Balanced set

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In linear algebra and related areas of mathematics a balanced set, circled set or disk in a vector space (over a field with an absolute value function ) is a set such that for all scalars satisfying

The balanced hull or balanced envelope of a set is the smallest balanced set containing The balanced core of a set is the largest balanced set contained in

Balanced sets are ubiquitous in functional analysis because every neighborhood of the origin in every topological vector space (TVS) contains a balanced neighborhood of the origin and every convex neighborhood of the origin contains a balanced convex neighborhood of the origin (even if the TVS is not locally convex). This neighborhood can also be chosen to be an open set or, alternatively, a closed set.

Definition

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Let   be a vector space over the field   of real or complex numbers.

Notation

If   is a set,   is a scalar, and   then let   and   and for any   let   denote, respectively, the open ball and the closed ball of radius   in the scalar field   centered at   where   and   Every balanced subset of the field   is of the form   or   for some  

Balanced set

A subset   of   is called a balanced set or balanced if it satisfies any of the following equivalent conditions:

  1. Definition:   for all   and all scalars   satisfying  
  2.   for all scalars   satisfying  
  3.   (where  ).
  4.  [1]
  5. For every    
    •   is a   (if  ) or   (if  ) dimensional vector subspace of  
    • If   then the above equality becomes   which is exactly the previous condition for a set to be balanced. Thus,   is balanced if and only if for every     is a balanced set (according to any of the previous defining conditions).
  6. For every 1-dimensional vector subspace   of     is a balanced set (according to any defining condition other than this one).
  7. For every   there exists some   such that   or  
  8.   is a balanced subset of   (according to any defining condition of "balanced" other than this one).
    • Thus   is a balanced subset of   if and only if it is balanced subset of every (equivalently, of some) vector space over the field   that contains   So assuming that the field   is clear from context, this justifies writing "  is balanced" without mentioning any vector space.[note 1]

If   is a convex set then this list may be extended to include:

  1.   for all scalars   satisfying  [2]

If   then this list may be extended to include:

  1.   is symmetric (meaning  ) and  

Balanced hull

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The balanced hull of a subset   of   denoted by   is defined in any of the following equivalent ways:

  1. Definition:   is the smallest (with respect to  ) balanced subset of   containing  
  2.   is the intersection of all balanced sets containing  
  3.  
  4.  [1]

Balanced core

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The balanced core of a subset   of   denoted by   is defined in any of the following equivalent ways:

  1. Definition:   is the largest (with respect to  ) balanced subset of  
  2.   is the union of all balanced subsets of  
  3.   if   while   if  

Examples

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The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular,   is always a balanced set.

Any non-empty set that does not contain the origin is not balanced and furthermore, the balanced core of such a set will equal the empty set.

Normed and topological vector spaces

The open and closed balls centered at the origin in a normed vector space are balanced sets. If   is a seminorm (or norm) on a vector space   then for any constant   the set   is balanced.

If   is any subset and   then   is a balanced set. In particular, if   is any balanced neighborhood of the origin in a topological vector space   then  

Balanced sets in   and  

Let   be the field real numbers   or complex numbers   let   denote the absolute value on   and let   denotes the vector space over   So for example, if   is the field of complex numbers then   is a 1-dimensional complex vector space whereas if   then   is a 1-dimensional real vector space.

The balanced subsets of   are exactly the following:[3]

  1.  
  2.  
  3.  
  4.   for some real  
  5.   for some real  

Consequently, both the balanced core and the balanced hull of every set of scalars is equal to one of the sets listed above.

The balanced sets are   itself, the empty set and the open and closed discs centered at zero. Contrariwise, in the two dimensional Euclidean space there are many more balanced sets: any line segment with midpoint at the origin will do. As a result,   and   are entirely different as far as scalar multiplication is concerned.

Balanced sets in  

Throughout, let   (so   is a vector space over  ) and let   is the closed unit ball in   centered at the origin.

If   is non-zero, and   then the set   is a closed, symmetric, and balanced neighborhood of the origin in   More generally, if   is any closed subset of   such that   then   is a closed, symmetric, and balanced neighborhood of the origin in   This example can be generalized to   for any integer  

Let   be the union of the line segment between the points   and   and the line segment between   and   Then   is balanced but not convex. Nor is   is absorbing (despite the fact that   is the entire vector space).

