Weighted Integral Means of Mixed Areas and Lengths under Holomorphic Mappings
aa r X i v : . [ m a t h . C V ] M a y WEIGHTED INTEGRAL MEANS OF MIXEDAREAS AND LENGTHS UNDER HOLOMORPHIC MAPPINGS
JIE XIAO AND WEN XUA bstract . This note addresses monotonic growths and logarithmic convexitiesof the weighted ((1 − t ) α dt , −∞ < α < ∞ , 0 < t <
1) integral means A α,β ( f , · )and L α,β ( f , · ) of the mixed area ( π r ) − β A ( f , r ) and the mixed length (2 π r ) − β L ( f , r )(0 ≤ β ≤ < r <
1) of f ( r D ) and ∂ f ( r D ) under a holomorphic map f from the unit disk D into the finite complex plane C .
1. I ntroduction
From now on, D represents the unit disk in the finite complex plane C , H ( D )denotes the space of holomorphic mappings f : D → C , and U ( D ) stands for allunivalent functions in H ( D ). For any real number α , positive number r ∈ (0 ,
1) andthe standard area measure dA , let dA α ( z ) = (1 − | z | ) α dA ( z ); r D = { z ∈ D : | z | < r } ; r T = { z ∈ D : | z | = r } . In their recent paper [11], Xiao and Zhu have discussed the following area 0 < p < ∞ -integral means of f ∈ H ( D ): M p ,α ( f , r ) = " A α ( r D ) Z r D | f | p dA α p , proving that r M p ,α ( f , r ) is strictly increasing unless f is a constant, and log r log M p ,α ( f , r ) is not always convex. This last result suggests a conjecture thatlog r log M p ,α ( f , r ) is convex or concave when α ≤ α >
0. But, moti-vated by [11, Example 10, (ii)] we can choose p = α = f ( z ) = z + c and c > p , p =
1. To understand this approach, let ustake a look at M ,α ( · , · ) from a di ff erential geometric viewpoint. Note that M ,α ( f ′ , r ) = R r D | f ′ | dA α A α ( r D ) = R r (cid:2) (2 π t ) − R t T | f ′ ( z ) || dz | (cid:3) (1 − t ) α dt R r (1 − t ) α dt . Mathematics Subject Classification.
Key words and phrases. monotonic growth, logarithmic convexity, mean mixed area, mean mixedlength, isoperimetric inequality, holomorphic map, univalent function.JX and WX were in part supported by NSERC of Canada and the Finnish Cultural Foundation,respectively.
So, if f ∈ U ( D ), then (2 π t ) − Z t T | f ′ ( z ) | | dz | is a kind of mean of the length of ∂ f ( t D ), and hence the square of this mean domi-nates a sort of mean of the area of f ( t D ) in the isoperimetric sense: Φ A ( f , t ) = ( π t ) − Z t D | f ′ ( z ) | dA ( z ) ≤ " (2 π t ) − Z t T | f ′ ( z ) | | dz | = (cid:2) Φ L ( f , t ) (cid:3) . According to the P ´olya-Szeg¨o monotone principle [9, Problem 309] (or [3, Propo-sition 6.1]) and the area Schwarz’s lemma in Burckel, Marshall, Minda, Poggi-Corradini and Ransford [3, Theorem 1.9], Φ L ( f , · ) and Φ A ( f , · ) are strictly in-creasing on (0 ,
1) unless f ( z ) = a z with a ,
0. Furthermore, log Φ L ( f , r ) andlog Φ A ( f , r ), equivalently, log L ( f , r ) and log A ( f , r ), are convex functions of log r for r ∈ (0 , c > D onto the annulus { e − c π < | z | < e c π } : f ( z ) = exp h ic log (cid:16) + z − z (cid:17)i enjoys the property that log r log A ( f , r ) is not convex; see [3, Example 5.1].In the above and below, we have used the following convention: Φ A ( f , r ) = A ( f , r ) π r & Φ L ( f , r ) = L ( f , r )2 π r , where under r ∈ (0 ,
1) and f ∈ H ( D ), A ( f , r ) and L ( f , r ) stand respectively for thearea of f ( r D ) (the projection of the Riemannian image of r D by f ) and the lengthof ∂ f ( r D ) (the boundary of the projection of the Riemannian image of r D by f )with respect to the standard Euclidean metric on C . For our purpose, we choose ashortcut notation d µ α ( t ) = (1 − t ) α dt & ν α ( t ) = µ α ([0 , t ]) ∀ t ∈ (0 , , and for 0 ≤ β ≤ Φ A ,β ( f , t ) = A ( f , t )( π t ) β & Φ L ,β ( f , t ) = L ( f , t )(2 π t ) β , and then A α,β ( f , r ) = R r Φ A ,β ( f , t ) d µ α ( t ) R r d µ α ( t ) & L α,β ( f , r ) = R r Φ L ,β ( f , t ) d µ α ( t ) R r d µ α ( t )which are called the weighted integral means of the mixed area and the mixedlength for f ( r D ) and ∂ f ( r D ), respectively.In this note, we consider two fundamental properties: monotonic growths andlogarithmic convexities of both A α,β ( f , r ) and L α,β ( f , r ), thereby producing two EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 3 specialities: (i) if r Φ L ( f , r ) is monotone increasing on (0 , r R r (cid:2) Φ L , ( f , t ) (cid:3) d µ α ( t ) R r d µ α ( t ) ≥ A α, ( f , r );(ii) the log-convexity for L α,β = ( f , r ) essentially settles the above-mentioned con-jecture. The details (results and their proofs) are arranged in the forthcoming twosections. 2. M onotonic G rowth In this section, we deal with the monotonic growths of A α,β ( f , r ) and L α,β ( f , r ),along with their associated Schwarz type lemmas. In what follows, N is used asthe set of all natural numbers.2.1. Two Lemmas.
