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Bounded operator

(Redirected from Bounded linear operator)

In functional analysis (a branch of mathematics), a bounded linear operator is a linear transformation L between normed vector spaces X and Y for which the ratio of the norm of L(v) to that of v is bounded by the same number, over all non-zero vectors v in X. In other words, there exists some M > 0 such that for all v in X,

\|L(v)\|_Y \le M \|v\|_X.\,

The smallest such M is called the operator norm \|L\|_{op} of L.

Let us note that a bounded linear operator is not necessarily a bounded function, the latter would require that the norm of L(v) is bounded for all v. Rather, a bounded linear operator is a locally bounded function.

It is quite easy to prove that a linear operator L is bounded if and only if it is a continuous function from X to Y.

Examples

  • Any linear operator between two finite-dimensional normed spaces is bounded, and such an operator may be viewed as multiplication by some fixed matrix.
K:[a, b]\times [c, d]\to {\mathbf R}
is a continuous function, then the operator L, defined on the space L1[a,b] of Lebesgue integrable functions with values in the space L1[c,d]
(Lf)(y)=\int_{a}^{b}\!K(x, y)f(x)\,dx,
is bounded.
\Delta:H^2({\mathbf R}^n)\to L^2({\mathbf R}^n)
(its domain is a Sobolev space and it takes values in a space of square integrable functions) is bounded.
  • The shift operator on the space of all sequences (x0, x1, x2...) of real numbers with x_0^2+x_1^2+x_2^2+\cdots < \infty,
L(x_0, x_1, x_2, \dots)=(x_1, x_2, x_3,\dots)
is bounded. Its norm is easily seen to be 1.

One can prove, by using the Baire category theorem, that if a linear operator L has as domain and range Banach spaces, then it will be bounded. Thus, to give an example of a linear operator which is not bounded, we need to pick some normed spaces which are not Banach. Let X be the space of all trigonometric polynomials defined on [−π, π], with the norm

\|P\|=\int_{-\pi}^{\pi}\!|P(x)|\,dx.

Define the operator L:XX which acts by taking the derivative, so it maps a polynomial P to its derivative P′. Then, for

v = einx

with n=1, 2, ...., we have \|v\|=2\pi,, while \|L (v)\|=2\pi n\to\infty as n→∞, so this operator is not bounded.

Further properties

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

See also

10-26-2009 08:16:03
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