Discontinuous linear map: Difference between revisions

Revision as of 01:44, 28 June 2021

Latest revision as of 10:22, 24 April 2025

(14 intermediate revisions by 7 users not shown)

Line 3:

== A linear map from a finite-dimensional space is always continuous ==

Let ''X'' and ''Y'' be two normed spaces and ''<math>f'' : X \to Y</math> a linear map from ''X'' to ''Y''. If ''X'' is [[finite-dimensional]], choose a basis (''e''<sub>1</submath>\left(e_1, ''e''₂e_2, …\ldots, ''e''<sub>''n''e_n\right)</submath>) in ''X'' which may be taken to be unit vectors. Then,

:<math display=block>f(x) = \sum^n_{i=1} x_if(e_i),</math>

and so by the [[triangle inequality]],

:<math display=block>\|f(x)\| = \left\|\sum^n_{i=1} x_if(e_i)\right\| \leleq \sum^n_{i=1} |x_i|\|f(e_i)\|.</math>

Letting

:<math display=block>M = \sup_i \{\|f(e_i)\|\},</math>

and using the fact that

:<math display=block>\sum^n_{i=1}|x_i|\leleq C \|x\|</math>

for some ''C''>0 which follows from the fact that [[Norm (mathematics)#Properties|any two norms on a finite-dimensional space are equivalent]], one finds

:<math display=block>\|f(x)\| \leleq \left(\sum^n_{i=1}|x_i|\right)M \leleq CM\|x\|.</math>

Thus, <math>f</math> is a [[bounded linear operator]] and so is continuous. In fact, to see this, simply note that ''f'' is linear,

and therefore <math>\|f(x)-f(x')\| = \|f(x-x')\| \leleq K \|x-x'\|</math> for some universal constant ''K''. Thus for any <math>\epsilon>0,</math>,

we can choose <math>\delta \leleq \epsilon/K</math> so that <math>f(B(x,\delta)) \subseteq B(f(x), \epsilon)</math> (<math>B(x, \delta)</math> and

<math>B(f(x), \epsilon)</math> are the normed balls around <math>x</math> and <math>f(x)</math>), which gives continuity.

Line 24:

Examples of discontinuous linear maps are easy to construct in spaces that are not complete; on any Cauchy sequence <math>e_i</math> of linearly independent vectors which does not have a limit, there is a linear operator <math>T</math> such that the quantities <math>\|T(e_i)\|/\|e_i\|</math> grow without bound. In a sense, the linear operators are not continuous because the space has "holes".

For example, consider the space ''<math>X''</math> of real-valued [[smooth function]]s on the interval [0, 1] with the [[uniform norm]], that is,

: <math display=block>\|f\| = \sup_{x\in [0, 1]}|f(x)|.</math>

The ''[[derivative]]-at-a-point'' map, given by

:<math display=block>T(f) = f'(0)\,</math>

defined on ''<math>X''</math> and with real values, is linear, but not continuous. Indeed, consider the sequence

:<math display=block>f_n(x)=\frac{\sin (n^2 x)}{n} </math>

for ''<math>n''≥1 \geq 1</math>. This sequence converges uniformly to the constantly zero function, but

:<math display=block>T(f_n) = \frac{n^2\cos(n^2 \cdot 0)}{n} = n\to \infty</math>

as ''<math>n''→∞ \to \infty</math> instead of <math>T(f_n)\to T(0)=0</math>, whichas would hold for a continuous map. Note that ''<math>T''</math> is real-valued, and so is actually a [[linear functional]] on ''<math>X''</math> (an element of the algebraic [[dual space]] ''X''<supmath>X^*</supmath>). The linear map ''<math>X'' →\to ''X''</math> which assigns to each function its derivative is similarly discontinuous. Note that although the derivative operator is not continuous, it is [[closed operator|closed]].

