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 Question concerning the Fourier Transform of $\mathbb{R}$
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.


