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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).