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Convex combination of analytic functions

  • Nak Eun Cho , Naveen Kumar Jain EMAIL logo and V. Ravichandran
Published/Copyright: March 29, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

Radii of convexity, starlikeness, lemniscate starlikeness and close-to-convexity are determined for the convex combination of the identity map and a normalized convex function F given by f(z) = α z+(1−α)F(z).

Keywords: Convex function; Starlike function; Close-to-convex function; Lemniscate of Bernoulli; Radius of starlikeness; Convolution; Convex combination
MSC 2010: 30C80; 30C45

1 Introduction

Let 𝒜 be the class of analytic functions f defined on the unit disc 𝔻 = {z ∈ ℂ : |z|<1}, and normalized by f(0) = 0 = f′(0) − 1. Let 𝒮, 𝒮𝒯, 𝒞𝒱 and 𝒞𝒞𝒱 denote the subclasses of 𝒜 consisting of functions univalent, starlike, convex and close-to-convex respectively. Recall that a function f ∈ 𝒜 is close-to-convex if there exists a convex function g such that Re(f′(z)/g′(z)) > 0 for all z ∈ 𝔻. The class 𝒮𝒯(β) of starlike functions of order β, 0 ≤β < 1, consists of f ∈ 𝒜 satisfying Re(zf′(z)/f(z))>β for all z ∈ 𝔻 and 𝒮𝒯 := 𝒮𝒯(0). The class 𝒞𝒱(β) of convex functions of order β is defined by 𝒞𝒱(β) = {f ∈ 𝒜 : zf′(z)∈ 𝒮𝒯(β)} and 𝒞𝒱 :=𝒞𝒱(0). The class 𝒮ℒ of lemniscate starlike functions, introduced by Sokół and Stankiewicz [1], consists of f ∈ 𝒜 satisfying |(zf′(z)/f(z))2−1| < 1 for all z ∈ 𝔻, or, equivalently, if zf′(z)/f(z) lies in the region bounded by the right-half of the lemniscate of Bernoulli given by | w2−1 | < 1. For recent investigation on the class 𝒮ℒ, see [1–5]. Another class of our interest is the class ℳβ, β > 1, consisting of f ∈ 𝒜 satisfying Re(zf′(z)/f(z)) < β for all z ∈ D. The class ℳβ was investigated by Uralegaddi et al. [6], while its subclass was investigated by Owa and Srivastava [7]. Related radius problem for this class can be found in [8] and [9].

Properties of linear combination, in particular, convex combination of functions belonging to various classes of functions were initially investigated by Rahmanov in 1952 and 1953 [10, 11]. The survey article of Campbell [12] provides several results concerning various combination of univalent functions as well as of locally univalent functions. Convex combination of univalent functions and the identity function were investigated by several authors (see Merkes [13] and references therein as well as [14]); in particular, Merkes [13] proved some results related to the present investigation. Obradovic and Nunokowa [15] investigated functions f ∈ 𝒜 satisfying the following condition

Re1+zf″(z)f(z)−α>0(z∈D) (1)

for some α∈[−1, 1) and obtained the following result.

Theorem 1.1

([15, Theorem 2]). If f ∈ 𝒜 satisfies the condition (1), then (i) f∈STfor−13≤α≤13, and (ii) f∈CCVfor−13≤α≤1.

If the function f is the convex combination f(z) = α z+(1−α)F(z), then the condition (1) is equivalent to the conditions that F ∈ 𝒞𝒱. If two subclasses 𝒢 and ℱ of 𝒜 are given, the 𝒢-radius of ℱ, denoted by R𝒢(ℱ), is the largest number R such that f(rz)/r ∈ 𝒢 for 0 < r < R, and for all f ∈ ℱ. Whenever 𝒢 is characterized by possesing a geometric property P, the number R is also referred to as the radius of property P for the class ℱ. In this paper, we investigate radius problem for functions f satisfying the condition (1) to belong to one of the classes introduced above. We also prove the correct results corresponding to [15, Theorem 1(a) and Theorem 2(a), p. 100] that f ∈ 𝒞𝒱 if f ∈ 𝒜 satisfies the condition (1) for some 0 ≤ α ≤ (12 2 −15)/9. Their result is correct only when α = 0. Unlike the radii problems associated with starlikeness and convexity, where a central feature is the estimate for the real part of the expressions zf′(z)/f(z) or 1+zf″(z)/f′(z) respectively, the 𝒮ℒ-radius problems are tackled by first finding the disc that contains the values of zf′(z)/f(z) or 1+zf″(z)/f′(z). The techniques used in this paper are earlier used for the class of uniformly convex functions investigated in [16–26].

