Bounded operator

In functional analysis, a bounded linear operator is a linear transformation $L : X \to Y$ 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 $M \geq 0$ such that for all $x$ in $X$ ,

| | Lx | | Y \leq M | | x | | X

.

The smallest such $M$ , denoted by $| | L | |$ , 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 $F : X \to Y$ between two topological vector spaces (TVSs) is locally bounded or just bounded if whenever $B \subseteq X$ 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.[1] 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

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 $F : X \to Y$ is continuous if and only if it is bounded.[2]

Every normed space is bornological.

Characterizations of bounded linear operators

Let $F : X \to Y$ be a linear operator between TVSs (not necessarily Hausdorff). The following are equivalent:

$F$ is (locally) bounded;[2]
(Definition): $F$ maps bounded subsets of its domain to bounded subsets of its codomain;[2]
$F$ maps bounded subsets of its domain to bounded subsets of its image $Im F := F (X)$ ;[2]
F maps every null sequence to a bounded sequence;[2]
- 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.[note 1]
- A sequence $x • = (x i) \infty i =1$ is said to be Mackey convergent to the origin in $X$ if there exists a divergent sequence $r • = (r i) \infty i =1 \to \infty$ of positive real number such that $(r i x i) \infty i =1$ is a bounded subset of $X.$

and if in addition $X$ and $Y$ are locally convex then the following may be add to this list:

$F$ maps bounded disks into bounded disks.[3]
$F -1$ maps bornivorous disks in $Y$ into bornivorous disks in $X$ .[3]

and if in addition $X$ is a bornological space and $Y$ is locally convex then the following may be added to this list:

$F$ is sequentially continuous.[4]
$F$ is sequentially continuous at the origin.

Bounded linear operators between normed spaces

A bounded linear operator is generally not a bounded function, as generally one can find a sequence $x • = (x i) \infty i =1$ in $X$ such that $\|Lx_{k}\|_{Y}\rightarrow \infty$ . Instead, all that is required for the operator to be bounded is that

{\frac {\|Lx\|_{Y}}{\|x\|_{X}}}\leq M<\infty

for all $x \neq 0$ . 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, $L(ax)=aL(x)$ 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 $x, h \in X$ with $h$ nonzero we have

\|L(x+h)-L(x)\|=\|L(h)\|\leq M\|h\|.\,

Letting ${\mathit {h}}\,$ go to zero shows that $L$ is continuous at $x$ . Moreover, since the constant $M$ does not depend on $x$ , this shows that in fact $L$ is uniformly continuous, and even Lipschitz continuous.

Conversely, it follows from the continuity at the zero vector that there exists a $\delta >0$ such that $\|L(h)\|=\|L(h)-L(0)\|\leq 1$ for all vectors $h \in X$ with $\|h\|\leq \delta$ . Thus, for all non-zero $x \in X$ , one has

\|Lx\|=\left\Vert {\|x\| \over \delta }L\left(\delta {x \over \|x\|}\right)\right\Vert ={\|x\| \over \delta }\left\Vert L\left(\delta {x \over \|x\|}\right)\right\Vert \leq {\|x\| \over \delta }\cdot 1={1 \over \delta }\|x\|.

This proves that $L$ is bounded.

Further properties

The condition for $L$ to be bounded, namely that there exists some $M$ such that for all x

\|Lx\|\leq M\|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.

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.
Any linear operator defined on a finite-dimensional normed space is bounded.
On the sequence space $c 00$ 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 $ℓ \infty$ norm, the same operator is not bounded.
Many integral transforms are bounded linear operators. For instance, if
$K:[a,b]\times [c,d]\to {\mathbb {R} }\,$

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

$(Lf)(y)=\int _{a}^{b}\!K(x,y)f(x)\,dx,\,$
is bounded. This operator is in fact compact. The compact operators form an important class of bounded operators.
The Laplace operator
$\Delta :H^{2}({\mathbb {R} }^{n})\to L^{2}({\mathbb {R} }^{n})\,$
(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 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 [−π, π], with the norm

\|P\|=\int _{-\pi }^{\pi }\!|P(x)|\,dx.

