How to choose $f\in C_{c}^{\infty}(\mathbb R)$ so that $ \hat{g}\in \ell^{1}(\mathbb Z)$, where $g(x)=f(x+2\pi)$?

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Suppose $K$ is compact proper subset of $[0, 2\pi]$ with the property $K\subset V \subset [0, 2\pi]$ where $V$ is open .

My Question: Is it possible to choose $f\in C_{c}^{\infty}(\mathbb R)$ such that $f=1$ on $K$ and $f(x)=0$ for all $x\in [0, \pi ]\setminus V$ with the property that $\hat{g}\in \ell^{1}(\mathbb Z)$ where $g(x):=f(x+2\pi), (x\in \mathbb R)$?

[We note that $g$ is periodic $2\pi-$ periodic on $\mathbb R$ and clearly it is possible to choose $f$ so that $g\in L^{1}(\mathbb T)$. We put $\hat{g}(n)= \int_0^{2\pi}g(t)e^{-int} dt, (n\in \mathbb Z)$, my question is how to choose $f$ so that $\sum_{n\in \mathbb Z}|\hat{g}(n)|<\infty$?]