The following two preliminary results are needed.
Lemma 1. [6, Theorems 1 & 2]
Let f ∈ H ( D ) be of the form f ( z ) = a + P ∞ k = n a k z k with n ∈ N . Then: (i) π r n h | f ( n ) (0) | n ! i ≤ A ( f , r ) ∀ r ∈ (0 , . (ii) 2 π r n h | f ( n ) (0) | n ! i ≤ L ( f , r ) ∀ r ∈ (0 , .Moreover, equality in (i) or (ii) holds if and only if f ( z ) = a + a n z n .Proof. This may be viewed as the higher order Schwarz type lemma for area andlength. See also the proofs of Theorems 1 & 2 in [6], and their immediate re-marks on equalities. Here it is worth noticing three matters: (a) f ( n ) (0) n ! is just a n ;(b) [5, Corollary 3] presents a di ff erent argument for the area case; (c) L ( f , r ) isgreater than or equal to the length l ( r , f ) of the outer boundary of f ( r D ) (defined in[6]) which is not less than the length l ( r , f ) of the exact outer boundary of f ( r D )(introduced in [12]). (cid:3) Lemma 2.
Let ≤ β ≤ . (i) If f ∈ H ( D ) , then r Φ A ,β ( f , r ) is strictly increasing on (0 , unlessf = ( constant when β < linear map when β = . (ii) If f ∈ U ( D ) or f ( z ) = a + a n z n with n ∈ N , then r Φ L ,β ( f , r ) is strictlyincreasing on (0 , unlessf = ( constant when β < linear map when β = . Proof.
It is enough to handle β < β = Φ A ,β ( f , r ) = ( π r ) − β Φ A , ( f , r ) & L ( f , r ) = (2 π r ) − β Φ L , ( f , r ) . To see the strictness, we consider two cases.
JIE XIAO AND WEN XU (i) Suppose that Φ A ,β ( f , · ) is not strictly increasing. Then there are r , r ∈ (0 , r < r , and Φ A ,β ( f , · ) is a constant on [ r , r ]. Hence ddr Φ A ,β ( f , r ) = ∀ r ∈ [ r , r ] . Equivalently, 2 β A ( f , r ) = r ddr A ( f , r ) ∀ r ∈ [ r , r ] . But, according to [3, (4.2)]:2 A ( f , r ) ≤ r ddr A ( f , r ) ∀ r ∈ (0 , . Since β <
1, we get A ( f , r ) = r ∈ [ r , r ], whence finding that f is constant.(ii) Now assume that Φ L ,β ( f , · ) is not strictly increasing. There are r , r ∈ (0 , r < r and0 = ddr Φ L ,β ( f , r ) = (2 π r ) − β h ddr L ( f , r ) − β r L ( f , r ) i = ∀ r ∈ [ r , r ] . If f ∈ U ( D ) then L ( f , r ) = Z r T | f ′ ( z ) | | dz | and hence one has the following “first variation formula” ddr L ( f , r ) = Z π | f ′ ( re i θ ) | d θ + r ddr Z π | f ′ ( re i θ ) | d θ ∀ r ∈ [ r , r ] . The previous three equations yield0 = (1 − β ) Z π | f ′ ( re i θ ) | d θ + r ddr Z π | f ′ ( re i θ ) | d θ ∀ r ∈ [ r , r ]and so Z π | f ′ ( re i θ ) | d θ = ∀ r ∈ [ r , r ] . This ensures that f is a constant, contradicting f ∈ U ( D ). Therefore, f ( z ) is ofthe form a + a n z n . But, since L ( z n , r ) = π r n is strictly increasing, f must beconstant. (cid:3) Monotonic Growth of A α,β ( f , · ) . This aspect is essentially motivated by thefollowing Schwarz type lemma.
