Prove that if locally compact group $G$ is discrete, then the group algebra $L^1(G)$ is unital.

Question

I am trying to prove that if $G$ is discrete, then the group algebra $L^1(G)$ is unital.

If $G$ is discrete, $\{1\}$ is an open set where 1 is the identity element of $G$. I am trying to show that the indicator function at $\{1\}$, $\chi_{\{1\}}$ is the unit element of $L^1(G)$, but I run into a problem. The convolution operation is for $f\in L^1$ and $\chi_{1}$ is given by:

$\int_G f(y)\chi_{\{1\}}(y^{-1}x)\,dy$. Replacing $y$ with $x$ in the above integral, we get: $\int_G f(x)\chi_{\{1\}}(x^{-1}x)\,dx=\int_G f(x)\,dx$, because $\chi_{{\1\}}(x^{-1}x)=\chi_{\{1\}})=1$.

So now we have 'isolated' f, but how could we possibly get from $\int_G f(x)\,dx$ to $f$, as needed? Clearly the integral of $f$ is not equal to $f$ in general.

Do we have that somewhere along the line, the integral is no longer over $G$ but over only $\{1\}$, because indicator function goes zero elsewhere? Then integrating over one element gives you the function itself back?

Any help is greatly appreciated. Thanks!

Answer 1

The convolution of $f$ and $\chi_{\{1\}}$, applied to $x$, is given by: $$ (f\star \chi _1)(x) = \int_G f(y)\chi_{\{1\}}(y^{-1}x)\,dy. \tag 1 $$

You cannot replace $y$ with $x$ in this integral, because $x$ and $y$ have very distinct roles. That would be the same as replacing $i$ by $n$ in the formula $$ \sum_{i=1}^n i = \frac{n(n+1)}2, $$ which I am sure you would never do!

Observing that Haar measure is actually counting measure, (1) becomes $$ (f\star \chi_{\{1\}})(x) = \sum_{y\in G} f(y)\chi_{\{1\}}(y^{-1}x) = $$$$ = f(x)\chi_{\{1\}}(x^{-1}x) = f(x), $$ so $f\star \chi _1 = f$, and a similar proof applies to give $\chi _1\star f = f$, as well.