Suppose $f, \hat{f} \in L^{p}(\mathbb R) \cap C(\mathbb R)\cap L^{\infty}(\mathbb R), (1<p<\infty).$
My Question: Can we expect $\lim_{|x|\to \infty} |f(x)|=0$ ? (In other words, If $f$ and its Fourier transform $\hat{f}$ both are in $L^{p} $ space and continuous , does it mean that $f$ vanishes at infinity)
[We note that for $p=1,$ the result follows from Riemann Lebsgue lemma, and inversion formula]
Edit: If needed, additionally, we assume $f$ is in $A(\mathbb T),$ that is, if we restrict to $f$ to finite interval, say $[0, 2\pi] \subset \mathbb R,$ then $\hat{f} \in \ell^{1}(\mathbb Z),$ that is, $\sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty.$
Edit again: We assume that $f\in \mathcal{F}L^{1}(\mathbb R)$ means that $\phi f \in A(\mathbb T)$ for all $\phi \in C^{\infty}(\mathbb R)$ with $\phi $ is compactly supported in $[0, 2\pi)$ (or in general case any interval I of length $2\pi$ ). With this assumption can we expect $|f(x)|\to 0 $ as $|x|\to \infty$?
No, this doesn't follow. Fix a smooth bump function $\varphi$ supported in $[-1,1]$ and let $$ f(x) = \sum_{n=1}^\infty \phi(2^n x-n) $$ The $n$th term is supported in $[n-2^{-n}, n+2^{-n}]$; these supports are disjoint. The series converges in every $L^p$ space. The function $f$ does not tend to zero at infinity, through.
On the Fourier side, $$ \hat f(x) = \sum_{n=1}^\infty 2^{-n}\hat \phi(2^{-n} \xi)e^{i \text{(something)}} $$ (The last term is unimodular, so I don't care what it is.) The $n$th term has $L^p$ norm $$2^{-np}2^n \|\hat \phi\|_p$$ so the series converges in every $L^p$ space with $p>1$. It also converges uniformly, so the sum is continuous.
The function $f$ in this example is $C^\infty$, and therefore belongs to any reasonable local function space that doesn't require analyticity.