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Let $G$ be a compact group and $C(G)$ be the set of continuous functions on $G$. Define the convolution of two functions $f_1,f_2 \in C(G)$ as $$ f_1 * f_2 := \int_{G} f_{1}(gh^{-1})f_2(h) dh$$ then with respect to above convolution $C(G)$ forms a ring.

Claim : $C(G)$ contains a unit element iff $G$ is finite.

My effort: I could prove the reverse direction by defining the unit as

$$ \delta_{e}(g)= \begin{cases} 1 & g=e\\ 0 & \text{otherwise} \end{cases} $$ Then it is easy to see that $f*\delta_{e}= f = \delta_{e}*f$.

For the forward direction (by contradiction): Suppose $G$ is a compact group and $\exists $ a unit $\delta$ in $C(G)$ such that $f*\delta= f = \delta*f$. Then, we would like to show that $\delta(g) =0 $ $\forall g \in G$. So, assume that $\exists$ $x \in G$ such that $\delta(x) \neq 0$. Since $\delta \in C(G)$ i.e. it is a continuous function implies that $\exists$ a symmetric neighbourhood $U_{x}$ of $x$ such that $\delta $ takes non-zero values on $U_{x}$. Now, by definition of convolution, we can write $$ |(f* \delta) (x) - \int_{U_{x}} f(xh^{-1}) \delta(h) dh| $$ After this step I am not understanding how to go further for showing that $\delta$ is identically zero. Thanks in advance! any help would be appreciated!

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(I will assume $G$ is Hausdorff.)

You can't just show $\delta$ is identically zero because that is false in finite groups! But you can show the following in any compact group: if $\delta$ is a unit element then $\delta(x)=0$ for all $x \ne e$. Since zero is not the unit element, we must have $\delta(e) \ne 0$, meaning $e$ is an isolated point, meaning $G$ is discrete, meaning (by compactness) that $G$ is finite.

To show the claim, take $x \ne e$ and suppose without loss of generality $\delta(x) > 0$. Using continuity, choose $\epsilon > 0$ and a neighborhood $V$ of $e$ such that $\delta(xy^{-1}) > \epsilon$ for all $y \in V$. Using Urysohn's lemma, produce a continuous nonnegative function $f$ supported inside $V \setminus \{x\}$ with $\int f(y)\,dy = 1$. Now verify that $(\delta \ast f)(x) \ge \epsilon \ne 0 = f(x)$ so that $\delta$ is not a unit.

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    Thank you! There was a small oversight on my part: I forgot to mention that the functions I am considering are complex-valued. Can you tell me how we can modify the proof of the claim to account for this (i.e. the $\delta(x) > 0$ part)? – 2017-02-21
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    @ShubhamNamdeo: Well, if you assume $\operatorname{Re} \delta(x) > 0$ then you get $\operatorname{Re} (\delta \ast f)(x) \ge \epsilon$. (Note that $f$ is real-valued.) So $\operatorname{Re} \delta(x) = 0$. The same logic applies if you look at imaginary parts instead. – 2017-02-21