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

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

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 ${\displaystyle X}$ be a vector space over the field ${\displaystyle \mathbb {K} }$ of real or complex numbers.

Notation

If ${\displaystyle S}$ is a set, ${\displaystyle a}$ is a scalar, and ${\displaystyle B\subseteq \mathbb {K} }$ then let ${\displaystyle aS=\{as:s\in S\))$ and ${\displaystyle BS=\{bs:b\in B,s\in S\))$ and for any ${\displaystyle 0\leq r\leq \infty ,}$ let

${\displaystyle B_{r}=\{a\in \mathbb {K} :|a|
denote, respectively, the open ball and the closed ball of radius ${\displaystyle r}$ in the scalar field ${\displaystyle \mathbb {K} }$ centered at ${\displaystyle 0}$ where ${\displaystyle B_{0}=\varnothing ,B_{\leq 0}=\{0\},}$ and ${\displaystyle B_{\infty }=B_{\leq \infty }=\mathbb {K} .}$ Every balanced subset of the field ${\displaystyle \mathbb {K} }$ is of the form ${\displaystyle B_{\leq r))$ or ${\displaystyle B_{r))$ for some ${\displaystyle 0\leq r\leq \infty .}$

Balanced set

A subset ${\displaystyle S}$ of ${\displaystyle X}$ is called a balanced set or balanced if it satisfies any of the following equivalent conditions:

1. Definition: ${\displaystyle as\in S}$ for all ${\displaystyle s\in S}$ and all scalars ${\displaystyle a}$ satisfying ${\displaystyle |a|\leq 1.}$
2. ${\displaystyle aS\subseteq S}$ for all scalars ${\displaystyle a}$ satisfying ${\displaystyle |a|\leq 1.}$
3. ${\displaystyle B_{\leq 1}S\subseteq S}$ (where ${\displaystyle B_{\leq 1}:=\{a\in \mathbb {K} :|a|\leq 1\))$).
4. ${\displaystyle S=B_{\leq 1}S.}$[1]
5. For every ${\displaystyle s\in S,}$ ${\displaystyle S\cap \mathbb {K} s=B_{\leq 1}(S\cap \mathbb {K} s).}$
• ${\displaystyle \mathbb {K} s=\operatorname {span} \{s\))$ is a ${\displaystyle 0}$ (if ${\displaystyle s=0}$) or ${\displaystyle 1}$ (if ${\displaystyle s\neq 0}$) dimensional vector subspace of ${\displaystyle X.}$
• If ${\displaystyle R:=S\cap \mathbb {K} s}$ then the above equality becomes ${\displaystyle R=B_{\leq 1}R,}$ which is exactly the previous condition for a set to be balanced. Thus, ${\displaystyle S}$ is balanced if and only if for every ${\displaystyle s\in S,}$ ${\displaystyle S\cap \mathbb {K} s}$ is a balanced set (according to any of the previous defining conditions).
6. For every 1-dimensional vector subspace ${\displaystyle Y}$ of ${\displaystyle \operatorname {span} S,}$ ${\displaystyle S\cap Y}$ is a balanced set (according to any defining condition other than this one).
7. For every ${\displaystyle s\in S,}$ there exists some ${\displaystyle 0\leq r\leq \infty }$ such that ${\displaystyle S\cap \mathbb {K} s=B_{r}s}$ or ${\displaystyle S\cap \mathbb {K} s=B_{\leq r}s.}$
8. ${\displaystyle S}$ is a balanced subset of ${\displaystyle \operatorname {span} S}$ (according to any defining condition of "balanced" other than this one).
• Thus ${\displaystyle S}$ is a balanced subset of ${\displaystyle X}$ if and only if it is balanced subset of every (equivalently, of some) vector space over the field ${\displaystyle \mathbb {K} }$ that contains ${\displaystyle S.}$ So assuming that the field ${\displaystyle \mathbb {K} }$ is clear from context, this justifies writing "${\displaystyle S}$ is balanced" without mentioning any vector space.[note 1]

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

1. ${\displaystyle aS\subseteq S}$ for all scalars ${\displaystyle a}$ satisfying ${\displaystyle |a|=1.}$[2]

