The truncated Fourier operator. I

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by  Victor Katsnelson

9 The trun
ated Fourier operator. I
0
0
2
Vi
tor Katsnelson Ronny Ma
hluf

n
a
J

6
1
F Abstra
t E
]
A Let be the one dimensional Fourier-Plan
herel operator and
C be a subsFet of the real aFxis.=TPheFtPrun
ated F(oPurixer)(to)p=eraχtor(ti)sx(tth)e
E E E E E E
h. operaχtEo(rt) of the form , wherEe ,
t and is the indi
ator fun
tion of the set . In thFe presented (cid:28)rst
a E
part of the work, the basi
properties of the operator a

ording to
m E
the set are dis
ussed.
[
F Among these properties there are the following one. The operator
1 E
: 1. has a not-trivial null-spa
e; 2. is stri
tly
ontra
tive; 3. is a
v
normal operator; 4. is a Hilbert-S
hmidt operator; 5. is a tra
e
lass
5
5 operator.
5
2
.
1 1 The trun
ated Fourier operator:
0
9
de(cid:28)nition, basi
properties
0
: E R E = R
v
Let be a measurable subset of the real axis . (The
ase is
i p 1 Lp(E)
X not ex
luded). For ≥ , let be the spa
e of
omplex valued
E x(t)pdt < .
r fun
tions on satisfying the
ondition | | ∞ We mainly
a E
p = 2 R p = 1
deal with the
ase , but episodi
the
ase is needed. The
L2(E)
spa
e , providedbythestandardlinearoperationsandthes
alar
x, y
E
produ
t h i :
x,y = x(t)y(t)dt, x,y L2(E) ,
E
h i ∈ (1.1)
Z
E (cid:0) (cid:1)
Mathemati
s Subje
t Classi(cid:28)
ation: (2000). Primary 47A38; Se
ondary 47B35,
47B06,47A10.
Keywords: Trun
ated Fourier-Plan
herel operator, normal operator,
ontra
tive opera-
tor, Hilbert-S
hmidt operator, tra
e
lass operator.
1
L2(E)
is a Hilbert spa
e. The norm in is
x = x,x .
E E
k k h i (1.2)
F p
The Fourier operator is de(cid:28)ned by the formula
1
( x)(t) = eitξx(ξ)dξ, (t R).
F √2π ∈ (1.3)
ZR
One of
entral fa
ts of the Fourier transform theory is the Parseval
equality: (Fx) 2R = x 2R ( x L2(R)).
k k k k ∀ ∈ (1.4)
F
TL2h(iRs )means that the Fourier operator is an isometri
operator in
. The next
entFral fa
t of the FourLie2r(Rtr)ansform theory is that
tLh2e(RF)ourier operatoFr maps the spa
e Lon2(tRo)the whole spa
e
, that is thFe 1 is an unitary operator in . Moreover, the
−
inverse operator is determined by the formula
1
(F 1x)(t) = e itξx(ξ)dξ, (t R).
− −
√2π ∈ (1.5)
ZR
Remark1.1. The integral in the rixght sLid1e(Rh)and oxf (1.L3)1(iRsa) LeLbe2s(gRu)e
integral. It is well de(cid:28)ned only if ∈ . If ∈ ∩ ,
then both the integral is weFll de(cid:28)ned and the Parseval equality (1.4)
holdxs. TLh1u(sR,)theLo2p(eRr)ator
an be de(cid:28)nxed originally Lby2(R(1).3) only
for ∈ F∩ . The set of su
h is dense in . Sin
e
the operator a
ts isometri
ally Lon2(Rth)is set, it
an be extended by
the
ontinuity on the whole spa
e . The same is related to the
operator whi
h appears in (1.5).
In this paper we deal with the trun
ated Fourier operator.
E 0 <
mD(eE(cid:28)n)ition.1.1. Let beFa m: eLa2s(uEra)ble Lsu2b(sEe)t of the real axis,
E
≤ ∞ The operator → , is de(cid:28)ned as
1
(F x)(t) = eitξx(ξ)dξ, (t E).
E
√2π ∈ (1.6)
Z
E
F x
E
The operator is said to be the trun
ated Fourier operator, or in
more detail, the Fourier operator trun
ated on the set E.
2
E dt <
Remark 1.2. If the set is a set of (cid:28)nite Lebesgue measure: E
L2(E) L1(E)
∞, then ⊂ and the integral in (1.6) is well de(cid:28)Rned for
x L2(E)
every ∈ .
