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{{Short description|Linear transformation between topological vector spaces}}
{{Distinguish|text=[[bounded function]] (set theory)}}

In [[functional analysis]] and [[operator theory]], a '''bounded linear operator''' is a [[linear transformation]] <math>L : X \to Y</math> between [[topological vector space]]s (TVSs) <math>X</math> and <math>Y</math> that maps [[Bounded set (topological vector space)|bounded]] subsets of <math>X</math> to bounded subsets of <math>Y.</math> 
If <math>X</math> and <math>Y</math> are [[normed vector space]]s (a special type of TVS), then <math>L</math> is bounded if and only if there exists some <math>M > 0</math> such that for all <math>x \in X,</math>
<math display=block>\|Lx\|_Y \leq M \|x\|_X.</math>
The smallest such <math>M</math> is called the [[operator norm]] of <math>L</math> and denoted by <math>\|L\|.</math> 
A bounded operator between normed spaces is [[Continuous linear operator|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 <math>f : X \to Y</math> is called "[[Bounded function|bounded]]" then this usually means that its [[Image of a function|image]] <math>f(X)</math> is a bounded subset of its codomain. A linear map has this property if and only if it is identically <math>0.</math> 
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 continuity|Lipschitz continuous]] at <math>0.</math>

===Equivalence of boundedness and continuity===

A linear operator between normed spaces is bounded if and only if it is [[Continuous linear operator|continuous]].

{{math proof|title=Proof|proof=
Suppose that <math>L</math> is bounded. Then, for all vectors <math>x, h \in X</math> with <math>h</math> nonzero we have
<math display=block>\|L(x + h) - L(x)\| = \|L(h)\| \leq M\|h\|.</math>
Letting <math>h </math> go to zero shows that <math>L</math> is continuous at <math>x.</math> 
Moreover, since the constant <math>M</math> does not depend on <math>x,</math> this shows that in fact <math>L</math> is [[Uniform continuity|uniformly continuous]], and even [[Lipschitz continuous]].

Conversely, it follows from the continuity at the zero vector that there exists a <math>\varepsilon > 0</math> such that <math>\|L(h)\| = \|L(h) - L(0)\| \leq 1</math> for all vectors <math>h \in X</math> with <math>\|h\| \leq \varepsilon.</math> 
Thus, for all non-zero <math>x \in X,</math> one has
<math display=block>\|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\|.</math>
This proves that <math>L</math> is bounded. [[Q.E.D.]]
}}

==In topological vector spaces==

A linear operator <math>F : X \to Y</math> between two [[topological vector space]]s (TVSs) is called a '''{{em|bounded linear operator}}''' or just '''{{em|bounded}}''' if whenever <math>B \subseteq X</math> is [[Bounded set (topological vector space)|bounded]] in <math>X</math> then <math>F(B)</math> is bounded in <math>Y.</math> 
A subset of a TVS is called bounded (or more precisely, [[Bounded set (topological vector space)|von Neumann bounded]]) if every neighborhood of the origin [[Absorbing set|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.{{sfn|Wilansky|2013|pp=47-50}} 
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 [[Metrizable topological vector space|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 space]]s, a weaker converse holds; any bounded linear map from an LF space is [[sequentially continuous]].

If <math>F : X \to Y</math> is a linear operator between two topological vector spaces and if there exists a neighborhood <math>U</math> of the origin in <math>X</math> such that <math>F(U)</math> is a bounded subset of <math>Y,</math> then <math>F</math> is continuous.{{sfn|Narici|Beckenstein|2011|pp=156-175}} 
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====
{{Main|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 <math>X</math> is a bornological space if and only if for every locally convex TVS <math>Y,</math> a linear operator <math>F : X \to Y</math> is continuous if and only if it is bounded.{{sfn|Narici|Beckenstein|2011|pp=441-457}}

Every normed space is bornological.

