Function from 1-dim. sphere to real line This is either trivial or wrong: 
Let $f:\mathbb S^1 \to \mathbb R$, where $\mathbb S^1 = [-1/2,1/2]/(-1/2\sim1/2)$ is the one-dimensional sphere, be such that $f\in L^1\big(\mathbb S^1\big)$; does it make sense to define $\tilde f:\mathbb R \to \mathbb R$ such that $\tilde f(x)= f(x)$ if $x\in (-1/2,1/2]$, and $\tilde f(x)= 0$ otherwise, and to say that for any $\kappa\in \mathbb Z$ $$\mathcal F \tilde f(u)|_{u=\kappa}:=\int_{\mathbb R}\tilde f(x)e^{2\pi i \kappa x }dx=\int_{\mathbb S^1} f(x) e^{2\pi i \kappa x }dx=:\mathcal F f(\kappa)?$$
I have some results about Fourier transforms of functions on $\mathbb R$ that I now need for functions on $\mathbb S^1$ and this argument would allow me to avoid reproving them. The definition of $\tilde f$ and the equality above seem to make sense computationally but for some reason my intuition tells me that they do not mathematically.
 A: The manipulation you described is correct, but it usually does not give an easy way to obtain results for the Fourier transform on $S^1$ from the results for the Fourier transform on $\mathbb{R}$. For example, $\tilde f$ is typically discontinuous even if $f$ is continuous, which is an obstacle to relating the smoothness of $f$ to the decay of its Fourier coefficients. 
What happens in such cases is: $\mathcal F\tilde f$ has slow decay at infinity, but dips toward zero at integer points, thus aligning with the quickly decaying Fourier coefficients $\hat f(n)$. A simple illustration is $f\equiv 1$, where all Fourier coefficients except one are zero. Since $\tilde f=\chi_{[-1/2,1/2]}$, its Fourier transform is the sinc function with some normalization: it is pretty heavy-tailed (not in $L^1(\mathbb{R})$) but duly vanishes at the integers. 
A relevant result on the relation between the Fourier transform of a function and the Fourier series of its periodization is the Poisson summation formula. 
