A pseudometric on is called aultrapseudometric or astrong pseudometric if it satisfies:
Strong/Ultrametric triangle inequality:
Pseudometric space
Apseudometric space is a pair consisting of a set and a pseudometric on such that's topology is identical to the topology on induced by We call a pseudometric space ametric space (resp.ultrapseudometric space) when is a metric (resp. ultrapseudometric).
If is a pseudometric on a set then collection ofopen balls: as ranges over and ranges over the positive real numbers,forms a basis for a topology on that is called the-topology or thepseudometric topology on induced by
Convention: If is a pseudometric space and is treated as atopological space, then unless indicated otherwise, it should be assumed that is endowed with the topology induced by
Pseudometrizable space
A topological space is calledpseudometrizable (resp.metrizable,ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric) on such that is equal to the topology induced by[1]
An additivetopological group is an additive group endowed with a topology, called agroup topology, under which addition and negation become continuous operators.
A topology on a real or complex vector space is called avector topology or aTVS topology if it makes the operations of vector addition andscalar multiplication continuous (that is, if it makes into atopological vector space).
Everytopological vector space (TVS) is an additive commutative topological group but not all group topologies on are vector topologies. This is because despite it making addition and negation continuous, a group topology on a vector space may fail to make scalar multiplication continuous. For instance, thediscrete topology on any non-trivial vector space makes addition and negation continuous but do not make scalar multiplication continuous.
If is an additive group then we say that a pseudometric on istranslation invariant or justinvariant if it satisfies any of the following equivalent conditions:
Theorem[2]—Suppose that is an additive commutative group. If is a translation invariant pseudometric on then the map is a value on calledthe value associated with, and moreover, generates a group topology on (i.e. the-topology on makes into a topological group).Conversely, if is a value on then the map is a translation-invariant pseudometric on and the value associated with is just
Theorem[2]—If is an additive commutativetopological group then the following are equivalent:
is induced by a pseudometric; (i.e. is pseudometrizable);
is induced by a translation-invariant pseudometric;
the identity element in has a countable neighborhood basis.
If is Hausdorff then the word "pseudometric" in the above statement may be replaced by the word "metric." A commutative topological group is metrizable if and only if it is Hausdorff and pseudometrizable.
An invariant pseudometric that doesn't induce a vector topology
Let be a non-trivial (i.e.) real or complex vector space and let be the translation-invarianttrivial metric on defined by and such that The topology that induces on is thediscrete topology, which makes into a commutative topological group under addition but doesnot form a vector topology on because isdisconnected but every vector topology is connected. What fails is that scalar multiplication isn't continuous on
This example shows that a translation-invariant (pseudo)metric isnot enough to guarantee a vector topology, which leads us to define paranorms andF-seminorms.
A collection of subsets of a vector space is calledadditive[5] if for every there exists some such that
Continuity of addition at 0—If is agroup (as all vector spaces are), is a topology on and is endowed with theproduct topology, then the addition map (i.e. the map) is continuous at the origin of if and only if the set ofneighborhoods of the origin in is additive. This statement remains true if the word "neighborhood" is replaced by "open neighborhood."[5]
All of the above conditions are consequently a necessary for a topology to form a vector topology. Additive sequences of sets have the particularly nice property that they define non-negative continuous real-valuedsubadditive functions. These functions can then be used to prove many of the basic properties of topological vector spaces and also show that a Hausdorff TVS with a countable basis of neighborhoods is metrizable. The following theorem is true more generally for commutative additivetopological groups.
Theorem—Let be a collection of subsets of a vector space such that and for all For all let
Define by if and otherwise let
Then issubadditive (meaning) and on so in particular If all aresymmetric sets then and if all are balanced then for all scalars such that and all If is a topological vector space and if all are neighborhoods of the origin then is continuous, where if in addition is Hausdorff and forms a basis of balanced neighborhoods of the origin in then is a metric defining the vector topology on
Proof
Assume that always denotes a finite sequence of non-negative integers and use the notation:
For any integers and
From this it follows that if consists of distinct positive integers then
It will now be shown by induction on that if consists of non-negative integers such that for some integer then This is clearly true for and so assume that which implies that all are positive. If all are distinct then this step is done, and otherwise pick distinct indices such that and construct from by replacing each with and deleting the element of (all other elements of are transferred to unchanged). Observe that and (because) so by appealing to the inductive hypothesis we conclude that as desired.