For two analytic functions f, g ∈ 𝒜 with f(z)=z+ ∑n=2∞ anzn and g(z) = z + ∑n=2∞ bnzn, their convolution or Hadamard product, denoted by f * g, is defined by (f*g)(z) := z+ ∑n=2∞ anbnzn. We need the following results.

Theorem 1.2

([28, Theorem 2.1-2.2, p. 125]). The classes 𝒮𝒯, 𝒞𝒱 and 𝒞𝒞𝒱 are closed under convolution with functions in 𝒞𝒱.

Theorem 1.3

The classes 𝒮𝒯(β), 𝒮ℒ and ℳ(β)∩ 𝒮𝒯 are closed under convolution with functions in 𝒞𝒱.

Lemma 1.4

([4, Lemma 2.2, p. 6559]). Let 0<a<2. If ra is given by

ra=(1−a2−(1−a2))1/2(0<a≤22/3)2−a(22/3≤a<2),

then {w : |w − a| < ra} ⊆ {w : |w2 − 1| < 1}.

Theorem 1.3 is a special case of results of Shanmugam [27, Theorem 3.3, p. 336; Theorem 3.5, p. 337] (see also Ma and Minda [18, Theorem 5, p. 167]).

2 Radii problems associated with convex combinations

For functions satisfying the condition (1), we determine, in the first part of the following theorem, the range of α so that the function is starlike of order β while the other parts of the theorem provide the radius of starlikeness of order β.

Theorem 2.1

Let −1 ≤ α < 1, 0 ≤ β < 1 and f ∈ 𝒜 satisfy the condition (1).

  1. If 0 ≤ β ≤ 1/2 and |α| ≤ (1 − 2β)/(3 − 2β), then f ∈ 𝒮𝒯(β).

  2. If either

    1. 0 ≤ β ≤ 1/2, and (1 − 2β)/(3 − 2β) < α < 1, or

    2. 1/2 < β < 1, and (−β + 2β2)/(6 − 7β + 2β2)< α < 1,

    then f(ρ1z)/(ρ1) ∈ 𝒮𝒯(β) where ρ1 = ρ1(α, β) is given by

    ρ 1 ( α , β ) = α ( 3 − 2 β ) 2 − ( 1 − 2 β ) 2 4 ( 1 − α ) α 2 ( α ( β − 2 ) − β ) ( β − 1 ) + α ( 1 + 4 β − 4 β 2 ) + α 2 ( 7 − 12 β + 4 β 2 ) 1 2 .

  3. If either

    1. 0 ≤ β ≤ 1/2, and −1 ≤ α < (1 − 2β)/(−3 + 2β), or

    2. 1/2 < β < 1, and −1 ≤ α < (−β + 2β2)/(6 − 7β + 2β2),

    then f(ρ0z)/ρ0 ∈ 𝒮𝒯(β) where ρ0 = ρ0(α, β) is given by

    ρ0(α,β)=2(1−β)α(β−2)+β+−4α(1−β)2+(α(β−2)+β)2.

Proof

Define the function g : 𝔻 → ℂ by

g(z)=z−αz21−z. (2)

For a fixed α ∈ [−1, 1), define the function f1 : 𝔻 → ℂ by

f1(z)=f(z)−αz1−α. (3)

If f satisfies the condition (1), then it follows that the function f1 ∈ 𝒞𝒱. With the function g defined by (2), the equation (3) shows that

f(z)=αz+(1−α)f1(z)=f1(z)∗g(z).

If g is starlike in the disc 𝔻ρ, then g(ρz)/ρ is starlike and hence, by Theorem 1.2, f1(z)*(g(ρz)/ρ) is starlike or equivalently f1*g is starlike in the disc 𝔻ρ. In view of this, it is enough to investigate the radius of starlikeness of the function g given by (2).

For the function g given by (2), we have

zg′(z)g(z)=11−z−αz1−αz.

