balanced function

In linear algebra and related areas of mathematics a balanced set, circled set or disk in a vector space (over a field $\mathbb$ with an absolute value function $, \cdot ,$) is a set $S$ such that $a S \subseteq S$ for all scalars $a$ satisfying $, a, \leq 1.$

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

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

Let $X$ be a vector space over the field $\mathbb$ of real or complex numbers.

Notation

If $S$ is a set, $a$ is a scalar, and $B \subseteq \mathbb$ then let $a S = \$ and $B S = \$ and for any $0 \leq r \leq \infty,$ let
$B_r = \ \qquad \text \qquad B_ = \.$
denote, respectively, the ''open ball'' and the ''closed ball'' of radius $r$ in the scalar field $\mathbb$ centered at $0$ where $B_0 = \varnothing, B_ = \,$ and $B_ = B_ = \mathbb.$
Every balanced subset of the field $\mathbb$ is of the form $B_$ or $B_r$ for some $0 \leq r \leq \infty.$

Balanced set

A subset $S$ of $X$ is called a ' or ''balanced'' if it satisfies any of the following equivalent conditions:

1. ''Definition'': $a s \in S$ for all $s \in S$ and all scalars $a$ satisfying $, a, \leq 1.$


2. $a S \subseteq S$ for all scalars $a$ satisfying $, a, \leq 1.$


3. $B_ S \subseteq S$ (where $B_ := \$).


4. $S = B_ S.$


5. For every $s \in S,$ $S \cap \mathbb s = B_ (S \cap \mathbb s).$
   * $\mathbb s = \operatorname \$ is a $0$ (if $s = 0$) or $1$ (if $s \neq 0$) dimensional vector subspace of $X.$
   * If $R := S \cap \mathbb s$ then the above equality becomes $R = B_ R,$ which is exactly the previous condition for a set to be balanced. Thus, $S$ is balanced if and only if for every $s \in S,$ $S \cap \mathbb s$ is a balanced set (according to any of the previous defining conditions).


6. For every 1-dimensional vector subspace $Y$ of $\operatorname S,$ $S \cap Y$ is a balanced set (according to any defining condition other than this one).


7. For every $s \in S,$ there exists some $0 \leq r \leq \infty$ such that $S \cap \mathbb s = B_r s$ or $S \cap \mathbb s = B_ s.$


8. $S$ is a balanced subset of $\operatorname S$ (according to any defining condition of "balanced" other than this one).
   * Thus $S$ is a balanced subset of $X$ if and only if it is balanced subset of every (equivalently, of some) vector space over the field $\mathbb$ that contains $S.$ So assuming that the field $\mathbb$ is clear from context, this justifies writing "$S$ is balanced" without mentioning any vector space.Assuming that all vector spaces containing a set $S$ are over the same field, when describing the set as being "balanced", it is not necessary to mention a vector space containing $S.$ That is, "$S$ is balanced" may be written in place of "$S$ is a balanced subset of $X$".

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

9. $a S \subseteq S$ for all scalars $a$ satisfying $, a, = 1.$

If $\mathbb = \R$ then this list may be extended to include:

10. $S$ is symmetric (meaning $- S = S$) and intersection of all balanced sets containing $S.$


11. $\operatorname S = \bigcup_ (a S).$


12. $\operatorname S = B_ S.$

Balanced core

$\operatorname S ~=~ \begin \displaystyle\bigcap_ a S & \text 0 \in S \\ \varnothing & \text 0 \not\in S \\ \end$

The ' of a subset $S$ of $X,$ denoted by $\operatorname S,$ is defined in any of the following equivalent ways:

1. ''Definition'': $\operatorname S$ is the largest (with respect to $\,\subseteq\,$) balanced subset of $S.$


2. $\operatorname S$ is the union of all balanced subsets of $S.$


3. $\operatorname S = \varnothing$ if $0 \not\in S$ while $\operatorname S = \bigcap_ (a S)$ if $0 \in S.$

Examples

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 $p$ is a seminorm (or norm) on a vector space $X$ then for any constant $c > 0,$ the set $\$ is balanced.

