On some Riesz and Carleman type inequalities for harmonic functions on the unit disk

0.1 MB·

by  David Kalaj

1
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR
HARMONIC FUNCTIONS ON THE UNIT DISK
7 DAVIDKALAJANDELVERBAJRAMI
1
0
2 ABSTRACT. We prove some isoperimetric type inequalities for real harmonic
functionsintheunitdiskbelongingtotheHardyspacehp,p> 1andforcom-
n plex harmonic functions in h4. The results extend some recent results on the
a
area.FurtherwediscussomeRiesztyperesultsforholomorphicfunctions.
J

2
1

] 1. INTRODUCTION AND STATEMENT OF MAIN RESULTS
V
C Throughout the paper we let U = z C : z < 1 be the open unit disk in
. the complex plane C, and let T = z{ C∈: z |=|1 be}the unit circle in C. The
th normalizedareameasureonUwill{be∈denote|db|ydσ}. Intermsofreal(rectangular
a
andpolar)coordinates, wehave
m
1 1
[ dσ = dxdy = rdrdθ, z = x+iy = reiθ.

π π
1
v For 0 < p < + let Lp(U,σ) = Lp denote the familiar Lebesgue space on U
9 withrespecttoth∞emeasureσ. TheBergmanspaceAp(U) = Ap isthespaceofall
2 holomorphic functions in Lp(U,σ). For a fixed 1 p < + , denote by Ap the
34 setofallfunctions f Ap forwhichf(0) = 0. Fo≤rafunctio∞n f inLp(U,σ)0, we
∈
0 write
. 1/p
1
f = f(z)pdσ .
0 p
k k U| |
7 (cid:18)Z (cid:19)
1 Bergmanspacebp ofharmonicfunctions isdefinedsimilarly.
: TheHardy space hp isdefined asthe space of (complex) harmonic functions f
v
i suchthat
X 2π dt
r f hp := sup ( f(reit)p )1/p < .
a k k 0≤r<1 Z0 | | 2π ∞
Iff hp,thenby[1,Theorem6.13],thereexists
∈
f(eit) = limf(reit),a.e.
r 1
→
andf Lp(T).Itcanbeshownthattherehold
∈
2π dt 2π dt
f p = lim f(reit)p = f(eit)p .
k khp r→1Z0 | | 2π Z0 | | 2π
SimilarlywedefinetheHardyspaceHp ofholomorphic functions.
File: kalbaj.tex, printed: 2017-1-13, 1.24
12010MathematicsSubjectClassification:Primary47B35
Keywordsandphrases. Harmonicfunctions,Riesinequality,Isoperimetricinequality.
1
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 2
The starting point of this paper is the well known isoperimetric inequality for
JordandomainsandisoperimetricinequalityforminimalsurfacesduetoCarleman
[3]. In that paper Carleman, among the other results proved that if u is harmonic
andsmoothinUthen
1 2π
e2udxdy ( eudt)2.
ZU ≤ 4π Z0
By using a similar approach as Carleman, Strebel ([13]) proved the isoperimetric
inequality forholomorphic functions; thatisiff H1(U)then
∈
1
(1.1) f(z)2dxdy ( f(eit)dt)2.
U| | ≤ 4π T| |
Z Z
Thisinequality has been proved independently by Mateljevic´ and Pavlovic´ ([11]).
In[6]havebeendonesomegeneralizations forthespace.
Inthispaperweprovethefollowingresults.
Theorem1.1. Letf bearealharmonic function. Forp > 1wehavethat
(1.2) f b2p Cp f hp
k k ≤ k k
where
cos π
4p, if1 < p 2;
C = M = cos2πp ≤
p 2p  cos4πp, ifp 2.
 sin2πp ≥
The inequality (1.2) extends the main result in [2], where Bajrami proved the

sameresultbutonlyforp =4. Butthewrongconstantappearin[2],duetoawrong
citation to the Duren approach [5, p. 67-68]. Howeverthe same method produces
the same constant namely C = 1 2.56292. We were not able to check
4 2sin1π6 ≈
if (1.2) is sharp. We are thankful to professor Hedenmalm who drawn attention
to his paper [7], where is treated a problem which suggests that the inequalities
from Theorem 2.1, which we use in order to prove (1.2), are maybe not sharp.
