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Research article Special Issues

The insiders' gaze: fieldworks, social media and visual methodologies in urban tourism

  • Received: 03 March 2022 Revised: 07 April 2022 Accepted: 08 May 2022 Published: 19 May 2022
  • This paper aims at scrutinizing the role of two overlapping dimensions in tourism geography teaching and learning: the real-world perspective, represented by the fieldwork as a long-established practice in geographical teaching; and the digital sphere, namely the use of visual methodologies and social media as tools to activate city image-making in tourist practices.

    In particular, the work deepens critical aspects and potentialities of a mixed-method approach combining both traditional methods and more innovative tools to carry out researches and teaching in the field of tourism geography and cultural heritage. Theoretically inserted at the interplay among tourism, digital geography and visual methods, it draws on a series of workshops organized with third-level students between 2016 and 2018. Although problematizing some crucial conceptual issues related to the role of visualities and new technologies, findings highlight the didactic effectiveness of mixed methods that entail the more traditional ones such as team working and fieldworks, and the most innovative such as visual methods and social media, to increase students' interest in geography learning and stimulating their own perceptions of ordinary landscapes through a gaze repositioning.

    Citation: Teresa Graziano. The insiders' gaze: fieldworks, social media and visual methodologies in urban tourism[J]. AIMS Geosciences, 2022, 8(3): 366-384. doi: 10.3934/geosci.2022021

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  • This paper aims at scrutinizing the role of two overlapping dimensions in tourism geography teaching and learning: the real-world perspective, represented by the fieldwork as a long-established practice in geographical teaching; and the digital sphere, namely the use of visual methodologies and social media as tools to activate city image-making in tourist practices.

    In particular, the work deepens critical aspects and potentialities of a mixed-method approach combining both traditional methods and more innovative tools to carry out researches and teaching in the field of tourism geography and cultural heritage. Theoretically inserted at the interplay among tourism, digital geography and visual methods, it draws on a series of workshops organized with third-level students between 2016 and 2018. Although problematizing some crucial conceptual issues related to the role of visualities and new technologies, findings highlight the didactic effectiveness of mixed methods that entail the more traditional ones such as team working and fieldworks, and the most innovative such as visual methods and social media, to increase students' interest in geography learning and stimulating their own perceptions of ordinary landscapes through a gaze repositioning.



    For real numbers a,b,c with cN{0}, the Gaussian hypergeometric function is defined by

    F(a,b;c;x)=n=0(a,n)(b,n)(c,n)xnn!,|x|<1,

    where (a,0)=1 for a0 and (a,n) is the shifted factorial function given by

    (a,n)=a(a+1)(a+2)(a+n1)

    for nN. It is well known that the Gaussian hypergeometric function, F(a,b;c;x), has a broad range of applications, including in geometric function theory, the theory of mean values, and numerous other areas within mathematics and related disciplines. Many elementary and special functions in mathematical physics are either particular cases or limiting cases. Specifically, F(a,b;c;x) is said to be zero-balanced if c=a+b. For the case of c=a+b, as x1, Ramanujan's asymptotic formula satisfies

    F(a,b;a+b,x)=R(a,b)ln(1x)B(a,b)+O((1x)ln(1x)), (1.1)

    where

    B(a,b)=Γ(a)Γ(b)Γ(a+b)

    is the classical beta function [1] and

    R(a,b)=2γΨ(a)Ψ(b),

    here Ψ(z)=Γ(z)/Γ(z), Re(x)>0 is the psi function, and γ is the Euler–Mascheroni constant [1].

    Throughout this paper, let a[12,1), and we denote r=1r2 for r(0,1). The generalized elliptic integrals of the first and second kind are defined on (0,1) as follows [2]:

    Ka=Ka(r)=π2F(a,1a;1,r2),Ka(0)=π2,Ka(1)=, (1.2)

    and

    Ea=Ea(r)=π2F(a1,1a;1,r2),Ea(0)=π2,Ea(1)=sin(πa)2(1a). (1.3)

    Set Ka(r)=Ka(r),Ea(r)=Ea(r). Note that when a=12, Ka(r) and Ea(r) reduce to the classical complete elliptic integrals K(r) and E(r) of the first and second kind, respectively

    K(r)=π2F(12,12;1;r2),E(r)=π2F(12,12;1;r2).