It is clear that and that so to prove that is subadditive, it suffices to prove that when are such that which implies that This is an exercise. If all are symmetric then if and only if from which it follows that and If all are balanced then the inequality for all unit scalars such that is proved similarly. Because is a nonnegative subadditive function satisfying as described in the article onsublinear functionals, is uniformly continuous on if and only if is continuous at the origin. If all are neighborhoods of the origin then for any real pick an integer such that so that implies If the set of all form basis of balanced neighborhoods of the origin then it may be shown that for any there exists some such that implies
If is a vector space over the real or complex numbers then aparanorm on is a G-seminorm (defined above) on that satisfies any of the following additional conditions, each of which begins with "for all sequences in and all convergent sequences of scalars":[6]
Continuity of multiplication: if is a scalar and are such that and then
If is a translation-invariant pseudometric on a vector space that induces a vector topology on (i.e. is a TVS) then the map defines a continuous paranorm on; moreover, the topology that this paranorm defines in is[8]
If is aHamel basis on a vector space then the real-valued map that sends (where all but finitely many of the scalars are 0) to is a paranorm on which satisfies for all and scalars[8]
The function is a paranorm on that isnot balanced but nevertheless equivalent to the usual norm on Note that the function is subadditive.[10]
Let be a complex vector space and let denote considered as a vector space over Any paranorm on is also a paranorm on[9]
If is a vector space over the real or complex numbers then anF-seminorm on (the stands forFréchet) is a real-valued map with the following four properties:[11]
Every positive scalar multiple of anF-seminorm (resp.F-norm, seminorm) is again anF-seminorm (resp.F-norm, seminorm).
The sum of finitely manyF-seminorms (resp.F-norms) is anF-seminorm (resp.F-norm).
If and areF-seminorms on then so is their pointwise supremum The same is true of the supremum of any non-empty finite family ofF-seminorms on[12]
The restriction of anF-seminorm (resp.F-norm) to a vector subspace is anF-seminorm (resp.F-norm).[9]
A non-negative real-valued function on is a seminorm if and only if it is aconvexF-seminorm, or equivalently, if and only if it is a convex balancedG-seminorm.[10] In particular, everyseminorm is anF-seminorm.
For any the map on defined by is anF-norm that is not a norm.
If is a linear map and if is anF-seminorm on then is anF-seminorm on[12]
Let be a complex vector space and let denote considered as a vector space over AnyF-seminorm on is also anF-seminorm on[9]
EveryF-seminorm is a paranorm and every paranorm is equivalent to someF-seminorm.[7] EveryF-seminorm on a vector space is a value on In particular, and for all
Theorem[11]—Let be anF-seminorm on a vector space Then the map defined by is a translation invariant pseudometric on that defines a vector topology on If is anF-norm then is a metric. When is endowed with this topology then is a continuous map on
The balanced sets as ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set. Similarly, the balanced sets as ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of open sets.
Suppose that is a non-empty collection ofF-seminorms on a vector space and for any finite subset and any let
The set forms a filter base on that also forms a neighborhood basis at the origin for a vector topology on denoted by[12] Each is abalanced andabsorbing subset of[12] These sets satisfy[12]
is the coarsest vector topology on making each continuous.[12]
is Hausdorffif and only if for every non-zero there exists some such that[12]
If is the set of all continuousF-seminorms on then[12]
If is the set of all pointwise suprema of non-empty finite subsets of of then is a directed family ofF-seminorms and[12]
Assume that is an increasing sequence of seminorms on and let be the Fréchet combination of Then is anF-seminorm on that induces the same locally convex topology as the family of seminorms.[13]
Since is increasing, a basis of open neighborhoods of the origin consists of all sets of the form as ranges over all positive integers and ranges over all positive real numbers.
If each is a paranorm then so is and moreover, induces the same topology on as the family of paranorms.[8] This is also true of the following paranorms on:
The Fréchet combination can be generalized by use of a bounded remetrization function.