The inequality (1.2) improves similar results for real harmonic functions proved
byKalajandMesˇtrovic´ in[10]andbyChenandPonnusamyandWangin[4]. We
expect that the inequality (1.2) is true for complex harmonic mappings for every
p > 1. KalajandMesˇtrovic´ in[10]proved itforp = 2,namelytheyobtained that
C = 1 1.3. Onthe same paper the example f (z) = z , when a 1
2 2sinπ8 ≈ a ℜ1−az ↑
produces the constant C = (5/2)1/4 1.257, so the constant C is not far from
0 2
≈
thesharpconstant. Inthispaperweextenditforp = 4.
Namelywehave
Theorem 1.2. Iff h4(U) isanonzero complex harmonic function thenf b8
∈ ∈
andthereholdtheinequality
1
kfkb8 ≤ 2sin π kfkh4.
16
The main ingredient of the proofs are some sharp M. Riesz type inequalities
provedforHardyspacebyVerbitsky([14]):
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 3
Proposition 1.3 (Riesz type inequality). Letp > 1. For every holomorphic map-
pingthereholdthesharpinequalities
Lp f hp f hp Rp f hp,
kℜ k ≤ k k ≤ kℜ k
provided argf(0) π π ,orf(0) = 0,wherep¯= max p, p and
| − 2| ≥ 2p¯ { p 1}
−
1 , if1 < p 2; 1 , if1< p 2;
R = cos2πp ≤ andL = sin2πp ≤
p 1 , ifp 2. p 1 , ifp 2.
( sin2πp ≥ ( cos2πp ≥
Inadditionwerefertothereferences [8]and[12]forrelated results.
By results of Verbitsky we prove some similar inequalities for Bergman space
(Theorem2.1).
FromProposition1.3,weobtainthat,iff = u+ivisholomorphicwithf(0) =
0,then
p
R
(1.3) f p p( u p + v p )
k kHp ≤ 2 k khp k khp
and
p
L
(1.4) p( u p + v p ) f p .
2 k khp k khp ≤ k kHp
Now we formulate a similar result, where Rpp and Lpp are replaced by smaller,
2 2
respectively biggerconstants forpcloseto4.
Theorem1.4. Letp 2andletf = u+iv beaholomorphicfunctionontheunit
≥
disksothatf(0)= 0andletf Hp. Thenforp [2,4]wehave
∈ ∈
1/p
p
(1.5) kfkHp ≤ p 1 kukphp +kvkphp 1/p
(cid:18) − (cid:19)
(cid:0) (cid:1)
andforp 4wehave
≥
1/p
p
(1.6) kfkHp ≥ p 1 kukphp +kvkphp 1/p.
(cid:18) − (cid:19)
(cid:0) (cid:1)
1/p
Remark1.5. IfC = p ,then2 1/pL C ifp 4andC 2 1/pR ,
p p 1 − p ≤ p ≥ p ≤ − p
−
if p1 p 4, so ine(cid:16)qualit(cid:17)ies (1.5) and (1.6) improve the inequalities (1.3) and
≤ ≤
(1.4), if p p 4, and if p 4, respectively. Here p 2.42484 is the only
1 1
≤ ≤ ≥ ≈
solutionoftheequation
1/p
p 1
= 2 1/p , p 2.
p 1 − sin π ≥
(cid:18) − (cid:19) 2p
Theproofsarepresented insections2and3and4.
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 4
2. PROOF OF THEOREM 1.1
FromProposition 1.3weobtain
Lp f(reit)pdt f(reit)pdt Rp f(reit)pdt.
p T|ℜ | ≤ T| | ≤ p T|ℜ |
Z Z Z
So
1 1
Lp rdr f(reit)pdt r f(reit)pdt
pZ0 ZT|ℜ | ≤ Z0 ZT | |
1
Rp rdr f(reit)pdt.
≤ pZ0 ZT|ℜ |
Thuswehavethefollowingtheorem
Theorem2.1. Letp > 2andf = f +i f bp(U). If argf(0) π π ,or
ℜ ℑ ∈ | − 2|≥ 2p
f(0)= 0,thenwehave
Lp f bp f bp Rp f bp
kℜ k ≤ k k ≤ kℜ k
and
Lp f bp f bp Rp f bp.
kℑ k ≤ k k ≤ kℑ k
Fromnowonwewillusetheshorthand notation
1 2π
f := f(eit)dt
ZT 2π Z0
and
1
f := f(z)dxdy, z = x+iy.