    It is well known that complete elliptic integrals play a crucial role in various areas of mathematics and physics. In particular, these integrals provide a foundation for investigating numerous special functions within conformal and quasiconformal mappings, including the Grötzsch ring function, Hübner's upper bound function, and the Hersch–Pfluger distortion function[3,4]. In 2000, Anderson, et al. [5] reintroduced the generalized elliptic integrals in geometric function theory. They discovered that the generalized elliptic integral of the first kind, denoted as Ka, originates from the Schwarz–Christoffel transformation [3] of the upper half–plane onto a parallelogram and established several monotonicity theorems for Ka and Ea. The generalized Grötzsch ring function in generalized modular equations and the generalized Hübner upper bound function can also be expressed in terms of generalized elliptic integrals[6]. Recently, generalized elliptic integrals have garnered significant attention from mathematicians. A wealth of properties and inequalities for these integrals can be found in the literature. Specifically, various properties of elliptic integrals and hypergeometric functions, including monotonicity, approximation, and discrete approximation, have been investigated in [7,8,9], with sharp inequalities derived for elliptic integrals. Additionally, studies in [10,11] primarily focus on inequalities between different means, such as the Toader mean, and Hölder mean, as well as their applications in elliptic integrals.

    For r(0,1), r=1r2, it is known that the arc-length of an ellipse with semiaxes 1 and r, denoted as L(1,r), is given by L(1,r)=4E(r). Muir indicated that L(1,r) can be approximated by 2π{(1+r322)23. Later, Vuorinen conjectured the following inequality for r(0,1):

    π2(1+r322)23<E(r),

    which was subsequently proven by Barnard et al.[12].

    The Hölder mean of positive numbers x,y>0 with order sR, is defined as

    Hs(x,y)={(xs+ys2)1s,s0,xy,s=0.

    It is easy to see Hs(x,y) is strictly increasing with respect to s. Alzer and Qiu [13] established the following inequalities:

    π2Hs1(1,r)<E(r)<π2Hs2(1,r) (1.4)

    with the best constants s1=3/2 and s2=log2/log(π/2)=1.5349, see [13,14] for details.

    The generalized weighted Hölder mean of positive numbers x,y, with weight ω and order s, is defined as [14]:

    Hs(x,y;ω)={[ωxs+(1ω)ys]1s,s0,xωy1ω,s=0. (1.5)

    Wang et al. [15] proved that for r(0,1), the following inequality holds:

    π2Hs1(1,r;α)<E(r)<π2Hs2(1,r;β), (1.6)

    and the best parameters α=α(s), β=β(s) satisfy

    α(s)={12,s(,32],1η,s(32,2),(2π)s,s[2,),β(s)={1,s(,0],(2π)s,s(0,s0),12,s[s0,),

    where s0=log2log(2/π)=1.5349, η=Fs(r0)>12, Fs=1[2E(r)/π]s1rs, and r0=r0(s)(0,1) is the value such that Fs(r) is strictly increasing on (0,r0) and strictly decreasing on (r0,1)) for s(32,2).

    The extension of the inequality (1.6) to the second kind of generalized elliptic integral Ea, where a[12,1), is a natural inquiry. This paper aims to address this question. One might wonder why the parameter a is restricted to the interval [12,1) rather than (0,1). For a(0,12), our analysis has revealed a lack of the expected monotonicity in the function Fa,p(x), as defined in Theorem 3.1. This monotonicity is crucial for establishing the desired inequalities.

    To achieve our purpose, we require some more properties of generalized elliptic integrals of the first and second kind. Therefore, Section 2 will introduce several lemmas that establish these properties. Section 3 will present our main results along with their corresponding proofs. In Section 4, we establish several functional inequalities involving Ea as applications. Finally, we give the conclusion of this article.

    In this section, several key formulas and lemmas are presented to support the proof of the main results. The derivative formulas of the generalized elliptic integrals are given.

    Lemma 2.1 ([5]). For a(0,1) and r(0,1), we have

    dKadr=2(1a)(Ear2Ka)rr2,dEadr=2(1a)(KaEa)r,ddr(KaEa)=2(1a)rEar2,ddr(Ear2Ka)=2arKa.

    The following lemma provides the monotonicity of some generalized elliptic integrals with respect to r, which can be found in [16].