Proposition 1.
Let −∞ < α < ∞ , ≤ β ≤ , and f ∈ H ( D ) be of the formf ( z ) = a + P ∞ k = n a k z k with n ∈ N . Then π − β h | f ( n ) (0) | n ! i ≤ A α,β ( f , r ) ν α ( r ) R r t n − β ) d µ α ( t ) ∀ r ∈ (0 , with equality if and only if f ( z ) = a + a n z n . EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 5
Proof.
The inequality follows from Lemma 1 (i) right away. When f ( z ) = a + a n z n , the last inequality becomes equality due to the equality case of Lemma 1(i). Conversely, suppose that the last inequality is an equality. If f does not havethe form a + a n z n , then the equality in Lemma 1 (i) is not true, then there are r , r ∈ (0 ,
1) such that r < r and A ( f , t ) > π t n h | f ( n ) (0) | n ! i ∀ t ∈ [ r , r ] . This strict inequality forces that for r ∈ [ r , r ], π − β h | f ( n ) (0) | n ! i Z r t n − β ) d µ α ( t ) = Z r ( π t ) − β A ( f , t ) d µ α ( t ) = Z r + Z r r + Z rr ! ( π t ) − β A ( f , t ) d µ α ( t ) > π − β h | f ( n ) (0) | n ! i Z r t n − β ) d µ α ( t ) , a contradiction. Thus f ( z ) = a + a n z n . (cid:3) Based on Proposition 1, we find the monotonic growth for A α,β ( · , · ) as follows. Theorem 1.
Let −∞ < α < ∞ , ≤ β ≤ , and f ∈ H ( D ) . Then r A α,β ( f , r ) isstrictly increasing on (0 , unlessf = ( constant when β < linear map when β = . Consequently, (i) lim r → A α,β ( f , r ) = ( when β < | f ′ (0) | when β = . (ii) If Φ A ,β ( f ,
0) : = lim r → Φ A ,β ( f , r ) & Φ A ,β ( f ,
1) : = lim r → Φ A ,β ( f , r ) < ∞ , then < r < s < ⇒ ≤ A α,β ( f , s ) − A α,β ( f , r )log ν α ( s ) − log ν α ( r ) ≤ Φ A ,β ( f , s ) − Φ A ,β ( f , with equality if and only iff = ( constant when β < linear map when β = . In particular, t A α,β ( f , t ) is Lipschitz with respect to log ν α ( t ) for t ∈ (0 , .Proof. Note that ν α ( r ) = R r d µ α ( t ). So d ν α ( r ), the di ff erential of ν α ( r ) with respectto r ∈ (0 , d µ α ( r ). By integration by parts we have Φ A ,β ( f , r ) ν α ( r ) − Z r Φ A ,β ( f , t ) d µ α ( t ) = Z r (cid:2) ddt Φ A ,β ( f , t ) (cid:3) ν α ( t ) dt . JIE XIAO AND WEN XU Di ff erentiating the function A α,β ( f , r ) with respect to r and using Lemma 2 (i), weget ddr A α,β ( f , r ) = Φ A ,β ( f , r )2 r (1 − r ) α ν α ( r ) − h R r Φ A ,β ( f , t ) d µ α ( t ) i r (1 − r ) α ν α ( r ) = r (1 − r ) α h Φ A ,β ( f , t ) ν α ( r ) − R r Φ A ,β ( f , t ) d µ α ( t ) i ν α ( r ) = r (1 − r ) α R r (cid:2) ddt Φ A ,β ( f , t ) (cid:3) ν α ( t ) dt ν α ( r ) ≥ . As a result, r A α,β ( f , r ) increases on (0 , r , r ∈ (0 ,
1) such that r < r and A α,β ( f , r ) = A α,β ( f , r ) = A α,β ( f , r ) ∀ r ∈ [ r , r ] . Consequently, ddr A α,β ( f , r ) = ∀ r ∈ [ r , r ]and so Z r (cid:2) ddt Φ A ,β ( f , t ) (cid:3) ν α ( t ) dt = ∀ r ∈ [ r , r ] . Then we must have ddt Φ A ,β ( f , t ) = ∀ t ∈ (0 , r ) with r ∈ [ r , r ] , whence getting that if β < f must be constant or if β = f must belinear, thanks to the argument for the strictness in Lemma 2 (i).It remains to check the rest of Theorem 1.(i) The monotonic growth of A α,β ( f , · ) ensures the existence of the limit. Anapplication of L’H ˆopital’s rule giveslim r → A α,β ( f , r ) = lim r → Φ A ,β ( f , r ) = ( β < | f ′ (0) | when β = . (ii) Again, the above monotonicity formula of A α,β ( f , · ) plus the given conditionyields that for s ∈ (0 , r ∈ (0 , s ) A α,β ( f , r ) = A α,β ( f , s ) < ∞ . EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 7
Integrating by parts twice and using the monotonicity of Φ A ,β ( f , · ), we obtain thatunder 0 < r < s < ≤ A α,β ( f , s ) − A α,β ( f , r ) = Z sr ddt A α,β ( f , t ) dt = Z sr Z t (cid:2) dd τ Φ A ,β ( f , τ ) (cid:3) ν α ( τ ) d τ ! h d ν α ( t ) ν α ( t ) i = Z sr ν α ( t ) Φ A ,β ( f , t ) − Z t Φ A ,β ( f , τ ) d ν α ( τ ) ! h d ν α ( t ) ν α ( t ) i ≤ h Φ A ,β ( f , s ) − Φ A ,β ( f , i Z sr d ν α ( t ) ν α ( t ) . This gives the desired inequality right away. Furthermore, the above argument plusLemma 2 (i) derives the equality case. (cid:3)
As an immediate consequence of Theorem 1, we get a sort of “norm” estimateassociated with Φ A ,β ( f , · ). Corollary 1.
