In functional analysis and operator theory, a bounded linear operator is a linear transformation 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
A bounded operator between normed spaces is continuous and vice versa.
The concept has been extended to certain "well-behaved" topological vector spaces.
In normed vector spaces
Equivalence of boundedness and continuity
A linear operator between normed spaces is bounded if and only if it is continuous.
Suppose that is bounded. Then, for all vectors with nonzero we have
Letting go to zero shows that is continuous at Moreover, since the constant does not depend on this shows that in fact is uniformly continuous, and even Lipschitz continuous.
Conversely, it follows from the continuity at the zero vector that there exists a such that for all vectors with Thus, for all non-zero one has
This proves that is bounded.
In topological vector spaces
A linear operator between two topological vector spaces (TVSs) is locally bounded or just bounded 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.
Every sequentially continuous linear operator between TVS is a bounded operator. This implies that every continuous linear operator 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. Clearly, this also means that boundedness is no longer equivalent to Lipschitz continuity in this context.
If the domain is a bornological space (e.g. 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.
Bornological spaces are exactly those locally convex spaces for 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 TVSs (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 [note 1]
- A sequence is said to be Mackey convergent to the origin in if there exists a divergent sequence of positive real number such that is a bounded subset of
and if in addition and are locally convex then the following may be add to this list:
and if in addition is a bornological space and is locally convex then the following may be added to this list:
- is sequentially continuous.
- is sequentially continuous at the origin.
- Any linear operator between two finite-dimensional normed spaces is bounded, and such an operator may be viewed as multiplication by some fixed matrix.
- Any linear operator defined on a finite-dimensional normed space is bounded.
- On the sequence space 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 compact. 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 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 and is a normed vector space.
- If is Banach, then so is
- from which it follows that dual spaces are Banach.
- For any the kernel of is a closed linear subspace of
- If is Banach and is nontrivial, then is Banach.
- Bounded set (topological vector space)
- Discontinuous linear map
- Continuous linear operator
- Norm (mathematics) – Length in a vector space
- Normed space
- Operator algebra – Branch of functional analysis
- Operator norm – Measure of the "size" of linear operators
- Operator theory
- Unbounded operator
- Topological vector space – Vector space with a notion of nearness
- Proof: Assume for the sake of contradiction that converges to but 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 for every positive integer The sequence is Mackey convergent to the origin (since is bounded in ) so by assumption, is bounded in So pick a real such that for every integer If is an integer then since is balanced, which is a contradiction. 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 "
- "Bounded operator", Encyclopedia of Mathematics, EMS Press, 2001 
- Kreyszig, Erwin: Introductory Functional Analysis with Applications, Wiley, 1989
- Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
- Wilansky, Albert (2013). Modern Methods in Topological Vector Spaces. Mineola, New York: Dover Publications, Inc. ISBN 978-0-486-49353-4. OCLC 849801114.