Balanced set - WikiHQ

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

The balanced hull or balanced envelope of a set <math>S</math> is the smallest balanced set containing <math>S.</math>

The balanced core of a set <math>S</math> is the largest balanced set contained in <math>S.</math>

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 <math>X</math> be a vector space over the field <math>\mathbb{K}</math> of real or complex numbers.

Notation

If <math>S</math> is a set, <math>a</math> is a scalar, and <math>B \subseteq \mathbb{K}</math> then let <math>a S = \{a s : s \in S\}</math> and <math>B S = \{b s : b \in B, s \in S\}</math> and for any <math>0 \leq r \leq \infty,</math> let

<math display=block>B_r = \{a \in \mathbb{K} : |a| < r\} \qquad \text{ and } \qquad B_{\leq r} = \{ a \in \mathbb{K} : |a| \leq r\}.</math>

denote, respectively, the open ball and the closed ball of radius <math>r</math> in the scalar field <math>\mathbb{K}</math> centered at <math>0</math> where <math>B_0 = \varnothing, B_{\leq 0} = \{0\},</math> and <math>B_{\infty} = B_{\leq \infty} = \mathbb{K}.</math>

Every balanced subset of the field <math>\mathbb{K}</math> is of the form <math>B_{\leq r}</math> or <math>B_r</math> for some <math>0 \leq r \leq \infty.</math>

Balanced set

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

<ol>

<li>Definition: <math>a s \in S</math> for all <math>s \in S</math> and all scalars <math>a</math> satisfying <math>|a| \leq 1.</math></li>

<li><math>a S \subseteq S</math> for all scalars <math>a</math> satisfying <math>|a| \leq 1.</math></li>

<li><math>B_{\leq 1} S \subseteq S</math> (where <math>B_{\leq 1} := \{a \in \mathbb{K} : |a| \leq 1\}</math>).</li>

<li>For every <math>s \in S,</math> <math>S \cap \mathbb{K} s = B_{\leq 1} (S \cap \mathbb{K} s).</math>

<math>\mathbb{K} s = \operatorname{span} \{s\}</math> is a <math>0</math> (if <math>s = 0</math>) or <math>1</math> (if <math>s \neq 0</math>) dimensional vector subspace of <math>X.</math>
If <math>R := S \cap \mathbb{K} s</math> then the above equality becomes <math>R = B_{\leq 1} R,</math> which is exactly the previous condition for a set to be balanced. Thus, <math>S</math> is balanced if and only if for every <math>s \in S,</math> <math>S \cap \mathbb{K} s</math> is a balanced set (according to any of the previous defining conditions).</li>

<li>For every 1-dimensional vector subspace <math>Y</math> of <math>\operatorname{span} S,</math> <math>S \cap Y</math> is a balanced set (according to any defining condition other than this one).</li>

<li>For every <math>s \in S,</math> there exists some <math>0 \leq r \leq \infty</math> such that <math>S \cap \mathbb{K} s = B_r s</math> or <math>S \cap \mathbb{K} s = B_{\leq r} s.</math></li>

<li><math>S</math> is a balanced subset of <math>\operatorname{span} S</math> (according to any defining condition of "balanced" other than this one).

Thus <math>S</math> is a balanced subset of <math>X</math> if and only if it is balanced subset of every (equivalently, of some) vector space over the field <math>\mathbb{K}</math> that contains <math>S.</math> So assuming that the field <math>\mathbb{K}</math> is clear from context, this justifies writing "<math>S</math> is balanced" without mentioning any vector space.</li>

</ol>

If <math>S</math> is a convex set then this list may be extended to include:

<li><math>a S \subseteq S</math> for all scalars <math>a</math> satisfying <math>|a| = 1.</math></li>

</ol>

If <math>\mathbb{K} = \R</math> then this list may be extended to include:

<li><math>S</math> is symmetric (meaning <math>- S = S</math>) and <math>[0, 1) S \subseteq S.</math></li>

</ol>

Balanced hull

<math display=block>\operatorname{bal} S ~=~ \bigcup_{|a| \leq 1} a S = B_{\leq 1} S</math>

The ' of a subset <math>S</math> of <math>X,</math> denoted by <math>\operatorname{bal} S,</math> is defined in any of the following equivalent ways:

<ol>

<li>Definition: <math>\operatorname{bal} S</math> is the smallest (with respect to <math>\,\subseteq\,</math>) balanced subset of <math>X</math> containing <math>S.</math></li>

<li><math>\operatorname{bal} S</math> is the intersection of all balanced sets containing <math>S.</math></li>

<li><math>\operatorname{bal} S = \bigcup_{|a| \leq 1} (a S).</math></li>

<li><math>\operatorname{bal} S = B_{\leq 1} S.</math></li>

</ol>

Balanced core

<math display=block>\operatorname{balcore} S ~=~ \begin{cases}

\displaystyle\bigcap_{|a| \geq 1} a S & \text{ if } 0 \in S \\

\varnothing & \text{ if } 0 \not\in S \\

\end{cases}</math>

The ' of a subset <math>S</math> of <math>X,</math> denoted by <math>\operatorname{balcore} S,</math> is defined in any of the following equivalent ways:

