A Question concerning the Fourier Transform of $\mathbb{R}$ Consider the classical Schwartz space $\mathcal{S}(\mathbb{R})$ together with the Fourier transform $\mathcal{F} : \mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}( \mathbb{R})$.
Consider the subspace $V$ of the even, smooth functions on the interval $[-1,1]$.
Can you construct a (bounded) operator $D:\mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}(\mathbb{R}) $ such that
$$ D \mathcal{F} v = 0, \quad Dv=v \qquad\forall v \in V ?$$
Observe that by Paley-Wiener, the intersection $\mathcal{F}V \cap V =0$ is trivial.
What is the associated Schwartz kernel?
 A: You don't need to do things the rough way; there is enough freedom for the smooth approach.
Take any even $C_0^\infty$ descent $\Phi$ from $[-1,1]$ and define $Pf=\Phi f$ and $Qf=\mathcal F^{-1}(\Phi\mathcal F f)$. Now take the standard $D=I-(I-PQ)^{-1}(1-P)$. This works in $L^2$ for the same reason as it does with orthogonal projections: $PQ$ is a contraction. The good news is that $PQ$ maps $L^2$ to $S$ continuously and $(I-PQ)^{-1}=I+PQ(I-PQ)^{-1}$, so the resulting operator is bounded from $S$ to $S$ as well. 
The kernel can be "found" by expanding $D$ into power series that converges geometrically but, since this construction involves an arbitrary smooth cutoff, to write an explicit formula seems quite hopeless.
A: I don't think that there is a continuous linear operator for the Schwartz space, but if you replace $\mathcal{S}$ by $L^2$ and define $V$ accordingly, then $V$ is a closed subspace of $L^2$. This implies that $\mathcal{F} V$ is closed, and thus $V+\mathcal{F}V\to L^2$, $v+\tilde v\mapsto v$ is a bounded operator (in the $L^2$-sense) and hence it can be extended to $L^2$, for example, by setting it to zero on the orthogonal complement of $V+\mathcal{F}V$.
A: Q1: I prefer taking $\mathcal{L}^2(\mathbb R)$. You have at least an 
(unbounded) operator with domain $V+\mathcal FV$ given by
$$D: V+ \mathcal{F}V \to \mathcal{L}^2(\mathbb R): x+y \mapsto x$$
which is well defined because $V\cap \mathcal{F}V=0$. 
