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Complex Anal. Oper. Theory (2013) 7:1167–1183 DOI 10.1007/s11785-011-0187-5

Complex Analysis and Operator Theory

Optimal Estimates for the Gradient of Harmonic Functions in the Unit Disk David Kalaj · Marijan Markovi´c

Received: 27 February 2011 / Accepted: 12 August 2011 / Published online: 28 August 2011 © Springer Basel AG 2011

Abstract Let U be the unit disk, p  1 and let h p (U) be the Hardy space of complex harmonic functions. We find the sharp constants C p and the sharp functions C p = C p (z) in the inequality |Dw(z)| ≤ C p (1 − |z|2 )−1−1/ p wh p (U) , w ∈ h p (U), z ∈ U, in terms of Gaussian hypergeometric and Euler functions. This generalizes some results of Colonna related to the Bloch constant of harmonic mappings of the unit disk into itself and improves some classical inequalities by Macintyre and Rogosinski. Keywords

Harmonic functions · Bloch functions · Hardy spaces

Mathematics Subject Classification (2000)

Primary 31A05; Secondary 42B30

1 Introduction and Statement of the Results A harmonic function w defined in the unit ball Bn belongs to the harmonic Hardy class h p = h p (Bn ), 1 ≤ p < ∞ if the following growth condition is satisfied

Communicated by Mihai Putinar. D. Kalaj (B) · M. Markovi´c Faculty of Natural Sciences and Mathematics, University of Montenegro, Cetinjski put b.b., 81000 Podgorica, Montenegro e-mail: [email protected] M. Markovi´c e-mail: [email protected]

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⎛ wh p := ⎝ sup



0 1 and let q be its conjugate. Let w ∈ h p be a z , and complex harmonic function defined in the unit disk and let z = 0. Define n = |z| z t = i |z| . (a) We have the following sharp inequalities |Dw(z)eiτ |  C p (z, eiτ )(1 − r 2 )−1/ p−1 wh p , 2 −1/ p−1

|Dw(z)|  C p (z)(1 − r )

(1.7)

wh p ,

(1.8)

where z = r eiα , ⎞1/q ⎛ π  q| |cos(s + τ − α) 1 C p (z, eiτ ) = ⎝ ds ⎠ π (1 + r 2 − 2r cos s)1−q −π

and C p (z) =

C p (z, n), if p < 2; C p (z, t), if p  2.

(1.9)

Moreover

C p (z, t)  C p (z, eiτ )  C p (z, n), if p < 2; C p (z, n)  C p (z, eiτ )  C p (z, t), if p  2.

(1.10)

(b) For p  2 the function C p (z) can be expressed as 21/q C p (z) = π

     1+q 1 3q q 2 1/q B , F 1− , 1 − q; 1 + ; r , 2 2 2 2 (1.11)

where B is the beta function and F is the Gauss hypergeometric function.

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(c) Finally

C p := sup C p (z) = z∈U

⎧ ⎪ ⎨ ⎪ ⎩

 π

1/q |cos s|q −π (2−2 cos s)1−q ds  1/q π |sin s|q 1 −π (2−2 cos s)1−q ds π 1 π

, if 1 < p < 2; , if p  2.

The constant C p is optimal for real harmonic functions as well. Theorem 1.2 Let p > 1 and let w ∈ h p , be a complex harmonic function defined in the unit disk. Then we have the following sharp inequalities ¯ |∂w(z)|, |∂w(z)|  c p (z)(1 − |z|2 )−1/ p−1 wh p , and ¯ |∂w(z)|, |∂w(z)|  c p (1 − |z|2 )−1/ p−1 wh p , where c p (z) = (2π )1/q−1 (F(1 − q, 1 − q; 1; r 2 ))1/q

(1.12)

and cp = 2

−1+q q

π

1 −1+ 2q



(−1/2 + q)

(q)

1/q .

