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Bounded Linear Operator
A linear operator that sends bounded subsets to bounded subsets
The smallest such M, denoted by ||||, is called the operator norm of L.
A linear operator that is sequentially continuous or continuous is a bounded operator and moreover, a linear operator between normed spaces is bounded if and only if it is continuous.
However, a bounded linear operator between more general topological vector spaces is not necessarily continuous.
In topological vector spaces
A linear operator F : X -> Y between two topological vector spaces (TVSs) is locally bounded or just bounded if whenever B ⊆ X 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.
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.
Bornological spaces are exactly those locally convex spaces for every bounded linear operator into another locally convex space is necessarily bounded.
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 -> Y is continuous if and only if it is bounded.
Every normed space is bornological.
Characterizations of bounded linear operators
Let F : X -> Y be a linear operator between TVSs (not necessarily Hausdorff).
The following are equivalent:
A bounded linear operator is generally not a bounded function, as generally one can find a sequence xo = (xi)? i=1 in X such that
Instead, all that is required for the operator to be bounded is that
for all x ? 0.
So, the operator L could only be a bounded function if it satisfied L(x) = 0 for all x, as is easy to understand by considering that for a linear operator,
for all scalars a.
Rather, a bounded linear operator is a locally bounded function.
A linear operator between normed spaces is bounded if and only if it is continuous, and by linearity, if and only if it is continuous at zero.
Equivalence of boundedness and continuity
As stated in the introduction, a linear operator L between normed spaces X and Y is bounded if and only if it is a continuous linear operator.
The proof is as follows.
Suppose that L is bounded. Then, for all vectors x, h ∈ X with h nonzero we have
Conversely, it follows from the continuity at the zero vector that there exists a such that for all vectors h ∈ X with .
Thus, for all non-zero x ∈ X, one has
This proves that L is bounded.
The condition for L to be bounded, namely that there exists some M such that for all x
is precisely the condition for L to be Lipschitz continuous at 0 (and hence everywhere, because L is linear).
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 its domain, such that it is locally bounded.
Then, extend the operator by continuity to a continuous linear operator on the whole domain.
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 spacec00 of eventually zero sequences of real numbers, considered with the l1 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 l? norm, the same operator is not bounded.
is 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
is bounded. This operator is in fact compact. The compact operators form an important class of bounded operators.
is bounded. Its operator norm is easily seen to be 1.
Unbounded linear operators
Not every linear operator between normed spaces is bounded.
Let X be the space of all trigonometric polynomials defined on [−?, ?], with the norm
Define the operator L : X -> X which acts by taking the derivative, so it maps a polynomial P to its derivative P′.
with n=1, 2, ...., we have while as n -> ?, so this operator is not bounded.
It turns out that this is not a singular example, but rather part of a general rule.
However, given any normed spaces X and Y with X infinite-dimensional and Y not being the zero space, one can find a linear operator which is not continuous from X to Y.
That such a basic operator as the derivative (and others) is not bounded makes it harder to study.
If, however, one defines carefully the domain and range of the derivative operator, one may show that it is a closed operator.
Closed operators are more general than bounded operators but still "well-behaved" in many ways.
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) and is a normed vector space.
^Proof: Assume for the sake of contradiction that xo = (xi)? i=1 converges to 0 but F(xo) = (F(xi))? 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(xo). Replacing xo with a subsequence if necessary, it may be assumed without loss of generality that F(xi) ? i2V for every positive integer i. The sequence zo := (xi / i)? i=1 is Mackey convergent to the origin (since (izi)? i=1 = (xi)? i=1 -> 0 is bounded in X) so by assumption, F(zo) = (F(zi))? i=1 is bounded in Y. So pick a real r > 1 such that F(zi) ? rV for every integer i. If i > r is an integer then since V is balanced, F(xi) ? riV ? i2V, which is a contradiction. ? This proof readily generalizes to give even stronger characterizations of "F is bounded." For example, the word "such that (rixi)? i=1 is a bounded subset of " in the definition of "Mackey convergent to the origin" can be replaced with "such that (rixi)? i=1 -> 0 in "