<ol>

<li>Definition: <math>\operatorname{balcore} S</math> is the largest (with respect to <math>\,\subseteq\,</math>) balanced subset of <math>S.</math></li>

<li><math>\operatorname{balcore} S</math> is the union of all balanced subsets of <math>S.</math></li>

<li><math>\operatorname{balcore} S = \varnothing</math> if <math>0 \not\in S</math> while <math>\operatorname{balcore} S = \bigcap_{|a| \geq 1} (a S)</math> if <math>0 \in S.</math></li>

</ol>

Examples

The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, <math>\{0\}</math> 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 <math>p</math> is a seminorm (or norm) on a vector space <math>X</math> then for any constant <math>c > 0,</math> the set <math>\{x \in X : p(x) \leq c\}</math> is balanced.

If <math>S \subseteq X</math> is any subset and <math>B_1 := \{a \in \mathbb{K} : |a| < 1\}</math> then <math>B_1 S</math> is a balanced set.

In particular, if <math>U \subseteq X</math> is any balanced neighborhood of the origin in a topological vector space <math>X</math> then <math display=block>\operatorname{Int}_X U ~\subseteq~ B_1 U ~=~ \bigcup_{0 < |a| < 1} a U ~\subseteq~ U.</math>

Balanced sets in <math>\R</math> and <math>\Complex</math>

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

The balanced subsets of <math>X = \mathbb{K}</math> are exactly the following:

<ol>

<li><math>\varnothing</math></li>

</ol>

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 <math>\Complex</math> 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, <math>\Complex</math> and <math>\R^2</math> are entirely different as far as scalar multiplication is concerned.

Balanced sets in <math>\R^2</math>

Throughout, let <math>X = \R^2</math> (so <math>X</math> is a vector space over <math>\R</math>) and let <math>B_{\leq 1}</math> is the closed unit ball in <math>X</math> centered at the origin.

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

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

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

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 <math>S := [-1, 1] \times \{1\},</math> which is a horizontal closed line segment lying above the <math>x-</math>axis in <math>X := \R^2.</math> The balanced hull <math>\operatorname{bal} S</math> is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles <math>T_1</math> and <math>T_2,</math> where <math>T_2 = - T_1</math> and <math>T_1</math> is the filled triangle whose vertices are the origin together with the endpoints of <math>S</math> (said differently, <math>T_1</math> is the convex hull of <math>S \cup \{(0,0)\}</math> while <math>T_2</math> is the convex hull of <math>(-S) \cup \{(0,0)\}</math>).

Sufficient conditions

A set <math>T</math> is balanced if and only if it is equal to its balanced hull <math>\operatorname{bal} T</math> or to its balanced core <math>\operatorname{balcore} T,</math> in which case all three of these sets are equal: <math>T = \operatorname{bal} T = \operatorname{balcore} T.</math>

The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field <math>\mathbb{K}</math>).

<ul>

<li>The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.</li>

<li>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).</li>

<li>Arbitrary unions of balanced sets are balanced, and the same is true of arbitrary intersections of balanced sets.</li>

<li>Scalar multiples and (finite) Minkowski sums of balanced sets are again balanced.</li>

<li>Images and preimages of balanced sets under linear maps are again balanced. Explicitly, if <math>L : X \to Y</math> is a linear map and <math>B \subseteq X</math> and <math>C \subseteq Y</math> are balanced sets, then <math>L(B)</math> and <math>L^{-1}(C)</math> are balanced sets.</li>

</ul>

Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced. The union of the origin <math>\{0\}</math> and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced. However, <math>\left\{(z, w) \in \Complex^2 : |z| \leq |w|\right\}</math> is a balanced subset of <math>X = \Complex^2</math> that contains the origin <math>(0, 0) \in X</math> but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set. Similarly for real vector spaces, if <math>T</math> denotes the convex hull of <math>(0, 0)</math> and <math>(\pm 1, 1)</math> (a filled triangle whose vertices are these three points) then <math>B := T \cup (-T)</math> is an (hour glass shaped) balanced subset of <math>X := \Reals^2</math> whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set <math>\{(0, 0)\} \cup \operatorname{Int}_X B</math> 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 <math>X</math> contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given <math>W \subseteq X,</math> the symmetric set <math>\bigcap_{|u|=1} u W \subseteq W</math> will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of <math>X</math>) whenever this is true of <math>W.</math> It will be a balanced set if <math>W</math> is a star shaped at the origin, which is true, for instance, when <math>W</math> is convex and contains <math>0.</math> In particular, if <math>W</math> is a convex neighborhood of the origin then <math>\bigcap_{|u|=1} u W</math> will be a convex neighborhood of the origin and so its topological interior will be a balanced convex neighborhood of the origin.