Bounded operator explained

L:X\toY

between topological vector spaces (TVSs)

X

and

Y

that maps bounded subsets of

X

to bounded subsets of

Y.

If

X

and

Y

are normed vector spaces (a special type of TVS), then

L

is bounded if and only if there exists some

M>0

such that for all

x\inX,

\|Lx\|_Y \leq M \|x\|_X.The smallest such

M

is called the operator norm of

L

and denoted by

\|L\|.

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

f:X\toY

is called "bounded" then this usually means that its image

f(X)

is a bounded subset of its codomain. A linear map has this property if and only if it is identically

0.

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

0.

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

F:X\toY

between two topological vector spaces (TVSs) is called a or just if whenever

B\subseteqX

is bounded in

X

then

F(B)

is bounded in

Y.

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

F:X\toY

is a linear operator between two topological vector spaces and if there exists a neighborhood

U

of the origin in

X

such that

F(U)

is a bounded subset of

Y,

then

F

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

X

is a bornological space if and only if for every locally convex TVS

Y,

a linear operator

F:X\toY

is continuous if and only if it is bounded.

Every normed space is bornological.

Characterizations of bounded linear operators

Let

F:X\toY

be a linear operator between topological vector spaces (not necessarily Hausdorff). The following are equivalent:

F

is (locally) bounded;
  1. (Definition):

F

maps bounded subsets of its domain to bounded subsets of its codomain;

F

maps bounded subsets of its domain to bounded subsets of its image

\operatorname{Im}F:=F(X)

;

F

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.

F

maps every Mackey convergent null sequence to a bounded subset of

Y.

[1]
    • A sequence

x\bull=\left(xi\right)

infty
i=1
is said to be Mackey convergent to the origin in

X

if there exists a divergent sequence

r\bull=\left(ri\right)

infty
i=1

\toinfty

of positive real number such that

r\bull=\left(rixi\right)

infty
i=1
is a bounded subset of

X.

if

X

and

Y

are locally convex then the following may be add to this list:
  1. F

    maps bounded disks into bounded disks.
  2. F-1

    maps bornivorous disks in

    Y

    into bornivorous disks in

    X.

if

X

is a bornological space and

Y

is locally convex then the following may be added to this list:
  1. F

    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

    X

    is also a sequential space, then

    F

    is sequentially continuous if and only if it is continuous.
  2. F

    is sequentially continuous at the origin.

Examples

c00

of eventually zero sequences of real numbers, considered with the

\ell1

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

\ellinfty

norm, the same operator is not bounded.
  • Many integral transforms are bounded linear operators. For instance, if K : [a, b] \times [c, d] \to \Ris a continuous function, then the operator

    L

    defined on the space

    C[a,b]

    of continuous functions on

    [a,b]

    endowed with the uniform norm and with values in the space

    C[c,d]

    with

    L

    given by the formula(Lf)(y) = \int_a^b\!K(x, y)f(x)\,dx, is bounded. This operator is in fact a compact operator. The compact operators form an important class of bounded operators.
  • The Laplace operator \Delta : H^2(\R^n) \to L^2(\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 Lp space

    \ell2

    of all sequences

    \left(x0,x1,x2,\ldots\right)

    of real numbers with
    2
    x
    0

    +

    2
    x
    1

    +

    2
    x
    2

    +<infty,

    L(x_0, x_1, x_2, \dots) = \left(0, x_0, x_1, x_2, \ldots\right) is bounded. Its operator norm is easily seen to be

    1.

  • Unbounded linear operators

    Let

    X

    be the space of all trigonometric polynomials on

    [-\pi,\pi],

    with the norm

    \|P\| = \int_^\!|P(x)|\,dx.

    The operator

    L:X\toX

    that maps a polynomial to its derivative is not bounded. Indeed, for

    vn=ei

    with

    n=1,2,\ldots,

    we have

    \|vn\|=2\pi,

    while

    \|L(vn)\|=2\pin\toinfty

    as

    n\toinfty,

    so

    L

    is not bounded.

    Properties of the space of bounded linear operators

    The space of all bounded linear operators from

    X

    to

    Y

    is denoted by

    B(X,Y)

    .

    B(X,Y)

    is a normed vector space.

    Y

    is Banach, then so is

    B(X,Y)

    ; in particular, dual spaces are Banach.

    A\inB(X,Y)

    the kernel of

    A

    is a closed linear subspace of

    X

    .

    B(X,Y)

    is Banach and

    X

    is nontrivial, then

    Y

    is Banach.

    Bibliography

    Notes and References

    1. Proof: Assume for the sake of contradiction that

      x\bull=\left(xi\right)

      infty
      i=1
      converges to

      0

      but

      F\left(x\bull\right)=\left(F\left(xi\right)\right)

      infty
      i=1
      is not bounded in

      Y.

      Pick an open balanced neighborhood

      V

      of the origin in

      Y

      such that

      V

      does not absorb the sequence

      F\left(x\bull\right).

      Replacing

      x\bull

      with a subsequence if necessary, it may be assumed without loss of generality that

      F\left(xi\right)\not\ini2V

      for every positive integer

      i.

      The sequence

      z\bull:=\left(xi/i\right)

      infty
      i=1
      is Mackey convergent to the origin (since

      \left(izi\right)

      infty
      i=1

      =\left(xi\right)

      infty
      i=1

      \to0

      is bounded in

      X

      ) so by assumption,

      F\left(z\bull\right)=\left(F\left(zi\right)\right)

      infty
      i=1
      is bounded in

      Y.

      So pick a real

      r>1

      such that

      F\left(zi\right)\inrV

      for every integer

      i.

      If

      i>r

      is an integer then since

      V

      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 "

      F

      is bounded." For example, the word "such that

      \left(rixi\right)

      infty
      i=1
      is a bounded subset of

      X.

      " in the definition of "Mackey convergent to the origin" can be replaced with "such that

      \left(rixi\right)

      infty
      i=1

      \to0

      in

      X.

      "