Bounded operator

In functional analysis and operator theory, 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 $M>0$ such that for all $x\in X,$ $\|Lx\|_{Y}\leq M\|x\|_{X}.$ 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 $f:X\to Y$ 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.

Proof

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 $h$ 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 $\varepsilon >0$ such that $\|L(h)\|=\|L(h)-L(0)\|\leq 1$ for all vectors $h\in X$ with $\|h\|\leq \varepsilon .$ Thus, for all non-zero $x\in X,$ one has $\|Lx\|=\left\Vert {\|x\| \over \varepsilon }L\left(\varepsilon {x \over \|x\|}\right)\right\Vert ={\|x\| \over \varepsilon }\left\Vert L\left(\varepsilon {x \over \|x\|}\right)\right\Vert \leq {\|x\| \over \varepsilon }\cdot 1={1 \over \varepsilon }\|x\|.$ This proves that $L$ is bounded. Q.E.D.

In topological vector spaces

A linear operator $F:X\to Y$ between two topological vector spaces (TVSs) is called a bounded linear operator 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.

Continuity and boundedness

Every sequentially continuous linear operator between TVS is a bounded operator.^[1] 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 $F:X\to Y$ 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.^[2] 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

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 $F:X\to Y$ is continuous if and only if it is bounded.^[3]

Every normed space is bornological.

Characterizations of bounded linear operators

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

$F$ is (locally) bounded;^[3]
(Definition): $F$ maps bounded subsets of its domain to bounded subsets of its codomain;^[3]
$F$ maps bounded subsets of its domain to bounded subsets of its image $\operatorname {Im} F:=F(X)$ ;^[3]
$F$ maps every null sequence to a bounded sequence;^[3]
- 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 $x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }$ is said to be Mackey convergent to the origin in $X$ if there exists a divergent sequence $r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty }\to \infty$ of positive real number such that $r_{\bullet }=\left(r_{i}x_{i}\right)_{i=1}^{\infty }$ is a bounded subset of $X.$

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

$F$ maps bounded disks into bounded disks.^[4]
$F^{-1}$ maps bornivorous disks in $Y$ into bornivorous disks in $X.$ ^[4]

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

$F$ is sequentially continuous at some (or equivalently, at every) point of its domain.^[5]
- A sequentially continuous linear map between two TVSs is always bounded,^[1] 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.
$F$ is sequentially continuous at the origin.

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 $\ell ^{1}$ 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 $\ell ^{\infty }$ 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 a compact operator. 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 Lp space $\ell ^{2}$ of all sequences $\left(x_{0},x_{1},x_{2},\ldots \right)$ of real numbers with $x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+\cdots <\infty ,\,$ $L(x_{0},x_{1},x_{2},\dots )=\left(0,x_{0},x_{1},x_{2},\ldots \right)$ 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

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

The operator $L:X\to X$ that maps a polynomial to its derivative is not bounded. Indeed, for $v_{n}=e^{inx}$ with $n=1,2,\ldots ,$ we have $\|v_{n}\|=2\pi ,$ while $\|L(v_{n})\|=2\pi n\to \infty$ as $n\to \infty ,$ 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.
If $Y$ is Banach, then so is $B(X,Y)$ ; in particular, 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 $x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }$ converges to $0$ but $F\left(x_{\bullet }\right)=\left(F\left(x_{i}\right)\right)_{i=1}^{\infty }$ is not bounded in $Y.$ Pick an open balanced neighborhood $V$ of the origin in $Y$ such that $V$ does not absorb the sequence $F\left(x_{\bullet }\right).$ Replacing $x_{\bullet }$ with a subsequence if necessary, it may be assumed without loss of generality that $F\left(x_{i}\right)\not \in i^{2}V$ for every positive integer $i.$ The sequence $z_{\bullet }:=\left(x_{i}/i\right)_{i=1}^{\infty }$ is Mackey convergent to the origin (since $\left(iz_{i}\right)_{i=1}^{\infty }=\left(x_{i}\right)_{i=1}^{\infty }\to 0$ is bounded in $X$ ) so by assumption, $F\left(z_{\bullet }\right)=\left(F\left(z_{i}\right)\right)_{i=1}^{\infty }$ is bounded in $Y.$ So pick a real $r>1$ such that $F\left(z_{i}\right)\in rV$ for every integer $i.$ If $i>r$ is an integer then since $V$ is balanced, $F\left(x_{i}\right)\in riV\subseteq i^{2}V,$ 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 $\left(r_{i}x_{i}\right)_{i=1}^{\infty }$ is a bounded subset of $X.$ " in the definition of "Mackey convergent to the origin" can be replaced with "such that $\left(r_{i}x_{i}\right)_{i=1}^{\infty }\to 0$ in $X.$ "

^ ^a ^b Wilansky 2013, pp. 47–50.
^ Narici & Beckenstein 2011, pp. 156–175.
^ ^a ^b ^c ^d ^e Narici & Beckenstein 2011, pp. 441–457.
^ ^a ^b 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.

[4] Proof: Assume for the sake of contradiction that $x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }$ converges to $0$ but $F\left(x_{\bullet }\right)=\left(F\left(x_{i}\right)\right)_{i=1}^{\infty }$ is not bounded in $Y.$ Pick an open balanced neighborhood $V$ of the origin in $Y$ such that $V$ does not absorb the sequence $F\left(x_{\bullet }\right).$ Replacing $x_{\bullet }$ with a subsequence if necessary, it may be assumed without loss of generality that $F\left(x_{i}\right)\not \in i^{2}V$ for every positive integer $i.$ The sequence $z_{\bullet }:=\left(x_{i}/i\right)_{i=1}^{\infty }$ is Mackey convergent to the origin (since $\left(iz_{i}\right)_{i=1}^{\infty }=\left(x_{i}\right)_{i=1}^{\infty }\to 0$ is bounded in $X$ ) so by assumption, $F\left(z_{\bullet }\right)=\left(F\left(z_{i}\right)\right)_{i=1}^{\infty }$ is bounded in $Y.$ So pick a real $r>1$ such that $F\left(z_{i}\right)\in rV$ for every integer $i.$ If $i>r$ is an integer then since $V$ is balanced, $F\left(x_{i}\right)\in riV\subseteq i^{2}V,$ 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 $\left(r_{i}x_{i}\right)_{i=1}^{\infty }$ is a bounded subset of $X.$ " in the definition of "Mackey convergent to the origin" can be replaced with "such that $\left(r_{i}x_{i}\right)_{i=1}^{\infty }\to 0$ in $X.$ "

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

[FOOTNOTENariciBeckenstein2011156–175-2] Narici & Beckenstein 2011, pp. 156–175.

[FOOTNOTENariciBeckenstein2011441–457-3] Narici & Beckenstein 2011, pp. 441–457.

[FOOTNOTENariciBeckenstein2011444-5] Narici & Beckenstein 2011, p. 444.

[FOOTNOTENariciBeckenstein2011451–457-6] Narici & Beckenstein 2011, pp. 451–457.

[1]

[2]

[3]

[note 1]

[4]

[5]

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