Bounded operator
From Academic Kids
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,
- <math>\|L(v)\|_Y \le M \|v\|_X.\,<math>
The smallest such M is called the operator norm <math>\|L\|_{op}<math> 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.
- Many integral transforms are bounded linear operators. For instance, if
- <math>K:[a, b]\times [c, d]\to {\mathbf R}<math>
- is a continuous function, then the operator <math>L,<math> defined on the space <math>L^1[a, b]<math> of Lebesgue integrable functions with values in the space <math>L^1[c, d]<math>
- <math>(Lf)(y)=\int_{a}^{b}\!K(x, y)f(x)\,dx,<math>
- is bounded.
- The Laplacian operator
- <math>\Delta:H^2({\mathbf R}^n)\to L^2({\mathbf R}^n)<math>
- (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 <math>x_0^2+x_1^2+x_2^2+\cdots < \infty,<math>
- <math>L(x_0, x_1, x_2, \dots)=(x_1, x_2, x_3,\dots)<math>
- 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
- <math>\|P\|=\int_{-\pi}^{\pi}\!|P(x)|\,dx.<math>
Define the operator L:X→X which acts by taking the derivative, so it maps a polynomial P to its derivative P′. Then, for
- <math>v=e^{in x}<math>
with n=1, 2, ...., we have <math>\|v\|=2\pi,<math>, while <math>\|L (v)\|=2\pi n\to\infty<math> 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).