(1.13)

Remark 1.3 (a) In particular, if in Theorem 1.1 we take p = 2, then we have the following estimate 1 (1 + |z|2 )1/2 |∇w(z)| ≤ √ wh 2 . π (1 − |z|2 )3/2

(1.14)

If we assume that w is a real harmonic function, i.e. w = g + g, where g is an analytic function, then this estimate is equivalent to the real part theorem 1 (1 + |z|2 )1/2  gh 2 . |g  (z)| ≤ √ π (1 − |z|2 )3/2

(1.15)

For the proof of (1.15) we refer to [9, pp. 87, 88]. See also a higher dimensional generalization of (1.14) by Maz’ya and Kresin in the recent paper [8, Corollary 3] for n  2. Also the relation (1.14) for real w can be deduced from the work of Macintyre and Rogosinski for analytic functions, see [10, p. 301]. (b) On the other hand, if we take p = ∞, then C p = π4 and therefore, the relation (1.8) coincides with the result of Colonna, while for real w, it is a real part theorem [4] which can be expressed as

Optimal Estimates for the Gradient of Harmonic Functions

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1.4

1.2

1.0

0.8

10

15

20

10

15

20

Fig. 1 The graph of C p 1.0

0.9

0.8

0.7

0.6

5

Fig. 2 The graph of c p

|g  (z)| ≤

1 4  g∞ . π 1 − |z|2

(1.16)

(c) Notice that c p < C p < 2c p if p > 1, and C1 = 2c1 = π4 . On the other hand, c∞ = 1 coincides with the constant of the Schwarz lemma for analytic functions. Notice also the interesting fact √ that the minimum of constants C p is achieved for p = 2 and it is equal to C2 = 2/π . The graphs of functions C p and c p with 1  p  20, are shown in Figs. 1 and 2. (d) From Theorem 1.1 we find out that the Khavinson–Kresin–Maz’ya hypothesis (see [8]) is not true for n = 2 and 2 < p < ∞. Namely, the maximum of the absolute value of the directional derivative of a harmonic function with a fixed L p -norm of its boundary values is attained at the radial direction for p  2 and at the tangential direction for 2 < p < ∞. In the classical paper [10, (8.3.8)] of Macintyre and Rogosinski they obtained the inequality

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D. Kalaj, M. Markovi´c

 | f (z)|  1 + 

|z|2 ( p − 1)2

1/q

(1 − |z|2 )−1−1/ p  f  H p .

(1.17)

As an immediate consequence of Theorem 1.2, we improve the inequality (1.17) as follows Corollary 1.4 Let w = f (z) be an analytic function from the Hardy class H p (U). Then there holds the following inequality | f  (z)|  c p (z)(1 − |z|2 )−1−1/ p  f  H p ,

(1.18)

where c p (z) is defined in (1.12). Remark 1.5 Corollary 1.4 is an improvement of the corresponding inequality [10, (8.3.8)] because it holds (2π )1−q F(1 − q, 1 − q; 1; r 2 ) < 1 +

r2 ( p − 1)2

√ 1+|z|2 for all q > 1. The function c2 (z) = √ in (1.18) is the best possible, because for 2π p = 2 the relation (1.18) coincides with the sharp inequality [10, p. 301, eq. (7.2.1)]. We expect that (1.18) is sharp for every p. On the other hand, the power −1 − 1/ p is optimal, see e.g. Garnett [3, p. 86]. The paper [10] contains some sharp estimates of the form | f (k) (z)|  c p  f  p for f ∈ H p (U) and k  1 but p depends on k and it seems that if k = 1 then p can be only equals to 1 or 2. 2 Proofs We need the following lemmas Lemma 2.1 Let aq (t), t ∈ [0, 2π ], q  1, 0 ≤ r ≤ 1, be a function defined by π aq (t) =

|cos(s − t)|q |r − eis |2q−2 ds. −π

Then max aq (t) =

0t 2π

aq ( π2 ), if q  2; aq (0), if q > 2

and min aq (t) =

0t 2π

aq (0), if q  2; aq ( π2 ), if q > 2.