F F
The operator E∗, whi
h is the adjoint operator to E with respe
t
to the s
alar produ
t (1.1), is
1
( x)(t) = e itξx(ξ)dξ, (t E).
FE∗ √2π − ∈ (1.7)
Z
E
F L2(E)
E
Remark 1.3. The operatoPr FP, a
ting in L2(R,)may be naturally
E E
identi(cid:28)ed with the operator , a
ting in , where
(P x)(t) = χ (t)x(t),
E E
(1.8)
1, t E,
χ (t) = ∈
E
(0, t / E. (1.9)
∈
E F
E
Lemma 1.1. For any , the operator is a
ontra
tive operator in
L2(E)
: F x 2 x 2 ( x L2(E)).
k E kE ≤ k kE ∀ ∈ (1.10)
Proof. Indeed, if
1
y(t)= eitξx(ξ)dξ, (t R),
√2π ∈ (1.11)
ZE
y(t)2 = x(t)2 y(t)2
thenbytheParsevalequality, R| | E| | ,hen
e E| | ≤
x(t)2
E| | . R R R
R
Thus, the inequalities hold
0 F F I 0 FF I .
≤ E∗ E ≤ E and ≤ ∗ ≤ E (1.12)
I
E
Here and in what follows is the identity operator.
Theorem 1E.1>. 0 F = 0 F = 0
1. If mes , then no-one of the operators E and E∗
x L2(E)
equals zero: there exists ∈ for whi
h both
F x= 0 F x= 0.
E 6 and E∗ 6 (1.13)
(R E) > 0 F F
2. If mes \ , then no-one of the operators E and E∗ is
x L2(E)
isometri
: there exists ∈ for whi
h both
F x < x F x < x .
k E k k k and k E∗ k k k (1.14)
3
t E
0
Proof. Let ∈ is su
h that
m E [t 1/n, t +1/n]
0 0
lim ∪ − = 1.
n m [t 1/n, t +1/n]
→∞ (cid:0) 0− 0 (cid:1)
(cid:0) (cid:1)
t E
0
(Almost every point ∈ possesses this property). Let us set
1 1, ξ E [t 1/n, t +1/n]
0 0
x (ξ) = ∈ ∩ − .
n
m [t0 1/n, t0+1/n] (0, ξ E [t0 1/n, t0+1/n]
− ∈ \ −
(cid:0) (cid:1)
x L2(E) n
n
It is
lear that ∈ ∀ , and that
x (ξ)e itξdξ e itt0 n , R.
n ± ±
→ as → ∞ the limit is lo
ally uniform on
Z
E
n Fx )(t)2dt > 0, (F x )(t)2dt > 0
n ∗ n
If islargeenough,then | | | | .
R E R E
x R\ E x (t)2dt =R\ x (t)2dt
n n n
Be
ause vanishes outside of ,E | | R | | . By the
Fx )(t)2dt =R (F x )(t)R2dt = x (t)2dt
n ∗ n n
Parseval equality, R | | R | | E | | .
R Rx= x n R
n
Thus, the inequalities (1.14) hold for if is large enough.
E
It is
lear that if the set is bounded, then then the inequalities
x L2(E), x = 0 x
(1.13), (1.14) hold for any ∈ 6 . Indeed, given ∈
L2(E), x= 0 y x
6 ,let isdetermined from a

ording to(1.11). Sin
ethe
E y(t) t
set isbounded, isanentirefun
tion of . Therefore thefun
tion
y(t) y(t)2 >
may vanish only in isolated points. In parti
ular, | |
E
0, y(t)2 > 0 FxR > 0
| | . The (cid:28)rst inequality means that k k , the
R E
R\ Fx < x
se
ond one (cid:22) that k k k k. In [AmBe, Proposition 5℄ it was
E < E
shown that if mes ∞ (the set may be unbounded), then the
x L2(E), x= 0.
inequalities (1.14) holds for arbitrary ∈ 6
A
tually, the mu
h more stronger statement takes pla
e.
E <
Theorem 1.2. If mes ∞, then the inequalities
F x 2 1 A 1e A( E)2 x 2, F x 2 1 A 1e A( E)2 x 2
k E k ≤ − − − mes k k k E∗ k ≤ − − − mes k k
(cid:0) (cid:1) (cid:0) (1.15(cid:1))
x L2(E) A, 1 A <
hold for every ∈ . Here ≤ ∞, is an absolute
x E

onstant: it does not depend neither on , nor on .