===Characterizations of bounded linear operators===

Let <math>F : X \to Y</math> be a linear operator between topological vector spaces (not necessarily Hausdorff). 
The following are equivalent:
#<math>F</math> is (locally) bounded;{{sfn|Narici|Beckenstein|2011|pp=441-457}}
#(Definition): <math>F</math> maps bounded subsets of its domain to bounded subsets of its codomain;{{sfn|Narici|Beckenstein|2011|pp=441-457}}
#<math>F</math> maps bounded subsets of its domain to bounded subsets of its [[Image of a function|image]] <math>\operatorname{Im} F := F(X)</math>;{{sfn|Narici|Beckenstein|2011|pp=441-457}}
#<math>F</math> maps every null sequence to a bounded sequence;{{sfn|Narici |Beckenstein| 2011|pp=441-457}}
#* 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.
#<math>F</math> maps every Mackey convergent null sequence to a bounded subset of <math>Y.</math><ref group=note>Proof: Assume for the sake of contradiction that <math>x_{\bull} = \left(x_i\right)_{i=1}^{\infty}</math> converges to <math>0</math> but <math>F\left(x_{\bull}\right) = \left(F\left(x_i\right)\right)_{i=1}^{\infty}</math> is not bounded in <math>Y.</math> Pick an open [[Balanced set|balanced]] neighborhood <math>V</math> of the origin in <math>Y</math> such that <math>V</math> does not absorb the sequence <math>F\left(x_{\bull}\right).</math> Replacing <math>x_{\bull}</math> with a subsequence if necessary, it may be assumed without loss of generality that <math>F\left(x_i\right) \not\in i^2 V</math> for every positive integer <math>i.</math> The sequence <math>z_{\bull} := \left(x_i/i\right)_{i=1}^{\infty}</math> is Mackey convergent to the origin (since <math>\left(i z_i\right)_{i=1}^{\infty} = \left(x_i\right)_{i=1}^{\infty} \to 0</math> is bounded in <math>X</math>) so by assumption, <math>F\left(z_{\bull}\right) = \left(F\left(z_i\right)\right)_{i=1}^{\infty}</math> is bounded in <math>Y.</math> So pick a real <math>r > 1</math> such that <math>F\left(z_i\right) \in r V</math> for every integer <math>i.</math> If <math>i > r</math> is an integer then since <math>V</math> is balanced, <math>F\left(x_i\right) \in r i V \subseteq i^2 V,</math> which is a contradiction. Q.E.D. This proof readily generalizes to give even stronger characterizations of "<math>F</math> is bounded." For example, the word "such that <math>\left(r_i x_i\right)_{i=1}^{\infty}</math> is a bounded subset of <math>X.</math>" in the definition of "Mackey convergent to the origin" can be replaced with "such that <math>\left(r_i x_i\right)_{i=1}^{\infty} \to 0</math> in <math>X.</math>"</ref><!---Old proof:<ref group=note>Proof if <math>Y</math> is locally convex: Assume for the sake of contradiction that <math>x_{\bull} = \left(x_i\right)_{i=1}^{\infty}</math> converges to <math>0</math> but <math>F\left(x_{\bull}\right) = \left(F\left(x_i\right)\right)_{i=1}^{\infty}</math> is not bounded in <math>Y.</math> Then there exists a continuous seminorm <math>p</math> on <math>Y</math> such that <math>p\left(F\left(x_{\bull}\right)\right) = \left(p\left(F\left(x_i\right)\right)\right)_{i=1}^{\infty}</math> is unbounded. Let <math>r_i = \sqrt{p\left(x_i\right)}</math> for every <math>i</math> where by going to a subsequence, it can be assumed without loss of generality that <math>r_i > 0</math> for all <math>i</math> and also <math>r_i \to \infty.</math> Then <math>\left(x_i/r_i\right)_{i=1}^{\infty}</math> Mackey convergent to the origin but its image under <math>F</math> is not bounded in <math>Y.</math> <math>\blacksquare</math></ref> End: old proof --->
#* A sequence <math>x_{\bull} = \left(x_i\right)_{i=1}^{\infty}</math> is said to be ''[[Mackey convergence|Mackey convergent to the origin]] in <math>X</math>'' if there exists a divergent sequence <math>r_{\bull} = \left(r_i\right)_{i=1}^{\infty} \to \infty</math> of positive real number such that <math>r_{\bull} = \left(r_i x_i\right)_{i=1}^{\infty}</math> is a bounded subset of <math>X.</math>

if <math>X</math> and <math>Y</math> are [[Locally convex topological vector space|locally convex]] then the following may be add to this list:
<ol start=6>
<li><math>F</math> maps bounded [[Absolutely convex set|disks]] into bounded disks.{{sfn|Narici|Beckenstein|2011|p=444}}</li>
<li><math>F^{-1}</math> maps [[Bornivorous set|bornivorous]] disks in <math>Y</math> into bornivorous disks in <math>X.</math>{{sfn|Narici|Beckenstein|2011|p=444}}</li>
</ol>

if <math>X</math> is a [[bornological space]] and <math>Y</math> is locally convex then the following may be added to this list:
<ol start=8>
<li><math>F</math> is [[Sequential continuity at a point|sequentially continuous at some]] (or equivalently, at every) point of its domain.{{sfn|Narici|Beckenstein|2011|pp=451-457}}
* A [[Sequential continuity|sequentially continuous]] linear map between two TVSs is always bounded,{{sfn|Wilansky|2013|pp=47-50}} but the converse requires additional assumptions to hold (such as the domain being bornological and the codomain being locally convex).
* If the domain <math>X</math> is also a [[sequential space]], then <math>F</math> is [[Sequential continuity|sequentially continuous]] if and only if it is continuous.</li>
<li><math>F</math> is [[Sequential continuity at a point|sequentially continuous at the origin]].</li>
</ol>