With z = reit and x = cos t, a calculation shows that

Re(zg′(z)g(z))=1−rcos⁡t1+r2−2rcos⁡t+rα(rα−cos⁡t)1+r2α2−2rαcos⁡t=1+2r2α2+r4α2−r(1+3α)(1+r2α)x+4r2αx2(1+r2−2rx)(1−2rxα+r2α2).

Therefore, g is starlike of order β in |z| < ρ1 if, for all 0 ≤ r < ρ1 and for all x ∈ [−1, 1], we have

1+2r2α2+r4α2−r(1+3α)(1+r2α)x+4r2αx2−β(1+r2−2rx)(1−2rxα+r2α2)≥0.

This inequality is equivalent to

h(r,x):=4r2α(1−β)x2+r(1+r2α)(−1−3α+2β+2αβ)x+1+2r2α2+r4α2−β−r2β−r2α2β−r4α2β≥0

for 0 ≤ r < ρ1 and for all x ∈ [−1, 1]. It follows that the derivative of function h(r, x) with respect to x vanishes for

x=x0=(1+r2α)(1−2β+α(3−2β))8rα(1−β)(α≠0).

It should be noted for later use that, for β ≤ 1/2 and α ≥ 0,

h(r,1)−h(r,−1)=(1−r)(1−rα)(1+r2α(1−β)−β+r(α(β−2)+β))−(1+r)(1+rα)(1+r2α(1−β)−β+r(α(2−β)−β))=2r(1+r2α)(−1−3α+2β+2αβ)≤2r(1+r2α)(−2α)≤0, (4)

and so we have h(r, 1) ≤ h(r, −1).

  1. Case (i) Let 0 ≤β ≤ 1/2 and 0 ≤ α ≤(1 − 2β)/(3 − 2β) . If β = 1/2, then α = 0, h(r, x) = (1 − r2)/2 and hence min|x|≤ 1h(r, x) > 0 for 0 ≤ r < 1. If 0 ≤ β < 1/2 and α = 0, then

    h(r,x)=1−β−r2β+(2β−1)rx

    and so

    min|x|≤1h(r,x)=h(r,1)=(1−r)(1−β+βr)>0for0≤r<1.

    If 0 ≤ β < 1/2 and 0 < α ≤ (1 − 2β)/(3 − 2β), then it can be verified that x0 > 1. In view of this and from (4), it follows that min|x|≤ 1h(r, x) = h(r, 1). Since h(r, 1) is a decreasing function of α, we have, for 0 ≤ r < 1,

    (1−r)(3−2β+r(−1+2β))min|x|≤1h(r,x)=h(r,1)≥×(3−5β+2β2+r(−2+8β−4β2)+r2(1−3β+2β2))(3−2β)2>0.

    Therefore, for 0 ≤ r < 1, min|x|≤ 1h(r, x) > 0 and so g is starlike of order β.

    Case (ii) Let 0 ≤ β < 1/2 and (1 − 2β)/(−3 + 2β) < α < 0. For fixed r, the second partial derivative of h(r, x) with respect to x is negative and so the minimum of h(r, x) in [−1, 1] is attained at the end points x = ± 1. Using (4) and the fact that h(r, 1) is a decreasing function of α, min|x|≤ 1h(r, x) > 0 for 0 ≤ r < 1, and min|x|≤ 1h(r, x) > 0. Therefore, g is starlike of order β.

  2. Case (i) Let 0 ≤ β ≤ 1/2 and (1 − 2β)/(3 − 2β)< α < 1. It can be seen that −1 ≤x0 ≤ 1 if

    r≥4α(1−β)−(1−α)α(α(3−2β)2−(1−2β)2)α(1−2β+α(3−2β)):=γ0.

    So, for γ0 ≤ r < ρ1, min|x|≤1h(r, x)= h(r, x0) > 0, where

    h(r,x0)=1−α16α(1−β)(9α−1+4β−12αβ−4β2+4αβ2−2(1+7α+4β−12αβ−4β2+4αβ2)αr2+(9α−1+4β−12αβ−4β2+4αβ2)α2r4).