4
Proof. Infa
t,Theorem1.2isaspe
ial
aseoftheNazarov un
ertainty
prin
iple. In [Naz℄, F.L. Nazarov prove the remarkable inequality
y(t)2dt AeA( E)( F) y(t)2dt+ x(ξ)2dξ
mes mes
| | ≤ | | | |
ZR (cid:0)RZE RZF (cid:1)
\ \
x L2(R) y (1.16)
where ∈ isanarbitraryfun
tions, istheFouriertransformof
x y(t)= 1 eitξx(ξ)dξ E F
: √2π R , and arearbitrary measurablesubsets
R F = ER x E
of . If and is an arbitrary fun
tion vanishing outside of ,
then the inequality (1.16) takes the form
y(t)2dt AeA( E)2 y(t)2dt y(t)2dt .
mes
| | ≤ | | − | |
ZR ZR ZE
(cid:0) (cid:1)
Invoking the Parseval identity, we rewrite this inequality in the form
x(t)2dt AeA( E)2 x(t)2dt y(t)2dt
mes
| | ≤ | | − | |
ZE ZE ZE
(cid:0) (cid:1)
The latter inequality
oin
ides with the (cid:28)rst of the inequalities
(1.15).
RS
ehmmiadrtkno1r.m4. kAFsEkitS2ishsotladtse:dkbFeElokwS,2 t=hemeeqsuEal.itySinfo
re tthhee HHiillbbeerrtt--
S
hmidt norm majorizes the operator norm, the inequalities hold
F x ( E) x , F x ( E) x ,
k E k ≤ mes k k k E∗ k ≤ mes k k (1.17)
E x
where is an arbitrary measurable set and is an arbitrary fun
tion
L2(E)
from .
Bothinequalities(1.15)and(1.17)aretrue. However(1.17)ismore
E
pre
ise for small values of mes , and (1.15) (cid:21) for large ones.
E
Remark 1.5. If the set is not just of a set of (cid:28)nite measure, but a
(cid:28)nite interval, the the estimate (1.15)
an be re(cid:28)ned for large values
E E = [ l,l] λ (l)
0
Fof mFes . For − , the largest eigenvaFlues2 Eof=the[ olp,le]rator
E∗ E
oiFn
iFdes with the squares of norm kL2E([k l.,lF])or − , the
operator E∗ E is the integral operator in − of the form
1 sinl(t τ)
FE∗FEx (t) = π t −τ x(τ)dτ . (1.18)
Z −
(cid:0) (cid:1) [ l.l]
−
5
l λ (l)
0
The asymptoti
behavior as → ∞ of the eigenvalue of the
integral operator (1.18) was found by W.H.J.Fu
hs in [Fu℄:
1 λ (l) 4√πl1/2e 2l (= 2√2π( E)1/2e E.
0 − −mes
− ≈ mes
E = [ l,l]
Thus, for − , the estimate holds whi
h is strongest then the
estimate (1.15):
F x (1 A(ε)e (1+ε) ) x , F x (1 A(ε)e (1+ε) ) x ,
k E k ≤ − − mesE k k k E∗ k ≤ − − mesE k k
(1.19)
x L2(E) ε > 0 A(ε) <
for every ∈ . Here is arbitrary, and ∞ for any
ε> 0 A(ε) l x
. The value does not depend on and .
E
Theorem 1.1
laims thatFifFtheFsetF is bounded, then the null-
spa
es of ea
h of operators E∗ E, E∗ E∗ are trivial(cid:22)they
onsist of
zero-ve
tor only.
E <
Theorem 1.2 imI plieFs thFat iIf meFs F ∞, then the null-spa
es of
ea
h of operators E − E∗ E, − E E∗ are trivial.
E =
Thefollowingexampleshowsthatifmes ∞,thenea
hofthese
null-spa
es
anbenotonlynon-trivial,butevenanin(cid:28)nitedimensional
one.
K R
Example 1.1. Let ⊂ be the interval:
π
K = [ a,a], 0 < a< .
− where 2 (1.20)
r
E
The set is a "periodi
" systems of intervals:
E = K +p√2π .