:<math>T(f)=f'(0)\,</math>

The fact that the ___domain is not complete here is important.: Discontinuousdiscontinuous operators on complete spaces require a little more work.

defined on ''X'' and with real values, is linear, but not continuous. Indeed, consider the sequence

:<math>f_n(x)=\frac{\sin (n^2 x)}{n} </math>

for ''n''≥1. This sequence converges uniformly to the constantly zero function, but

:<math>T(f_n)=\frac{n^2\cos(n^2 \cdot 0)}{n}=n\to \infty</math>

as ''n''→∞ instead of <math>T(f_n)\to T(0)=0</math> which would hold for a continuous map. Note that ''T'' is real-valued, and so is actually a [[linear functional]] on ''X'' (an element of the algebraic [[dual space]] ''X''^*). The linear map ''X'' → ''X'' which assigns to each function its derivative is similarly discontinuous. Note that although the derivative operator is not continuous, it is [[closed operator|closed]].

The fact that the ___domain is not complete here is important. Discontinuous operators on complete spaces require a little more work.

== A nonconstructive example ==

An algebraic basis for the [[real number]]s as a vector space over the [[rationals]] is known as a [[Hamel basis]] (note that some authors use this term in a broader sense to mean an algebraic basis of ''any'' vector space). Note that any two [[commensurability (mathematics)|noncommensurable]] numbers, say 1 and π<math>\pi</math>, are linearly independent. One may find a Hamel basis containing them, and define a map ''<math>f'' from: '''\R''' \to '''\R'''</math> so that ''<math>f''(π\pi) = 0,</math> ''f'' acts as the identity on the rest of the Hamel basis, and extend to all of '''<math>\R'''</math> by linearity. Let {''r''_''n''}_''n'' be any sequence of rationals which converges to π<math>\pi</math>. Then lim_''n'' ''f''(''r''_''n'') = π, but ''<math>f''(π\pi) = 0.</math> By construction, ''f'' is linear over '''<math>\Q'''</math> (not over '''<math>\R'''</math>), but not continuous. Note that ''f'' is also not [[measurable function|measurable]]; an [[Additive map|additive]] real function is linear if and only if it is measurable, so for every such function there is a [[Vitali set]]. The construction of ''f'' relies on the axiom of choice.

This example can be extended into a general theorem about the existence of discontinuous linear maps on any infinite-dimensional normed space (as long as the codomain is not trivial).

Line 50 ⟶ 45:

== General existence theorem ==

Discontinuous linear maps can be proven to exist more generally, even if the space is complete.{{clarify|reason=A general "constructive" proof in the incomplete case was not given above, so this contrast seems kind of hand-wavey.|date=May 2015}} Let ''X'' and ''Y'' be [[normed space]]s over the field ''K'' where ''<math>K'' = '''\R'''</math> or ''<math>K'' = '''C'''\Complex.</math> Assume that ''X'' is infinite-dimensional and ''Y'' is not the zero space. We will find a discontinuous linear map ''f'' from ''X'' to ''K'', which will imply the existence of a discontinuous linear map ''g'' from ''X'' to ''Y'' given by the formula ''<math>g''(''x'') = ''f''(''x'')''y''<sub>0 y_0</submath> where ''y''<submath>0y_0</submath> is an arbitrary nonzero vector in ''Y''.

If ''X'' is infinite-dimensional, to show the existence of a linear functional which is not continuous then amounts to constructing ''f'' which is not bounded. For that, consider a [[sequence]] (''e''_''n'')_''n'' (''<math>n'' ≥\geq 1</math>) of [[linearly independent]] vectors in ''X'', which we normalize. DefineThen, we define

:<math display=block>T(e_n) = n\|e_n\|\,</math>

for each ''<math>n'' = 1, 2, ...\ldots</math> Complete this sequence of linearly independent vectors to a [[basis (vector space)|vector space basis]] of ''X'', andby definedefining ''T'' at the other vectors in the basis to be zero. ''T'' so defined will extend uniquely to a linear map on ''X'', and since it is clearly not bounded, it is not continuous.

Notice that by using the fact that any set of linearly independent vectors can be completed to a basis, we implicitly used the axiom of choice, which was not needed for the concrete example in the previous section but one.