Let −∞ < α < ∞ , ≤ β ≤ , and f ∈ H ( D ) . (i) If −∞ < α ≤ − , then Z Φ A ,β ( f , t ) d µ α ( t ) = sup r ∈ (0 , Z r Φ A ,β ( f , t ) d µ α ( t ) < ∞ if and only if f is constant. Moreover, sup r ∈ (0 , A α,β ( f , r ) = Φ A ,β ( f , . (ii) If − < α < ∞ , then A α,β ( f , r ) ≤ A α,β ( f ,
1) : = sup s ∈ (0 , A α,β ( f , s ) ∀ r ∈ (0 , , where the inequality becomes an equality for all r ∈ (0 , if and only iff = ( constant when β < linear map when β = . (iii) The following function α A α,β ( f , is strictly decreasing on ( − , ∞ ) unlessf = ( constant when β < linear map when β = . Proof. (i) By Theorem 1, we have A α,β ( f , r ) ≤ R s Φ A ,β ( f , t ) d µ α ( t ) ν α ( s ) ∀ r ∈ (0 , s ) . Note thatlim s → ν α ( s ) = ∞ & lim s → Z s Φ A ,β ( f , t ) d µ α ( t ) = Z Φ A ,β ( f , t ) d µ α ( t ) . So, the last integral is finite if and only if Φ A ,β ( f , r ) = ∀ r ∈ (0 , , JIE XIAO AND WEN XU equivalently, A ( f , r ) = r ∈ (0 , f is constant.For the remaining part of (i), we may assume that f is not a constant map. Dueto lim r → ν α ( r ) = ∞ , we obtainlim r → Z r Φ A ,β ( f , t ) d µ α ( t ) = Z Φ A ,β ( f , t ) d µ α ( t ) = ∞ . So, an application of L’H ˆopital’s rule yieldssup < r < A α,β ( f , r ) = lim r → R r Φ A ,β ( f , t ) d µ α ( t ) ν α ( r ) = lim r → Φ A ,β ( f , r ) r (1 − r ) α r (1 − r ) α = Φ A ,β ( f , . (ii) Under − < α < ∞ , we havelim r → ν α ( r ) = ν α (1) & lim r → Z r Φ A ,β ( f , t ) d µ α ( t ) = Z Φ A ,β ( f , t ) d µ α ( t ) . Thus, by Theorem 1 it follows that for r ∈ (0 , A α,β ( f , r ) ≤ lim s → A α,β ( f , s ) = (cid:2) ν α (1) (cid:3) − Z Φ A ,β ( f , t ) d µ α ( t ) = sup s ∈ (0 , A α,β ( f , s ) . The equality case just follows from a straightforward computation and Theorem 1.(iii) Suppose − < α < α < ∞ and A α ,β ( f , < ∞ , then integrating by partstwice, we obtain A α ,β ( f , = (cid:2) ν α (1) (cid:3) − Z Φ A ,β ( f , r ) d µ α ( r ) = (cid:2) ν α (1) (cid:3) − Z (1 − r ) α − α ddr "Z r Φ A ,β ( f , t ) d µ α ( t ) dr = (cid:2) ν α (1) (cid:3) − " − Z Z r Φ A ,β ( f , t ) d µ α ( t ) ! d (1 − r ) α − α ≤ (cid:2) ν α (1) (cid:3) − A α ,β ( f , Z ν α ( r ) d (cid:2) − (1 − r ) α − α (cid:3) = A α ,β ( f , (cid:2) ν α (1) (cid:3) − "Z (1 − r ) α − α d µ α ( r ) = A α ,β ( f , , thereby establishing A α ,β ( f , ≤ A α ,β ( f , Z r Φ A ,β ( f , t ) d µ α ( t ) = A α ,β ( f , ν α ( r ) ∀ r ∈ (0 , , whence yielding (via the just-verified (ii)) f = ( constant when β < β = . (cid:3) EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 9
Monotonic Growth of L α,β ( f , · ) . Correspondingly, we first have the follow-ing Schwarz type lemma.
