Prove that if $f(x) \sim \sum c_k e^{ikx}$, then $f(x+t) \sim \sum c_k e^{ikt} e^{ikx}$.
Replacing the instance of $x$ with $x + t$, we have that $$f(x + t) \sim \sum c_k e^{ik(x+t)} = \sum c_k e^{ikx + ikt} = \sum c_k e^{ikt} e^{ikx},$$ but this seems too simple. Is there something I'm missing?
On
Your claim that we can find the Fourier series for $f_t(x):=f(x+t)$ (which, for each fixed $t$, is a function of $x$, and thus we can talk about its Fourier series) by plugging in $x+t$ in the Fourier series for $f(x)$ is false. The Fourier series for $f$ does not necessarily converge to $f$, if it even converges at all, without additional hypotheses. That is, we don't know that $f(x)=\sum c_k e^{ikx}$ for any $x\in S^1$, so replacing $x$ by $x+t$ on both sides of $f_t(x)\sim\sum c_k e^{ikt}e^{ikx}$ doesn't make sense. The symbol "$\sim$" is not equality. It means that the function on the left has the Fourier series on the right.
However, the statement $f_t(x)\sim\sum c_k e^{ikt}e^{ikx}$ is still true. By hypothesis, we are given $$ f(x)\sim\sum c_k e^{ikx}, $$ which means by definition that for all $k\in\mathbb{Z}$, $$ c_k=\int_{S^1}f(x)e^{-ikx}dx. $$ Then the Fourier coefficients $c_k'$ for $f_t$ are gotten by computing \begin{align*} c_k' &= \int_{S^1}f_t(x)e^{-ikx}dx \\ &= \int_{S^1}f(x+t)e^{-ikx}dx \\ &= \int_{S^1}f(y)e^{-ik(y-t)}dy \\ &= e^{ikt}\int_{S^1}f(y)e^{-iky}dy \\ &= e^{ikt}c_k \end{align*} using the change of variables $y=x+t$. Thus, \begin{align*} f(x+t)=f_t(x)&\sim \sum c_k'e^{ikx} \\ &= \sum c_k e^{ikt}e^{ikx} \end{align*} since $c_k'=e^{ikt}c_k$.
Well, it depends on what $\sim$ means here. Ususally it just means that the $c_k$ are the Fourier coefficients of $f$. It does not necessarily mean that the series converges by any means, so your approach is not correct in these cases. But if, e.g., $f\in L^2(0,2\pi)$, then it is correct because the series converges in terms of the $L^2$ norm and the operator of translation by $t$ (which you apply here) is bounded (even unitary). But if the convergence of the series is not guaranteed, you will have to show that the $k$th Fouriercoefficient of $f(\cdot + t)$ equals $e^{ikt}c_k$, where $c_k$ is the $k$th Fourier coefficient of $f$. And this is done via a simple substitution in the integral defining the Fourier coefficient.