Anomalous localized resonance using a folded geometry in three dimensions

0.2 MB·English

by  Habib Ammari

Anomalous localized resonance using a folded geometry in
three dimensions
∗
Habib Ammari† Giulio Ciraolo‡ Hyeonbae Kang§ Hyundae Lee§
Graeme W. Milton¶
January 25, 2013
3
1
Abstract
0
2 If a body of dielectric material is coated by a plasmonic structure of negative dielectric
n materialwithnonzerolossparameter,thencloakingbyanomalouslocalizedresonance(CALR)
a may occur as the loss parameter tends to zero. It was proved in [1, 2] that if the coated
J structure is circular (2D) and dielectric constant of theshell is a negative constant (with loss

4 parameter),thenCALRoccurs,andifthecoatedstructureisspherical(3D),thenCALRdoes
2 notoccur. TheaimofthispaperistoshowthattheCALRtakesplaceifthesphericalcoated
structure has a specially designed anisotropic dielectric tensor. The anisotropic dielectric

] tensor is designed by unfolding a folded geometry.
h
p
- 1 Introduction
h
t
a
m If a body of dielectric material (core) is coated by a plasmonic structure of negative dielectric
constant with nonzero loss parameter (shell), then anomalous localized resonance may occur as
[
the lossparametertends tozero. To be precise,letΩ be abounded domaininRd, d=2,3,andD

1 be a domain whose closure is contained in Ω. In other words, D is the core and Ω D is the shell.
v For a given loss parameter δ >0, the permittivity distribution in Rd is given by \
2
1 1 in Rd Ω,
7 \
5 ǫδ =ǫs+iδ in Ω D, (1.1)
\
1. ǫc in D.
0
3 Here ǫc is a positive constant, but ǫs isa negative constant representing the negative dielectric
1 constant of the shell. For a given function f compactly supported in Rd Ω satisfying
\
:
v
i fdx=0 (1.2)
X ZRd
r
(which is required by conservation of charge), we consider the following dielectric problem:
a
ǫ V =f in Rd, (1.3)
δ δ
∇· ∇
∗This work was supported by the ERC Advanced Grant Project MULTIMOD–267184, by Korean Ministry of
Education, Sciences and Technology through NRF grants No. 2010-0004091 and 2010-0017532, and by the NSF
throughgrants DMS-0707978andDMS-1211359.
†Department of Mathematics and Applications, Ecole Normale Sup´erieure, 45 Rue d’Ulm, 75005 Paris, France
([email protected]).
‡Dipartimento di Matematica e Informatica, Universit`a di Palermo Via Archirafi 34, 90123, Palermo, Italy
([email protected]).
§DepartmentofMathematics, InhaUniversity,Incheon 402-751,Korea([email protected], [email protected]).
¶DepartmentofMathematics, UniversityofUtah,SaltLakeCity,UT84112,USA([email protected]).
1
with the decay condition V (x) 0 as x . The equation (1.3) is known as the quasistatic
δ
→ | | → ∞
equationandthe realpartof V (x)e−iωt,whereω is the frequencyandt isthe time, represents
δ
−∇
an approximation for the physical electric field in the vicinity of Ω, when the wavelength of the
electromagnetic radiation is large compared to Ω.
Let
E := ǫ V 2dx= δ V 2dx (1.4)
δ δ δ δ
ℑZRd |∇ | ZΩ\D |∇ |
( for the imaginary part), which, within a factor proportional to the frequency, approximately
ℑ
represents the time averaged electromagnetic power produced by the source dissipated into heat.
Also for any region Υ let
E0(Υ)= V 2dx (1.5)
δ |∇ δ|
ZΥ
which when Υ is outside Ω approximately represents, within a proportionality constant, the time
averagedelectricalenergystoredintheregionΥ. Anomalouslocalizedresonanceisthephenomenon
of field blow-up in a localized region. It may (and may not) occur depending upon the structure
and the location of the source. Quantitatively, it is characterized by E0(Υ) as δ 0 for all
δ →∞ →
regions Υ that overlap the region of anomalous resonance, and this defines that region. Cloaking
due to anomalous localized resonance (CALR) may occur when the support of the source, or part
of it, lies in the anomalously resonant region. Physically the enormous fields in the anomalously
resonant region interact with the source to create a sort of optical molasses, against which the
source has to do a tremendous amount of work to maintain its amplitude, and this work tends to
infinity as δ 0. Quantitatively it is characterized by E as δ .