For every   let   be any positive real number and let   be the (open or closed) line segment in   between the points   and   Then the set   is a balanced and absorbing set but it is not necessarily convex.

The balanced hull of a closed set need not be closed. Take for instance the graph of   in  

The next example shows that the balanced hull of a convex set may fail to be convex (however, the convex hull of a balanced set is always balanced). For an example, let the convex subset be   which is a horizontal closed line segment lying above the  axis in   The balanced hull   is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles   and   where   and   is the filled triangle whose vertices are the origin together with the endpoints of   (said differently,   is the convex hull of   while   is the convex hull of  ).

Sufficient conditions

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A set   is balanced if and only if it is equal to its balanced hull   or to its balanced core   in which case all three of these sets are equal:  

The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field  ).

  • The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.[4]
  • The convex hull of a balanced set is convex and balanced (that is, it is absolutely convex). However, the balanced hull of a convex set may fail to be convex (a counter-example is given above).
  • Arbitrary unions of balanced sets are balanced, and the same is true of arbitrary intersections of balanced sets.
  • Scalar multiples and (finite) Minkowski sums of balanced sets are again balanced.
  • Images and preimages of balanced sets under linear maps are again balanced. Explicitly, if   is a linear map and   and   are balanced sets, then   and   are balanced sets.

Balanced neighborhoods

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In any topological vector space, the closure of a balanced set is balanced.[5] The union of the origin   and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced.[5][proof 1] However,   is a balanced subset of   that contains the origin   but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set.[6] Similarly for real vector spaces, if   denotes the convex hull of   and   (a filled triangle whose vertices are these three points) then   is an (hour glass shaped) balanced subset of   whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set   formed by adding the origin is balanced, it is neither an open set nor a neighborhood of the origin).

Every neighborhood (respectively, convex neighborhood) of the origin in a topological vector space   contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given   the symmetric set   will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of  ) whenever this is true of   It will be a balanced set if   is a star shaped at the origin,[note 2] which is true, for instance, when   is convex and contains   In particular, if   is a convex neighborhood of the origin then   will be a balanced convex neighborhood of the origin and so its topological interior will be a balanced convex open neighborhood of the origin.[5]

Proof

Let   and define   (where   denotes elements of the field   of scalars). Taking   shows that   If   is convex then so is   (since an intersection of convex sets is convex) and thus so is  's interior. If   then   and thus   If   is star shaped at the origin[note 2] then so is every   (for  ), which implies that for any     thus proving that   is balanced. If   is convex and contains the origin then it is star shaped at the origin and so   will be balanced.

Now suppose   is a neighborhood of the origin in   Since scalar multiplication   (defined by  ) is continuous at the origin   and   there exists some basic open neighborhood   (where   and  ) of the origin in the product topology on   such that   the set   is balanced and it is also open because it may be written as   where   is an open neighborhood of the origin whenever   Finally,   shows that   is also a neighborhood of the origin. If   is balanced then because its interior   contains the origin,   will also be balanced. If   is convex then   is convex and balanced and thus the same is true of    

Suppose that   is a convex and absorbing subset of   Then   will be convex balanced absorbing subset of   which guarantees that the Minkowski functional   of   will be a seminorm on   thereby making   into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples   as   ranges over   (or over any other set of non-zero scalars having   as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If   is a topological vector space and if this convex absorbing subset   is also a bounded subset of   then the same will be true of the absorbing disk   if in addition   does not contain any non-trivial vector subspace then   will be a norm and   will form what is known as an auxiliary normed space.[7] If this normed space is a Banach space then   is called a Banach disk.

Properties

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Properties of balanced sets

A balanced set is not empty if and only if it contains the origin. By definition, a set is absolutely convex if and only if it is convex and balanced. Every balanced set is star-shaped (at 0) and a symmetric set. If   is a balanced subset of   then:

  • for any scalars   and   if   then   and   Thus if   and   are any scalars then  
  •   is absorbing in   if and only if for all   there exists   such that  [2]
  • for any 1-dimensional vector subspace   of   the set   is convex and balanced. If   is not empty and if   is a 1-dimensional vector subspace of   then   is either   or else it is absorbing in  
  • for any   if   contains more than one point then it is a convex and balanced neighborhood of   in the 1-dimensional vector space   when this space is endowed with the Hausdorff Euclidean topology; and the set   is a convex balanced subset of the real vector space   that contains the origin.