Proposition 2.
Let −∞ < α < ∞ , ≤ β ≤ , and f ∈ H ( D ) be of the formf ( z ) = a + P ∞ k = n a k z k with n ∈ N . Then (2 π ) − β h | f ( n ) (0) | n ! i ≤ L α,β ( f , r ) ν α ( r ) R r t n − β d µ α ( t ) ∀ r ∈ (0 , with equality when and only when f = a + a n z n .Proof. This follows from Lemma 1 (ii) and its equality case. (cid:3)
The coming-up-next monotonicity contains a hypothesis stronger than that forTheorem 1.
Theorem 2.
Let −∞ < α < ∞ , ≤ β ≤ , and f ∈ U ( D ) or f ( z ) = a + a n z n withn ∈ N . Then r L α,β ( f , r ) is strictly increasing on (0 , unlessf = ( constant when β < linear map when β = . Consequently, (i) lim r → L α,β ( f , r ) = ( when β < | f ′ (0) | when β = . (ii) If Φ L ,β ( f ,
0) : = lim r → Φ L ,β ( f , r ) & Φ L ,β ( f ,
1) : = lim r → Φ L ,β ( f , r ) < ∞ , then < r < s < ⇒ ≤ L α,β ( f , s ) − L α,β ( f , r )log ν α ( s ) − log ν α ( r ) ≤ Φ L ,β ( f , s ) − Φ L ,β ( f , with equality if and only iff = ( constant when β < linear map when β = . In particular, t L α,β ( f , t ) is Lipschitz with respect to log ν α ( t ) for t ∈ (0 , .Proof. Similar to that for Theorem 1, but this time by Lemma 2 (ii). (cid:3)
Naturally, we can establish the so-called “norm” estimate associated to Φ L ,β ( f , · ). Corollary 2.
Let ≤ β ≤ and f ∈ U ( D ) or f ( z ) = a + a n z n with n ∈ N . (i) If −∞ < α ≤ − , then Z Φ L ,β ( f , t ) d µ α ( t ) = sup r ∈ (0 , Z r Φ L ,β ( f , t ) d µ α ( t ) < ∞ if and only if f is constant. Moreover, sup r ∈ (0 , L α,β ( f , r ) = Φ L ,β ( f , . (ii) If − < α < ∞ , then L α,β ( f , r ) ≤ L α,β ( f ,
1) : = sup s ∈ (0 , L α,β ( f , s ) ∀ r ∈ (0 , , where the inequality becomes an equality for all r ∈ (0 , if and only iff = ( constant when β < linear map when β = . (iii) α L α,β ( f , is strictly decreasing on ( − , ∞ ) unlessf = ( constant when β < linear map when β = . Proof.
The argument is similar to that for Corollary 1, but via Lemma 2 (ii). (cid:3) logarithmic convexity In this section, we treat the convexities of the functions: log r log A α,β ( f , r )and log r log L α,β ( f , r ) for r ∈ (0 , Two More Lemmas.
The following are two technical preliminaries.
Lemma 3. [10, Corollaries 2-3 & Proposition 7]
Suppose f ( x ) and { h k ( x ) } ∞ k = are positive and twice di ff erentiable for x ∈ (0 , such that the function H ( x ) = P ∞ k = h k ( x ) is also twice di ff erentiable for x ∈ (0 , . Then: (i) log x log f ( x ) is convex if and only if log x log f ( x ) is convex. (ii) The function log x log f ( x ) is convex if and only if the D-notation of fD ( f ( x )) : = f ′ ( x ) f ( x ) + x f ′ ( x ) f ( x ) ! ′ ≥ ∀ x ∈ (0 , . (iii) If for each k the function log x log h k ( x ) is convex, then log x log H ( x ) isalso convex. Lemma 4.
Let f ∈ H ( D ) . Then f belongs to U ( D ) provided that one of the follow-ing two conditions is valid: (i) [8] or [1, Lemma 2.1] f (0) = f ′ (0) − = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z f ′ ( z ) f ( z ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ∀ z ∈ D . (ii) [7, Theorem 1] or [4, Theorem 8.12] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)" f ′′ ( z ) f ′ ( z ) ′ − " f ′′ ( z ) f ′ ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ − | z | ) − ∀ z ∈ D . EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 11
Log-convexity for A α,β ( f , · ) . Such a property is given below.
Theorem 3.