U π U
Z Z
Thusforp > 2wehave
1
( f p)1/p ( f p)1/p
U|ℜ | ≤ Lp U| |
Z Z
1
( f p/2)2/p
≤ Lp T| |
Z
R
p/2( f p/2)2/p
≤ Lp T|ℜ |
Z
= M ( f p/2)2/p,
p
T|ℜ |
Z
where
cos π
2p, if2 < p 4;
M = cosπp ≤
p  cos2πp, ifp 4.
 sinπp ≥
ThisfinishestheproofofTheorem1.1.

ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 5
3. PROOF OF THEOREM 1.2
Weassumethatf(z)= g(z)+h(z),whereh(0) = 0,andgandhareholomor-
phicontheunitdisk.
Lemma3.1. Thefunction a 2+ b 2islog-subharmonic, providedthataandbare
| | | |
analytic.
Proof. Weneed toshowthat f(z) = log(a 2 + b 2)issubharmonic. Bycalcula-
| | | |
tionwefind
aa¯+b¯b
′ ′
f =
z a 2+ b 2
| | | |
andso
aa¯ +b¯b aa¯+b¯b aa¯ +b¯b
′ ′ ′ ′ ′ ′ ′ ′
f =
zz¯ a 2+ b 2 − a 2+ b 2 a 2+ b 2
| | | | | | | | | | | |
(a 2+ b 2)(a 2 + b 2) a¯a +¯bb 2
′ ′ ′ ′
= | | | | | | | | −| |
(a 2+ b 2)2
| | | |
whichisclearlypositive.
(cid:3)
Now from the isoperimetric inequality for log-subharmonic functions (e.g. [9,
Lemma2.2]),wehave
Lemma 3.2. For every positive number p and analytic functions a and b defined
ontheunitdiskU wehavethat
2
(a 2 + b 2)2p (a 2 + b 2)p .
U | | | | ≤ T | | | |
Z (cid:18)Z (cid:19)
Furtherlet
L = (g+¯h2)4 = (g 2 + h2+2 (gh))4.
U | | U | | | | ℜ
Z Z
Then
4
4
L = (g 2 + h2)k(2 (gh))4 k
−
k U | | | | ℜ
k=0(cid:18) (cid:19)Z
X
4
4
((g 2 + h2)4)k/4( 2 (gh)4)(4 k)/4.
−
≤ k U | | | | U| ℜ |
k=0(cid:18) (cid:19)Z Z
X
LetE = cos π. FromLemma3.2andTheorem2.1wehave
4 8
4
4
L E4 k( (g 2 + h2)4)k/4( (2gh)4)(4 k)/4
≤ k 4− U | | | | U | | −
k=0(cid:18) (cid:19) Z Z
X
4
4
E4 k( (g 2 + h2)4/2)2k/4( (2gh)4/2)2(4 k)/4.
≤ k 4− T | | | | T | | −
k=0(cid:18) (cid:19) Z Z
X
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 6
Furtherwehave
X := g+h¯ 4
T| |
Z
= (g 2 + h2+2 (gh))2
T | | | | ℜ
Z
= (g 2 + h2)2+4(g 2 + h2) (gh)2 +4( (gh))2
T | | | | | | | | ℜ ℜ
Z
= (g 2 + h2)2+4(g 2 + h2) (gh)+2gh2.
T | | | | | | | | ℜ | |
Z
Let
1/2
A= (g 2 + h2)2
T | | | |
(cid:18)Z (cid:19)
and
1/2
B = 4g 2 h2 .
T | | | |
(cid:18)Z (cid:19)
Then
(g 2 + h2) (gh) ( ((g 2 + h2))2)1/2( gh2/2)1/2
T | | | | ℜ ≤ | T | | | | T| |
(cid:12)Z (cid:12) Z Z
(cid:12)(cid:12) (cid:12)(cid:12) √2
(cid:12) (cid:12) = AB .
· 4
Furtherwehave
B2 √2
X A2+ √2AB = (A B)2.
≥ 2 − − 2
Furthermore
A2 B2 = (g 2 h2)2 0
− T | | −| | ≥
Z
andthus
2 √2
(3.1) X ( − )2B2
≥ 2
andsimilarly
2
2 √2
(3.2) X − A2.
≥ 2
!
