# 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,

$\|L(v)\|_{Y}\leq M\|v\|_{X}.\,$

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 $L^{1}[a,b]$ of Lebesgue integrable functions with values in the space $L^{1}[c,d]$
$(Lf)(y)=\int _{{a}}^{{b}}\!K(x,y)f(x)\,dx,$
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 l2 space of all sequences (x0, x1, x2...) of real numbers with $x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+\cdots <\infty ,$
$L(x_{0},x_{1},x_{2},\dots )=(0,x_{0},x_{1},x_{2},\dots )$
is bounded. Its norm is easily seen to be 1.

Not every linear operator between normed spaces is bounded. 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

$v=e^{{inx}}$

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).