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Proof Since the case q = 1 is trivial, assume that q > 1 and π a(t) := aq (t) =

|cos(t − s)|q (1 + r 2 − 2r cos s)q−1 ds. −π

Note that a is π -periodic. Because sub-integral expression is 2π -periodic with respect to s, we obtain 2π a(t) = |cos s|q (1 + r 2 − 2r cos(t + s))q−1 ds, 0

and therefore a  (t) = 2(q − 1)r

2π |cos s|q sin(t + s)(1 + r 2 − 2r cos(t + s))q−2 ds. 0

Again by using the periodicity of sub-integral expression, we find that 2π a (t) = 2(q − 1)r |cos(t − s)|q sin s(1 + r 2 − 2r cos s)q−2 ds. 

0

Next we need the following transformations π



a (t) = 2(q − 1)r

|cos(t − s)|q sin s(1 + r 2 − 2r cos s)q−2 ds 0

2π + 2(q − 1)r |cos(t − s − π )|q sin(s + π )(1 + r 2 − 2r cos(s + π ))q−2 ds π

π = 2(q − 1)r

|cos(t − s)|q sin s Q (r, s − π/2) ds 0

π/2 |sin(t − s)|q cos s Q(r, s)ds,

= 2(q − 1)r −π/2

where Q(r, s) = (1 + r 2 + 2r sin s)q−2 − (1 + r 2 − 2r sin s)q−2 .

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Thus, the derivative is a  (t) = 2r (q − 1)

π/2 h(t, s) cos sds,

−π/2

where h(t, s) = |sin(t − s)|q Q(r, s). Also a  (t) is π -periodic and a  (0) = a  (π/2) = 0. Further, we have h(t, s) + h(t, −s) = (|sin(t − s)|q − |sin(t + s)|q )Q(r, s). If 1 < q < 2, then for 0 < t < π/2 we have h(t, s) + h(t, −s) > 0, 0 < s < π/2 and for π/2 < t < π h(t, s) + h(t, −s) > 0, 0 < s < π/2. We claim that π/2 a (t) = 2r (q − 1) (h(t, s) + h(t, −s)) cos sds > 0, 0 < t < π/2 

0

and π/2 a (t) = 2r (q − 1) (h(t, s) + h(t, −s)) cos sds < 0, π/2 < t < π. 

0

This means that the minimum of a is achieved in 0 and the maximum in π2 . Similarly, it can be treated the case q > 2. For q = 2 the function a(t) is a constant. The proof of Lemma 2.1 is completed.   Lemma 2.2 Let λ  0, 0  r  1 and q ≥ 1. For all t there exists t  ∈ [0, 2π ] such that 2π

λ

2π

| cos(s − t)| |r − e | 0

is 2q−2

ds 

| cos(s − t  )|λ |1 − eis |2q−2 ds.

0

Proof In order to prove Lemma 2.2, we need the following result.

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Proposition 2.3 [6, Lemma 3.2] Let U ⊂ C be the open unit disk and let (A, μ) be a measured space with μ(A) < ∞. Let f (z, ω) be a holomorphic function for z ∈ U and measurable for ω ∈ A. Let b > 0 and assume in addition that there exists an integrable function χ ∈ L max{b,2} (A, dμ) such that | f (0, ω)| + | f  (z, ω)|  χ (ω),

(2.1)

for (z, ω) ∈ U × A, where by f  (z, ω) we mean the complex derivative of f with respect to z. Then the function  | f (z, ω)|b dμ(ω)

φ(z) = log A

is subharmonic in U. Corollary 2.4 Assume together with the assumptions of the previous proposition that z → f (z, ω) is a continuous map up to the boundary T. Then we have the following inequality 

φ(z)  max φ(eiτ ) = φ(eiτ ). τ ∈[0,2π )

In order to apply Corollary 2.4, we take dμ(s) = | cos(s − t)|λ ds,

f (z, s) = z − eis and b = 2q − 2

and observe that 2π max τ

| cos(s − t)|λ |eiτ − eis |2q−2 ds

0

2π =



| cos(s − t)|λ |eiτ − eis |2q−2 ds

0

2π =

| cos(s − t  )|λ |1 − eis |2q−2 ds.

