Bounded operator explained
between
topological vector spaces (TVSs)
and
that maps
bounded subsets of
to bounded subsets of
If
and
are
normed vector spaces (a special type of TVS), then
is bounded if and only if there exists some
such that for all
The smallest such
is called the
operator norm of
and denoted by
A bounded operator between normed spaces is
continuous and vice versa.
The concept of a bounded linear operator has been extended from normed spaces to all topological vector spaces.
Outside of functional analysis, when a function
is called "
bounded" then this usually means that its
image
is a bounded subset of its codomain. A linear map has this property if and only if it is identically
Consequently, in functional analysis, when a linear operator is called "bounded" then it is never meant in this abstract sense (of having a bounded image).
In normed vector spaces
Every bounded operator is Lipschitz continuous at
Equivalence of boundedness and continuity
A linear operator between normed spaces is bounded if and only if it is continuous.
In topological vector spaces
A linear operator
between two
topological vector spaces (TVSs) is called a
or just
if whenever
is
bounded in
then
is bounded in
A subset of a TVS is called bounded (or more precisely,
von Neumann bounded) if every neighborhood of the origin
absorbs it. In a normed space (and even in a seminormed space), a subset is von Neumann bounded if and only if it is norm bounded. Hence, for normed spaces, the notion of a von Neumann bounded set is identical to the usual notion of a norm-bounded subset.
Continuity and boundedness
Every sequentially continuous linear operator between TVS is a bounded operator. This implies that every continuous linear operator between metrizable TVS is bounded. However, in general, a bounded linear operator between two TVSs need not be continuous.
This formulation allows one to define bounded operators between general topological vector spaces as an operator which takes bounded sets to bounded sets. In this context, it is still true that every continuous map is bounded, however the converse fails; a bounded operator need not be continuous. This also means that boundedness is no longer equivalent to Lipschitz continuity in this context.
If the domain is a bornological space (for example, a pseudometrizable TVS, a Fréchet space, a normed space) then a linear operators into any other locally convex spaces is bounded if and only if it is continuous. For LF spaces, a weaker converse holds; any bounded linear map from an LF space is sequentially continuous.
If
is a linear operator between two topological vector spaces and if there exists a neighborhood
of the origin in
such that
is a bounded subset of
then
is continuous. This fact is often summarized by saying that a linear operator that is bounded on some neighborhood of the origin is necessarily continuous. In particular, any linear functional that is bounded on some neighborhood of the origin is continuous (even if its domain is not a
normed space).
Bornological spaces
See main article: Bornological space.
Bornological spaces are exactly those locally convex spaces for which every bounded linear operator into another locally convex space is necessarily continuous. That is, a locally convex TVS
is a bornological space if and only if for every locally convex TVS
a linear operator
is continuous if and only if it is bounded.
Every normed space is bornological.
Characterizations of bounded linear operators
Let
be a linear operator between topological vector spaces (not necessarily Hausdorff). The following are equivalent:
is (locally) bounded;
- (Definition):
maps bounded subsets of its domain to bounded subsets of its codomain;
maps bounded subsets of its domain to bounded subsets of its
image
;
maps every null sequence to a bounded sequence;
- A null sequence is by definition a sequence that converges to the origin.
- Thus any linear map that is sequentially continuous at the origin is necessarily a bounded linear map.
maps every Mackey convergent null sequence to a bounded subset of
[1] -
is said to be
Mackey convergent to the origin in
if there exists a divergent sequence
r\bull=\left(ri\right)
\toinfty
of positive real number such that
is a bounded subset of
if
and
are
locally convex then the following may be add to this list:
-
maps bounded disks into bounded disks.
-
maps bornivorous disks in
into bornivorous disks in
if
is a
bornological space and
is locally convex then the following may be added to this list:
-
is sequentially continuous at some (or equivalently, at every) point of its domain.
- A sequentially continuous linear map between two TVSs is always bounded, but the converse requires additional assumptions to hold (such as the domain being bornological and the codomain being locally convex).
- If the domain
is also a sequential space, then
is sequentially continuous if and only if it is continuous. -
is sequentially continuous at the origin.
Examples
of eventually zero sequences of real numbers, considered with the
norm, the linear operator to the real numbers which returns the sum of a sequence is bounded, with operator norm 1. If the same space is considered with the
norm, the same operator is not bounded.
Many integral transforms are bounded linear operators. For instance, if is a continuous function, then the operator
defined on the space
of continuous functions on
endowed with the uniform norm and with values in the space
with
given by the formulais bounded. This operator is in fact a compact operator. The compact operators form an important class of bounded operators.The Laplace operator (its domain is a Sobolev space and it takes values in a space of square-integrable functions) is bounded.The shift operator on the Lp space
of all sequences \left(x0,x1,x2,\ldots\right)
of real numbers with
is bounded. Its operator norm is easily seen to be
Unbounded linear operators
Let
be the space of all
trigonometric polynomials on
with the norm
The operator
that maps a polynomial to its
derivative is not bounded. Indeed, for
with
we have
while
as
so
is not bounded.
Properties of the space of bounded linear operators
The space of all bounded linear operators from
to
is denoted by
.
is a normed vector space.
is Banach, then so is
; in particular,
dual spaces are Banach.
the kernel of
is a closed linear subspace of
.
is Banach and
is nontrivial, then
is Banach.
Bibliography
- Kreyszig, Erwin: Introductory Functional Analysis with Applications, Wiley, 1989
Notes and References
- Proof: Assume for the sake of contradiction that
converges to
but
F\left(x\bull\right)=\left(F\left(xi\right)\right)
is not bounded in
Pick an open balanced neighborhood
of the origin in
such that
does not absorb the sequence
Replacing
with a subsequence if necessary, it may be assumed without loss of generality that F\left(xi\right)\not\ini2V
for every positive integer
The sequence z\bull:=\left(xi/i\right)
is Mackey convergent to the origin (since \left(izi\right)
=\left(xi\right)
\to0
is bounded in
) so by assumption, F\left(z\bull\right)=\left(F\left(zi\right)\right)
is bounded in
So pick a real
such that
for every integer
If
is an integer then since
is balanced, F\left(xi\right)\inriV\subseteqi2V,
which is a contradiction. Q.E.D. This proof readily generalizes to give even stronger characterizations of "
is bounded." For example, the word "such that
is a bounded subset of
" in the definition of "Mackey convergent to the origin" can be replaced with "such that
in
"