If ${\displaystyle \mathbb {K} =\mathbb {R} }$ then this list may be extended to include:

1. ${\displaystyle S}$ is symmetric (meaning ${\displaystyle -S=S}$) and ${\displaystyle [0,1)S\subseteq S.}$

### Balanced hull

${\displaystyle \operatorname {bal} S~=~\bigcup _{|a|\leq 1}aS=B_{\leq 1}S}$

The balanced hull of a subset ${\displaystyle S}$ of ${\displaystyle X,}$ denoted by ${\displaystyle \operatorname {bal} S,}$ is defined in any of the following equivalent ways:

1. Definition: ${\displaystyle \operatorname {bal} S}$ is the smallest (with respect to ${\displaystyle \,\subseteq \,}$) balanced subset of ${\displaystyle X}$ containing ${\displaystyle S.}$
2. ${\displaystyle \operatorname {bal} S}$ is the intersection of all balanced sets containing ${\displaystyle S.}$
3. ${\displaystyle \operatorname {bal} S=\bigcup _{|a|\leq 1}(aS).}$
4. ${\displaystyle \operatorname {bal} S=B_{\leq 1}S.}$[1]

### Balanced core

${\displaystyle \operatorname {balcore} S~=~{\begin{cases}\displaystyle \bigcap _{|a|\geq 1}aS&{\text{ if ))0\in S\\\varnothing &{\text{ if ))0\not \in S\\\end{cases))}$

The balanced core of a subset ${\displaystyle S}$ of ${\displaystyle X,}$ denoted by ${\displaystyle \operatorname {balcore} S,}$ is defined in any of the following equivalent ways:

1. Definition: ${\displaystyle \operatorname {balcore} S}$ is the largest (with respect to ${\displaystyle \,\subseteq \,}$) balanced subset of ${\displaystyle S.}$
2. ${\displaystyle \operatorname {balcore} S}$ is the union of all balanced subsets of ${\displaystyle S.}$
3. ${\displaystyle \operatorname {balcore} S=\varnothing }$ if ${\displaystyle 0\not \in S}$ while ${\displaystyle \operatorname {balcore} S=\bigcap _{|a|\geq 1}(aS)}$ if ${\displaystyle 0\in S.}$

## Examples

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

If ${\displaystyle S\subseteq X}$ is any subset and ${\displaystyle B_{1}:=\{a\in \mathbb {K} :|a|<1\))$ then ${\displaystyle B_{1}S}$ is a balanced set. In particular, if ${\displaystyle U\subseteq X}$ is any balanced neighborhood of the origin in a topological vector space ${\displaystyle X}$ then

${\displaystyle \operatorname {Int} _{X}U~\subseteq ~B_{1}U~=~\bigcup _{0<|a|<1}aU~\subseteq ~U.}$

Balanced sets in ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} }$

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

The balanced subsets of ${\displaystyle X=\mathbb {K} }$ are exactly the following:[3]

1. ${\displaystyle \varnothing }$
2. ${\displaystyle X}$
3. ${\displaystyle \{0\))$
4. ${\displaystyle \{x\in X:|x| for some real ${\displaystyle r>0}$
5. ${\displaystyle \{x\in X:|x|\leq r\))$ for some real ${\displaystyle r>0.}$

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 ${\displaystyle \mathbb {C} }$ 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, ${\displaystyle \mathbb {C} }$ and ${\displaystyle \mathbb {R} ^{2))$ are entirely different as far as scalar multiplication is concerned.

Balanced sets in ${\displaystyle \mathbb {R} ^{2))$

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

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

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

For every ${\displaystyle 0\leq t\leq \pi ,}$ let ${\displaystyle r_{t))$ be any positive real number and let ${\displaystyle B^{t))$ be the (open or closed) line segment in ${\displaystyle X:=\mathbb {R} ^{2))$ between the points ${\displaystyle (\cos t,\sin t)}$ and ${\displaystyle -(\cos t,\sin t).}$ Then the set ${\displaystyle B=\bigcup _{0\leq t<\pi }r_{t}B^{t))$ 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 ${\displaystyle xy=1}$ in ${\displaystyle X=\mathbb {R} ^{2}.}$

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

### Sufficient conditions

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

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

• 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 ${\displaystyle L:X\to Y}$ is a linear map and ${\displaystyle B\subseteq X}$ and ${\displaystyle C\subseteq Y}$ are balanced sets, then ${\displaystyle L(B)}$ and ${\displaystyle L^{-1}(C)}$ are balanced sets.

### Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced.[5] The union of the origin ${\displaystyle \{0\))$ 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, ${\displaystyle \left\{(z,w)\in \mathbb {C} ^{2}:|z|\leq |w|\right\))$ is a balanced subset of ${\displaystyle X=\mathbb {C} ^{2))$ that contains the origin ${\displaystyle (0,0)\in X}$ but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set.[6] Similarly for real vector spaces, if ${\displaystyle T}$ denotes the convex hull of ${\displaystyle (0,0)}$ and ${\displaystyle (\pm 1,1)}$ (a filled triangle whose vertices are these three points) then ${\displaystyle B:=T\cup (-T)}$ is an (hour glass shaped) balanced subset of ${\displaystyle X:=\mathbb {R} ^{2))$ whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set ${\displaystyle \{(0,0)\}\cup \operatorname {Int} _{X}B}$ 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 ${\displaystyle X}$ contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given ${\displaystyle W\subseteq X,}$ the symmetric set ${\displaystyle \bigcap _{|u|=1}uW\subseteq W}$ will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of ${\displaystyle X}$) whenever this is true of ${\displaystyle W.}$ It will be a balanced set if ${\displaystyle W}$ is a star shaped at the origin,[note 2] which is true, for instance, when ${\displaystyle W}$ is convex and contains ${\displaystyle 0.}$ In particular, if ${\displaystyle W}$ is a convex neighborhood of the origin then ${\displaystyle \bigcap _{|u|=1}uW}$ 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 ${\displaystyle 0\in W\subseteq X}$ and define ${\displaystyle A=\bigcap _{|u|=1}uW}$ (where ${\displaystyle u}$ denotes elements of the field ${\displaystyle \mathbb {K} }$ of scalars). Taking ${\displaystyle u:=1}$ shows that ${\displaystyle A\subseteq W.}$ If ${\displaystyle W}$ is convex then so is ${\displaystyle A}$ (since an intersection of convex sets is convex) and thus so is ${\displaystyle A}$'s interior. If ${\displaystyle |s|=1}$ then

${\displaystyle sA=\bigcap _{|u|=1}suW\subseteq \bigcap _{|u|=1}uW=A}$
and thus ${\displaystyle sA=A.}$ If ${\displaystyle W}$ is star shaped at the origin[note 2] then so is every ${\displaystyle uW}$ (for ${\displaystyle |u|=1}$), which implies that for any ${\displaystyle 0\leq r\leq 1,}$
${\displaystyle rA=\bigcap _{|u|=1}ruW\subseteq \bigcap _{|u|=1}uW=A}$
thus proving that ${\displaystyle A}$ is balanced. If ${\displaystyle W}$ is convex and contains the origin then it is star shaped at the origin and so ${\displaystyle A}$ will be balanced.

Now suppose ${\displaystyle W}$ is a neighborhood of the origin in ${\displaystyle X.}$ Since scalar multiplication ${\displaystyle M:\mathbb {K} \times X\to X}$ (defined by ${\displaystyle M(a,x)=ax}$) is continuous at the origin ${\displaystyle (0,0)\in \mathbb {K} \times X}$ and ${\displaystyle M(0,0)=0\in W,}$ there exists some basic open neighborhood ${\displaystyle B_{r}\times V}$ (where ${\displaystyle r>0}$ and ${\displaystyle B_{r}:=\{c\in \mathbb {K} :|c|) of the origin in the product topology on ${\displaystyle \mathbb {K} \times X}$ such that ${\displaystyle M\left(B_{r}\times V\right)\subseteq W;}$ the set ${\displaystyle M\left(B_{r}\times V\right)=B_{r}V}$ is balanced and it is also open because it may be written as