Define the operator $L : X \to X$ 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 $n \to \infty$ , 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.
If $Y$ is Banach, then so is $B(X, Y)$ .
from which it follows that dual spaces are Banach.
For any $A \in B(X, Y)$ , the kernel of $A$ is a closed linear subspace of $X$ .
If $B(X, Y)$ is Banach and $X$ is nontrivial, then $Y$ is Banach.

References

Proof: Assume for the sake of contradiction that $x • = (x i) \infty i =1$ converges to $0$ but $F (x •) = (F (x i)) \infty i =1$ is not bounded in $Y$ . Pick an open balanced neighborhood $V$ of the origin in $Y$ such that $V$ does not absorb the sequence $F (x •)$ . Replacing $x •$ with a subsequence if necessary, it may be assumed without loss of generality that $F (x i) \notin i 2 V$ for every positive integer $i$ . The sequence $z • := (x i / i) \infty i =1$ is Mackey convergent to the origin (since $(i z i) \infty i =1 = (x i) \infty i =1 \to 0$ is bounded in $X$ ) so by assumption, $F (z •) = (F (z i)) \infty i =1$ is bounded in $Y$ . So pick a real $r > 1$ such that $F (z i) \in r V$ for every integer $i$ . If $i > r$ is an integer then since $V$ is balanced, $F (x i) \in r i V \subseteq i 2 V$ , which is a contradiction. ∎ This proof readily generalizes to give even stronger characterizations of " $F$ is bounded." For example, the word "such that $(r i x i) \infty i =1$ is a bounded subset of $X.$ " in the definition of "Mackey convergent to the origin" can be replaced with "such that $(r i x i) \infty i =1 \to 0$ in $X.$ "

Wilansky 2013, pp. 47-50.
Narici & Beckenstein 2011, pp. 441-457.
Narici & Beckenstein 2011, p. 444.
Narici & Beckenstein 2011, pp. 451-457.

Bibliography

"Bounded operator", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
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.

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[3] Proof: Assume for the sake of contradiction that $x • = (x i) \infty i =1$ converges to $0$ but $F (x •) = (F (x i)) \infty i =1$ is not bounded in $Y$ . Pick an open balanced neighborhood $V$ of the origin in $Y$ such that $V$ does not absorb the sequence $F (x •)$ . Replacing $x •$ with a subsequence if necessary, it may be assumed without loss of generality that $F (x i) \notin i 2 V$ for every positive integer $i$ . The sequence $z • := (x i / i) \infty i =1$ is Mackey convergent to the origin (since $(i z i) \infty i =1 = (x i) \infty i =1 \to 0$ is bounded in $X$ ) so by assumption, $F (z •) = (F (z i)) \infty i =1$ is bounded in $Y$ . So pick a real $r > 1$ such that $F (z i) \in r V$ for every integer $i$ . If $i > r$ is an integer then since $V$ is balanced, $F (x i) \in r i V \subseteq i 2 V$ , which is a contradiction. ∎ This proof readily generalizes to give even stronger characterizations of " $F$ is bounded." For example, the word "such that $(r i x i) \infty i =1$ is a bounded subset of $X.$ " in the definition of "Mackey convergent to the origin" can be replaced with "such that $(r i x i) \infty i =1 \to 0$ in $X.$ "

[FOOTNOTEWilansky201347-50-1] Wilansky 2013, pp. 47-50.

[FOOTNOTENariciBeckenstein2011441-457-2] Narici & Beckenstein 2011, pp. 441-457.

[FOOTNOTENariciBeckenstein2011444-4] Narici & Beckenstein 2011, p. 444.

[FOOTNOTENariciBeckenstein2011451-457-5] Narici & Beckenstein 2011, pp. 451-457.

Boundedness and Bornology
Basic concepts	Barrelled space Bounded set Bornological space (Vector) Bornology
Operators	(Un)Bounded operator
Subsets	Barrelled set Bornivorous set Saturated family
Operators	(Countably) Barrelled space (Countably) Quasi-barrelled space Ultrabarrelled space Ultrabornological space