$f\in M(\mathbb{R})$ but $\hat{f}$ is not I am studying Fourier analysis. I noted some problems state $f,\hat{f}\in M(\mathbb{R})$ as assumption, where $M(\mathbb{R})$ denote the collection of all continuous and of moderate decrease functions on $\mathbb{R}$, which means $$\exists A \in \mathbb{R}\, \text{such that}\, \forall x\in \mathbb{R}, \ |f(x)| \lt \frac{A}{1 + |x|^{1+\epsilon}}$$ 
Hence I think there must be some function in $M(\mathbb{R})$ whose Fourier transform doesn't. Since I only know the Fourier transform for a few functions, may you give me an example? 
 A: $M(\mathbb{R})\subset L^1(\mathbb{R})$. This post shows the existence of a continuous function $f$ with compact support (hence $f\in M(\mathbb{R})$) with $\hat f\notin L^1(\mathbb{R})$, and hence $\hat f\notin M(\mathbb{R})$.
An explicit example of such an $f$ is
$$
f(x)=\bigl(1-\log(1-x^2)\bigr)^{-\alpha}\cdot\mathbb{1}_{[-1,1]},\quad0<\alpha\le1.
$$
According to the answer of T. Tao to a question of mine, the behavior at infinity of $\hat f$ is like
$$
\frac{1}{|\xi|(\log|\xi|)^\alpha}.
$$
A: The Schwartz Space is mapped onto itself by the Fourier Transform.  This is a big source of examples of $f$ and $\hat{f}$ both being in $\mathcal{M}$.  
For an example where this fails, look at the function $f(x) = 1$ if $x \in [0,2\pi]$ and $0$ otherwise.
A: The smoother the function $f$ is, the smaller is $\hat f$  at $\infty$. Take a discontinuous function $f$ like the characteristic function of the interval $[-1,1]$, for which
$$
\hat f(\xi)=\text{some constant}\cdot\frac{\sin \xi}{\xi}.
$$
Note (added after ZHANG's comment): The example above is not continuous, so it is not an answer to the OP question.
