A function with support of finite measure and "small" spectrum

0.07 MB·

by  Fedor Nazarov

A FUNCTION WITH SUPPORT OF FINITE MEASURE
AND “SMALL” SPECTRUM
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1
FEDOR NAZAROV, ALEXANDER OLEVSKII
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2
n Abstract. We construct a function on R supported on a set of
a finite measure whose spectrum has density zero.
J

1
3

] 1. The result
A
C Let F be a function in L2(R). We say that it is supported on S if
h. F = 0 almost everywhere on R\S.
t
a Suppose the set S ⊂ R is of finite Lebesgue measure. Then the Fourier
m
transform F of F is a continuous function, so the spectrum of F is nat-
[
urally defined as the closure of the set where F takes non-zero values.
1 b
v According to the uncertianty principle, the support and the spec-
b
4 trum of a (non-trivial) function F cannot be both “small sets” . This
1
principle has various versions (see e.g. [HAJ94]).
1
9 In particular, the classic uniqueness theorem for analytic functions
0 implies that if F is supported on an interval and it has a “spectral gap”
.
1 (that is, F = 0 on an interval ) then F = 0.
0
7 AnotherimportantresultsaysthatifthesupportS andthespectrum
b
1 Q of F are both of finite measure then F = 0 [Ben74/85], [AB77].
:
v On the other hand, F may have a support of finite measure and a
i
X spectral gap; see [Kr82], where such an example was constructed with
r F = 1S.
a
Answering a question posed by Benedicks, Kargaev and Volberg
[KV92] constructed an example of a function F such that
|S| < +∞,|R\Q| = +∞
(here and below by |A| we denote the Lebesgue measure of the set A).
The goal of this note is to prove the following
Theorem. There is a function F ∈ L2(R) supported by a set S of
finite measure, such that
|Q∩(−R,R)| = o(R) as R → ∞.
In addition, F can be chosen as the indicator function of S.
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2 FEDOR NAZAROV,ALEXANDEROLEVSKII
The proof below is based on a simple construction, completely dif-
ferent from the ones in the cited papers.
2. Proof
2.1. Take a Schwartz function F such that
0
0 6 F (t) 6 1 (t ∈ R)
0
and its Fourier transform F is positive on (−1,1) and vanishes outside
0
that interval. Define a sequence of functions F recursively by
b n
F := F +G (n = 1,2,...),
n n−1 n
where
(1) G (t) := F (t)[1−F (t)]cosk t
n n−1 n−1 n
We are going to prove that if the numbers k grow sufficiently fast,
n
then the sequence F converges to a function F satisfying the require-
n
ments of the theorem.
2.2. Clearly, F and G are Schwartz functions.
n n
A simple induction shows that for every t ∈ R, we have
|G (t)| 6 max{F (t),1−F (t)}
n n−1 n−1
and
0 6 F (t) 6 1.
n
TheFouriertransformsofF [1−F ],F2 [1−F ],andF2 [1−
n−1 n−1 n−1 n−1 n−1
F ]2 vanish outside a compact interval, so for each n > 1, we have:
n−1
G = F G = 0
Z n Z n−1 n
R R
and
1
G2 = F2 [1−F ]2,
Z n 2 Z n−1 n−1
R R
provided that k is chosen sufficiently large. It follows that
n
(2) F = F =: C
Z n Z 0
R R
and, thereby,
I := F (1−F ) 6 C
n Z n n
R
(here, as usual, by C we denote a positive constant that may vary from
line to line).
Observe also that
I = [F +G ][1−F −G ] = I − G2,
n Z n−1 n n−1 n n−1 Z n
R R
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which implies that
G2 ≤ I −I 6 C,
Z n 0 N
X R
n∈[1,N]
and so
(3) G2 6 C
Z n
X R
n
2.3. Define the sequence Q of intervals on (another copy of) R re-
n
cursively as follows:
Q := [−1,1],
0
Q := conv(Q ∪[k +2Q ]∪[−k +2Q ])
n n−1 n n−1 n n−1
(here convE denotes the convex hull of a set E ⊂ R). Clearly, for
every n,
specF ⊂ Q ;
n−1 n−1
specG ⊂ [k +2Q ]∪[−k +2Q ].
n n n−1 n n−1
Set Q := Q ∪ ([k +2Q ]∪[−k +2Q ]).
0 n n n−1 n n−1
Choosing k Sgrowing sufficiently fast we can ensure that the spectra
n
of G are parewise disjoint and
n
(4) |Q∩(−R,R)| = o(R) as R → ∞.
2.4. Consider the series F +G +G +... Since the spectra of the
0 1 2
terms are pairwise disjoint, this series is orthogonal in L2(R). Then (3)
implies that it converges in L2(R) to some non-trivial function F. The
partial sums of this series are F . Take a subsequence F such that
n nℓ
F → F almost everywhere on R as ℓ → ∞.
nℓ
Recall that all F are non-negative functions, so (2) implies that
n
(5) F > 0 almost everywhere and F < ∞.
Z
R
It follows from (1) and (3) that
[F (1−F )]2 = 2 G2 < +∞,
Z n n Z n
X R X R
n n
so we must have
F(1−F) = lim F (1−F ) = 0 almost everywhere,
ℓ→∞ nℓ nℓ
4 FEDOR NAZAROV,ALEXANDEROLEVSKII
which implies that F is the indicator-function of a set S. According to
(5), this set has finite measure. Clearly the spectrum of F is a subset
of Q. Due to (4) it has density zero. This finishes the proof.
Remark. Consider the function
h(R) := |Q∩(−R,R)|.
In the conditions of the Theorem, it can not be bounded. However the
proof above shows that it may increase arbitrarily slowly. It remains
an open question, however, if Q can have uniform density 0, i.e., if it
is possible that
1
lim sup |Q∩(x−R,x+R)| = 0.
R→∞x∈R 2R
References
[AB77] Amrein,W.O.,Bertier,A.M. Onsupport properties of Lp-functions and
their Fourier transfroms. J.Funct.Anal. 24 (1977), 258-267.
[Ben74/85] Benedicks M., On Fourier transforms of functions supported on sets
of finite Lebesgue measure. - Royal Institute of Technology, Stockholm
(1974), preprint; - J.of Math.Anal.and Appl. 106 (1985), 180-183.
[HAJ94] Havin, V.P., J¨orickeB., The Uncertainty Principle in Harmonic Anal-
ysis, Springer-Verlag, Berlin, Heidelberg, 1994.
[Kr82] Kargaev, P.P. The Fourier transform of the characteristic function of
a set vanishing on an interval (Russian), Mat. Sb. (N.S.) 117 (1982),
397411.English translation in Math. USSR-Sb 45 (1983), 397-411.
[KV92] Kargaev,P.P., Volberg A.L., Three results concerning the support of
functions and their Fourier transforms, Indiana Univ. Math. J. 41
(1992), 1143-1164.