    Lemma 2.2. Let a(0,1). Then the following function:

    (1) rEar2Kar2 is increasing from (0,1) to (πa2,sin(πa)2(1a));

    (2) rEar2Kar2Ka is decreasing from (0,1) to (0,a);

    (3) rr2(KaEa)r2Ea is decreasing from (0,1) to (0,1a);

    (4) rKaEa)r2Ka is increasing from (0,1) to (1a,1);

    (5) rrc(KaEa)r2 is decreasing on (0,1) if and only if ca(2a).

    Lemmas 2.3 and 2.4 are important tools for proving the monotonicity of the related functions.

    Lemma 2.3 ([17]). Let α(x)=n=0anxn and β(x)=n=obnxn be real power series that converge on (r,r)(r>0), and bn>0 for all n. If the sequence {anbn}n0 is increasing (or decreasing) on (0,r), then so is α(x)β(x).

    Lemma 2.4 ([3]). For a,b(,) and a<b, let f,g:[a,b] be continuous on [a,b] and be differentiable on (a,b), and g(x)0 for all x(a,b). If f(x)g(x) is increasing (or decreasing) on (a,b), then so are

    f(x)f(a)g(x)g(a)andf(x)f(b)g(x)g(b).

    In particular, if f(a)=g(a)=0(orf(b)=g(b)=0), then the monotonicity of f(x)g(x) is the same as f(x)g(x).

    However, f(x)g(x) is not always monotonic; it is sometimes piecewise monotonic. An auxiliary function Hf,g [8] is defined as

    Hf,g:=fggf, (2.1)

    where f and g are differentiable on (a,b) and g0 on (a,b) for <a<b<. If f and g are twice differentiable on (a,b), the function Hf,g satisfies the following identities:

    (fg)=fgfgg2=gg2(fggf)=gg2Hf,g, (2.2)
    Hf,g=(fg)g. (2.3)

    Here, Hf,g establishes a connection between fg and fg.

    Lemma 2.5. Define the function f1(x) on [12,1) by

    f1(x)=2(1x)logxlog(sin(πx)/(π(1x))).

    Then 2x<f1(x)<2.

    Proof. To establish the right-hand side of the inequality, it suffices to prove that

    (1x)logxlogsin(πx)π(1x)>0.

    Denote

    g1(x)=(1x)logxlogsin(πx)π(1x).

    By differentiation, we obtain

    g1(x)=logx+1xxπcos(πx)sin(πx)11x,g1(x)=1x1x2+π2sin2(πx)1(1x)2.

    Observe that g1(12)=π210=0.130...<0, and limx1g1(x)=+. This implies that there exists x0[12,1) such that g1(x) is decreasing on [12,x0) and increasing on (x0,1). Since g1(12)=log21=0.306, and g1(1)=0, it is clear that

    g1(x)max{g1(12),g1(1)}=0,

    which implies that g1(x) is decreasing on [12,1). Consequently,

    g1(x)>g1(1)=0.

    In order to establish the left-hand side of the inequality, we define

    g2(x)=2(1x)logx(2x)logsin(πx)π(1x).

    Note

    g2(12)=log1232log2π=0.015...,g2(1)=0. (2.4)

    Differentiating g2(x) yields

    g2(x)=2logx+2(1x)x(2x)πcos(πx)sin(πx)2x1x+logsin(πx)π(1x).

    Observe that

    g2(12)=log8π1=0.065...<0,g2(34)=log3229π+5π4133=4.166...>0.

    Based on these observations and the intermediate value theorem, there exists x2[12,1) such that g2(x2)=0 and g2(x) is decreasing on [12,x2) and increasing on (x2,1). Therefore, together with (2.4), we conclude that

    g2(x)<0.

    This completes the proof.

    Lemma 2.6. For each a[12,1), the function

    f2(r)=r2a(2a)[a(KaEa)(1a)(Ear2Ka)]Ear2Kaar2Ea

    is decreasing from (0,1) to (0,a2a).

    Proof. Following from (1.2) and (1.3), we deduce that

    a(KaEa)(1a)(Ear2Ka)=π4a2(1a)r4F(a+1,2a;3;r2),Ear2Kaar2Ea=a(1a)(2a)π4r4F(a,2a;3;r2).