δ
→ →∞ →∞
This phenomena of anomolous resonance was first discovered by Nicorovici, McPhedran and
Milton [15] and is related to invisibility cloaking [11]: the localized resonant fields created by a
sourcecanactbackonthe sourceandmaskit(assumingthe sourceisnormalizedtoproducefixed
power). It is also related to superlenses [16, 17] since, as shown in [15], the anomalous resonance
can create apparent point sources. For these connections and further developments tied to this
formofinvisibilitycloaking,wereferto[1,2,3,4,10]andreferencestherein. Anomalousresonance
is also presumably responsible for cloaking due to complementary media [8, 18, 14], although we
do not study this here.
The problem of cloaking by anomalous localized resonance (CALR) can be formulated as the
problem of identifying the sources f such that first
E := δ V 2dx as δ 0, (1.6)
δ δ
|∇ | →∞ →
ZΩ\D
and secondly, V /√E goes to zero outside some radius a, as δ 0:
δ δ
→
V (x)/ E 0 as δ 0 when x >a. (1.7)
δ δ
| |→ → | |
p
SincethequantityE isproportionaltotheelectromagneticpowerdissipatedintoheatbythetime
δ
harmonic electrical field averaged over time, (1.6) implies an infinite amount of energy dissipated
per unit time in the limit δ 0 which is unphysical. If we rescale the source f by a factor of
→
1/√E then the source will produce the same power independently of δ and the new associated
δ
potential V /√E will, by (1.7), approach zero outside the radius a. Hence, cloaking due to
δ δ
anomalouslocalizedresonance(CALR) occurs. The normalizedsourceis essentially invisiblefrom
the outside, yet the fields inside are very large.
In the recent papers [1, 2] the authors developed a spectral approach to analyze the CALR
phenomenon. Inparticular,theyshowthatifD andΩareconcentricdisksinR2 ofradiir andr ,
i e
respectively, and ǫ = 1, then there is a critical radius r such that for any source f supported
s ∗
−
outside r CALR does not occur, and for sources f satisfying a mild (gap) condition CALR takes
∗
place. Thecriticalradiusr isgivenbyr = r3/r ifǫ =1,andbyr =r2/r ifǫ =1. Itisalso
∗ ∗ e i c ∗ e i c 6
p
2
provedthatifǫ = 1,thenCALRdoesnotoccur: E isboundedregardlessofδ andthe location
s δ
6 −
of the source. It is worth mentioning that these results (when ǫ = ǫ =1) were extended in [7]
c s
−
to the case when the core D is not radial by a different method based on a variational approach.
There the source f is assumed to be supported on circles.
Thesituationinthreedimensionsiscompletelydifferent. IfD andΩareconcentricballsinR3,
CALR does not occur whatever ǫ and ǫ are, as long as they are constants. We emphasize that
s c
this discrepancycomes fromthe convergencerateofthe singularvalues ofthe Neumann-Poincar´e-
type operator associated with the structure. In 2D, they converge to 0 exponentially fast, but in
3D they converge only at the rate of 1/n. See [2]. The absence of CALR in such coated sphere
geometries is also linked with the absence of perfect plasmon waves: see the appendix in [7]. On
the other hand, in a slab geometry CALR is known to occur in three dimensions with a single
dipolar source [11]. (CALR is also known to occur for the full time-harmonic Maxwell equations
with a single dipolar source outside the slab superlens [6, 11, 19].)