${\displaystyle B_{r}V=\bigcup _{|a|
where ${\displaystyle aV}$ is an open neighborhood of the origin whenever ${\displaystyle a\neq 0.}$ Finally,
${\displaystyle A=\bigcap _{|u|=1}uW\supseteq \bigcap _{|u|=1}uB_{r}V=\bigcap _{|u|=1}B_{r}V=B_{r}V}$
shows that ${\displaystyle A}$ is also a neighborhood of the origin. If ${\displaystyle A}$ is balanced then because its interior ${\displaystyle \operatorname {Int} _{X}A}$ contains the origin, ${\displaystyle \operatorname {Int} _{X}A}$ will also be balanced. If ${\displaystyle W}$ is convex then ${\displaystyle A}$ is convex and balanced and thus the same is true of ${\displaystyle \operatorname {Int} _{X}A.}$ ${\displaystyle \blacksquare }$

Suppose that ${\displaystyle W}$ is a convex and absorbing subset of ${\displaystyle X.}$ Then ${\displaystyle D:=\bigcap _{|u|=1}uW}$ will be convex balanced absorbing subset of ${\displaystyle X,}$ which guarantees that the Minkowski functional ${\displaystyle p_{D}:X\to \mathbb {R} }$ of ${\displaystyle D}$ will be a seminorm on ${\displaystyle X,}$ thereby making ${\displaystyle \left(X,p_{D}\right)}$ into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples ${\displaystyle rD}$ as ${\displaystyle r}$ ranges over ${\displaystyle \left$$(\tfrac {1}{2)),{\tfrac {1}{3)),{\tfrac {1}{4)),\ldots \right$$)$ (or over any other set of non-zero scalars having ${\displaystyle 0}$ as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If ${\displaystyle X}$ is a topological vector space and if this convex absorbing subset ${\displaystyle W}$ is also a bounded subset of ${\displaystyle X,}$ then the same will be true of the absorbing disk ${\displaystyle D:={\textstyle \bigcap \limits _{|u|=1))uW;}$ if in addition ${\displaystyle D}$ does not contain any non-trivial vector subspace then ${\displaystyle p_{D))$ will be a norm and ${\displaystyle \left(X,p_{D}\right)}$ will form what is known as an auxiliary normed space.[7] If this normed space is a Banach space then ${\displaystyle D}$ is called a Banach disk.

## 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 ${\displaystyle B}$ is a balanced subset of ${\displaystyle X}$ then:

• for any scalars ${\displaystyle c}$ and ${\displaystyle d,}$ if ${\displaystyle |c|\leq |d|}$ then ${\displaystyle cB\subseteq dB}$ and ${\displaystyle cB=|c|B.}$ Thus if ${\displaystyle c}$ and ${\displaystyle d}$ are any scalars then ${\displaystyle (cB)\cap (dB)=\min _{}\{|c|,|d|\}B.}$
• ${\displaystyle B}$ is absorbing in ${\displaystyle X}$ if and only if for all ${\displaystyle x\in X,}$ there exists ${\displaystyle r>0}$ such that ${\displaystyle x\in rB.}$[2]
• for any 1-dimensional vector subspace ${\displaystyle Y}$ of ${\displaystyle X,}$ the set ${\displaystyle B\cap Y}$ is convex and balanced. If ${\displaystyle B}$ is not empty and if ${\displaystyle Y}$ is a 1-dimensional vector subspace of ${\displaystyle \operatorname {span} B}$ then ${\displaystyle B\cap Y}$ is either ${\displaystyle \{0\))$ or else it is absorbing in ${\displaystyle Y.}$
• for any ${\displaystyle x\in X,}$ if ${\displaystyle B\cap \operatorname {span} x}$ contains more than one point then it is a convex and balanced neighborhood of ${\displaystyle 0}$ in the 1-dimensional vector space ${\displaystyle \operatorname {span} x}$ when this space is endowed with the Hausdorff Euclidean topology; and the set ${\displaystyle B\cap \mathbb {R} x}$ is a convex balanced subset of the real vector space ${\displaystyle \mathbb {R} x}$ that contains the origin.

