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

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!

• Have you noticed that the integral is actually a sum? This is because Haar measure is counting measure.
– Ruy
Commented Apr 11, 2021 at 20:18
• Is the group supposed to be abelian? If so, recall that the convolution is commutative and thus $\int_G f(y)\chi_{\{1\}}(y^{-1}x)dy=\int_Gf(y^{-1}x)\chi(y)dy=f(x)$
– user840639
Commented Apr 11, 2021 at 20:21

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.