ThepurposeofthispaperistoshowthatweareabletomakeCALRoccurinthreedimensions
by using a shell with a specially designed anisotropic dielectric constant. In fact, let D and Ω be
concentric balls in R3 of radii r and r , and choose r so that r >r . For a givenloss parameter
i e 0 0 e
δ >0, define the dielectric constant ǫ by
δ
I, x >r ,
e
| |
b(b 2x)
ǫδ(x)=ǫ(ǫcs+r0iδI),a−1(cid:18)I+ |−x|2| | xb⊗xb(cid:19), rxi <<|xri|,<re, (1.8)
r | |
where I is the 3 3 identityrmatirix, ǫ and ǫ constants, x= x , and
× s c |x|
r r
a:= e− i >0, b:=(1b+a)r . (1.9)
e
r r
0 e
−
Note that ǫ is anisotropic and variable in the shell. This dielectric constant is obtained by push-
δ
forwarding(unfolding)thatofafoldedgeometryasinFigure1. (Seethenextsectionfordetails.) It
isworthmentioningthatthisideaofafoldedgeometryhasbeenusedin[12]toproveCALRinthe
analogoustwo-dimensionalcylinder structure for a finite set of dipolar sources. Folded geometries
were first introduced in [9] to explain the properties of superlenses, and their unfolding map was
generalized in [12] to allow for three different fields, rather than a single one, in the overlapping
regions. Folded cylinder structures were studied as superlensesin [20] and foldedgeometriesusing
bipolarcoordinateswereintroducedin[5]toobtainnewcomplementarymediacloakingstructures.
More general folded geometries were rigorously investigated in [14].
Figure 1: unfolding map
3
For a given source f supported outside B let V be the solution to
re δ
(ǫ V )=f in R3,
δ δ
∇· ∇ (1.10)
(Vδ(x) 0 as x ,
→ | |→∞
and define
E = ǫ V V dx, (1.11)
δ δ δ δ
ℑZR3 ∇ ·∇
where V is the complex conjugate of V . Let F be the Newtonian potential of the source f, i.e.,
δ δ
F(x):= G(x y)f(y)dy, (1.12)
ZR3 −
with G(x y) = 1 . Since f is supported in R3 B , F is harmonic in x < R for some
− −4π|x−y| \ re | |
R>r and can be expressed there as
e
∞ n
F(x)= fk xnYk(x), (1.13)
n| | n
n=0k=−n
X X
b
where Yk(x) is the (real) spherical harmonic of degree n and order k.
n
The following is the main result of this paper.
b
Theorem 1.1 (i) Ifǫ = ǫ =1,thenweakCALRoccursandthecriticalradiusisr =√r r ,
c s ∗ e 0
−
i.e., if the source function f is supported inside the sphere of radius r (and its Newtonian
∗
potential does not extend harmonically to R3), then
limsupE = , (1.14)
δ
∞
δ→0
and there exists a constant C such that
V (x) <C (1.15)
δ
| |
for all x with x > r2r −1. If, in addition, the Fourier coefficients fk of F satisfy the
| | 0 e n
following gap condition:
[GC1]: There exists a sequence n with n <n < such that
j 1 2
{ } ···
nj
lim ρnj+1−nj n r2nj fk 2 =
j→∞ j ∗ | nj| ∞
k=X−nj
where ρ:=r /r , then CALR occurs, i.e.,
e 0
limE = , (1.16)
δ
δ→0 ∞
and V /√E goes to zero outside the radius r2/r , as implied by (1.15).
δ δ 0 e
(ii) If ǫ = ǫ =1, then weak CALR occurs and the critical radius is r =r . If, in addition,
c s ∗∗ 0
6 −
the Fourier coefficients fk of F satisfy
n
[GC2]: There exists a sequence n with n <n < such that
j 1 2
{ } ···
nj
lim ρ2(nj+1−nj) n r2nj fk 2 = ,
j→∞ j 0 | nj| ∞
k=X−nj
4
then CALR occurs.
(iii) If ǫ =1, then CALR does not occur.
s
− 6
We emphasize that [GC1] and [GC2] are mild conditions on the Fourier coefficients of the
Newtonian potential of the source function. For example, if the source function is a dipole in
B B ,i.e., f(x)=a δ (x)foravectoraandy B B whereδ istheDiracdeltafunction
r∗\ e ·∇ y ∈ r∗\ e y
aty,[GC1]and[GC2]holdandCALRtakesplace. Aproofofthisfactisprovidedintheappendix.