Hence
4 4 2
4 2
(g+¯h)8 E4 k g+h¯ 4
U | | ≤ k 4− 2 √2 T| |
Z Xk=0(cid:18) (cid:19) (cid:18) − (cid:19) (cid:18)Z (cid:19)
4 2
2(1+E )
= 4 g+h¯ 4 .
2 √2 T| |
(cid:18) − (cid:19) (cid:18)Z (cid:19)
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 7
So
2(1+E )
kg+h¯kb8 ≤ s 2 √24 kg+h¯kh4,
−
whereE = cos π = √2+√2. Finallywehavethat
4 8 2
1
(cid:2) (cid:3) kg+h¯kb8 ≤ 2sin π kg+h¯kh4.
16
Here 1 2.56292. ThisfinishestheproofofTheorem1.2.
2sin1π6 ≈
4. PROOF OF THEOREM 1.4
WeusethefollowingformofGreenformula
2π ∂G(reit)
(4.1) r dt = ∆G(z)dxdy.
∂r
Z0 Z|z|≤r
Let p > 2 and let f = u+iv be an analytic function. Let q = p and ǫ > 0
p 1
anddefine −
F (z) = qǫ+ f(z)2 p/2
ǫ
| | | |
U (z) = ǫ+u(z)2 p/2
ǫ
| |
V (z) = ǫ+v(z)2 p/2
ǫ
| |
Thenbydirectcalculation weobtain
(4.2) ∆F = p(qǫ+ f 2)p/2 2(2qǫ+p f 2)f 2,
ǫ − ′
| | | | | |
(4.3) ∆U = p(u2+ǫ)p/2 2 f 2(ǫ+(p 1)u2),
ǫ − ′
| | −
and
(4.4) ∆V = p(v2+ǫ)p/2 2 f 2(ǫ+(p 1)v2).
ǫ − ′
| | −
Ifp= 4,then
3
∆U +∆V = ∆F .
ǫ ǫ ǫ
4
Lemma 4.1. Let f be a holomorphic function defined on the unit disk U. For
p > 4andz Uandǫ > 0wehave
∈
p 1
∆(U +V ) − ∆F .
ǫ ǫ ǫ
≤ p
(cid:18) (cid:19)
For1 p 4andz Uandǫ > 0wehave
≤ ≤ ∈
p 1
∆(U +V ) − ∆F .
ǫ ǫ ǫ
≥ p
(cid:18) (cid:19)
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 8
Proof. Letf = u+iv = reis anddefine
∆(U +V )
ǫ ǫ
Q(s) = .
∆F
ǫ
Thenfrom(4.2),(4.3)and(4.4)wehave
ǫ+r2cos2s −2+p2 ǫ+(p 1)r2cos2s
Q(s)= −
2+p
(cid:0) (2ǫ+(p (cid:1)1)r2)(cid:0) ǫp +r2 − 2 (cid:1)
− 1+p
−
ǫ+r2sin2s −2+p2 ǫ+(cid:16)(p 1)r2s(cid:17)in2s
+ − .
2+p
(cid:0) (2ǫ+(p (cid:1)1)r2)(cid:0) ǫp +r2 − 2 (cid:1)
− 1+p
−
Weshouldprovethat (cid:16) (cid:17)
ifp < 4;
Q(s) ≥
ifp 4.
(cid:26) ≤ ≥
Firstofall
2 p
( 2+p)r2 ǫp +r2 −2 cosssins
− p 1
Q′(s)= 2(cid:16)ǫ+− (p 1(cid:17))r2 T
−
where
T = ǫ+r2cos2s −3+p2 3ǫ+(1 p)r2cos2s
− −
+ (cid:0)ǫ+r2sin2s(cid:1)−3+p2 (cid:0)3ǫ+(p 1)r2sin2s .(cid:1)
−
ThenT = 0,ifandonlyif
(cid:0) (cid:1) (cid:0) (cid:1)
1 3ǫ+(p 1)r2sin2s
L = S2(6−p) = R := − ,
3ǫ+(p 1)r2cos2s
−
where
ǫ+r2sin2s
S = .
ǫ+r2cos2s
Ifp = 6,thencos2s = sin2s. Ifp > 6thenwealsohavecos2s= sin2s,because
ifforexamplecos2s > sin2s,thenL > 1 > R. Similarlycos2s < sin2simplies
L < 1< R. If4 < p < 6andcos2s > sin2s,then
ǫ(4 p)r2cos(2s)
R S = − < 0.
