Balanced set

{{Short description|Construct in functional analysis}} {{use mdy dates|date=September 2021}} {{Use American English|date = March 2019}} 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 ''{{visible anchor|balanced set}}'' 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><math>S = B_{\leq 1} S.</math>{{sfn|Swartz|1992|pp=4-8}}</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.<ref group=note>Assuming that all vector spaces containing a set <math>S</math> are over the same field, when describing the set as being "balanced", it is not necessary to mention a vector space containing <math>S.</math> That is, "<math>S</math> is balanced" may be written in place of "<math>S</math> is a balanced subset of <math>X</math>".</ref></li> </ol>

If <math>S</math> is a convex set then this list may be extended to include: <ol start=9> <li><math>a S \subseteq S</math> for all scalars <math>a</math> satisfying <math>|a| = 1.</math>{{sfn|Narici|Beckenstein|2011|pp=107-110}}</li> </ol>

If <math>\mathbb{K} = \R</math> then this list may be extended to include: <ol start=10> <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 ''{{visible anchor|balanced hull|Balanced hull}}'' 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>{{sfn|Swartz|1992|pp=4-8}}</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 ''{{visible anchor|balanced core|Balanced core}}'' 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:{{sfn|Jarchow|1981|p=34}} <ol> <li><math>\varnothing</math></li> <li><math>X</math></li> <li><math>\{0\}</math></li> <li><math>\{x \in X : |x| < r\}</math> for some real <math>r > 0</math></li> <li><math>\{x \in X : |x| \leq r\}</math> for some real <math>r > 0.</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 {{em|any}} 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.{{sfn|Narici|Beckenstein|2011|pp=156-175}}</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.{{sfn|Rudin|1991|pp=10-14}} 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.{{sfn|Rudin|1991|pp=10-14}}<ref group=proof>Let <math>B \subseteq X</math> be balanced. If its topological interior <math>\operatorname{Int}_X B</math> is empty then it is balanced so assume otherwise and let <math>|s| \leq 1</math> be a scalar. If <math>s \neq 0</math> then the map <math>X \to X</math> defined by <math>x \mapsto s x</math> is a homeomorphism, which implies <math>s \operatorname{Int}_X B = \operatorname{Int}_X (s B) \subseteq s B \subseteq B;</math> because <math>s \operatorname{Int}_X B</math> is open, <math>s \operatorname{Int}_X B \subseteq \operatorname{Int}_X B</math> so that it only remains to show that this is true for <math>s = 0.</math> However, <math>0 \in \operatorname{Int}_X B</math> might not be true but when it is true then <math>\operatorname{Int}_X B</math> will be balanced. <math>\blacksquare</math></ref> 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.{{sfn|Rudin|1991|p=38}} 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,<ref group=note name=DefStarShapedAtOrigin><math>W</math> being star shaped at the origin means that <math>0 \in W</math> and <math>r w \in W</math> for all <math>0 \leq r \leq 1</math> and <math>w \in W.</math></ref> 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 {{em|balanced}} convex neighborhood of the origin and so its topological interior will be a balanced convex {{em|open}} neighborhood of the origin.{{sfn|Rudin|1991|pp=10-14}}

{{math proof|proof= Let <math>0 \in W \subseteq X</math> and define <math>A = \bigcap_{|u|=1} u W</math> (where <math>u</math> denotes elements of the field <math>\mathbb{K}</math> of scalars). Taking <math>u := 1</math> shows that <math>A \subseteq W.</math> If <math>W</math> is convex then so is <math>A</math> (since an intersection of convex sets is convex) and thus so is <math>A</math>'s interior. If <math>|s| = 1</math> then <math display=block>s A = \bigcap_{|u|=1} s u W \subseteq \bigcap_{|u|=1} u W = A</math> and thus <math>s A = A.</math> If <math>W</math> is star shaped at the origin<ref group=note name=DefStarShapedAtOrigin /> then so is every <math>u W</math> (for <math>|u| = 1</math>), which implies that for any <math>0 \leq r \leq 1,</math> <math display=block>r A = \bigcap_{|u|=1} r u W \subseteq \bigcap_{|u|=1} u W = A</math> thus proving that <math>A</math> is balanced. If <math>W</math> is convex and contains the origin then it is star shaped at the origin and so <math>A</math> will be balanced.