    Notice that the number ρ1(α, β) is the root of h(r, x0) = 0. For 0 ≤ r < γ0, since

    s(α)=1+r2α(1−β)−β+r(α(β−2)+β)

    is a decreasing function of α,

    min|x|≤1h(r,x)=h(r,1)=(1−r)(1−rα)(1+r2α(1−β)−β+r(α(β−2)+β))≥(1−r)4(1−β)>0. (5)

    Therefore, min|x|≤ 1h(r, x) ≥ 0 for 0 ≤ r<ρ1 and hence g is starlike of order β in |z| < ρ1.

    Case (ii) Let 1/2 < β < 1 and (1 − 2β)/(−3 + 2β) < α < 1. Let r < ρ1. It can be shown that

    −1≤x0≤1forr≥4α(1−β)−(1−α)α(α(3−2β)2−(1−2β)2)α(1−2β+α(3−2β)):=γ1.

    This shows that for γ1 ≤ r < ρ1, min|x|≤ 1h(r, x) = h(r, x0) > 0.

    On the other hand if 0 < r < γ1, then x0 > 1. Now proceeding as in case (i) we get min|x|≤ 1h(r, x) = h(r, 1) > 0. Thus g is starlike of order β in |z| < ρ1.

    Case (iii) Let 1/2 < β < 1 and (−β + 2β2)/(6 − 7β + 2β2) < α ≤ (1 − 2β)/(−3 + 2β). Let r2 be the root of the equation x0 = −1. Then r2 ≤ ρ1. Let r < ρ1. If r ≥ r2 then −1 ≤x0 ≤ 1. Thus min|x|≤ 1h(r, x) = h(r, x0) > 0. On the other hand if r < r2, then x0 < − 1. Also ρ1< ρ0. Thus min|x|≤ 1h(r, x)= h(r, −1) > 0 in |z|< ρ1 and hence g is starlike of order β in |z|< ρ1.

  3. Case (i) Let 0 ≤ β ≤ 1/2 and −1 ≤ α < (1 − 2β)/(−3 + 2β). For fixed r, the second derivative test shows that the minimum of h(r, x) in [−1, 1] is attained at the end points x = ±1. It follows from (4) that min|x|≤ 1h(r, x) = h(r, −1). Notice that

    h(r,−1)=(1+r)(1+rα)(1−β−r(β+α(−2+r(−1+β)+β)))

    and the number ρ0(α, β) is the root of h(r, −1) = 0. Since h(r, −1) > 0 for r < ρ0, min|x|≤ 1h(r, x) > 0 in |z| < ρ0. Thus g is starlike of order β in |z| < ρ0.

    Case (ii) Let 1/2 < β < 1 and 0 < α ≤ (−β + 2β2)/(6 − 7β + 2β2). Then r2≥ρ0. Let r < ρ0. Since x0 < − 1, we have min|x|≤ 1h(r, x) = h(r, −1) > 0 in |z| < ρ0. Thus g is starlike of order β in |z| < ρ0.

    Case (iii) Let 1/2 < β < 1 and −1 ≤ α ≤ 0. For fixed r, the minimum of h(r, x) in [−1, 1] is attained at the end points x = ±1. Also h(r, 1) − h(r, −1) > 0. Thus min|x|≤ 1h(r, x) = h(r, −1) > 0 in |z| < ρ0. Thus g is starlike of order β in |z| < ρ0.

To prove the sharpness, consider the functions f and g given by

f(z)=g(z)=z−αz21−z. (6)

For the function f given in (6), we have

Re(1+zf″(z)f′(z)−α)=Re(1+z1−z)>0.

The function f satisfies the condition (1) and since radius is sharp for the function g, the sharpness follows. □

Remark 2.2

If −1/3 ≤ α ≤ 1/3 and f ∈ 𝒜 satisfy the condition (1), then f ∈ 𝒮𝒯(β) where

β=1−3|α|2(1−|α|)≥0.

In particular we have the following corollary.

Corollary 2.3

([15, Theorem 2(b), p. 100]). If f satisfies the condition (1) for some α with |α|≤13, then f ∈ 𝒮𝒯.

For other ranges of α, we have the following corollary.

Corollary 2.4

The radius of starlikeness of the class of functions f ∈ 𝒜 satisfying the condition (1) for α ∈ (−1, −1/3] ∪ [1/3, 1) is given by

r=(9α−1α+7α2+42(1−α)α3)12if13≤α<1,1α(α−1)−αif−1<α≤−13.