    To establish the desired monotonicity of f2(r), it suffices to prove that the function f3(x), defined on (0, 1) by

    f3(x)=(1x)p(a)F(a+1,2a;3;x)F(a,2a;3;x),

    is decreasing on (0,1), where p(a)=2a(2a)2. Using the power series expansion, the function can be expressed as

    xn=1Unxnn=1Vnxn,

    where the coefficients Un and Vn satisfy the recursive relations, as detailed in [18]:

    U0=1,V0=1,Un+1=anUnbnUn1,Vn=(a)n(2a)n(3)nn!, (2.5)

    with

    an=4n2+2(3a2+2a)n+(5a2+8a2)2(n+1)(n+3),bn=(2n+4a2a2)(2na2)4(n+1)(n+3).

    By Lemma 2.3, we aim to prove that the sequence {UnVn}n0 is decreasing. Note that

    Un>0,Vn>0,

    and

    U0V0=1,U1V1=5a2+8a+22a(2a),U2V2=8a3+10a2+a2a(3a)(1+a).

    Observe that

    U0V0>U1V1>U2V2,

    which implies

    U1V1V0U0<0,U2V2V1U1<0.

    Assuming that UkVkVk1Uk1<0 for all 1kn, we prove by induction that Un+1Vn+1VnUn<0. According to (2.5), we have

    Un+1Vn+1VnUn=(anUnbnUn1)Vn+1VnUn=(anVn+1Vn)Un+(anVn+1Vn)VnVn1Un1(anVn+1Vn)VnVn1Un1bnUn1=(anVn+1Vn)(UnVnVn1Un1)+[(anVn+1Vn)VnVn1bn]Un1.

    Since a[12,1), it is easy to know that

    6+4a2a2=2(1a)2+8152,5a2+8a+2=5(a4/5)2+26/5194,

    and

    anVn+1Vn=2(n1)2+(6+4a2a2)(n1)+(5a2+8a+2)2(n+1)(n+3)

    is positive for a[12,1) when n1. For a[12,1) and n1, we have that

    (anVn+1Vn)VnVn1bn=δ(n)4n(n+1)(n+2)(n+3)<0,

    where

    δ(n)=a2(a2)2n2+2(a44a3+6a22)n+2(1a)2(3a24a+2).

    In fact, δ(n) is a quadratic function of n and is decreasing on (1,), it follows that

    2(a44a3+6a22)2(a2(a2)2)=1+2a22a2(a2)2<1,δ(n)δ(2)=2(a1)(3a37a2+10a+2)<0forn2, (2.6)

    which implies that

    (anVn+1VnUn)VnVn1bn<0.

    By induction, we conclude that Un+1Vn+1VnUn<0 for all n1. Therefore, the sequence {UnVn}n0 is decreasing. Consequently, the function f2(r) is decreasing on (0,1). Moreover,

    limr0+f2(r)=a2a,limr1f2(r)=0.

    This completes the proof.

    Lemma 2.7. For each a[12,1), we define the function h(r) on (0,1) by

    h(r)=2Ea(KaEa)2(1a)r2E2a2(1a)r2(KaEa)2(KaEa)(Ear2Ka).

    Then, 2a<h(r)<2.

    Proof. First of all, we prove the right-hand side inequality. To establish the desired result, we need to show the following inequality:

    2Ea(KaEa)2(1a)r2E2a2(1a)r2(KaEa)2<2(KaEa)(Ear2Ka),

    which is equivalent to

    2(1a)Ea(Ear2Ka)+2ar2Ka(KaEa)<0.

    Denote that

    h1(r)=2(1a)Ea(Ear2Ka)+2ar2Ka(KaEa).

    By differentiation, we obtain

    h1(r)=2(1a)[2(1a)(EaKa)r(Ear2Ka)+2arEaKa]+2a[2rKa(KaEa)+2(1a)(Ear2Ka)r(KaEa)+2(1a)rEaKa]=KaEar[4(1a)(EaKa)+(48a)r2Ka]<0.

    Therefore, h1(r) is decreasing on (0,1) and

    h1(r)<limr0+h(r)=0,

    which implies h(r)<2.

    Next, we prove h(r)>2a. This is equivalent to the following inequality.

    Ea[a(KaEa)(1a)(Ear2Ka)][(1a)(Ear2Ka)ar2Ka(KaEa)]>0.

    Denote

    F(r)=Ea[a(KaEa)(1a)(Ear2Ka)][(1a)(Ear2Ka)ar2Ka(KaEa)].