(1.21)
p Z
[∈ (cid:0) (cid:1)
u(t) 0 R
Let 6≡ be a (smooth) fun
tion on su
h that
u K.
supp ⊆ (1.22)
u(t)
The fun
tion is representable in the form
∞
u(t) = eitξv(ξ)dξ,
(1.23)
Z
−∞
6
v(ξ)
where is a fast de
aying fun
tion. Let
y(t) = c u(t+p√2π),
p
(1.24)
p Z
X∈
cp p Z
where { } ∈ be a summable sequen
e. From (1.21)-(1.24) it follows
that
y E.
supp ⊆ (1.25)
Moreover
∞
y(t)= eitξv(ξ)ϕ(ξ)dξ,
(1.26)
Z
−∞
where ϕ(ξ) = c eipξ√2π, < ξ < .
p
−∞ ∞ (1.27)
p Z
X∈
ϕ
The fun
tion is a periodi
one:
ϕ(ξ +√2π) ϕ(ξ), < ξ < .
≡ −∞ ∞ (1.28)
u(t)
Let us invert the order of reasoning. Starting from a fun
tion
K √2π ϕ(ξ)
supported on , (1.22), and -periodi
fun
tion , (1.28), we
y(t) v(ξ) u
de(cid:28)ne the fun
tion by (1.26), where is determined from by
cp p Z
(1.23). Then the equality (1.24) holds, where { } ∈ is the sequen
e
ϕ
of the Fourier
oe(cid:30)
ient by the originally given fun
tion :(1.27). Let
√2π ϕ 0
a periodi
fun
tion 6≡ satisfy the
ondition
ϕ [ √π/2, √π/2] K,
supp ∩ − − ⊆ (1.29)
K
where is the same as before. Then
v(ξ)ϕ(ξ) E.
supp ⊆ (1.30)
ϕ cp p Z
If moreover the fun
tion is smooth, then the sequen
e { } ∈ ,
y(t)
(1.27), is summable. Thus the fun
tion is representable in the
x(ξ) = √2πv(ξ)ϕ(ξ) x E, y E
form (1.11), where , supp ⊆ supp ⊆ ,
therefore
y(t)2 = y(t)2 = x(ξ)2dξ,
| | | | | |
EZ ZR EZ
i.e. F x 2 = x 2.
E
k k k k (1.31)
7
u(t) ϕ(ξ)
Bxe∈
aLus2e(Eof) the freedom in the
hoi
e of and , the set of
L2(sEat)isfying the
ondition (1.31) is an in(cid:28)nite dimensional sub-
spa
e ofx (ξ) =.x(ξ)e ihξ h R
1 −
Let , where ∈ , and
1
y (t) = eitξx (ξ)dξ, t R,
1 1
√2π ∈
ZE
y (t) = y(t h), y = h+ y a < 1 π h
Then 1 − supp 1 supp . If 2 2,then
an
(E +h) E = . y (t) = 0 t E
1
be
hosen su
h that ∩ ∅ In this
ase, p ∀ ∈ ,
thus F x = 0.
E 1
(1.32)
x1 ∈ L2(E)
As before, the set of L2(Esa)tisfying the
ondition (1.32) is an
in(cid:28)nite dimensional subspa
e of .
In this example, both
(E) = , (R E) = .
mes ∞ mes \ ∞ (1.33)
(cid:3)
E
Remark 1.6. In [AmBe, P(RropoEsi)tio<n 6℄ it was shown thatxif aLs2e(tE)
satis(cid:28)es the
ondition mes \ ∞, then the set of ∈
satisfying the equality (1.31) is an in(cid:28)nite dimensional subspa
e of
L2(E)
.
A
We re
all that the operator a
ting in a Hilbert spa
e is said to
be normal if
A A= AA ,
∗ ∗
A A
∗
where is the operator adjoint to the operator .
Here and further
E = t R : t E
− { ∈ − ∈ } (1.34)
.
F
E
Lemma 1.2. The trun
ated Fourier operator is normal if and only
if the equality
y(t)2dt = y(t)2dt,
| | | | (1.35)
Z Z
E ( E) ( E) E
\ − − \
y(t) y(t) = 1 eitξx(ξ)dξ t R
holds for every of the form √2π E , ∈ , where
x L2(E)
R
runs over the whole spa
e .
