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Question:
Suppose $E,F$ subsets of $\Bbb R^n$ have finite measure. Show that for any $g_1,g_2 \in L^2(\Bbb R^n)$ there exists $f \in L^2(\Bbb R^n)$ with $f=g_1$ on $E$ and $\mathscr{F}(f)=g_2$ on $F$,where $\mathscr{F}(f)$ is the Fourier transform of $f$.
My attempt:
This is a problem from Schlag's book Classical and multilinear harmonic analysis, chapter 10 problem 10.3. I try to prove it by contradiction and Amrein-Berthier thm but failed.
Any solution or hint is highly appreciated.

A-B thm: Let E,F be sets of finite measure in $\Bbb R^n$.Then $ \left\|f\right\|_{L^2(\Bbb R^n)} \leq C( \left\|f\right\|_{L^2(E^c)}+ \left\|{F}(f)\right\|_{L^2(F^c)} )$ for some $C=C(E,F,n)$

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    $\begingroup$ I looked up your 'Amrein-Berthier Theorem' but found it difficult to find the statement. So if you could at least state the statement or provide some link... $\endgroup$
    – Brozovic
    Dec 21, 2020 at 9:51
  • $\begingroup$ Here's a hint, admittedly with a lot to fill in: if $V$ is the subspace of $f \in L^2$ where $f = 0$ on $E$ and $W$ is the subspace of $f \in L^2$ where $\mathscr{F}(f) = 0$ on $F$, and $P_V$ and $P_W$ denote the projections onto these subspaces, use the fact that the norm of $P_V P_W$ is less than $1$ to show that $V + W = L^2$. $\endgroup$
    – Jason
    Dec 26, 2020 at 22:43
  • $\begingroup$ This has been posted on MO, with the notice "The question has been posted here but had no response." $\endgroup$ Dec 29, 2020 at 5:12
  • $\begingroup$ @Jason Thanks for the hint.Can you explain why the norm is less than 1 is able to ensure V+W=L2 $\endgroup$
    – mathdogcmf
    Dec 29, 2020 at 11:01
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    $\begingroup$ @mathdogcmf Ack, I meant to write that $P_{V^{\perp}} P_{W^{\perp}}$ has norm less than $1$, not $P_V P_W$, apologies. My thinking was to do a series expansion based on the fact that $I = P_{W} + P_V P_{W^{\perp}} + P_{V^{\perp}} P_{W^{\perp}}$. $\endgroup$
    – Jason
    Dec 29, 2020 at 20:26

1 Answer 1

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Let $V$ be the closed subspace of $L^2(\mathbb{R}^n)$ consisting of those functions that vanish almost everywhere on $E$, let $W \subseteq L^2$ be the closed subspace of functions whose Fourier transforms vanish almost everywhere on $F$, and let $V^{\perp}$ and $W^{\perp}$ be their orthogonal complements.

It suffices to show that there exists $f \in L^2$ with $f = g_1$ on $E$ and $\mathscr{F}(f) = 0$ on $F$. Or, put another way, we'd like to show that an arbitrary coset of $V$ intersects $W$.

Now, if we knew that $V + W = L^2$, the result would follow from the (purely algebraic) isomorphism $(V + W) / V \cong W / (V \cap W)$. To see this, note that the natural isomorphism is given by $h + (V \cap W) \mapsto h + V$. So given an arbitrary coset $g + V$ ($g \in L^2 = V + W$), we could find a $\tilde{g} \in W$ so that $\tilde{g} + V = g + V$. In particular, $\tilde{g} \in g + V$, so this function demonstrates that the coset $g + V$ intersects $W$.

To finish, we show that $V + W = L^2$. We rely on the fact that the $L^2 \to L^2$ operator norm of $P_{V^{\perp}} P_{W^{\perp}}$ is strictly less than $1$, where $P_{\cdot}$ represents the orthogonal projection onto the relevant subspace. (This norm statement is connected to the Amrein-Berthier theorem, and follows from Theorem 10.4 and Lemma 10.5 in Muscalu and Schlag's book.) Using the fact that $I = P_W + P_V P_{W^{\perp}} + P_{V^{\perp}} P_{W^{\perp}}$, we can write an arbitrary $f \in L^2$ as $f = g_1 + h_1 + P_{V^{\perp}} P_{W^{\perp}} f$, where $g_1 \in W$ and $h_1 \in V$. Iterating this expansion on the remainder term and using the norm bound, we obtain convergent series $\sum_j g_j$ in $W$ and $\sum_j h_j$ in $V$ satisfying $f = (\sum_j g_j) + (\sum_j h_j)$. This completes the proof.

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