    The derivative of F(r) yields

    F(r)=2(1a)KaEar[a(KaEa)(1a)(Ear2Ka)]+Ea[2a(1a)r(Ear2Ka)r2]2r(KaEa)[a(KaEa)(1a)(Ear2Ka)r2+aEar2Kar2]=r(Ear2Kaar2Ea)r2[2aEar2Kar2(2a)r2(KaEa)r2a(KaEa)(1a)(Ear2Ka)Ear2Kaar2Ea].

    Note that (Ear2Kaar2Ea)/r2 is increasing from (0, 1) to (0,). In fact, by differentiation, we know

    (Ear2Kaar2Ear2)=2a(2a)r(KaEa)r4>0.

    According to Lemma 2.2(1)(5) and Lemma 2.6, we have that F(r) is increasing on (0, 1) and F(r)>limr0+F(r)=0, which implies that F(r) is increasing on (0,1). Moreover,

    F(r)>limr0+F(r)=0.

    Thus, h(r)>2a. The proof is completed.

    For a[12,1), it is also found that h(r) is strictly increasing on (0,1).

    Lemma 2.8. For each a[12,1), r(0,1), we define the function f4(r) by

    f4(r)=r2a(KaEa)22Ea2ar2Ea2r2Ka.

    Then f4(r) is strictly decreasing from (0,1) to (0,(1a)π2a(2a)).

    Proof. Let

    f41(r)=r2a(KaEa)2,f42(r)=2Ea2ar2Ea2r2Ka.

    With Lemma 2.4 and f41(0+)=f42(0+)=0, we only prove the monotonicity of f41(r)/f42(r). Then we have

    f41(r)=rr22a(KaEa)[(42a)Ea2aKa],
    f42(r)=4a(2a)r(KaEa),
    4a(2a)f41(r)f42(r)=(42a)EaKar22af43(r).

    By differentiation, we see

    f43(r)=2(1a)rKar42a[(44a)Ear2Kar2Ka2a].

    With Lemma 2.2(2), we obtain

    (44a)Ear2Kar2Ka2a<a(44a)2a=2a(12a)0.

    Thus, f43(r) is strictly decreasing on (0,1), which shows f4(r) is strictly decreasing. And

    limr0+f4(r)=limr0+f41(r)f42(r)=(1a)π2a(2a),limr1f4(r)=0.

    The proof is completed.

    Lemma 2.9. For each a[12,1), r(0,1), we define the function f5(r) by

    f5(r)=Ea(Ear2Ka)+r2Ka(KaEa)r2r22aKa.

    Then f5(r) is strictly increasing from (0,1) to (π2,+).

    Proof. Let

    f51(r)=Ea(Ear2Ka)+r2Ka(KaEa),f52(r)=r2r22aKa.

    Taking the derivative, we have

    f51(r)=2rKa(2EaKa),f52(r)=rr2a[2r2Ka2(1a)(KaEa)],
    f5(r)=f51(r)f52(r)f51(r)f52(r)f252(r)=f53(r)r3r42aK2a,

    where

    f53(r)=(KaEa)[2a(E2ar2K2a)(4a2)Ea(Ear2Ka)].

    In fact, we see

    (2a(E2ar2K2a)(4a2)Ea(Ear2Ka))=KaEar[4a(KaEa)+2(4a2)(Ear2Ka)]>0,

    which demonstrates f5(r)>0 for r(0,1) and f5(r) is increasing on (0,1). Moreover,

    limr0+f5(r)=Ea(Ear2Ka)/r2+r2Ka(KaEa)/r2r22aKa=π2,limr1f5(r)=+.

    The proof is completed.

    Lemma 2.10. For each, a[12,1), r(0,1), h(r) is given as in Lemma 2.7. Then, h(r) is strictly increasing from (0,1) to (2a,2).

    Proof. Let

    h2(r)=2Ea(KaEa)2(1a)r2E2a2(1a)r2(KaEa)2KaEa,h3(r)=Ear2Ka.

    Clearly, h(r)=h2(r)h3(r) and h2(0+)=h3(0+)=0. By differentiations,

    h2(r)=2(1a)2r2(KaEa)2(Ear2Ka)+r2Ea[2(1a)Ea2+(4a2)r2EaKa2ar2K2a]rr2(KaEa)2,h3(r)=2arKa.