:<math>T(e_n)=n\|e_n\|\,</math>

for each ''n'' = 1, 2, ... Complete this sequence of linearly independent vectors to a [[basis (vector space)|vector space basis]] of ''X'', and define ''T'' at the other vectors in the basis to be zero. ''T'' so defined will extend uniquely to a linear map on ''X'', and since it is clearly not bounded, it is not continuous.

Notice that by using the fact that any set of linearly independent vectors can be completed to a basis, we implicitly used the axiom of choice, which was not needed for the concrete example in the previous section but one.

== Role of the axiom of choice ==

Line 73 ⟶ 66:

| volume = 92

| year = 1970

| issue = 1 | doi=10.2307/1970696| jstor = 1970696 }}.</ref> This implies that there are no discontinuous linear real functions. Clearly AC does not hold in the model.

Solovay's result shows that it is not necessary to assume that all infinite-dimensional vector spaces admit discontinuous linear maps, and there are schools of analysis which adopt a more [[constructivism (mathematics)|constructivist]] viewpoint. For example, H. G. Garnir, in searching for so-called "dream spaces" (topological vector spaces on which every linear map into a normed space is continuous), was led to adopt ZF + [[dependent choice|DC]] + [[Baire property|BP]] (dependent choice is a weakened form and the [[Baire property]] is a negation of strong AC) as his axioms to prove the [[Garnir–Wright closed graph theorem]] which states, among other things, that any linear map from an [[F-space]] to a TVS is continuous. Going to the extreme of [[Constructivism (mathematics)|constructivism]], there is [[Ceitin's theorem]], which states that ''every'' function is continuous (this is to be understood in the terminology of constructivism, according to which only representable functions are considered to be functions).<ref>{{citation|title=Handbook of Analysis and Its Foundations|first=Eric|last=Schechter|publisher=Academic Press|year=1996|isbn=9780080532998|page=136|url=https://books.google.com/books?id=eqUv3Bcd56EC&pg=PA136}}.</ref> Such stances are held by only a small minority of working mathematicians.

Line 82 ⟶ 75:

Many naturally occurring linear discontinuous operators are [[closed operator|closed]], a class of operators which share some of the features of continuous operators. It makes sense to ask which linear operators on a given space are closed. The [[closed graph theorem]] asserts that an ''everywhere-defined'' closed operator on a complete ___domain is continuous, so to obtain a discontinuous closed operator, one must permit operators which are not defined everywhere.

To be more concrete, let <math>T</math> be a map from <math>X</math> to <math>Y</math> with ___domain <math>\operatorname{Dom}(T)</math>, written <math>T: \operatorname{Dom}(T)\subseteq X\to Y</math>. We don't lose much if we replace ''X'' by the closure of <math>\operatorname{Dom}(T)</math>. That is, in studying operators that are not everywhere-defined, one may restrict one's attention to [[densely defined operator]]s without loss of generality.

To be more concrete, let <math>T</math> be a map from <math>X</math> to <math>Y</math> with ___domain <math>\operatorname{Dom}(T),</math>, written <math>T : \operatorname{Dom}(T) \subseteq X \to Y.</math>. We don't lose much if we replace ''X'' by the closure of <math>\operatorname{Dom}(T).</math>. That is, in studying operators that are not everywhere-defined, one may restrict one's attention to [[densely defined operator]]s without loss of generality.

If the graph <math>\Gamma(T)</math> of <math>T</math> is closed in ''X'' ×''Y'', we call ''T'' ''closed''. Otherwise, consider its closure <math>\overline{\Gamma(T)}</math> in ''X'' ×''Y''. If <math>\overline{\Gamma(T)}</math> is itself the graph of some operator <math>\overline{T}</math>, <math>T</math> is called ''closable'', and <math>\overline{T}</math> is called the ''closure'' of <math>T</math>.