Properties of balanced hulls and balanced cores

For any collection   of subsets of    

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If   is a Hausdorff topological vector space and if   is a compact subset of   then the balanced hull of   is compact.[8]

If a set is closed (respectively, convex, absorbing, a neighborhood of the origin) then the same is true of its balanced core.

For any subset   and any scalar    

For any scalar     This equality holds for   if and only if   Thus if   or   then   for every scalar  

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A function   on a real or complex vector space is said to be a balanced function if it satisfies any of the following equivalent conditions:[9]

  1.   whenever   is a scalar satisfying   and  
  2.   whenever   and   are scalars satisfying   and  
  3.   is a balanced set for every non-negative real  

If   is a balanced function then   for every scalar   and vector   so in particular,   for every unit length scalar   (satisfying  ) and every  [9] Using   shows that every balanced function is a symmetric function.

A real-valued function   is a seminorm if and only if it is a balanced sublinear function.

See also

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References

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  1. ^ a b Swartz 1992, pp. 4–8.
  2. ^ a b Narici & Beckenstein 2011, pp. 107–110.
  3. ^ Jarchow 1981, p. 34.
  4. ^ Narici & Beckenstein 2011, pp. 156–175.
  5. ^ a b c Rudin 1991, pp. 10–14.
  6. ^ Rudin 1991, p. 38.
  7. ^ Narici & Beckenstein 2011, pp. 115–154.
  8. ^ Trèves 2006, p. 56.
  9. ^ a b Schechter 1996, p. 313.
  1. ^ Assuming that all vector spaces containing a set   are over the same field, when describing the set as being "balanced", it is not necessary to mention a vector space containing   That is, "  is balanced" may be written in place of "  is a balanced subset of  ".
  2. ^ a b   being star shaped at the origin means that   and   for all   and  

Proofs

  1. ^ Let   be balanced. If its topological interior   is empty then it is balanced so assume otherwise and let   be a scalar. If   then the map   defined by   is a homeomorphism, which implies   because   is open,   so that it only remains to show that this is true for   However,   might not be true but when it is true then   will be balanced.  

Sources

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  • Bourbaki, Nicolas (1987) [1981]. Topological Vector Spaces: Chapters 1–5. Éléments de mathématique. Translated by Eggleston, H.G.; Madan, S. Berlin New York: Springer-Verlag. ISBN 3-540-13627-4. OCLC 17499190.
  • Conway, John (1990). A course in functional analysis. Graduate Texts in Mathematics. Vol. 96 (2nd ed.). New York: Springer-Verlag. ISBN 978-0-387-97245-9. OCLC 21195908.
  • Dunford, Nelson; Schwartz, Jacob T. (1988). Linear Operators. Pure and applied mathematics. Vol. 1. New York: Wiley-Interscience. ISBN 978-0-471-60848-6. OCLC 18412261.
  • Edwards, Robert E. (1995). Functional Analysis: Theory and Applications. New York: Dover Publications. ISBN 978-0-486-68143-6. OCLC 30593138.
  • Jarchow, Hans (1981). Locally convex spaces. Stuttgart: B.G. Teubner. ISBN 978-3-519-02224-4. OCLC 8210342.
  • 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. ISBN 978-3-642-64988-2. MR 0248498. OCLC 840293704.
  • Köthe, Gottfried (1979). Topological Vector Spaces II. Grundlehren der mathematischen Wissenschaften. Vol. 237. New York: Springer Science & Business Media. ISBN 978-0-387-90400-9. OCLC 180577972.
  • Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
  • Robertson, Alex P.; Robertson, Wendy J. (1980). Topological Vector Spaces. Cambridge Tracts in Mathematics. Vol. 53. Cambridge England: Cambridge University Press. ISBN 978-0-521-29882-7. OCLC 589250.
  • Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.
  • Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
  • Schechter, Eric (October 24, 1996). Handbook of Analysis and Its Foundations. Academic Press. ISBN 978-0-08-053299-8.
  • Swartz, Charles (1992). An introduction to Functional Analysis. New York: M. Dekker. ISBN 978-0-8247-8643-4. OCLC 24909067.
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