Properties of balanced hulls and balanced cores

For any collection ${\displaystyle {\mathcal {S))}$ of subsets of ${\displaystyle X,}$

${\displaystyle \operatorname {bal} \left(\bigcup _{S\in {\mathcal {S))}S\right)=\bigcup _{S\in {\mathcal {S))}\operatorname {bal} S\quad {\text{ and ))\quad \operatorname {balcore} \left(\bigcap _{S\in {\mathcal {S))}S\right)=\bigcap _{S\in {\mathcal {S))}\operatorname {balcore} S.}$

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If ${\displaystyle X}$ is a Hausdorff topological vector space and if ${\displaystyle K}$ is a compact subset of ${\displaystyle X}$ then the balanced hull of ${\displaystyle K}$ 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 ${\displaystyle S\subseteq X}$ and any scalar ${\displaystyle c,}$ ${\displaystyle \operatorname {bal} (c\,S)=c\operatorname {bal} S=|c|\operatorname {bal} S.}$

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

${\displaystyle \operatorname {balcore} (c\,S)=c\operatorname {balcore} S=|c|\operatorname {balcore} S}$
for every scalar ${\displaystyle c.}$

## Related notions

A function ${\displaystyle p:X\to [0,\infty )}$ 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. ${\displaystyle p(ax)\leq p(x)}$ whenever ${\displaystyle a}$ is a scalar satisfying ${\displaystyle |a|\leq 1}$ and ${\displaystyle x\in X.}$
2. ${\displaystyle p(ax)\leq p(bx)}$ whenever ${\displaystyle a}$ and ${\displaystyle b}$ are scalars satisfying ${\displaystyle |a|\leq |b|}$ and ${\displaystyle x\in X.}$
3. ${\displaystyle \{x\in X:p(x)\leq t\))$ is a balanced set for every non-negative real ${\displaystyle t\geq 0.}$

If ${\displaystyle p}$ is a balanced function then ${\displaystyle p(ax)=p(|a|x)}$ for every scalar ${\displaystyle a}$ and vector ${\displaystyle x\in X;}$ so in particular, ${\displaystyle p(ux)=p(x)}$ for every unit length scalar ${\displaystyle u}$ (satisfying ${\displaystyle |u|=1}$) and every ${\displaystyle x\in X.}$[9] Using ${\displaystyle u:=-1}$ shows that every balanced function is a symmetric function.

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

## References

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 ${\displaystyle S}$ are over the same field, when describing the set as being "balanced", it is not necessary to mention a vector space containing ${\displaystyle S.}$ That is, "${\displaystyle S}$ is balanced" may be written in place of "${\displaystyle S}$ is a balanced subset of ${\displaystyle X}$".
2. ^ a b ${\displaystyle W}$ being star shaped at the origin means that ${\displaystyle 0\in W}$ and ${\displaystyle rw\in W}$ for all ${\displaystyle 0\leq r\leq 1}$ and ${\displaystyle w\in W.}$

Proofs

1. ^ Let ${\displaystyle B\subseteq X}$ be balanced. If its topological interior ${\displaystyle \operatorname {Int} _{X}B}$ is empty then it is balanced so assume otherwise and let ${\displaystyle |s|\leq 1}$ be a scalar. If ${\displaystyle s\neq 0}$ then the map ${\displaystyle X\to X}$ defined by ${\displaystyle x\mapsto sx}$ is a homeomorphism, which implies ${\displaystyle s\operatorname {Int} _{X}B=\operatorname {Int} _{X}(sB)\subseteq sB\subseteq B;}$ because ${\displaystyle s\operatorname {Int} _{X}B}$ is open, ${\displaystyle s\operatorname {Int} _{X}B\subseteq \operatorname {Int} _{X}B}$ so that it only remains to show that this is true for ${\displaystyle s=0.}$ However, ${\displaystyle 0\in \operatorname {Int} _{X}B}$ might not be true but when it is true then ${\displaystyle \operatorname {Int} _{X}B}$ will be balanced. ${\displaystyle \blacksquare }$

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