If $S \subseteq X$ is any subset and $B_1 := \$ then $B_1 S$ is a balanced set.
In particular, if $U \subseteq X$ is any balanced neighborhood of the origin in a topological vector space $X$ then $\operatorname_X U ~\subseteq~ B_1 U ~=~ \bigcup_ a U ~\subseteq~ U.$

Balanced sets in $\R$ and $\Complex$

Let $\mathbb$ be the field real numbers $\R$ or complex numbers $\Complex,$ let $, \cdot,$ denote the absolute value on $\mathbb,$ and let $X := \mathbb$ denotes the vector space over $\mathbb.$ So for example, if $\mathbb := \Complex$ is the field of complex numbers then $X = \mathbb = \Complex$ is a 1-dimensional complex vector space whereas if $\mathbb := \R$ then $X = \mathbb = \R$ is a 1-dimensional real vector space.

The balanced subsets of $X = \mathbb$ are exactly the following:

1. $\varnothing$


2. $X$


3. $\$


4. $\$ for some real $r > 0$


5. $\$ for some real $r > 0.$

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 $\Complex$ 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, $\Complex$ and $\R^2$ are entirely different as far as scalar multiplication is concerned.

Balanced sets in $\R^2$

Throughout, let $X = \R^2$ (so $X$ is a vector space over $\R$) and let $B_$ is the closed unit ball in $X$ centered at the origin.

If $x_0 \in X = \R^2$ is non-zero, and $L := \R x_0,$ then the set $R := B_ \cup L$ is a closed, symmetric, and balanced neighborhood of the origin in $X.$ More generally, if $C$ is closed subset of $X$ such that $(0, 1) C \subseteq C,$ then $S := B_ \cup C \cup (-C)$ is a closed, symmetric, and balanced neighborhood of the origin in $X.$ This example can be generalized to $\R^n$ for any integer $n \geq 1.$

Let $B \subseteq \R^2$ be the union of the line segment between the points $(-1, 0)$ and $(1, 0)$ and the line segment between $(0, -1)$ and $(0, 1).$ Then $B$ is balanced but not convex. Nor is $B$ is absorbing (despite the fact that $\operatorname B = \R^2$ is the entire vector space).

For every $0 \leq t \leq \pi,$ let $r_t$ be any positive real number and let $B^t$ be the (open or closed) line segment in $X := \R^2$ between the points $(\cos t, \sin t)$ and $- (\cos t, \sin t).$ Then the set $B = \bigcup_ r_t B^t$ 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 $x y = 1$ in $X = \R^2.$

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 S := 1, 1\times \, which is a horizontal closed line segment lying above the $x-$axis in $X := \R^2.$ The balanced hull $\operatorname S$ is a non-convex subset that is " hour glass shaped" and equal to the union of two closed and filled isosceles triangles $T_1$ and $T_2,$ where $T_2 = - T_1$ and $T_1$ is the filled triangle whose vertices are the origin together with the endpoints of $S$ (said differently, $T_1$ is the convex hull of $S \cup \$ while $T_2$ is the convex hull of $(-S) \cup \$).

Sufficient conditions

A set $T$ is balanced if and only if it is equal to its balanced hull $\operatorname T$ or to its balanced core $\operatorname T,$ in which case all three of these sets are equal: $T = \operatorname T = \operatorname T.$

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

• The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.


• 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 $L : X \to Y$ is a linear map and $B \subseteq X$ and $C \subseteq Y$ are balanced sets, then $L(B)$ and $L^(C)$ are balanced sets.

Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced. 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.Let $B \subseteq X$ be balanced. If its topological interior $\operatorname_X B$ is empty then it is balanced so assume otherwise and let $, s, \leq 1$ be a scalar. If $s \neq 0$ then the map $X \to X$ defined by $x \mapsto s x$ is a homeomorphism, which implies $s \operatorname_X B = \operatorname_X (s B) \subseteq s B \subseteq B;$ because $s \operatorname_X B$ is open, $s \operatorname_X B \subseteq \operatorname_X B$ so that it only remains to show that this is true for $s = 0.$ However, $0 \in \operatorname_X B$ might not be true but when it is true then $\operatorname_X B$ will be balanced. $\blacksquare$ However, $\left\$ is a balanced subset of $X = \Complex^2$ that contains the origin $(0, 0) \in X$ but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set. Similarly for real vector spaces, if $T$ denotes the convex hull of $(0, 0)$ and $(\pm 1, 1)$ (a filled triangle whose vertices are these three points) then $B := T \cup (-T)$ is an ( hour glass shaped) balanced subset of $X := \Reals^2$ whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set $\ \cup \operatorname_X B$ 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 $X$ contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given $W \subseteq X,$ the symmetric set $\bigcap_ u W \subseteq W$ will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of $X$) whenever this is true of $W.$ It will be a balanced set if $W$ is a star shaped at the origin,$W$ being star shaped at the origin means that $0 \in W$ and $r w \in W$ for all $0 \leq r \leq 1$ and $w \in W.$ which is true, for instance, when $W$ is convex and contains $0.$ In particular, if $W$ is a convex neighborhood of the origin then $\bigcap_ u W$ will be a convex neighborhood of the origin and so its topological interior will be a balanced convex neighborhood of the origin.

