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Let $G$ be a compact group. Let $C(G)$ denote the set of all continuous functions $G\to \mathbb{C}$ and let $\mu$ denote the normed Haar measure on $G$. Convolution on $G$, $*:C(G)\times C(G)\to C(G)$, is defined by $$(f*g)(a)=\int_G f(ab^{-1})g(b)d\mu (b),\quad f,g,\in C(G).$$

I know that $C(G)$ with convolution is a Banach algebra.

Allegedly, this algebra is unital iff G is finite. How does one prove this?

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up vote 2 down vote accepted

One direction is easy, so suppose the algebra has a unit $\delta$. We'll show that $\delta(a) = 0$ for any $a \ne e$ (where $e \in G$ is the identity of the group). If $G$ was infinite and hence not discrete, this would imply $\delta(e) = 0$, which can't happen.

If $a \ne e$, suppose $\delta(a) = z \ne 0$; then we can pick a symmetric neighbourhood $U$ of $e$ such that $\lvert \delta(x) - z \rvert < \lvert z \rvert/2$ for $x \in aU$. Then by picking, say, a positive real-valued function $f \in C(G)$ supported in $U$ with $f(a) = 0$, it should be straightfoward to show that $(\delta * f)(a)$ is nonzero, contradicting $\delta * f = f$. Hint: look at $$\left\lvert (\delta * f)(a) - \int_U zf(b) \,d\mu(b) \right\rvert$$

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I get it! I am new to topology, especially to topological groups, and this problem was a mountain to me! Thank you so much! – sonjcy Dec 9 '12 at 10:42

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