Continuous linear operator: Difference between revisions

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{{Short description|Function between topological vector spaces}}

In [[functional analysis]] and related areas of [[mathematics]], a '''continuous linear operator''' or '''continuous linear mapping''' is a [[Continuous function (topology)|continuous]] [[linear transformation]] between [[topological vector space]]s.

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Throughout, <math>F : X \to Y</math> is a [[linear map]] between [[topological vector space]]s (TVSs).

'''Bounded on a setsubset'''

{{See also|Bounded set (topological vector space)}}

The notion of a "bounded set" for a topological vector space is that of being a [[Bounded set (topological vector space)|von Neumann bounded set]].

If the space happens to also be a [[normed space]] (or a [[seminormed space]]), such as the scalar field with the [[absolute value]] for instance, then a subset <math>S</math> is von Neumann bounded if and only if it is {{em|[[Norm (mathematics)|norm]] bounded; that is}}, ifmeaning and only ifthat <math>\sup_{s \in S} \|s\| < \infty.</math>

A subset of a normed (or seminormed) space is called {{em|bounded}} if it is norm-bounded (or equivalently, von Neumann bounded).

For example, the scalar field (<math>\Reals</math> or <math>\Complex</math>) with the [[absolute value]] <math>|\cdot|</math> is a normed space, so a subset <math>S</math> is bounded if and only if <math>\sup_{s \in S} |s|</math> is finite, which happens if and only if <math>S</math> is contained in some open (or closed) ball centered at the origin (zero).

Any translation, scalar multiple, and subset of a bounded set is again bounded.

'''Function bounded on a set'''

If <math>S \subseteq X</math> is a set then <math>F : X \to Y</math> is said to be {{em|{{visible anchor|function bounded on a set|bounded on a set|text=bounded on <math>S</math>}}}} if <math>F(S)</math> is a [[Bounded set (topological vector space)|bounded subset]] of <math>Y,</math> which if <math>(Y, \|\cdot\|)</math> is a normed (or seminormed) space happens if and only if <math>\sup_{s \in S} \|F(s)\| < \infty.</math>

A linear map <math>F</math> is bounded on a set <math>S</math> if and only if it is bounded on <math>x + S := \{x + s : s \in S\}</math> for every <math>x \in X</math> (because <math>F(x + S) = F(x) + F(S)</math> and any translation of a bounded set is again bounded) if and only if it is bounded on <math>c S := \{c s : s \in S\}</math> for every non-zero scalar <math>c \neq 0</math> (because <math>F(c S) = c F(S)</math> and any scalar multiple of a bounded set is again bounded).

Consequently, if <math>(X, \|\cdot\|)</math> is a normed or seminormed space, then a linear map <math>F : X \to Y</math> is bounded on some (equivalently, on every) non-degenerate open or closed ball (not necessarily centered at the origin, and of any radius) if and only if it is bounded on the closed unit ball centered at the origin <math>\{x \in X : \|x\| \leq 1\}.</math>

'''Bounded linear maps'''

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By definition, a linear map <math>F : X \to Y</math> between [[Topological vector space|TVS]]s is said to be {{em|[[Bounded linear operator|bounded]]}} and is called a {{em|{{visible anchor|bounded linear operator|text=[[bounded linear operator]]}}}} if for every [[Bounded set (topological vector space)|(von Neumann) bounded subset]] <math>B \subseteq X</math> of its ___domain, <math>F(B)</math> is a bounded subset of it codomain; or said more briefly, if it is bounded on every bounded subset of its ___domain. When the ___domain <math>X</math> is a normed (or seminormed) space then it suffices to check this condition for the open or closed unit ball centered at the origin. Explicitly, if <math>B_1</math> denotes this ball then <math>F : X \to Y</math> is a bounded linear operator if and only if <math>F\left(B_1\right)</math> is a bounded subset of <math>Y;</math> if <math>Y</math> is also a (semi)normed space then this happens if and only if the [[operator norm]] <math>\|F\| := \sup_{\|x\| \leq 1} \|F(x)\| < \infty</math> is finite. Every [[sequentially continuous]] linear operator is bounded.{{sfn|Wilansky|2013|pp=47-50}}

'''BoundedFunction bounded on a neighborhood and local boundedness'''

{{See also|Local boundedness}}

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</li>

<li><math>f</math> is continuous at the origin.