If the graph <math>\Gamma(T)</math> of <math>T</math> is closed in ''<math>X'' ×''\times Y'',</math> we call ''T'' ''closed''. Otherwise, consider its closure <math>\overline{\Gamma(T)}</math> in ''<math>X'' ×''\times Y''.</math> If <math>\overline{\Gamma(T)}</math> is itself the graph of some operator <math>\overline{T},</math>, <math>T</math> is called ''closable'', and <math>\overline{T}</math> is called the ''closure'' of <math>T.</math>.

So the natural question to ask about linear operators that are not everywhere-defined is whether they are closable. The answer is, "not necessarily"; indeed, every infinite-dimensional normed space admits linear operators that are not closable. As in the case of discontinuous operators considered above, the proof requires the axiom of choice and so is in general nonconstructive, though again, if ''X'' is not complete, there are constructible examples.

In fact, there is even an example of a linear operator whose graph has closure ''all'' of ''<math>X'' ×''\times Y''.</math> Such an operator is not closable. Let ''X'' be the space of [[polynomial function]]s from [0,1] to '''<math>\R'''</math> and ''Y'' the space of polynomial functions from [2,3] to '''<math>\R'''</math>. They are subspaces of ''C''([0,1]) and ''C''([2,3]) respectively, and so normed spaces. Define an operator ''T'' which takes the polynomial function ''x'' ↦ ''p''(''x'') on [0,1] to the same function on [2,3]. As a consequence of the [[Stone–Weierstrass theorem]], the graph of this operator is dense in ''<math>X''×'' \times Y'',</math> so this provides a sort of maximally discontinuous linear map (confer [[nowhere continuous function]]). Note that ''X'' is not complete here, as must be the case when there is such a constructible map.

== Impact for dual spaces ==

Line 97 ⟶ 90:

== Beyond normed spaces ==

The argument for the existence of discontinuous linear maps on normed spaces can be generalized to all metrisablemetrizable topological vector spaces, especially to all Fréchet- spaces, but there exist infinite-dimensional locally convex topological vector spaces such that every functional is continuous.<ref>For example, the weak topology w.r.t. the space of all (algebraically) linear functionals.</ref> On the other hand, the [[Hahn–Banach theorem]], which applies to all locally convex spaces, guarantees the existence of many continuous linear functionals, and so a large dual space. In fact, to every convex set, the [[Minkowski gauge]] associates a continuous [[linear functional]]. The upshot is that spaces with fewer convex sets have fewer functionals, and in the worst-case scenario, a space may have no functionals at all other than the zero functional. This is the case for the [[Lp space|''L''<supmath>''L^p''</sup>('''\R''','' dx'')</math>]] spaces with <math>0&nbsp; <&nbsp;'' p''&nbsp; <&nbsp; 1,</math> from which it follows that these spaces are nonconvex. Note that here is indicated the [[Lebesgue measure]] on the real line. There are other ''L''<supmath>''L^p''</supmath> spaces with <math>0&nbsp; <&nbsp;'' p''&nbsp; <&nbsp; 1</math> which do have nontrivial dual spaces.

Another such example is the space of real-valued [[measurable function]]s on the unit interval with [[quasinorm]] given by

:<math display=block>|\|f|\| = \int_I \frac{|f(x)|}{1+|f(x)|}dx.</math>

This non-locally convex space has a trivial dual space.

One can consider even more general spaces. For example, the existence of a homomorphism between complete separable metric [[groupGroup (mathematics)|group]]s can also be shown nonconstructively.

==See also==

* {{annotated link|Finest locally convex topology}}

* {{annotated link|Sublinear function}}

<!--

==Notes==

{{reflist|group=note}}

-->

== References ==

{{reflist}}

== References ==

{{Reflist}}

* Constantin Costara, Dumitru Popa, ''Exercises in Functional Analysis'', Springer, 2003. {{isbn|1-4020-1560-7}}.

* [[Eric Schechter|Schechter, Eric]], ''Handbook of Analysis and its Foundations'', Academic Press, 1997. {{isbn|0-12-622760-8}}.

{{Functional Analysisanalysis}}

{{Topological vector spaces}}

{{TopologicalVectorSpaces}}

[[Category:Functional analysis]]

[[Category:Axiom of choice]]

[[Category:Functions and mappings]]