We define Fourier transform on $L^1(\mathbb{R})$ and extend the definition to $L^2(\mathbb{R})$. But for a compact group $G$, we define Fourier transform for $L^2(G)$ and there is no such thing of defining for $L^1(G)$.
My question is why such unusual thing happens for groups? Is this because $L^2(G)$ is Hilbert space? I want an answer from very deep point.
If $G$ is a compact group and $\mu$ is the left-invariant Haar measure on $G$, then $\mu(G)$ is finite, so $L^2(G)\subset L^1(G)$ by Holder's inequality and we can directly define the Fourier transform on $L^2(G)$ since all the integrals in question are defined.
However $L^2(\mathbb{R})$ is not a subset of $L^1(\mathbb{R})$, which is why it is necessary to extend the Fourier transform to $L^2(\mathbb{R})$ using some dense subset.