Similarlyonecanshowthatiff isa quadrupole,f(x)=A: δ (x)= 2 a ∂2 δ (x) for
∇∇ y i,j=1 ij∂xi∂xj y
a 3 3 matrix A=(a ) and y B B , then [GC1] and [GC2] hold.
× ij ∈ r∗ \ e P
2 Proof of Theorem 1.1
Let r , r and r be positive constants satisfying r < r < r , as before. For given constants κ ,
i e 0 i e 0 c
κ and κ , and a source f supported outside B , let u , u and u be the functions satisfying
s m re c s m
u =0 in B ,
△ c r0
u =0 in B B ,
△ s r0 \ re
△ucu=mu=s,f κicn∂∂Rurc3\=Bκrse∂,∂urs on ∂Br0, (2.1)
∂u ∂u
s m
u =u , κ =κ on ∂B ,
We emphasize that the domauismns(xo)fm→uc,0ussa,s∂a|rnxd|→um∞ma.r∂eroverlappinrgeon re x r0 so that the
≤ | | ≤
solutions combined may be considered as the solution of the transmission problem with dielectric
constants κ , κ and κ in the folded geometry as shownin Figure 1. We unfold the solutioninto
c s m
one whose domain is not overlapping, following the idea in [12].
In terms of spherical coordinates (r,θ,φ), the unfolding map Φ= Φ ,Φ ,Φ is given by
c s m
{ }
Φ (r,θ,φ)=(r,θ,φ), r r ,
m e
≥
r r
Φ (r,θ,φ)= r e− i(r r ),θ,φ , r r r ,
 s (cid:18)rei− r0−re − e (cid:19) e ≤ ≤ 0 (2.2)
Φ (r,θ,φ)= r,θ,φ , r r .
c 0
r ≤
Then the folding map can be writ(cid:18)ten0(with(cid:19)an abuse of notation) as
x, x >r ,
e
| |
Φ−1(x)= ax+bx, ri < x <re, (2.3)
−r | |
 0x, x <ri,
r | |
i b
where a and b are constants defined in(1.9).
Let κ(x) be the push-forward by the unfolding map Φ, namely,
κ det Φ (y)−1 Φ (y) Φ (y)t, x >r ,
m m m m e
| ∇ | ∇ ∇ | |
κ(x)= κ det Φ (y)−1 Φ (y) Φ (y)t, r < x <r , (2.4)
 s s s s i e
− | ∇ | ∇ ∇ | |
κc det Φc(y)−1 Φc(y) Φc(y)t, x <ri,
| ∇ | ∇ ∇ | |
where x=Φ(y). By straight-forwardcomputations one can see that κ=ǫ given in (1.8) if we set
κ =1, κ = (ǫ +iδ), κ =ǫ . (2.5)
m s s c c
−
5
Moreover,the solution V to (1.10) is given by
δ
u Φ−1(x) if x >r ,
m e
◦ | |
V (x)= u Φ−1(x) if r < x <r , (2.6)
δ  s i e
◦ | |
uc Φ−1(x) if x <ri,
◦ | |
and by the change of variables x=Φ(y) and (2.4), we have
s
E = ǫ(x) V (x) V (x)=δ u (y)2. (2.7)
δ δ δ s
ℑZR3 ∇ ·∇ Zre<|y|<r0|∇ |
Suppose that the source f is supported in x > R for some R > r . Then the solution u to
e
| |
(2.1) can be expressed in x <R as follows:
| |
∞ n
u (x)= ak xnYk(x), if x <r ,
c n| | n | | 0
 n=0k=−n
us(x)=nXX∞=∞0kXX=n−nn(bkn|x|n+cbkn|x|−n−1)Ynk(x), if re <|x|<r0, (2.8)
u (x)= (ek xn+dk x−n−1)Ykb(x), if r < x <R,
where the coefficiemnts satisnXfy=0tkhX=e−fnollonw|in|g relant|ion|s resultningbfrom theeinte|rfa|ce conditions:
akrn =bkrn+ckr−n−1,
n 0 n 0 n 0
ekrn+dkr−n−1 =bkrn+ckr−n−1,
n e n e n e n e
κ aknrn =κ (bknrn ck(n+1)r−n−1),
c n 0 s n 0 − n 0
κ (bknrn ck(n+1)r−n−1)=κ (eknrn dk(n+1)r−n−1).