Abounded remetrization function[15] is a continuous non-negative non-decreasing map that has a bounded range, issubadditive (meaning that for all), and satisfies if and only if
Examples of bounded remetrization functions include and[15] If is a pseudometric (respectively, metric) on and is a bounded remetrization function then is a bounded pseudometric (respectively, bounded metric) on that is uniformly equivalent to[15]
Suppose that is a family of non-negativeF-seminorm on a vector space is a bounded remetrization function, and is a sequence of positive real numbers whose sum is finite. Thendefines a boundedF-seminorm that is uniformly equivalent to the[16] It has the property that for any net in if and only if for all[16] is anF-norm if and only if the separate points on[16]
A pseudometric (resp. metric) is induced by a seminorm (resp. norm) on a vector space if and only if is translation invariant andabsolutely homogeneous, which means that for all scalars and all in which case the function defined by is a seminorm (resp. norm) and the pseudometric (resp. metric) induced by is equal to
If is a complete pseudometrizable TVS and is a closed vector subspace of then is complete.[11]
If is metrizable TVS and is a closed vector subspace of then is metrizable.[11]
If is anF-seminorm on then the map defined byis anF-seminorm on that induces the usualquotient topology on[11] If in addition is anF-norm on and if is a closed vector subspace of then is anF-norm on[11]
Everyseminormed space is pseudometrizable with a canonical pseudometric given by for all[19].
If is pseudometricTVS with a translation invariant pseudometric then defines a paranorm.[20] However, if is a translation invariant pseudometric on the vector space (without the addition condition that ispseudometric TVS), then need not be either anF-seminorm[21] nor a paranorm.
If a TVS has a bounded neighborhood of the origin then it is pseudometrizable; the converse is in general false.[14]
If a Hausdorff TVS has a bounded neighborhood of the origin then it is metrizable.[14]
If is Hausdorff locally convex TVS then with thestrong topology, is metrizable if and only if there exists a countable set of bounded subsets of such that every bounded subset of is contained in some element of[22]
In particular, if a metrizable locally convex space (such as aFréchet space) isnot normable then itsstrong dual space is not aFréchet–Urysohn space and consequently, thiscomplete Hausdorff locally convex space is also neither metrizable nor normable.
Another consequence of this is that if is areflexivelocally convex TVS whose strong dual is metrizable then is necessarily a reflexive Fréchet space, is aDF-space, both and are necessarilycomplete Hausdorffultrabornologicaldistinguishedwebbed spaces, and moreover, is normable if and only if is normable if and only if is Fréchet–Urysohn if and only if is metrizable. In particular, such a space is either aBanach space or else it is not even a Fréchet–Urysohn space.
Suppose that is a pseudometric space and The set ismetrically bounded or-bounded if there exists a real number such that for all; the smallest such is then called thediameter or-diameter of[14] If isbounded in a pseudometrizable TVS then it is metrically bounded; the converse is in general false but it is true forlocally convex metrizable TVSs.[14]
Every completepseudometrizable TVS is abarrelled space and aBaire space (and hence non-meager).[31] However, there exist metrizable Baire spaces that are notcomplete.[31]
If is a metrizable locally convex space, then the strong dual of isbornological if and only if it isbarreled, if and only if it isinfrabarreled.[26]
If is a complete pseudometrizable TVS and is a closed vector subspace of then is complete.[11]
If and are complete metrizable TVSs (i.e.F-spaces) and if is coarser than then;[33] this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete.[34] Said differently, if and are bothF-spaces but with different topologies, then neither one of and contains the other as a subset. One particular consequence of this is, for example, that if is aBanach space and is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of (i.e. if or if for some constant), then the only way that can be a Banach space (i.e. also be complete) is if these two norms and areequivalent; if they are not equivalent, then can not be a Banach space. As another consequence, if is a Banach space and is aFréchet space, then the map is continuous if and only if the Fréchet spaceis the TVS (here, the Banach space is being considered as a TVS, which means that its norm is "forgetten" but its topology is remembered).
Everytopological vector space (and more generally, atopological group) has a canonicaluniform structure, induced by its topology, which allows the notions of completeness anduniform continuity to be applied to it. If is a metrizable TVS and is a metric that defines's topology, then its possible that is complete as a TVS (i.e. relative to its uniformity) but the metric isnot acomplete metric (such metrics exist even for). Thus, if is a TVS whose topology is induced by a pseudometric then the notion of completeness of (as a TVS) and the notion of completeness of the pseudometric space are not always equivalent. The next theorem gives a condition for when they are equivalent:
Theorem—If is a pseudometrizable TVS whose topology is induced by atranslation invariant pseudometric then is a complete pseudometric on if and only if is complete as a TVS.[36]
Theorem[37][38](Klee)—Let beany[note 2] metric on a vector space such that the topology induced by on makes into a topological vector space. If is a complete metric space then is a complete-TVS.