* By definition, <math>f</math> said to be continuous at the origin if for every open (or closed) ball <math>B_r</math> of radius <math>r > 0</math> centered at <math>0</math> in the codomain <math>\mathbb{F},</math> there exists some [[Neighbourhood (mathematics)|neighborhood]] <math>U</math> of the origin in <math>X</math> such that <math>f(U) \subseteq B_r.</math>
* If <math>B_r</math> is a closed ball then the condition <math>f(U) \subseteq B_r</math> holds if and only if <math>\sup_{u \in U} |f(u)| \leq r.</math>

** However,It assumingis important that <math>B_r</math> be a closed ball in this [[supremum]] characterization. Assuming that <math>B_r</math> is instead an open ball, then <math>\sup_{u \in U} |f(u)| < r</math> is a sufficient but {{em|not necessary}} condition for <math>f(U) \subseteq B_r</math> to be true (consider for example when <math>f = \operatorname{Id}</math> is the identity map on <math>X = \mathbb{F}</math> and <math>U = B_r</math>), whereas the non-strict inequality <math>\sup_{u \in U} |f(u)| \leq r</math> is instead a necessary but {{em|not sufficient}} condition for <math>f(U) \subseteq B_r</math> to be true (consider for example <math>X = \R, f = \operatorname{Id},</math> and the closed neighborhood <math>U = [-r, r]</math>). This is one of several reasons why many definitions involving linear functionals, such as [[polar set]]s for example, involve closed (rather than open) neighborhoods and non-strict <math>\,\leq\,</math> (rather than strict<math>\,<\,</math>) inequalities.

</li>

<li><math>f</math> is [[#bounded on a neighborhood|bounded on a neighborhood]] (of some point). Said differently, <math>f</math> is a [[#locally bounded at|locally bounded at some point]] of its ___domain.

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<li>There exists some neighborhood <math>U</math> of the origin such that <math>\sup_{u \in U} |f(u)| \leq 1</math>

* This inequality holds if and only if <math>\sup_{x \in r U} |f(x)| \leq r</math> for every real <math>r > 0,</math> which shows that the positive scalar multiples <math>\{r U : r > 0\}</math> of this single neighborhood <math>U</math> will satisfy the definition of [[Continuity at a point|continuity at the origin]] given in (4) above.

* By definition of the set <math>U^\circ,</math> which is called the [[Polar set|(absolute) polar]] of <math>U,</math> the inequality <math>\sup_{u \in U} |f(u)| \leq 1</math> holds if and only if <math>f \in U^\circ.</math> Polar sets, and thusso also this particular inequality, play important roles in [[duality theory]].

</li>

<li><math>f</math> is a [[#locally bounded|locally bounded at every point]] of its ___domain.</li>

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</ol>

Thus, ifIf <math>X</math> is a complex then either all three of <math>f,</math> <math>\operatorname{Re} f,</math> and <math>\operatorname{Im} f</math> are [[Continuous linear map|continuous]] (resp.respectively, [[Bounded linear operator|bounded]]), or else all three are [[Discontinuous linear functional|discontinuous]] (resp.respectively, unbounded).

==Examples==

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Every linear map whose ___domain is a finite-dimensional Hausdorff [[topological vector space]] (TVS) is continuous. This is not true if the finite-dimensional TVS is not Hausdorff.

Every (constant) map <math>X \to Y</math> between TVS that is identically equal to zero is a linear map that is continuous, bounded, and bounded on the neighborhood <math>X</math> of the origin. In particular, every TVS has a non-empty [[continuous dual space]] (although it is possible for the constant zero map to be its only continuous linear functional).

Suppose <math>X</math> is any Hausdorff TVS. Then {{em|every}} [[linear functional]] on <math>X</math> is necessarily continuous if and only if every vector subspace of <math>X</math> is closed.{{sfn|Wilansky|2013|p=55}} Every linear functional on <math>X</math> is necessarily a bounded linear functional if and only if every [[Bounded set (topological vector space)|bounded subset]] of <math>X</math> is contained in a finite-dimensional vector subspace.{{sfn|Wilansky|2013|p=50}}

Suppose <math>X</math> is any Hausdorff TVS. Then {{em|every}} [[linear functional]] on <math>X</math> is necessarily continuous if and only if every vector subspace of <math>X</math> is closed.{{sfn|Wilansky|2013|p=55}} Every linear functional on <math>X</math> is necessarily a bounded linear functional if and only if every [[Bounded set (topological vector space)|bounded subset]] of <math>X</math> is contained in a finite-dimensional vector subspace.{{sfn|Wilansky|2013|p=50}}

==Properties==

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* {{Köthe Topological Vector Spaces I}} <!-- {{sfn|Köthe|1969|p=}} -->

* {{Narici Beckenstein Topological Vector Spaces|edition=2}} <!-- {{sfn|Narici|Beckenstein|2011|p=}} -->

* {{Cite book|last1=Rudin|first1=Walter|author-link1=Walter Rudin|isbn=978-0-07-054236-5|title=Functional analysis|yeardate=January 1991|publisher=McGraw-Hill Science/Engineering/Math|url-access=registration|url=https://archive.org/details/functionalanalys00rudi}} <!-- Rudin, Walter (1991) Functional Analysis -->

* {{Schaefer Wolff Topological Vector Spaces|edition=2}} <!-- {{sfn|Schaefer|1999|p=}} -->

* {{Swartz An Introduction to Functional Analysis}} <!-- {{sfn|Swartz|1992|p=}} -->