s n e − n e m n e − n e
By solving this system of linear equations one can see that
ak =a ek, bk =b ek, ck =c ek, dk =d ek,
n n n n n n n n n n n n
where
ρ2n+1(2n+1)2κ κ
m s
a = − , (2.9)
n (n2+n)(κ κ )(κ κ ) ρ2n+1((n+1)κ +nκ )((n+1)κ +nκ )
s c s m s c m s
− − −
ρ2n+1κ (2n+1)((n+1)κ +nκ )
m s c
b = − , (2.10)
n (n2+n)(κ κ )(κ κ ) ρ2n+1((n+1)κ +nκ )((n+1)κ +nκ )
s c s m s c m s
− − −
r2n+1κ n(2n+1)(κ κ )
c = − e m s− c , (2.11)
n (n2+n)(κ κ )(κ κ ) ρ2n+1((n+1)κ +nκ )((n+1)κ +nκ )
s c s m s c m s
− − −
nr2n+1[ρ2n+1(κ κ )((n+1)κ +nκ )+(κ κ )(nκ +(n+1)κ )]
d = e m− s s c s− c m s . (2.12)
n −(n2+n)(κ κ )(κ κ ) ρ2n+1((n+1)κ +nκ )((n+1)κ +nκ )
s c s m s c m s
− − −
Here ρ is defined to be r /r
e 0
Let F be the Newtonian potential of f, as before. Since u F is harmonic in x > r and
e
− | |
tends to 0 as x , we have
| |→∞ ek =fk. (2.13)
n n
So u (the solution in the matrix) is given by
m
∞ n
u (x)=F(x)+ fkd x−n−1Yk(x). (2.14)
m n n| | n
n=0k=−n
X X
b
6
Since d Cr2n, we have
| n|≤ 0
∞ n
u (x) F(x) C fk r2n x−n−1 < (2.15)
| m − |≤ | n| 0 | | ∞
n=0k=−n
X X
if x =r >r2r −1. This proves (1.15).
| | 0 e
The solution in the shell u is given by
s
∞ n
u (y)= fk(b yn+c y−n−1)Yk(y). (2.16)
s n n| | n| | n
n=0k=−n
X X
Using Green’s identity and the orthogonality of Yk, we obtain that b
n
∂u ∂u
u (y)2 = u s u s
s s s
|∇ | ∂r − ∂r
Zre<|y|<r0 Z|y|=r0 Z|y|=re
∞ n
= fk 2 (b rn+c r−n−1)(nb rn (n+1)c r−n−1)r
| n| n 0 n 0 n 0 − n 0 0
n=0k=−n
X X (cid:2) (cid:3)
∞ n
fk 2 (b rn+c r−n−1)(nb rn (n+1)c r−n−1)r
− | n| n e n e n e − n e e
n=0k=−n
X X (cid:2) (cid:3)
∞ n
= fk 2 nb 2(r2n+1 r2n+1) (n+1)c 2(r−2n−1 r−2n−1)
| n| | n| 0 − e − | n| 0 − e
n=0k=−n
X X (cid:2) (cid:3)
∞ n
nfk 2 b 2r2n+1+ c 2r−2n−1 .
≈ | n| | n| 0 | n| e
n=0k=−n
X X (cid:0) (cid:1)
The estimate (2.7) yields
∞ n
E δ nfk 2 b 2r2n+1+ c 2r−2n−1 . (2.17)
δ ≈ | n| | n| 0 | n| e
n=0k=−n
X X (cid:0) (cid:1)
(i) Suppose that ǫ = ǫ =1. With the notation in (2.5), we have
c s
−
(n2+n)(κ κ )(κ κ ) ρ2n+1((n+1)κ +nκ )((n+1)κ +nκ ) n2(δ2+ρ2n+1),
s c s m s c m s
| − − − |≈
and hence
ρ2n δr2n
b , c e . (2.18)
| n|≈ δ2+ρ2n | n|≈ δ2+ρ2n
It then follows from (2.17) that
∞ n δnr2n fk 2
E e | n| . (2.19)
δ ≈ δ2+ρ2n
n=0k=−n
X X
Let
logδ
N = . (2.20)
δ
logρ
If n N , then we have that δ ρ|n|, and hence
δ
≤ ≤
∞ n δnr2n fk 2 n δnr2n fk 2
e | n| e | n|
δ2+ρ2n ≥ δ2+ρ2n
nX=0kX=−n nX≤NδkX=−n
m m 2
δm
δmr2m fk 2 r2m fk
≥ 0 | m| ≥ 2m+1 0 | m|!