Theorem 2.5

For −1 ≤ α < 1, the 𝒮ℒ-radius of the class of functions f ∈ 𝒜 satisfying the condition (1) is given by

ρ2(α)=2−21+α−2α+(1−α)(1−3α+22α).

Proof

First we observe that

|(1−αz)(1−z)|≥(1−αr)(1−r),|z|=r<1. (7)

For 0 ≤ α < 1, (7) is trivial. For −1 ≤ α < 0, the inequality (7) holds as the function

h(r,x):=|(1−αz)(1−z)|2=(1−2rαx+r2α2)(1−2rx+r2),

where x = cos t, is a decreasing function. Since the function g given by (2) satisfies

|zg′(z)g(z)−1|=|(1−α)z(1−z)(1−αz)|≤r(1−α)(1−r)(1−αr), (8)

Equation 8 and Lemma 1.4 yield

|(zg′(z)g(z))2−1|<1(|z|<r)

provided

1−2+2(1−2α+α)r+α(1−2)r2≤0orr≤ρ2.

Thus g(ρ2z)/ρ2 ∈ 𝒮ℒ. As the function f satisfies (1), the function f1 defined by (3) is convex. Hence, by Theorem 1.3, we have

f(ρ2z)ρ2=f1(z)∗g(ρ2z)ρ2∈SL

or, equivalently

|(zf′(z)f(z))2−1|<1(|z|<ρ2).

For z = ρ2, the function g given by (2) satisfies

(zg′(z)g(z))2−1=(11−ρ2−αρ21−αρ2)2−1=(ρ2(1−α)(1−ρ2)(1−αρ2)+1)2−1=1.

Thus, the result is sharp for the function

f(z)=g(z)=z−αz21−z.

 □

Corollary 2.6

The 𝒮ℒ-radius of the class 𝒞𝒱 of convex functions is 2−2.

Theorem 2.7

For −1/3 ≤ α ≤ 1/3 and β > 1, the ℳβ ∩ 𝒮𝒯 radius of the class of functions f ∈ 𝒜 satisfying the condition (1) is given by

ρ3(α,β)=2(β−1)β−2α+αβ+(α−1)(4α−4αβ−β2+αβ2).

Proof

From the inequality (8) for the function g given by (2), we have

Rezg′(z)g(z)≤1+r(1−α)(1−r)(1−αr)≤β

provided

1−β+(−2α+β+αβ)r+α(1−β)r2≤0.

The last inequality holds if 0 < r ≤ ρ3(α, β) . Thus g(ρ3z)/ρ3 ∈ ℳβ. Since the function f satisfies (1), the function f1 defined by (3) is convex. Hence, by Theorem 1.3, we have

f(ρ3z)ρ3=f1(z)∗g(ρ3z)ρ3∈Mβ

or

Re(zf′(z)f(z))<β(|z|<ρ3).

Also, by Corollary 2.3,

0<Re(zf′(z)f(z))(|z|<1).

Thus f(ρ3z)/ρ3 ∈ ℳβ ∩ 𝒮𝒯. For z = ρ3, the function g given by (2) satisfies

Rezg′(z)g(z)=1+ρ3(1−α)(1−ρ3)(1−αρ3)=β.

Thus the result is sharp for the function

f(z)=g(z)=z−αz21−z.

 □

Corollary 2.8

For β > 1, the ℳβ ∩ 𝒮𝒯-radius of the class 𝒞𝒱 of convex functions is 1 − β−1.

The results [15, Theorem 1(a) and Theorem 2(a), p. 100] of Obradovic and Nunokawa are incorrect. The correct version of these results are given in the following theorem.

Theorem 2.9

For −1 ≤ α < 1, the radius of convexity of the class of functions f ∈ 𝒜 satisfying the condition (1) is

ρ4(α)=r1,if0≤α<1,r2if−1≤α≤0, (9)

where r1 ∈ (0, 1] is the root of the equation in r:

1−r2(1+α)−r4α(1−2α)−2r3α(1−α)(1−r2α)=0

and r2 ∈ (0, 1] is the root of the equation in r:

1+6αr−(1−8α−8α2)r2+2α(1+9α)r3+15α2r4+6α2r5+α2r6=0.