0

This finishes the proof of Lemma 2.2. The Poisson kernel for the disk can be expressed as P(z, eθ ) =

  1 − |z|2 e−iθ eiθ . = − 1 + + |z − eiθ |2 z − e−iθ z − eiθ

 

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Then we have grad(P) = (Px , Py ) = Px + i Py = 2∂¯ P = ∂P = 

eiθ

2e−iθ , (z − e−iθ )2

2

z − eiθ

and ∂¯ P = 

e−iθ z¯ − e−iθ

2 .

Proof of Theorem 1.1 (a) Let l = eiτ . Then for p > 1 1 Dw(z)l = 2π =

1 π

2π grad(P), l f (eiθ )dθ

0 2π 

 0

e−i(θ+τ ) iθ 2 f (e )dθ . z¯ − e−iθ

By applying (2.2) and Hölder inequality we obtain ⎞1/ p ⎛ 2π  q ⎞1/q ⎛2π    −i(θ+τ ) e 1   |Dw(z)l|  ⎝    dθ ⎠ ⎝ | f (eiθ )| p dθ ⎠ .  z¯ − e−iθ 2  π 0

0

We should consider the integral q 2π  e−i(θ+τ )   I q =    dθ.  z¯ − e−iθ 2  0

First of all q 2π  2π  i(θ+τ −α) q ei(θ+τ )    e  Iq =    dθ =    dθ.  z − eiθ 2   r − eiθ 2  0

0

Take the substitution eiθ =

r − eis . 1 − r eis

(2.2)

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Then deiθ =

1 − r2 deis , (1 − r eis )2

and thus 1 − r 2 eis ds (1 − r eis )2 eiθ 1 − r 2 1 − r eis ds = eis (1 − r eis )2 r − eis 1 − r2 ds. = 1 + r 2 − 2r cos s

dθ =

On the other hand, we easily find that ei(θ+τ −α) (1 + r 2 − 2r cos s) cos(s + τ − α)

 . 2 = (1 − r 2 )2 r − eiθ Therefore, finally we have the relation q 2π  π e−i(θ+τ )  |cos(s + τ − α)|q  2 1−2q dθ = (1 − |z| ) ds,      z¯ − e−iθ 2  (1 + r 2 − 2r cos s)1−q

(2.3)

−π

0

which together with the first relation gives |Dw(z)l| ≤ C p (z, l)(1 − |z|2 )−1−1/ p wh p . Now by using Lemma 2.1, we conclude that C p (z) =

C p (z, n), if p < 2; C p (z, t), if p  2,

which coincides with (1.9). This implies (1.8). Lemma 2.1 implies at once (1.10). (b) By using the following formula π 0

    1−μ 1+μ 2 sinμ−1 t μ 1 (2.4) , F ν, ν + ; , r dt = B (1 + r 2 − 2r cos t)ν 2 2 2 2

(see, e.g., Prudnikov et al. [11, 2.5.16(43)]), where B(u, v) is the Beta-function, and F(a, b; c; x) is the hypergeometric Gauss function; taking μ = q + 1 and ν = 1 − q, because |cos(s + τ − α)|q = |sin s|q , for τ = α + π2 , we obtain (1.11).

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(c) By Lemma 2.2 and Lemma 2.1, we obtain C p (z, l) ≤ C p (1, l  ) ≤ C p for some l  with |l  | = 1, and we have the second conclusion of the main theorem. Let us now show that the constant C p is sharp. We will show the sharpness of the result for p  2. A similar analysis works for p > 2. Let 0 < ρ < 1 and take eis =

ρ − eit , 1 − ρeit

eit =

ρ − eis . 1 − ρeis

i.e.,

Define f ρ (eit ) = (1 − ρ 2 )−1/ p |cos s(1 − cos s)|q−1 sign(cos s) and take wρ = P[ f ρ ]. Then 1 − ρ2 ds, 1 + ρ 2 − 2ρ cos s eit (1 + r 2 − 2r cos s) cos(s)