8
F F = F F
Proof. The
oFndxitio2n= E∗FEx 2 E E∗ is equivalexnt toLt2h(eE)
onditxion:
tLh2e(Eeq)uality k(FExk)(t) =ky(E∗t),kt hoElds for e(vFeryx)(t)∈= y( t)., tIf E∈
, then E F x 2 = ∈F x, 2and E∗ −y(t)2∈dt =.
Thus, the equality k E k k E∗ k takes the form | |
E
y( t)2dt R
| − | . The latter equality is equivalent to the equality (1.35).
E
R
E
De(cid:28)nition 1.2. The set is said to be symmetri
if
∆(E, E) = 0,
mes − (1.36)
∆(E, E) E E
where − is the symmetri
di(cid:27)eren
e of the sets and − :
∆(E, E) = (E ( E)) (( E) E).
− \ − ∪ − \
(E ( E)) (( E) E) = (E ( E)) = (( E)
Sin
e \ − ∩ − \ ∅,andmes \ − mes − \
E)
, the
ondition (1.36)
an be expressed in asymmetri
form:
∆(E, E) = 0.
mes −
E F
E
Theorem 1.3. If the set is symmetri
, then the operator is a
normal operator.
Proof. The theorem is an evident
onsequen
e of Lemma 1.2: the ex-
pressions in both sides of (1.35) are equal be
ause both of them van-
ish.
F E
E
Question 1.1. Let the operator be normal. Is the set symmet-
ri
?
We
an not answer this question in full generality. However, under
E
someextra
ondition imposed on the set the answer to thisquestion
is a(cid:30)rmativeS., S R
The set ⊂ , is said to be bounded, bounded from below,
S
and bounded from above respe
tively, if is
ontained respe
tively
[a,b] [a,+ )
in some bounded interval , bounded from above interval ∞
( ,b] a,b
or bounded from below intearv<al b−∞ , whSe,reS Rare some (cid:28)nite
numbers. (In the (cid:28)rst
ase, .) The set ⊂ , is said to be
S
semi-bounded, if is either bounded from above, or is bounded from
below. (In parti
ular, every bounded set is semi-bounded).
9
Theorem 1.4. AssFume that the following two
onditions are satis(cid:28)ed:
E
1. The operator is normal;
E ( E)
2. The set \ − is semi-bounded.
E
Then the set is symmetri
.
E, E R (E) > 0,
Lemma 1.3S.,LSet R ⊂ be a set of positive measure: mes
and the set ⊂ , is semi-bounded. Then the set of all fun
tions
y(t)= 1 eitξx(ξ)dξ, t S x L2(E)
of the form √2π E ∈ , where runs over ,
L2(S)
R
is dense in .
S
Proof. Assume for de(cid:28)niteness that the set is bounded from above,
S ( ,b] b < y
say ⊆ −∞ , where ∞. If the set of all su
h is not dense in
L2(S) v L2(S), v = 0, v(t)y(t)dt = 0
, then there exists ∈ 6 su
h that
S
y(t) v(t)e itξdt = 0R ξ E
−
for all . From this follow that ∀ ∈ . Sin
e
S
S ( ,b] f(ξ) = Reibξ v(t)e itξdt, ξ R
−
⊆ −∞ , the fun
tion ∈ , belongs
S
H2 v L2(SR) f
to the Hardy
lass +. Sin
e ∈ is non-zero, is a non-zero
H2 f(ξ) = 0 ξ E
fun
tion from +. Moreover, for ∈ . However, the non-
zerofun
tionfromtheHardy
lass
annotvanishonthesetofpositive
measure.
Remark 1.7.
S
Proof of Theorem 1.4. We show that if the set is not symmetri
,
(E ( E)) > 0
that is if mes \ − , then the
ondition (1.35) is violated for
y(t) = 1 eitξx(ξ)dξ x L2(E)
some √2π E , where ∈ . Then by Lemma
F
1.2, the operator RE is not normal.
E ( E)
We(cid:28)rstpresenttheproofassumingthattheset \ − isbounded.
E ( E)
If the set \ − is bounded, then the set
S d=ef (E ( E)) (( E) E)
\ − ∪ − \ (1.37)
g(t)
is bounded as well. We de(cid:28)ne the fun
tion as
g(t) = 1, t (E ( E)), g(t) = 0, t (( E) E).
∈ \ − ∈ − \
(E ( E)) (( E) E)
Sin
ethesets \ − and − \ donotinterse
t,thede(cid:28)nition
g
of the fun
tion is not
ontradi
tory. A

ording to Lemma, 1.3, for
10