Operator norm

In functional analysis, a linear transformation L between normed vector spaces is said to be bounded, or to be a bounded linear operator, if the ratio of the norms of L(v) and v is bounded above, over all non-zero vectors v.

It is simple to prove that this is the same condition on L as continuity, for the topologies induced from the norms.

In the case of a matrix A acting as a linear transformation, from R^m to Rⁿ, or from C^m to Cⁿ, one can prove directly that A must be bounded. In fact the function

f(v) = ||A(v)||

is continuous as a function of v, for any norm ||.||; and the set of v with ||v|| = 1 is compact, being a closed, bounded subset. The matrix norm of A is by definition the supremum of f. In this case it is attained somewhere, again by the compactness of the domain.

In general the operator norm of a bounded linear transformation L from V to W, where V and W are both normed real vector spaces (or both normed complex) vector spaces is defined as the supremum of the ||L(v)|| taken over all v in V of norm 1. This definition uses the property ||c.v|| = |c|.||v|| where c is a scalar, to restrict attention to v with ||v|| = 1. Geometrically we need (for real scalars) to look at one vector only on each ray out from the origin 0.

Note that there are two different norms here: that in V and that in W. Even if V = W we might wish to take distinct norms on it. In fact given two norms ||.|| and |||.||| on V, the identity operator on V will have an operator norm, in passing from V with ||.|| as norm to V with |||.|||, only if we can say

|||v||| < C.||v||

for some absolute constant C, for all v. When V is finite-dimensional, we can be sure of this: for example in the case of two dimensions the conditions ||v|| = 1 and |||v||| may define a rectangle and an ellipse respectively, centred at 0. Whatever their proportions and orientations, we can magnify the rectangle so that the the ellipse fits inside the enlarged rectangle; and vice versa.

This is, however, a phenomenon of finite dimensions. That all norms will turn out to be equivalent: they stay within constant multiples of each other, and from a topological point of view they give the same open sets. This all fails for infinite-dimensional spaces. This can be seen, for example, by considering the differentiation operator D, as applied to trigonometric polynomials. We can take the root mean square as norm: since D(e^inx) = ine^inx, the norms of D applied to the finite-dimensional subspaces of the Hilbert space grow beyond any bounds. Therefore an operator as fundamental as D can fail to have an operator norm.

A basic theorem applies the Baire category theorem to show that if L has as domain and range Banach spaces, it will be bounded. That is, in the example just given, D must not be defined on all square-integrable Fourier series; and indeed we know that they ca represent continuous but nowhere differentiable functions. The intuition is that if L magnifies norms of some vectors by as large a number as we choose, we should be able to condense singularities - choose a vector v that sums up others for which it would be contradictory for ||L(v)|| to be finite - showing that the domain of L cannot be the whole of V.