\lvert\psi(x)\rvert\colon \lVert x\rVert\leq 1\}.$$ 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: $$\lVert A\rVert=\sup\{\lVert Ax\rVert\colon \lVert x\rVert\leq 1\}.$$ 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 $$\lVert AB\rVert \leq \lVert A\rVert\cdot\lVert B\rVert,\quad\lVert I\rVert=1,$$ 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 $$\lvert x\rvert=\lim_{n\rightarrow\infty}\lVert x^n\rVert^{1/n}$$ 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.

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.
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 $E$.