Let ≤ β ≤ and < r < . (i) If α ∈ ( −∞ , − , then there exist f , g ∈ H ( D ) such that log r log A α,β ( f , r ) isnot convex and log r log A α,β ( g , r ) is not concave. (ii) If α ∈ [ − , , then log r log A α, ( a n z n , r (cid:1) is convex for a n , with n ∈ N .Consequently, log r log A α, (cid:0) f , r (cid:1) is convex for all f ∈ U ( D ) . (iii) If α ∈ (0 , ∞ ) , then log r log A α,β ( a n z n , r ) is not convex for a n , and n ∈ N .Proof. The key issue is to check whether or not log r log A α,β ( z n , r ) is convexfor r ∈ (0 , λ ≥ < x < f λ ( x ) = Z x t λ (1 − t ) α dt and ∆ ( λ, x ) = f ′ λ ( x ) f λ ( x ) + x f ′ λ ( x ) f λ ( x ) ! ′ − " f ′ ( x ) f ( x ) + x f ′ ( x ) f ( x ) ! ′ . Given n ∈ N . A simple calculation shows Φ A ,β ( z n , t ) = π − β t n − β ) , and then achange of variable derives A α,β ( z n , r ) = R r Φ A ,β ( z n , t ) d µ α ( t ) ν α ( r ) = π − β R r t n − β (1 − t ) α dt R r (1 − t ) α dt = π − β f n − β ( r ) f ( r ) . In accordance with Lemma 3 (i)-(ii), it is readily to work out that log r log A α,β ( z n , r ) is convex for r ∈ (0 ,
1) if and only if ∆ ( n − β, x ) ≥ x ∈ (0 , α ∈ ( −∞ , − x → ∆ ( λ, x ) = λ ( α + λ + + α )( α + ( α + . Choosing f ( z ) = z n = ( z when β < z when β = λ = n − β , we find lim x → ∆ ( λ, x ) <
0, whence deriving that log r log A α ( f , r )is not convex.In the meantime, picking n ∈ N such that n > β − (2 + α ) and putting g ( z ) = z n ,we obtain lim x → ∆ ( n − β, x ) = ( n − β )( α + n − β + + α )( α + ( α + > , whence deriving that log r log A α,β ( g , r ) is not concave.(ii) Under α ∈ [ − , Situation 1 : f ∈ U ( D ). Upon writing f ( z ) = P ∞ n = a n z n , we compute Φ A , (cid:0) f ( z ) , t (cid:1) = ( π t ) − A ( f , t ) = ∞ X n = n | a n | t n − , and consequently, A α, ( f , r ) = P ∞ n = n | a n | R r ( π t ) − A ( z n , t ) d µ α ( t ) ν α ( r ) = ∞ X n = n | a n | A α, ( z n , r ) . So, by Lemma 3 (iii), we see that the convexity oflog r log A α, ( f , r ) under f ∈ U ( D )follows from the convexity oflog r log A α, ( z n , r ) under n ∈ N . So, it remains to verify this last convexity via the coming-up-next consideration.
Situation 2 : f ( z ) = a n z n with a n ,
0. Three cases are required to control.
Case 1 : α =
0. An easy computation shows A , ( z n , r ) = n − r n − and so log r log A , ( z n , r ) is convex. Case 2 : − ≤ α <
0. Under this condition, we see from the arguments for [10,Propositions 4-5] that ∆ ( n − , x ) ≥ ∀ n − ≥ < x < , and so that log r log A α, ( z n , r ) is convex. Case 3 : − ≤ α < −
2. With the assumption, we also get from the arguments for[10, Propositions 4-5] that ∆ ( n − , x ) ≥ ∆ ( − − α, x ) > ∀ x ∈ (0 ,
1) & n − ∈ [ − − α, ∞ )and so that log r log A α, ( z n , r ) is convex when n ≥
2. Here it is worth notingthat the convexity of log r log A α, ( z , r ) = < α < ∞ , from the argument for [10, Proposition 6] we know that ∆ ( n − β, x ) < x is su ffi ciently close to 1. Thus log r log A α,β ( a n z n , r ) is notconvex under a n , (cid:3) The following illustrates that the function log r log A α,β ( f , r ) is not alwaysconcave for α >
0, 0 ≤ β ≤
1, and f ∈ U ( D ). Example 1.