    Then,

    h2(r)h3(r)=2(1a)2a2r2(KaEa)2(Ear2Ka)+r2Ea[2(1a)Ea2+(4a2)r2EaKa2ar2K2a]r2r2Ka(KaEa)2=1aa[2Ea2ar2Ea2r2Kar2a(KaEa)2][Ea(Ear2Ka)+r2Ka(KaEa)r2r22aKa]=1aaf5(r)f4(r).

    With Lemmas 2.8 and 2.9, we obtain that h(r) is strictly increasing on (0,1). Furthermore,

    limr0+h(r)=2a,limr1h(r)=2.

    This completes the proof.

    In this section, we present some of the main results of Ea(r).

    Theorem 3.1. Let a[12,1), pR{0}, and for r(0,1), define

    Fa,p(r)=1[2Ea(r)/π]p2(1a)1rp.

    The monotonicity of Fa,p(r) is as follows:

    (1) Fa,p(r) is strictly increasing from (0,1) to (1a,1b) if and only if p2, where

    b=(sin(πa)(1a)π)p2(1a).

    (2) Fa,p(r) is strictly decreasing on (0,1) if and only if p2a. Moreover, if p(0,2a], the range of Fa,p(r) is (1b,1a), and the range is (0,1a) if p(,0).

    (3) If p(2a,2), Fa,p(r) is piecewise monotonic. To be precise, there exsists r0=r0(a,p)(0,1) such that Fa,p(r) is strictly increasing on (0,r0) and strictly decreasing on (r0,1). Furthermore, for r(0,1), if p(2a,p0), the range of Fa,p(r) satisifies

    1b<Fa,p(r)σ0, (3.1)

    while

    1a<Fa,p(r)σ0, (3.2)

    if p[p0,2), where

    p0=2(1a)logalog(sin(πa)/(1a)π)(2a,2),σ0=Fa,p(r0)>1a.

    Proof. For r(0,1),

    Fa,p(r)=1[2Ea(r)/π]p2(1a)1rp=:φ1(r)φ2(r).

    Clearly, we have φ1(0)=φ2(0)=0. By differentiation,

    φ1(r)=p2(1a)(2π)p2(1a)Ep2(1a)1a2(1a)(KaEa)r,φ2(r)=prrp2,

    and

    φ1(r)φ2(r)=(2π)p2(1a)Ep2(1a)1a(KaEa)r2rp2=:φ3(r).

    By differentiating logφ3(r), we obtain

    φ3(r)φ3(r)=p2(1a)2(a1)(KaEa)rEa+prr22r+2(1a)rEar2(KaEa)+2(1a)(KaEa)rEa2rr2=pEar2Karr2Ea+2(1a)r2E2a2Ea(KaEa)+2(1a)r2(KaEa)2rr2Ea(KaEa)=Ear2Karr2Ea[p2Ea(KaEa)2(1a)r2E2a2(1a)r2(KaEa)2(KaEa)(Ear2Ka)]=Ear2Karr2Ea(ph(r)), (3.3)

    where h(r) is defined as in Lemma 2.7. By Lemmas 2.2(2), 2.7, and 2.10, there are three cases to consider.

    (i) If p2. It follows from (3.3) that φ3(r) is strictly increasing on (0,1), and so is Fa,p(r). Furthermore, in this case,

    Fa,p(0+)=limr0+φ1(r)φ2(r)=1a,Fa,p(1)=1(sin(πa)(1a)π)p2(1a).

    (ii) If p2a, as in the proof of case (i), we know that φ3(r) is strictly decreasing on (0,1), and so is Fa,p(r). Also, Fa,p(0+)=1a, and

    Fa,p(1)={0,forp<0,1(sin(πa)(1a)π)p2(1a),for0<p2a.

    (iii) If 2a<p<2. According to Ramanujan's approximation (1.1), it shows that rcKa0 (r1) if c0. With Lemma 2.2(4) and the equation

    Hφ1,φ2(r)=φ1φ2φ2φ1=φ2φ3φ1,

    we obtain

    limr0+Hφ1,φ2(r)=0,limr1Hφ1,φ2(r)=(sin(πa)(1a)π)p2(1a)1<0. (3.4)

    Together with (3.3), (3.4), Lemmas 2.7 and 2.10, and the formulas

    Fa,p(r)=(φ1φ2)=φ2φ22Hφ1,φ2(r),Hφ1,φ2(r)=(φ1φ2)φ2=φ3(r)φ2(r),

    which follows from (2.2) and (2.3), it shows that there exists r0(0,1) such that Hφ1,φ2(r)>0 for r(0,r0) and Hφ1,φ2(r)<0 for r(r0,1). Thus, Fa,p(r) is strictly increasing on (0,r0) and strictly decreasing on (r0,1). Therefore, for all r(0,1), we get

    Fa,p(r)Fa,p(r0)=σ0.