 , 2 = (1 − r 2 )2 r − eit

dt =

and 2π

2π | f ρ (eit )| p dt =

0

| f ρ (eit )| p 0 2π

=

1 ds 1 + ρ 2 − 2ρ cos s

|cos s(1 − cos s)|q 0

1 + ρ2

1 ds. − 2ρ cos s

Thus

lim

ρ→1

p  fρ  p

2π =

| f ρ (eit )| p dt = 0

πq q C p. 2q

(2.5)

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Taking r = ρ, we obtain (1 − ρ )

2 1+1/ p

(1 − ρ 2 )1+1/ p |Dwρ (ρ)1| = π =

(1−ρ 2 )1+1/ p π

2π

 0

2π 0

eit

ρ

2 f ρ (e − eiθ

=

)dt

(1 + ρ 2 −2ρ cos s) cos(s) (1−ρ 2 )−1/ p (1−ρ 2 )2

× |cos s(1 − cos s)|q−1 sign(cos s) 1 = π

it

1 − ρ2 ds 1 + ρ 2 − 2ρ cos s

2π |cos s|q (1 − cos s)q−1 ds

0 q−1 π

2q−1

q

Cp

From (2.5) it follows that (1 − ρ 2 )1+1/ p |Dwρ (ρ)1| = C p. ρ→1  fρ  p lim

 

This shows that the constant C p is sharp. Proof of Theorem 1.2 First of all 2π ∂w =

 0

eiθ z − eiθ

iθ 2 f (e )

dθ . 2π

By applying Hölder inequality, we have ⎞1/q ⎛ 2π ⎞1/ p ⎛ 2π   1 ⎝ 1 iθ p |∂w|    dθ ⎠ ⎝ | f (e )| dt ⎠ 2π z − eiθ 2q 0

0

⎞1/q ⎛ 2π ⎞1/ p ⎛ 2π   2 )2q−1 1 (1 − |z| ⎝ = (1 − |z|2 )1/q−2 dθ ⎠ ⎝ | f (eiθ )| p dt ⎠ .   2π z − eiθ 2q 0

0

It remains to estimate the integral 2π Jq = 0

(1 − |z|2 )2q−1 dθ =   z − eiθ 2q

2π 0

(1 − r 2 )2q−1 dθ. |r − eiθ |2q

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By making use again of the change eiθ =

r − eis , 1 − r eis

we obtain dθ =

1 − r2 ds |1 − r eis |2

and (1 − r 2 )eis . 1 − r eis

r − eiθ =

Therefore, by using Lemma 2.2 for λ = 0, we obtain 2π Jq = 0

(1 − r 2 )2q−1 dθ = (1 − r 2 )1−q |r − eiθ |2q

2π |1 − r eis |2q−2 ds 0 2π

= (1 − r 2 )1−q

|1 + r 2 − 2r cos s|q−1 ds 0

2π ≤2

q−1

(1 − r )

|1 − cos s|q−1 ds.

2 1−q 0

Thus |∂w|  c p (1 − |z|2 )−1−1/ p  f  L p (T) , where cp = 2

−1+q q

π

1 −1+ 2q



(−1/2 + q)

(q)

1/q .

This proves (1.13). By formula (2.4), for μ = 1, ν = 1 − q we have 2π

π |1 + r − 2r cos s| 2

0

q−1

ds = 2

|1 + r 2 − 2r cos s|q−1 ds 0

  = 2π F 1 − q, 1 − q; 1, r 2 .

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This implies (1.12). The sharpness of constant c p can be verified by taking f ρ± (eit ) = (1 − ρ 2 )−1/ p |cos s(1 − cos s)|q−1 e±is and following the proof of sharpness of C p .

 

Acknowledgments After we wrote the first version of this paper, we had useful discussion about this subject with Professor Vladimir Maz’ya. We thank Professor Romeo Meštrovi´c for some language remarks.

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