k=−m k=−m
X X
7
for any m N . By taking δ to be ρn, n=1,2,..., we see that if the following holds
δ
≤
n
limsup(r r )n/2 fk = , (2.21)
e 0 | n| ∞
n→∞
k=−n
X
then there is a sequence n such that
k
{ }
lim E = , (2.22)
k→∞ ρ|nk| ∞
i.e., weak CALR occurs.
Suppose that the source function f is supported inside the critical radius r∗ = √rer0 (and
outside r ) and its Newtonian potential F cannot be extended harmonically in x <r . Then
e ∗
| |
n 1/n
limsup fk >1/√r r . (2.23)
n→∞ | n|! e 0
k=−n
X
and consequently, (2.21) holds. This proves that if the source function f is supported inside the
sphere of critical radius r , then weak CALR occurs.
∗
If the source function f is supported outside the sphere of critical radius r∗ = √rer0, then its
Newtonian potential F can be extended harmonically in x <r +2η for η >0 and
∗
| |
∞ n δnr2n fk 2 ∞ n
e | n| nr2n fk 2 C F 2 < . (2.24)
δ2+ρ2n ≤ ∗ | n| ≤ k kL2(∂Br∗+η) ∞
n=0k=−n n=0k=−n
X X X X
So E is bounded regardless of δ and CALR does not occur.
δ
Suppose that f is supported inside r and [GC1] holds. Let n be the sequence satisfying
∗ j
{ }
nj
lim ρnj+1−nj n r2nj fk 2 = .
j→∞ j ∗ | nj| ∞
k=X−nj
If δ =ρα for some α, let j(α) be the number in the sequence such that
n α<n .
j(α) j(α)+1
≤
Then, we have
n δnr2n fk 2 n nr2n fk 2
E e | n| ρα e | n|
δ ≈ δ2+ρ2n ≥ ρ2n
nX≤NδkX=−n nX≤NδkX=−n
nj(α)
≥ρ|nj(α)+1|−|nj(α)| nj(α)r∗2nj(α)|fnkj(α)|2 →∞
k=−Xnj(α)
as α . So CALR takes place. This completes the proof of (i).
→∞
To prove (ii) assume that ǫ = ǫ =1. In this case, we have
c s
6 −
ρ2n r2n
b , c e ,
| n|≈ δ+ρ2n | n|≈ δ+ρ2n
and
∞ n δnr2n fk 2
E e | n| .
δ ≈ δ2+ρ4n
n=0k=−n
X X
8
The rest of proof of (ii) is the same as that for (i).
Suppose now that ǫ =1. If ǫ =1, then we have
s c
− 6
ρ2n δr2n
b , c e ,
| n|≈ δ+ρ2n | n|≈ δ+ρ2n
and
∞ n δ(δ2+ρ2n)nr2n fk 2 ∞ n ∂F 2
E e | n| nr2n fk 2 .
δ ≈ (δ+ρ2n)2 ≤ e | n| ≤ ∂r
n=0k=−n n=0k=−n (cid:13) (cid:13)L2(∂Be)
X X X X (cid:13) (cid:13)
(cid:13) (cid:13)
Since the source function f is supported outside the radius re, we have (cid:13) (cid:13)
∂F
∂r ≤CkfkL2(R3),
(cid:13) (cid:13)L2(∂Be)
(cid:13) (cid:13)
(cid:13) (cid:13)
and Eδ is bounded independently o(cid:13)f δ. (cid:13)The case when ǫc =1 can be treated similarly.
