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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?

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    $\begingroup$ Did you try to solve the problem for symmetric rectangular functions supported in $[-1,1]$? They are not smooth, but you know the Fourier transforms (a sinc function) explicitly. $\endgroup$ Commented Nov 12, 2010 at 9:04
  • $\begingroup$ How should that help? $\endgroup$
    – Marc Palm
    Commented Nov 14, 2010 at 12:52
  • $\begingroup$ @Marcel and Florian: I modified the question a little bit to force an analytic construction. I would definitely prefer a kernel transformantion, since the operator is needed for a construction. $\endgroup$
    – Marc Palm
    Commented Nov 14, 2010 at 13:01
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    $\begingroup$ @Marc, the idea above was just that it allows you to make (maybe) explicit calculations. Because I didn't made these, I cannot say if there is any hope of helping solving the problem, but the heuristic idea was you can approx. any function in $V$ by this step functions. I would be interested for what kind of construction. $\endgroup$ Commented Nov 15, 2010 at 11:27
  • $\begingroup$ It is a question in the context of the explicit formula of Andr\'{e} Weil. $\endgroup$
    – Marc Palm
    Commented Nov 17, 2010 at 8:51

3 Answers 3

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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.

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  • $\begingroup$ Do I understand you correctly? You choose a smooth compactly supported function $\Theta$, which is equal to $1$ on $[-1,1]$. Why do you say $D$ is the standard operator? Do you have a good reference for this argument? I have to check the details, but if it works like that this ... amazing construction ...Where can I study further details for this! $\endgroup$
    – Marc Palm
    Commented Nov 14, 2010 at 15:15
  • $\begingroup$ Sorry, I meant of course $\Theta =\Phi$. $\endgroup$
    – Marc Palm
    Commented Nov 14, 2010 at 15:17
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    $\begingroup$ Also you require that $0\le\Phi\le 1$ everywhere (this is what I meant by "descent") to ensure contractivity. The formulae are taken directly from Havin and Joricke's (I wish the Uppsala aministration to be kicked out of their jobs the next year the same way they kicked out her...) book "The Uncertainty Principle in Harmonic Analysis" but you do not need to read the book to see how it works (one case is $Pf=f$ and the other is $Qf=f$). $\endgroup$
    – fedja
    Commented Nov 14, 2010 at 15:57
  • $\begingroup$ I understand. We assume $f \in V$, then we have $Pf = f$, hence (1-P)f=0 and Df =f. On the other hand if $g = \mathcal{F}^{-1} f$, then we have $Qg=g$, which gives with geometric series expansion $ Dg = g - \sum\limits_{k=0}^{\infty} (PQ)^{k} (g -Pg) = \sum\limits_{k=0}^{\infty} (PQ)^{k} (g -PQg) = 0$. Hence it works formally, but on which page in this book I find the analysis for that. $\endgroup$
    – Marc Palm
    Commented Nov 14, 2010 at 16:15
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    $\begingroup$ Chapter 3 "Hilbert space methods"; pp. 87-102 are most relevant. $\endgroup$
    – fedja
    Commented Nov 14, 2010 at 17:00
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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 $V+\mathcal{F}V$.

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    $\begingroup$ Why is this operator bounded ? This is he same as saying that the angle between any two unit vectors $u\in\mathcal{F}V$ and $v\in V$ is bounded away from $0$ and $\pi$. Is this clear ? $\endgroup$
    – BS.
    Commented Nov 14, 2010 at 13:27
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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 $V\cap \mathcal{F}V=0$.

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