If X is a finite set, what does the Hilbert space $L^2(X)$ means? - saw this notion on The Princeton Companion to Mathematics.
Hilbert space on a finite set
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abstract-algebra
hilbert-spaces
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0Which bit of the definition of Hilbert space is causing confusion here? – 2011-09-27
1 Answers
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If $X$ contains, say, $n$ elements then $L^2(X)$ reduces to the ordinary Euclidean space $\mathbb{R}^n$. EDIT: As commenter points out, here we have implicitly assumed that $X$ is equipped with the counting measure (call it $m$), that is, every subset of $X$ has a measure equal to its cardinality. This way the integral
$\int_X\lvert f(x) \rvert^2\, dm$
reduces to the sum
$\sum_{k=1}^n \lvert f(k)\rvert^2.$
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0I think what's missing is the explicit mention of a measure. So you should say that you take counting measure, for example. Just give *some* positive weight to each point in $X$, otherwise your $L^2$-space won't have dimension $n$. – 2011-09-27