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For $n > 1$ an integer, there are well-known formulas for volume of the balls.

What is the analogous statement in a Banach space/Hilbert Space?

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There is no analogue of the volume form for an infinite-dimensional space, since there is no top exterior power. So what does "volume" mean here? – Qiaochu Yuan Jul 27 '10 at 21:37
See the answer of John Cook. – user218 Jul 27 '10 at 23:15

3 Answers 3

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Maybe this is helpful, a formula for the volume of Lp balls in R^n.

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I don't understand how this answers the question. The L^p ball is not the L^p analogue of a simplex. – Qiaochu Yuan Jul 28 '10 at 0:00
@Qiaochu: I was seeding the site with questions and forgot what exact answer I wanted to bait; so I was too vague. Sorry. Now I modified the question. – user218 Jul 28 '10 at 13:04

In an infinite-dimensional normed space with a translation-invariant measure, the measure of a ball must be either $0$ or $\infty$. This fact is sometimes summarized as "there is no infinite-dimensional Lebesgue measure." So unless you have some other notion of "volume" in mind, only the finite-dimensional case (i.e. $\mathbb{R}^n$ with some other norm) has any content. And with regard to Hilbert spaces, the only finite-dimensional Hilbert spaces are $\mathbb{R}^n$ with the Euclidean inner product, so of course we know about this.

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A finite measure $\mu$ on a separable complete metric space has the property of tightness; i.e. for any $\epsilon > 0$ there is some compact subset $K$ so that $\mu(K^c) < \epsilon$. A Banach space that is locally compact is necessarily finite dimensional. Hence, a compact subset is a very small subset that must have a void interior.

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