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Convolution on groups

  1. Apr 24, 2015 #1
    Let ##G## be a group and let ##R## be the set of reals.
    Consider the set ## R(G) = \{ f : G \rightarrow R \, | f(a) \neq 0 ## for finitely many ## a \in G \} ##.
    For ## f, g \in R(G) ##, define ## (f+g)(a) = f(a) + g(a) ## and ## (f * g)(a) = \sum_{b \in G} f(b)g(b^{-1}a) ##.

    Prove that ## R(G) ## is a ring.
    I'm not even sure if this is true, so I'm posting it here. I proved everything except closure under product, which is giving me some trouble. To show closure under product, I'd need to show

    ## \sum_{b \in G} f(b)g(b^{-1}a) \neq 0 ## is true for finitely many ## a \in G ##. But is this even true??

    Thanks!

    BiP
     
  2. jcsd
  3. Apr 24, 2015 #2

    jbunniii

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    First, let's check that the sum defining the convolution is well defined. Since ##G## is presumably infinite (otherwise the problem is not very interesting), and ##f,g## are not constrained to be nonnegative, the sum ##\sum_{b \in G}## is not well defined unless only finitely many of the terms are nonzero.

    We're given that ##f(b)## is nonzero for only finitely many ##b \in G##. Also, for a fixed value of ##a##, we are also given that ##g(b^{-1}a)## is nonzero for only finitely many ##b \in G##.

    Therefore, for fixed ##a##, the product ##f(b)g(b^{-1}a)## is nonzero for only finitely many ##b \in G##. So the sum ##\sum_{b \in G}f(b)g(b^{-1}a)## makes sense for every ##a \in G##, and the convolution is well defined.

    Let's define ##N(f) = \{x \in G \mid f(x) \neq 0\}## and ##N(g) = \{x \in G \mid g(x) \neq 0\}##. Both sets are finite since ##f,g \in R(G)##.

    Now, for a given ##a \in G##, if ##(f * g)(a) = \sum_{b \in G}f(b)g(b^{-1}a)## is nonzero, then there must be at least one ##b \in G## such that both ##f(b)## and ##g(b^{-1}a)## are nonzero. So ##b \in N(f)## and ##b^{-1}a \in N(g)##. The condition ##b^{-1}a \in N(g)## is equivalent to ##a \in bN(g)##.

    The two conditions ##b\in N(f)## and ##a \in bN(g)## together imply that ##a \in N(f)N(g)##, and the latter is a finite set.
     
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