Theorem—If is a TVS whose topology is induced by a paranorm then is complete if and only if for every sequence in if then converges in[39]
If is a closed vector subspace of a complete pseudometrizable TVS then the quotient space is complete.[40] If is acomplete vector subspace of a metrizable TVS and if the quotient space is complete then so is[40] If is not complete then but not complete, vector subspace of
Let be aseparable locally convex metrizable topological vector space and let be its completion. If is a bounded subset of then there exists a bounded subset of such that[41]
Everytotally bounded subset of a locally convex metrizable TVS is contained in the closedconvex balanced hull of some sequence in that converges to
In a pseudometrizable TVS, everybornivore is a neighborhood of the origin.[42]
If is a translation invariant metric on a vector space then for all and every positive integer[43]
If is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence of positive real numbers diverging to such that[43]
A subset of a complete metric space is closed if and only if it is complete. If a space is not complete, then is a closed subset of that is not complete.
Banach-Saks theorem[45]—If is a sequence in alocally convex metrizable TVS that convergesweakly to some then there exists a sequence in such that in and each is a convex combination of finitely many
Mackey's countability condition[14]—Suppose that is a locally convex metrizable TVS and that is a countable sequence of bounded subsets of Then there exists a bounded subset of and a sequence of positive real numbers such that for all
Generalized series
As describedin this article's section on generalized series, for any-indexed family family of vectors from a TVS it is possible to define their sum as the limit of thenet of finite partial sums where the domain isdirected by If and for instance, then the generalized series converges if and only ifconverges unconditionally in the usual sense (which for real numbers,is equivalent toabsolute convergence). If a generalized series converges in a metrizable TVS, then the set is necessarilycountable (that is, either finite orcountably infinite);[proof 1] in other words, all but at most countably many will be zero and so this generalized series is actually a sum of at most countably many non-zero terms.
If is a pseudometrizable TVS and maps bounded subsets of to bounded subsets of then is continuous.[14] Discontinuous linear functionals exist on any infinite-dimensional pseudometrizable TVS.[46] Thus, a pseudometrizable TVS is finite-dimensional if and only if its continuous dual space is equal to itsalgebraic dual space.[46]
If is a linear map between TVSs and is metrizable then the following are equivalent:
Theorem: If is a surjective linear operator from a TVS onto aBaire space then is almost open.[47]
Theorem: Suppose is a continuous linear operator from a complete pseudometrizable TVS into a Hausdorff TVS If the image of is non-meager in then is a surjective open map and is acomplete metrizable space.[47]
A vector subspace of a TVS hasthe extension property if any continuous linear functional on can be extended to a continuous linear functional on[22] Say that a TVS has theHahn-Banach extension property (HBEP) if every vector subspace of has the extension property.[22]
TheHahn-Banach theorem guarantees that every Hausdorff locally convex space has the HBEP. For complete metrizable TVSs there is a converse:
Theorem(Kalton)—Every complete metrizable TVS with the Hahn-Banach extension property is locally convex.[22]
If a vector space has uncountable dimension and if we endow it with thefinest vector topology then this is a TVS with the HBEP that is neither locally convex or metrizable.[22]
^Suppose the net converges to some point in a metrizable TVS where recall that this net's domain is thedirected set Like every convergent net, this convergent net of partial sums is aCauchy net, which for this particular net means (by definition) that for every neighborhood of the origin in there exists a finite subset of such that for all finite supersetsthis implies that for every (by taking and). Since is metrizable, it has a countable neighborhood basis at the origin, whose intersection is necessarily (since is a Hausdorff TVS).For every positive integer pick a finite subset such that for every If belongs to then belongs to Thus for every index that does not belong to the countable set
Berberian, Sterling K. (1974).Lectures in Functional Analysis and Operator Theory. Graduate Texts in Mathematics. Vol. 15. New York: Springer.ISBN978-0-387-90081-0.OCLC878109401.
Köthe, Gottfried (1983) [1969].Topological Vector Spaces I. Grundlehren der mathematischen Wissenschaften. Vol. 159. Translated by Garling, D.J.H. New York: Springer Science & Business Media.ISBN978-3-642-64988-2.MR0248498.OCLC840293704.