In mathematics, the operator norm is a norm defined on the space of bounded operators between two Banach spaces.

Given a linear operator , where X and Y are Banach spaces with norms , respectively, we define

Further analysis

In functional analysis, a bounded linear operator is a linear transformation L between normed vector spaces for which the ratio of the norm of L(v) to that of 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 (or 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.

The operator norm indeed satisfies the conditions for the norm, so the space of all bounded linear transformations from V to W is itself a normed vector space. It is complete if W is complete.

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||| = 1 may define a rectangle and an ellipse respectively, centred at 0. Whatever their proportions and orientations, we can magnify the rectangle so that 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 differential 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 can 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.

A common procedure for defining a bounded linear operator between two given complete normed spaces is as follows. First, define the operator on a dense subset of the domain. Then extend the operator by continuity to a continuous linear operator on the whole domain (see continuous linear extension).

See also:

• operator algebra
• Topologies on the set of operators on a Hilbert space