Let α = , β ∈ { , } , and f ( z ) = z + z . Then the function log r log A α,β ( f , r ) is neither convex nor concave for r ∈ (0 , .Proof. A direct computation shows (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z f ′ ( z ) f ( z ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z (1 + z )( z + z ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = | z | | z + | < EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 13 since | z | < < − | z | ≤ | z + | ∀ z ∈ D . So, f ∈ U ( D ) owing to Lemma 4 (i). By f ′ ( z ) = z + A ( f , t ) = Z t D | z + | dA ( z ) = π (cid:16) t + t (cid:17) , plus Z r Φ A ,β ( f , t ) d µ ( t ) = π (cid:16) r − r − r (cid:17) when β = r − r − r when β = ν ( r ) = Z r (1 − t ) dt = r − r . So, we get A ,β ( f , r ) = π (12 r − r − r )12(2 − r ) when β = − r − r − r ) when β = h β ( x ) = x − x − x − x when β = − x − x − x when β = x ∈ (0 , D ( h β ( x )) = ( D (12 x − x − x ) − D (2 − x ) when β = D (12 − x − x ) − D (2 − x ) when β = x ∈ (0 , D (12 x − x − x ) = − − x + x (12 − x − x ) D (2 − x ) = − − x ) D (12 − x − x ) = − − x + x (12 − x − x ) . Consequently, D ( h β ( x )) = g β ( x )(12 − x − x ) (2 − x ) when β = g β ( x )(12 − x − x ) (2 − x ) when β = , where g β ( x ) = ( − x + x − x + x when β = − x + x − x + x when β = . Now, under x ∈ (0 ,
1) we find g ′ ( x ) = − + x − x + x & g ′′ ( x ) = − x + x . Clearly, g ′′ ( x ) is an open-upward parabola with the axis of symmetry x = > g ′′ (1) = > g ′′ on (0 , g ′′ ( x ) > x ∈ (0 , g ′ is increasing on (0 , g ′ (0) = − < g ′ (1) = > x ∈ (0 ,
1) such that g ′ ( x ) < x ∈ (0 , x ) and g ′ ( x ) > x ∈ ( x , g (0) =
48 and g (1) = −
65, there exists an x ∈ (0 ,
1) such that g ( x ) > x ∈ (0 , x ) and g ( x ) < x ∈ ( x , x log h ( x ) isneither convex nor concave.Similarly, under x ∈ (0 ,
1) we have g ′ ( x ) = − + x − x + x & g ′′ ( x ) = − x + x . Obviously, g ′′ ( x ) is an open-upward parabola with the axis of symmetry x = >
1. By g ′′ (1) = > g ′′ on (0 , g ′′ ( x ) > x ∈ (0 , g ′ is increasing on (0 , g ′ (0) = − < g ′ (1) = − < g ′ ( x ) < x ∈ (0 , g (0) =
72 and g (1) = −
14, there exists an x ∈ (0 ,
1) such that g ( x ) > x ∈ (0 , x ) and g ( x ) < x ∈ ( x , x log h ( x ) is neither convex nor concave. (cid:3) Log-convexity for L α,β ( f , · ) . Analogously, we can establish the expected con-vexity for the mixed lengths.
Theorem 4.
Let ≤ β ≤ and < r < . (i) If α ∈ ( −∞ , − , then there exist f , g ∈ H ( D ) such that log r log L α,β ( f , r ) isnot convex and log r log L α,β ( g , r ) is not concave. (ii) If α ∈ [ − , , then log r log L α, ( a n z n , r (cid:1) is convex for a n , with n ∈ N .Consequently, log r log L α, ( f , r ) is convex for f ∈ U ( D ) . (iii) If α ∈ (0 , ∞ ) , then log r log L α,β ( a n z n , r ) is not convex for a n , and n ∈ N .Proof. The argument is similar to that for Theorem 3 except using the followingstatement for α ∈ [ − ,
0] – If f ∈ U ( D ), then there exists g ( z ) = P ∞ n = b n z n suchthat g is the square root of the zero-free derivative f ′ on D and f ′ (0) = g (0), andhence Φ L , ( f , t ) = (2 π t ) − Z t T | f ′ ( z ) || dz | = (2 π t ) − Z t T | g ( z ) | | dz | = ∞ X n = | b n | t n . (cid:3) Our concluding example shows that under 0 < α < ∞ and 0 ≤ β ≤ L α,β ( f , r ) is convex or concave in log r for all functions f ∈ U ( D ). EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 15
Example 2.
