Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $B \subset X$ where $X$ is a vector space. $B$ is called balanced if $\alpha B \subset B$ for every $\alpha \in \Phi$ with $|\alpha| \leq 1$. Note that $\Phi = \textbf{R}$ or $\Phi = \textbf{C}$. What is the intuition behind balanced sets? Why do we define them? Is it "good" for a vector space to have more balanced sets? Why do we require $| \alpha| \leq 1$?

share|improve this question
Think about this: Look at $\mathbb{R}^2$ as a vector space over $\mathbb{R}$ ($\mathbb{C}$), what is the smallest balanced set containing $A=\{(1,1)\}$, $B=\{(1,1),(0,1)\}$, $C=\{(x,y):x^2+y^2\le 1\}$? –  AD. Oct 27 '10 at 18:11
The unit circle? –  PEV Oct 28 '10 at 1:44
Do you adopt $|\alpha|\le 1$ (KennyTM edited the text)? Think it over again (using pen+paper), $\alpha$ is real over $\mathbb{R}$ and complex over $\mathbb{C}$. In one occasion you WILL get the unit disc (usually unit circle is the boundary of the unit disc)! –  AD. Oct 28 '10 at 4:37

3 Answers 3

up vote 5 down vote accepted

[Added in response to the comments below:] Despite what is written below, balanced needn't imply convex; it simply doesn't follow from the definition. So one should take the following discussion as rather an explanation about convex balanced sets, for whatever that is worth.

[Original answer follows:]

Balanced means that the set is symmetric around the origin (i.e. invariant under $v \mapsto -v$) and is convex. These are nice properties, generalizing some useful features of balls around the origin in $\mathbb R^n$. (Incidentally, if you didn't have the condition $|\alpha| = 1$, then any balanced subset with non-empty interior (in $\mathbb R^n$ say) would be all of $\mathbb R^n$, so that wouldn't be such an interesting condition.)

Also, any vector space will have plenty of such sets; you can form them just by taking the convex hull of a collection of points invariant under $v \mapsto -v$. What is more relevant in functional analysis is whether the origin of a topological vector space contains a basis of balaced neighbourhoods. The topological vector space is then called locally convex and this has important consequences for the space (especially for its duality theory).

Note that balls around the origin in a normed space are automatically balanced, so normed spaces are locally convex; so again, in the topological setting, one sees that this is a way of capturing some the geometric aspects of normed vector spaces, without insisting on the existence of a norm. (So, while normed spaces are not closed under certain operations, such as forming infinite direct sums, locally convex spaces are --- any direct sum (just to take one example) of a collection of locally convex spaces has a natural locally convex topology of its own.)

share|improve this answer
Is a balanced set necessary convex? If over a real vector space, I thought it suffices that the set is star-shaped w.r.t. 0. –  Willie Wong Oct 27 '10 at 18:49
I also don't see why if you didn't have $\lvert\alpha\rvert \le 1$, a balanced subset couldn't be an infinite star-shaped region, like a double cone, and not all of $\mathbb{R}^n$. –  Rahul Oct 27 '10 at 19:17
@Willie, @Rahul: Dear Willie and Rahul, Thanks a lot, you are both correct; I was thinking too much about the locally convex set-up, rather than what was literally under discussion. I have added a big caveat at the start of my answer. –  Matt E Oct 27 '10 at 20:24

It's like balancing a tire, you spin the set around! (A bit tongue in cheek, but that's how I've always felt about it...)

share|improve this answer

In Complex analysis in locally convex spaces by Seán Dineen, it states that:

Balanced open sets arise as the natural domain of convergence of Taylor series expansion at the origin of holomorphic functions.
share|improve this answer
@Chandru1: If you are going to copy from a google books search, you might want to try cut-n-paste instead, and thus avoid introducing incorrect punctuation. –  Arturo Magidin Oct 27 '10 at 18:36
@Arturo: The thing is i couldn't do that, –  anonymous Oct 27 '10 at 18:37
@Chandru: No, I'm not asking anything. I'm pointing out that you are trying (or claiming) to quote verbatim from a book, but you introduced incorrect punctuation (a comma after "arise"; capital letter in "Holomorphic"; a missing period at the end of the sentence). If you are going to quote, then quote accurately. If you cannot quote accurately when copying, then use cut-n-paste so you don't introduce errors that will then be attributed to the original author. –  Arturo Magidin Oct 27 '10 at 18:41
@Chandru1: You couldn't do what? You couldn't cut and paste? Or are you physically unable to quote without introducing errors? Learn to quote accurately, or stop trying to quote other people. "Quoting accurately" is not about giving appropriate citations, it's about quoting: repeating. If you are claiming to quote, then your quote should be exactly what the other person wrote/said, and any changes need to be explicitly marked (in sundry typographical ways). If you are merely paraphrasing, then don't present it as a quote. –  Arturo Magidin Oct 27 '10 at 18:46
@Chandru1: If you look up quickly, you might catch a glimpse of my point flying by. –  Arturo Magidin Oct 27 '10 at 18:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.