Suppose that $W$ is a convex and absorbing subset of $X.$ Then $D := \bigcap_ u W$ will be convex balanced absorbing subset of $X,$ which guarantees that the Minkowski functional $p_D : X \to \R$ of $D$ will be a seminorm on $X,$ thereby making $\left(X, p_D\right)$ into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples $r D$ as $r$ ranges over $\left\$ (or over any other set of non-zero scalars having $0$ as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If $X$ is a topological vector space and if this convex absorbing subset $W$ is also a bounded subset of $X,$ then the same will be true of the absorbing disk $D := u W;$ if in addition $D$ does not contain any non-trivial vector subspace then $p_D$ will be a norm and $\left(X, p_D\right)$ will form what is known as an auxiliary normed space. If this normed space is a Banach space then $D$ is called a .

Properties

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 $B$ is a balanced subset of $X$ then:

• for any scalars $c$ and $d,$ if $, c, \leq , d,$ then $c B \subseteq d B$ and $c B = , c, B.$ Thus if $c$ and $d$ are any scalars then $(c B) \cap (d B) = \min_ \ B.$


• $B$ is absorbing in $X$ if and only if for all $x \in X,$ there exists $r > 0$ such that $x \in r B.$


• for any 1-dimensional vector subspace $Y$ of $X,$ the set $B \cap Y$ is convex and balanced. If $B$ is not empty and if $Y$ is a 1-dimensional vector subspace of $\operatorname B$ then $B \cap Y$ is either $\$ or else it is absorbing in $Y.$


• for any $x \in X,$ if $B \cap \operatorname x$ contains more than one point then it is a convex and balanced neighborhood of $0$ in the 1-dimensional vector space $\operatorname x$ when this space is endowed with the Hausdorff Euclidean topology; and the set $B \cap \R x$ is a convex balanced subset of the real vector space $\R x$ that contains the origin.

Properties of balanced hulls and balanced cores

For any collection $\mathcal$ of subsets of $X,$
$\operatorname \left(\bigcup_ S\right) = \bigcup_ \operatorname S \quad \text \quad \operatorname \left(\bigcap_ S\right) = \bigcap_ \operatorname S.$

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

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 $S \subseteq X$ and any scalar $c,$ $\operatorname (c \, S) = c \operatorname S = , c, \operatorname S.$

For any scalar $c \neq 0,$ $\operatorname (c \, S) = c \operatorname S = , c, \operatorname S.$ This equality holds for $c = 0$ if and only if $S \subseteq \.$ Thus if $0 \in S$ or $S = \varnothing$ then $\operatorname (c \, S) = c \operatorname S = , c, \operatorname S$ for every scalar $c.$

Related notions

A function $p : X \to [0, \infty)$ on a real or complex vector space is said to be a if it satisfies any of the following equivalent conditions:

1. $p(a x) \leq p(x)$ whenever $a$ is a scalar satisfying $, a, \leq 1$ and $x \in X.$


2. $p(a x) \leq p(b x)$ whenever $a$ and $b$ are scalars satisfying $, a, \leq , b,$ and $x \in X.$


3. $\$ is a balanced set for every non-negative real $t \geq 0.$

If $p$ is a balanced function then $p(a x) = p(, a, x)$ for every scalar $a$ and vector $x \in X;$
so in particular, $p(u x) = p(x)$ for every unit length scalar $u$ (satisfying $, u, = 1$) and every $x \in X.$
Using $u := -1$ shows that every balanced function is a symmetric function.

A real-valued function $p : X \to \R$ is a seminorm if and only if it is a balanced sublinear function.

See also

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References

Proofs

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Linear algebra