Let α = , β ∈ { , } , and f ( z ) = ( z + . Then the function log r log L α,β ( f , r ) is neither convex nor concave for r ∈ (0 , .Proof. Clearly, we have f ′ ( z ) = z + & f ′′ ( z ) = z + " f ′′ ( z ) f ′ ( z ) ′ − " f ′′ ( z ) f ′ ( z ) = − z + . It is easy to see that √ − | z | ) ≤ − | z | ∀ z ∈ D . So, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)" f ′′ ( z ) f ′ ( z ) ′ − " f ′′ ( z ) f ′ ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = | z + | ≤ − | z | ) ≤ − | z | ) . By Lemma 4 (ii), f belongs to U ( D ). Consequently, L ( f , t ) = Z π | f ′ ( te i θ ) | t d θ = π t ( t + Z r Φ L ,β ( f , t ) d µ ( t ) = π (cid:16) r − r − r (cid:17) when β = r − r − r when β = . Note that ν ( r ) = r − r . So, L ,β ( f , r ) = π (140 r − r − r )105(2 − r ) when β = − r − r − r when β = . To gain our conclusion, we only need to consider the logarithmic convexity ofthe function h β ( x ) = x − x − x − x when β = − x − x − x when β = . Case 1 : β =
0. Applying the definition of D -notation, we obtain D (140 x − x − x ) = − x − x + x (140 − x − x ) and D (2 − x ) = − x (2 − x ) , whence reaching D (cid:0) h ( x ) (cid:1) = D (140 x − x − x ) − D (2 − x ) = xg ( x )(140 − x − x ) (2 − x ) , where g ( x ) = − x + x − x + x . Obviously, g (0) = > g (1) = − < . Now letting s = x , we get g ( x ) = G ( s ) = − s + s − s + s , and G ′ ( s ) = − + s − s + s & G ′′ ( s ) = − s + s . Since the axis of symmetry of G ′′ is s = > G ′′ is decreasing on (0 , G ′′ (1) = >
0, we have G ′′ ( s ) > s ∈ (0 , G ′ ( s ) is increasingon (0 , G ′ (0) = − < G ′ (1) = > , we conclude that there exists an s ∈ (0 ,
1) such that G ′ ( s ) < s ∈ (0 , s )and G ′ ( s ) > s ∈ ( s , x ∈ (0 ,
1) such that g ( x ) isdecreasing for x ∈ (0 , x ) and g ( x ) is increasing for x ∈ ( x , x ∈ (0 ,
1) such that g ( x ) > x ∈ (0 , x ) and g ( x ) < x ∈ ( x , r log L α, ( f , r ) is neither concave nor convex. Case 2 : β =
1. Again using the D -notation, we obtain D (24 − x − x ) = − − x + x (24 − x − x ) and D (2 − x ) = − − x ) , whence deriving D (cid:0) h ( x ) (cid:1) = D (24 − x − x ) − D (2 − x ) = g ( x )(24 − x − x ) (2 − x ) , where g ( x ) = − x + x − x + x . Now we have g ′ ( x ) = − + x − x + x & g ′′ ( x ) = − x + x . Since the axis of symmetry of g ′′ ( x ) is x = > g ′′ ( x ) is decreasing on (0 , g ′′ (1) = >
0, we have g ′′ ( x ) > x ∈ (0 , g ′ ( x ) isincreasing on (0 , g ′ (0) = − < g ′ (1) = − < , we conclude that g ′ ( x ) < x ∈ (0 , g (0) = > g (1) = − < . Hence there exists an x ∈ (0 ,
1) such that g ( x ) > x ∈ (0 , x ) and g ( x ) < x ∈ ( x , r log L α,β = ( f , r ) is neither concavenor convex. (cid:3) EIGHTED INTEGRAL MEANS OF MIXED AREAS AND LENGTHS 17 R eferences [1] M. H. Al-Abbadi and M. Darus, Angular estimates for certain analytic univalent functions, Int. J. Open Problems Complex Analysis (2010), 212–220.[2] J. Arazy, S. D. Fisher and J. Peetre, M¨obius invariant function spaces, J. Reine Angew. Math. (1985), 110–145.[3] R. B. Burckel, D. E. Marshall, D. Minda, P. Poggi-Corradini and T. J. Ransford, Area, capacityand diameter versions of Schwarz’s Lemma,
Conform. Geom. Dyn. (2008), 133–152.[4] P. L. Duren, Univalent functions , Springer-verlag, NewYork, 1983.[5] P. J¨arvi, On some function-theoretic extremal problems,
Complex Variables Theory Appl. (1994), 267–270.[6] T. H. Macgregor, Length and area estimates for analytic functions, Michigan Mathe. J. (1964), 317–320.[7] Z. Nehari, The Schwarzian derivative and schlicht functions, Bull. Amer. Math. Soc. (1949),545–551.[8] M. Nunokawa, On some angular estimates of analytic functions, Math. Japon. (1995), no.2, 447–452.[9] G. P´olya and G. Szeg¨o, Problems and Theorems in Analysis, I. , Springer 1978.[10] C. Wang and K. Zhu, Logarithmic convexity of area integral means for analytic functions,arXiv:1101.2998v1[math.CV]15 Jan 2011.[11] J. Xiao and K. Zhu, Volume integral means of holomorphic functions,
Proc. Amer. Math. Soc. (2011), 1455–1465.[12] S. Yamashita, Length estimates for holomorphic functions,
Proc. Amer. Math. Soc. (1981),250–252.D epartment of M athematics and S tatistics , M emorial U niversity , NL A1C 5S7, C anada E-mail address ::