6
A Gap property of dipoles
In this appendix we show that the Newtonian potentials of dipole source functions satisfy the gap
conditions [GC1] and [GC2]. We only prove [GC1] since the other can be proved similarly.
Let f be a dipole in B B , i.e., f(x) = a δ (x) for a vector a and y B B . Then
r∗ \ e ·∇ y ∈ r∗ \ e
its Newtonian potential is given by F(x) = a G(x y). It is well-known (see, for example,
y
− ·∇ −
[13]) that the fundamental solution G(x y) admits the following expansion if y > x:
− | | | |
∞ n 1 xn
G(x y)= Yk(x)Yk(y) | | .
− − 2n+1 n n yn+1
n=0k=−n | |
X X
b b
So we have
∞ n
1 1
F(x)= xnYk(x)a Yk(y) ,
2n+1| | n ·∇ yn+1 n
n=0k=−n (cid:18)| | (cid:19)
X X
and hence b b
1 1
fk = a Yk(y) . (A.1)
n 2n+1 ·∇ yn+1 n
(cid:18)| | (cid:19)
We show that
b
n
nr2n fk 2 as n , (A.2)
∗ | n| →∞ →∞
k=−n
X
and hence [GC1] holds. The following lemma is needed.
Lemma A.1 For any a and y on S2 and for any positive integer n there is a homogeneous
harmonic polynomial h of degree n such that
b
a h(y)=1 (A.3)
·∇
and
b
√3
max h(x) . (A.4)
|xb|=1| |≤ n
b
9
Proof. After rotationif necessary,we may assume that y=(1,0,0). We introduce three homoge-
neous harmonic polynomials of degree n:
1 b
h (x):= [(x +ix )n+(x ix )n],
1 1 2 1 2
2n −
1
h (x):= [(x +ix )n (x ix )n],
2 1 2 1 2
2ni − −
1
h (x):= [(x +ix )n (x ix )n].
3 1 3 1 3
2ni − −
Then one can easily see that
h (y)=(1,0,0), h (y)=(0,1,0), h (y)=(0,0,1).
1 2 3
∇ ∇ ∇
So if we define
b h=a hb+a h +a h , b
1 1 2 2 3 3
then (A.3) holds.
Since
1
max h (x) for j =1,2,3,
j
|xb|=1| |≤ n
we obtain (A.4) using the Cauchy-Schwartzinequality. This completes the proof. (cid:3)
b
Let a and y be two unit vectors, and let h be a homogeneous harmonic polynomial of degree
n satisfying (A.3) and (A.4). Then h can be expressed as
b n
h(x)= α xnYk(x),
k| | n
k=−n
X
where b
1
α = h(x)Yk(x)dS. (A.5)
k 4π n
ZS2
Because of (A.3), we have
b b
n
1=a h(y) α a xnYk(x) .
·∇ ≤ | k| ·∇ | | n
k=−n
X (cid:12) (cid:0) (cid:1)(cid:12)
So there is k, say k , between n andbn such that (cid:12) b (cid:12)
n
−
1
α a xnYkn(x) . (A.6)
| kn| ·∇ | | n ≥ 2n+1
On the other hand, from (A.4) and ((cid:12)A.5),(cid:0)it follows by(cid:1)(cid:12)using Jensen’s inequality that
(cid:12) b (cid:12)
1 3
α 2 h(x)2 Ykn(x)2dS .
| kn| ≤ 4π | | | n | ≤ n2
ZS2
Thus we have
b bn
a xnYkn(x) C (A.7)
·∇ | | n ≥ √3(2n+1) ≥
for some C independent of n.(cid:12)(cid:12) (cid:0) b (cid:1)(cid:12)(cid:12)
Now one can see from (A.1) that
C
fkn (A.8)
| n |≥ nyn+1
| |
for some C independent of n. Since y <r , we obtain that
∗
| |
n 2n
C r
nr2n fk 2 nr2n fkn 2 ∗ as n ,
∗ | n| ≥ ∗ | n | ≥ n y →∞ →∞
k=−n (cid:18)| |(cid:19)
X
as desired. It is worth mentioning that the constants C in the above may be different at each
occurrence, but are independent of n.
10