Let $C$ is a compact subset of $\mathbb R,$ $V\subset \mathbb R,$ and $0<m(V)<\infty,$ where $m$ is a Lebsgue measure on $\mathbb R.$
My Question is: Can we expect to find $k\in \mathcal{S(\mathbb R)}$ (=Schwartz space) such that (a) $\hat{k}(x)=1$ for all $x \in C,$ $\hat{k}(x)=0,$ outside $C+V\setminus V,$ and $0\leq \hat{k}(x)\leq 1, $ for all $x\in \mathbb R.$ (b)What can we say about $\|k\|_{L^{1}(\mathbb R)}$ ?
Motivation:
Fact. Let $C$ is a compact subset of $\mathbb R,$ $V\subset \mathbb R,$ and $0<m(V)<\infty,$ where $m$ is a Lebsgue measure on $\mathbb R.$ Then there exists $k\in L^{1}(\mathbb R)$ such that (a) $\hat{k}(x)=1$ for all $x \in C,$ $\hat{k}(x)=0,$ outside $C+V\setminus V,$ and $0\leq \hat{k}(x)\leq 1, $ for all $x\in \mathbb R.$ (b) $\|k\|_{L^{1}(\mathbb R)}\leq \{m(C\setminus V)/m(V)\}^{\frac{1}{2}}.$
Proof. Let $g$ and $h$ be the functions in $L^{2}(\mathbb R)$ whose Plancherel Fourier transforms are the characteristic functions of $V$ and $C\setminus V$ respectively, and define
$$k(y):=\frac{g(y) h(y)}{m(v)}, (y\in \mathbb R).$$
Then $\hat{k}= m(V)^{-1} (\hat{g}\ast \hat{h}),$ or
$\hat{k}(x)=\frac{1}{m(V)} \int_{V} \hat{h}(x-y) dy, (x\in \mathbb R).$ If $x\in C, $ then $\hat{h}(x-y)=1, $ for all $y\in V,$ hence $\hat{k}(x)= 1.$ If $x\notin C+V\setminus V,$ then
$\hat{h}(x-y)=0$ for all $y\in V.$ This proves (a); and (b) follows from Schwartz inequality.
Thanks,
The Fourier transform ${\cal S}(\mathbb{R})\to {\cal S}(\mathbb{R})$ is a linear isometry (Fourier inversion). If $C\subseteq W$ where $C$ is compact and $W$ is open, then we can find a smooth function $g$ such that $g\equiv 1$ on $C$ and $g\equiv 0$ outside $W$ (why?). Let us take $W=\{x\in\mathbb{R}: d(x,C)<1\}$ for concreteness and observe now that $g$ is evidently Schwartz. Therefore, there exists $f\in {\cal S}(\mathbb{R})$ such that $\widehat{f}=g$ (you can take $f(x)=\widehat{g}(-x)$ for $x\in\mathbb{R}$; again this is simply Fourier inversion).
Hope this helps!