==Examples==

<ul>
<li>Any linear operator between two finite-dimensional normed spaces is bounded, and such an operator may be viewed as multiplication by some fixed [[matrix (mathematics)|matrix]].</li>
<li>Any linear operator defined on a finite-dimensional normed space is bounded.</li>
<li>On the [[sequence space]] <math>c_{00}</math> of eventually zero sequences of real numbers, considered with the <math>\ell^1</math> 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 <math>\ell^{\infty}</math> norm, the same operator is not bounded.</li>
<li>Many [[integral transform]]s are bounded linear operators. For instance, if 
<math display=block>K : [a, b] \times [c, d] \to \R</math>
is a continuous function, then the operator <math>L</math> defined on the space <math>C[a, b]</math> of continuous functions on <math>[a, b]</math> endowed with the [[uniform norm]] and with values in the space <math>C[c, d]</math> with <math>L</math> given by the formula
<math display=block>(Lf)(y) = \int_a^b\!K(x, y)f(x)\,dx, </math>
is bounded. This operator is in fact a [[compact operator]]. The compact operators form an important class of bounded operators.</li>
<li>The [[Laplace operator]] 
<math display=block>\Delta : H^2(\R^n) \to L^2(\R^n) \,</math>
(its [[domain of a function|domain]] is a [[Sobolev space]] and it takes values in a space of [[square-integrable function]]s) is bounded.</li>
<li>The [[shift operator]] on the [[Lp space]] <math>\ell^2</math> of all [[sequence]]s <math>\left(x_0, x_1, x_2, \ldots\right)</math> of real numbers with <math>x_0^2 + x_1^2 + x_2^2 + \cdots < \infty, \,</math> 
<math display=block>L(x_0, x_1, x_2, \dots) = \left(0, x_0, x_1, x_2, \ldots\right) </math> 
is bounded. Its operator norm is easily seen to be <math>1.</math></li>
</ul>

===Unbounded linear operators===

Let <math>X</math> be the space of all [[trigonometric polynomial]]s on <math>[-\pi, \pi],</math> with the norm 

<math display=block>\|P\| = \int_{-\pi}^{\pi}\!|P(x)|\,dx.</math>

The operator <math>L : X \to X</math> that maps a polynomial to its [[derivative]] is not bounded. Indeed, for <math>v_n = e^{i n x}</math> with <math>n = 1, 2, \ldots,</math> we have <math>\|v_n\| = 2\pi,</math> while <math>\|L(v_n)\| = 2 \pi n \to \infty</math> as <math>n \to \infty,</math> so <math>L</math> is not bounded.

===Properties of the space of bounded linear operators===
The space of all bounded linear operators from <math>X</math> to <math>Y</math> is denoted by <math>B(X, Y)</math>.
* <math>B(X, Y)</math> is a normed vector space.
* If <math>Y</math> is Banach, then so is <math>B(X, Y)</math>; in particular, [[dual space]]s are Banach.
* For any <math>A \in B(X, Y)</math> the kernel of <math>A</math> is a closed linear subspace of <math>X</math>.
* If <math>B(X, Y)</math> is Banach and <math>X</math> is nontrivial, then <math>Y</math> is Banach.

==See also==

* {{annotated link|Bounded set (topological vector space)}}
* {{annotated link|Contraction (operator theory)}}
* {{annotated link|Discontinuous linear map}}
* {{annotated link|Continuous linear operator}}
* {{annotated link|Local boundedness}}
* {{annotated link|Norm (mathematics)}}
* {{annotated link|Operator algebra}}
* {{annotated link|Operator norm}}
* {{annotated link|Operator theory}}
* {{annotated link|Seminorm}}
* {{annotated link|Unbounded operator}}

==References==

{{reflist|group=note}}
{{reflist}}

==Bibliography==

* {{springer|title=Bounded operator|id=p/b017420}}
* Kreyszig, Erwin: ''Introductory Functional Analysis with Applications'', Wiley, 1989
* {{Narici Beckenstein Topological Vector Spaces|edition=2}} <!-- {{sfn|Narici|Beckenstein|2011|p=}} -->
* {{Wilansky Modern Methods in Topological Vector Spaces}} <!-- {{sfn|Wilansky|2013|p=}} -->

{{Banach spaces}}
{{Functional Analysis}}
{{BoundednessAndBornology}}

[[Category:Linear operators]]
[[Category:Operator theory]]
[[Category:Theory of continuous functions]]