Now suppose <math>W</math> is a neighborhood of the origin in <math>X.</math> Since scalar multiplication <math>M : \mathbb{K} \times X \to X</math> (defined by <math>M(a, x) = a x</math>) is continuous at the origin <math>(0, 0) \in \mathbb{K} \times X</math> and <math>M(0, 0) = 0 \in W,</math> there exists some basic open neighborhood <math>B_r \times V</math> (where <math>r > 0</math> and <math>B_r := \{c \in \mathbb{K} : |c| < r\}</math>) of the origin in the product topology on <math>\mathbb{K} \times X</math> such that <math>M\left(B_r \times V\right) \subseteq W;</math> the set <math>M\left(B_r \times V\right) = B_r V</math> is balanced and it is also open because it may be written as <math display=block>B_r V = \bigcup_{|a| < r} a V = \bigcup_{0 < |a| < r} a V \qquad \text{ (since } 0 \cdot V = \{0\} \subseteq a V \text{ )}</math> where <math>a V</math> is an open neighborhood of the origin whenever <math>a \neq 0.</math> Finally, <math display=block>A = \bigcap_{|u|=1} u W \supseteq \bigcap_{|u|=1} u B_r V = \bigcap_{|u|=1} B_r V = B_r V</math> shows that <math>A</math> is also a neighborhood of the origin. If <math>A</math> is balanced then because its interior <math>\operatorname{Int}_X A</math> contains the origin, <math>\operatorname{Int}_X A</math> will also be balanced. If <math>W</math> is convex then <math>A</math> is convex and balanced and thus the same is true of <math>\operatorname{Int}_X A.</math> <math>\blacksquare</math> }}

Suppose that <math>W</math> is a convex and absorbing subset of <math>X.</math> Then <math>D := \bigcap_{|u|=1} u W</math> will be convex balanced absorbing subset of <math>X,</math> which guarantees that the Minkowski functional <math>p_D : X \to \R</math> of <math>D</math> will be a seminorm on <math>X,</math> thereby making <math>\left(X, p_D\right)</math> into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples <math>r D</math> as <math>r</math> ranges over <math>\left\{\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \ldots\right\}</math> (or over any other set of non-zero scalars having <math>0</math> as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If <math>X</math> is a topological vector space and if this convex absorbing subset <math>W</math> is also a bounded subset of <math>X,</math> then the same will be true of the absorbing disk <math>D := {\textstyle\bigcap\limits_{|u|=1}} u W;</math> if in addition <math>D</math> does not contain any non-trivial vector subspace then <math>p_D</math> will be a norm and <math>\left(X, p_D\right)</math> will form what is known as an auxiliary normed space.{{sfn|Narici|Beckenstein|2011|pp=115-154}} If this normed space is a Banach space then <math>D</math> is called a {{em|Banach disk}}.

==Properties== {{See also|Topological vector space#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 <math>B</math> is a balanced subset of <math>X</math> then: <ul> <li>for any scalars <math>c</math> and <math>d,</math> if <math>|c| \leq |d|</math> then <math>c B \subseteq d B</math> and <math>c B = |c| B.</math> Thus if <math>c</math> and <math>d</math> are any scalars then <math>(c B) \cap (d B) = \min_{} \{|c|, |d|\} B.</math></li> <li><math>B</math> is absorbing in <math>X</math> if and only if for all <math>x \in X,</math> there exists <math>r > 0</math> such that <math>x \in r B.</math>{{sfn|Narici|Beckenstein|2011|pp=107-110}}</li> <li>for any 1-dimensional vector subspace <math>Y</math> of <math>X,</math> the set <math>B \cap Y</math> is convex and balanced. If <math>B</math> is not empty and if <math>Y</math> is a 1-dimensional vector subspace of <math>\operatorname{span} B</math> then <math>B \cap Y</math> is either <math>\{0\}</math> or else it is absorbing in <math>Y.</math></li> <li>for any <math>x \in X,</math> if <math>B \cap \operatorname{span} x</math> contains more than one point then it is a convex and balanced neighborhood of <math>0</math> in the 1-dimensional vector space <math>\operatorname{span} x</math> when this space is endowed with the Hausdorff Euclidean topology; and the set <math>B \cap \R x</math> is a convex balanced subset of the real vector space <math>\R x</math> that contains the origin.</li> </ul>