Proof

If the function f satisfies (1) with α = 0, then the function f is convex and so ρ4(0) = 1 as claimed. Now, assume that α ≠ 0. For the function g given by (2), a calculation shows that, with x = cos t,

Re(1+zg″(z)g′(z))=1−r21+r2−2rcos⁡t+2rα(−(1+3r2α)cos⁡t+r((2+r2)α+cos⁡(2t)))1+r2(4+r2)α2+2rα(−2(1+r2α)cos⁡t+rcos⁡(2t))=ϕ(r,x)h(r,x)

where

h(r,x)=(1+r2−2rx)(1+r2(4+r2)α2−2rα(r+2x−2rx2+2r2xα))

and

ϕ(r,x)=1−r2−4r2α+8r2α2+3r4α2+r6α2−6rα(1−r2+3r2α+r4α)x+12r2α1+r2αx2−8r3αx3 (10)

Then

∂∂xϕ(r,x)=−6rα1−r2+4r2x2+3r2α+r4α+24r2xα1+r2α.

A calculation shows that

∂∂xϕ(r,x)=0ifx=x0=1+r2α−r(1−α)(1−r2α)2r

and

∂2∂x2ϕ(r,x0)=24r3α(1−α)(1−r2α)>0for0<α<1.

So for 0 < α < 1,

min|x|≤1ϕ(r,x)=ϕ(r,x0)=(1−α)(1−r2α)(1−r2(1+α)−r4α(1−2α)−2r3α(1−α)(1−r2α))>0.

On the other hand, if −1 < α < 0, then, for r < r2, it can be verified that

min|x|≤1ϕ(r,x)=ϕ(r,−1)=1+6αr−(1−8α−8α2)r2+2α(1+9α)r3+15α2r4+6α2r5+α2r6>0.

Thus g(ρ4z)/ρ4 ∈ 𝒞𝒱. Since the function f satisfies (1), the function f1 defined by (3) is convex. Hence

f(ρ4z)ρ4=f1(z)∗g(ρ4z)ρ4∈CV

or

Re(1+zf″(z)f′(z))>0(|z|<ρ4).

The result is sharp for the function f given by (6). □

Theorem 2.10

For −1 ≤ α < −1/3, the radius of close-to-convexity of the class of functions f ∈ 𝒜 satisfying the condition (1) is given by

ρ5(α)=1α(α−1)−α.

Proof

The function g1 : 𝔻 → ℂ by

g1(z)=−ln⁡(1−z)(z∈D)

is clearly convex in 𝔻. For the functions g given by (2) and g1 above, we have

Re(g′(reit)g1′(reit))=1+αr2−(2α+1+αr2)rx+2αr2x21+r2−2rx (11)

where x := cos t. Let h : [−1, 1] → ℝ be defined by

h(r,x)=1+αr2−(2α+1+αr2)rx+2αr2x2.

Then

∂∂xh(r,x)=0ifx=x0=1+2α+r2α4rα

and

∂2∂x2h(r,x)=4r2α<0.

Therefore, for a fixed r, the minimum of h(r, x) is attained at x = ± 1. Since

h(r,−1)−h(r,1)=2r(1+2α+r2α)<0,

it follows that

min|x|≤1h(r,x)=h(r,−1)=(1+r)(1+2rα+r2α)>0forr<ρ5.

Thus g(ρ5z)/ρ5 ∈ 𝒞𝒞𝒱. Since the function f1 defined by (3) is convex as f satisfies (1), we have, by Theorem 1.2,

f(ρ5z)ρ5=f1(z)∗g(ρ5z)ρ5∈CCV

or

Re(g′(z)g1′(z))>0(|z|<ρ5).

The result is sharp for the function

f(z)=g(z)=z−αz21−z.

 □

Acknowledgement

This research was supported by the Basic Science Research Program (No. 2016R1D1A1A09916450) through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology. The authors are thankful to the referees for their comments.

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Received: 2016-8-18
Accepted: 2017-1-26
Published Online: 2017-3-29

© 2017 Cho et al.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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https://doi.org/10.1515/math-2017-0021
Keywords for this article
Convex function; Starlike function; Close-to-convex function; Lemniscate of Bernoulli; Radius of starlikeness; Convolution; Convex combination
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