− (ǫ+r2cos2s)(3ǫ+(p 1)r2cos2s)
−
Thus
S < R < 1.
Since0 < 6−p < 1,itfollowsthat
2
6−p
S < R 2 < 1.
SoL = R. If4 < p < 6andcos2s < sin2s,then
6
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 9
S > R > 1.
So
6−p
S > R 2 .
ThusL = R.
6
Ifp< 4,thencos2s > sin2simpliesthat
6−p
1 > S > R > R 2 .
Finallyifp < 4andcos2s >sin2s. Then
6−p
1 < S < R < R 2 .
Weprovedthattheonlystationary pointsofQare
πj
s = , j = 0,...,7.
j
4
SoweeneedtoshowthatQ(s ) 1forp 4andQ(s ) 1forp 4.
j j
≤ ≤ ≥ ≥
Lets = 0. Showthat
ǫ−1+p2p+p ǫ+r2 −2+p2 ǫ+(p 1)r2 1, ifp 4;
−
Q(0) = ≥ ≤
(cid:16) p(2ǫ+((cid:0)p 1)r(cid:1)2) ǫp(cid:0) +r2 p2−2 (cid:1)(cid:17)(cid:26) ≤ 1, ifp ≥ 4.
− p 1
−
Letǫ = tr2,t > 0. Then (cid:16) (cid:17)
1+ ppt1 2−p2 t−1+p2 +(1+t)−2+p2(p−1+t)
Q(0) = − .
(cid:16)(cid:16) (cid:17)(cid:17) (cid:16) p 1+2t (cid:17)
−
Byusingconvexity ofthefunction k(x) = xp/2 2,forp < 4wehave
−
t−1+p2 (1+t)−2+p2(p 1+t) p 1+pt+2t2 −2+p2
+ − > − .
p 1+2t p 1+2t p 1+2t
− − (cid:18) − (cid:19)
Further
p 1+pt+2t2 −2+p2 p 1+pt −2+p2
− > − .
p 1+2t p 1
(cid:18) − (cid:19) (cid:18) − (cid:19)
SoQ(0) 1.
≥
For4 pweneedtoshowthat
≤
t−1+p2 (1+t)−2+p2(p 1+t) p 1+pt −2+p2
+ − < − ,
p 1+2t p 1+2t p 1
− − (cid:18) − (cid:19)
i.e.
t−1+p2 (1+t)−2+p2(p 1+t) t −2+p2
+ − < 1+t+ .
p 1+2t p 1+2t p 1
− − (cid:18) − (cid:19)
Thelastinequality isequivalent withthetrivialinequalities
t(t−2+p2 −(t+1)−2+p2) < 0< 1+t+ t −2+p2 (1+t)−2+p2.
p 1+2t p 1 −
− (cid:18) − (cid:19)
ONSOMERIESZANDCARLEMANTYPEINEQUALITIESFORHARMONICFUNCTIONS 10
Fors = π/4,andǫ = r2t,
2 p
Q(s)= 22−p2 p−1+pt −2 .
(p 1)(1+2t)
(cid:18) − (cid:19)
So
1, ifp 4;
Q(s) ≥ ≤
1, ifp 4.
(cid:26) ≤ ≥
Theothercasescanbetreatedinasimilarway. (cid:3)
ProofofTheorem1.4. Assumethat p 4. Byusing theGreen formula (4.1),and
≤
Lemma4.1weobtain
2π ∂ r 2π
r F dt = ρ∆F dρdt.
ǫ ǫ
∂r
Z0 Z0 Z0
p r 2π
ρ(∆U +∆V )dρdt
ǫ ǫ
≤ p 1
− Z0 Z0
p 2π ∂
= r (U +V )dt.
ǫ ǫ
p 1 ∂r
− Z0
Dividingbyrandintegrating in[0,1]withrespecttor weobtain
2π
[F (eit) F (0)]dt
ǫ ǫ
−
Z0
p 2π 2π
[U (eit) U (0)]dt+ [V (eit) V (0)] dt.
ǫ ǫ ǫ ǫ
≤ p 1 − −
− (cid:18)Z0 Z0 (cid:19)
Lettingǫ 0weobtain
→
2π p 2π 2π
f(eit)pdt u(eit)pdt+ v(eit)p dt.
| | ≤ p 1 | | | |
Z0 − (cid:18)Z0 Z0 (cid:19)
Similarlyweprovetherelatedinequality forp 4. (cid:3)
≥
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