'''Properties of balanced hulls and balanced cores'''

For any collection <math>\mathcal{S}</math> of subsets of <math>X,</math> <math display=block>\operatorname{bal} \left(\bigcup_{S \in \mathcal{S}} S\right) = \bigcup_{S \in \mathcal{S}} \operatorname{bal} S \quad \text{ and } \quad \operatorname{balcore} \left(\bigcap_{S \in \mathcal{S}} S\right) = \bigcap_{S \in \mathcal{S}} \operatorname{balcore} S.</math>

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If <math>X</math> is a Hausdorff topological vector space and if <math>K</math> is a compact subset of <math>X</math> then the balanced hull of <math>K</math> is compact.{{sfn|Trèves|2006|p=56}}

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 <math>S \subseteq X</math> and any scalar <math>c,</math> <math>\operatorname{bal} (c \, S) = c \operatorname{bal} S = |c| \operatorname{bal} S.</math>

For any scalar <math>c \neq 0,</math> <math>\operatorname{balcore} (c \, S) = c \operatorname{balcore} S = |c| \operatorname{balcore} S.</math> This equality holds for <math>c = 0</math> if and only if <math>S \subseteq \{0\}.</math> Thus if <math>0 \in S</math> or <math>S = \varnothing</math> then <math display=block>\operatorname{balcore} (c \, S) = c \operatorname{balcore} S = |c| \operatorname{balcore} S</math> for every scalar <math>c.</math>

==Related notions==

A function <math>p : X \to [0, \infty)</math> on a real or complex vector space is said to be a {{em|{{visible anchor|balanced function}}}} if it satisfies any of the following equivalent conditions:{{sfn|Schechter|1996|p=313}} <ol> <li><math>p(a x) \leq p(x)</math> whenever <math>a</math> is a scalar satisfying <math>|a| \leq 1</math> and <math>x \in X.</math></li> <li><math>p(a x) \leq p(b x)</math> whenever <math>a</math> and <math>b</math> are scalars satisfying <math>|a| \leq |b|</math> and <math>x \in X.</math></li> <li><math>\{x \in X : p(x) \leq t\}</math> is a balanced set for every non-negative real <math>t \geq 0.</math></li> </ol> If <math>p</math> is a balanced function then <math>p(a x) = p(|a| x)</math> for every scalar <math>a</math> and vector <math>x \in X;</math> so in particular, <math>p(u x) = p(x)</math> for every unit length scalar <math>u</math> (satisfying <math>|u| = 1</math>) and every <math>x \in X.</math>{{sfn|Schechter|1996|p=313}} Using <math>u := -1</math> shows that every balanced function is a symmetric function.

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

==See also==

* {{annotated link|Absolutely convex set}} * {{annotated link|Absorbing set}} * {{annotated link|Bounded set (topological vector space)}} * {{annotated link|Convex set}} * {{annotated link|Star domain}} * {{annotated link|Symmetric set}} * {{annotated link|Topological vector space}}

==References==

'''Proofs'''

===Sources=== * {{Bourbaki Topological Vector Spaces}}  * {{Conway A Course in Functional Analysis}}  * {{Dunford Schwartz Linear Operators Part 1 General Theory}}  * {{Edwards Functional Analysis Theory and Applications}}  * {{Jarchow Locally Convex Spaces}}  * {{Köthe Topological Vector Spaces I}}  * {{Köthe Topological Vector Spaces II}}  * {{Narici Beckenstein Topological Vector Spaces|edition=2}}  * {{Robertson Topological Vector Spaces| page=4}}  * {{Rudin Walter Functional Analysis|edition=2}}  * {{Schaefer Wolff Topological Vector Spaces|edition=2}}  * {{cite book |last1=Schechter |first1=Eric |title=Handbook of Analysis and Its Foundations |date=24 October 1996 |publisher=Academic Press |isbn=978-0-08-053299-8 |url=https://books.google.com/books?id=eqUv3Bcd56EC&pg=PA313 |language=en}} * {{Swartz An Introduction to Functional Analysis}}  * {{Trèves François Topological vector spaces, distributions and kernels}}  * {{Wilansky Modern Methods in Topological Vector Spaces|edition=1}}

Category:Linear algebra