    In fact, Fa,p(r0)Fa,p(r)>max{Fa,p(0+),Fa,p(1)}. It follows from Lemma 2.5 that

    p0=2(1a)logalog(sin(πa)/(1a)π)(2a,2),

    which makes p0 the unique root of

    1(sin(πa)(1a)π)p2(1a)=1a

    on (2a,2) and p1(sin(πa)(1a)π)p2(1a) is strictly increasing on (,). Hence we have Fa,p(0+)Fa,p(1) if p(2a,p0] and Fa,p(0+)<Fa,p(1) if p(p0,2). Consequently, the range of Fa,p(r) in case 3 is valid. The proof is completed.

    Figure 1 shows the monotonicity of Fa,p with a=0.7 as an example.

    Figure 1.  Monotonicity of Fa,p with a=0.7 as an example.

    Applying the property of Fa,p(r) from Theorem 3.1, we obtain our main result.

    Theorem 3.2. For a[12,1), let μ,ν[0,1] and p0,σ0 be given as in Theorem 3.1. Then for any fixed pR, the double inequality

    π2H2(1a)p(1,r;μ)Eaπ2H2(1a)p(1,r;ν) (3.5)

    holds for all r(0,1) with the equality only for certain values of r if and only if μμ(a,p) and νν(a,p), where μ(a,p) and ν(a,p) satisfy

    μ(a,p)={a,p(,0)(0,2a],1σ0,p(2a,2),b,p[2,+),ν(a,p)={1,p(,0),b,p(0,p0),a,p[p0,+), (3.6)

    where

    b=(sin(πa)(1a)π)p2(1a).

    Particularly, for p=0, (3.5) holds if and only if μ12(1a)2 and ν1.

    Proof. First we consider the case of p0, by (1.5), the inequality (3.5) is equivalent to

    μ<1Fa,p(r)<ν, (3.7)

    where Fa,p(r) is defined as in Theorem 3.1. It follows from Theorem 3.1 that we immediately conclude the best possible constants μ=μ(a,p) and ν=ν(a,p) in (3.6).

    For p=0, we define the function T(r) on (0, 1) by

    T(r)=log(2Ea/π)logr=:T1(r)T2(r).

    Obviously, we see that T1(0+)=T2(0+)=0. By differentiation, we have

    T1(r)T2(r)=2(1a)r2(KaEa)r2Ea.

    Together with Lemma 2.2(3), this implies T1(r)T2(r) is strictly decreasing on (0, 1), and by Lemma 2.4, T(r) shares the same monotonicity. Clearly, T(1)=0 and

    T(0+)=limr0+T1(r)T2(r)=2(1a)2,

    which indicates 12(1a)2<1T(r)<1 for r(0,1). As a result, Eq (1.5) demonstrates that the inequality

    π2H2(1a)p(1,r;μ)<Ea(r)<π2H2(1a)p(1,r;ν)

    holds for all r(0,1) if and only if μ12(1a)2 and ν1.

    This completes the proof.

    Figure 2 shows the sharpness of the bound with a=0.7 as an example.

    Figure 2.  Sharp bound for Ea with a=0.7 as an example.

    Remark 3.1. For a=12, we see that (3.5) holds if the parameters satisfy the conditions given in Theorem 3.2. This conclusion has been proved in [15].

    In this section, by applying Theorem 3.2, we present several sharp bounds of weighted Hölder mean for Ea.

    Note that for the case of μ(a,p)=ν(a,p)=a, the best bounds of Ea are attained at p=2a and p=p0, which will be proved in the following corollary.

    Corollary 4.1. Let a[12,1) and p1,p2R. Then the inequality

    π2H2(1a)p1(1,r;a)<Ea(r)<π2H2(1a)p2(1,r;a) (4.1)

    holds for all r(0,1) with the best possible constants p1=2a and p2=p0, where p0 is given as in Theorem 3.1.

    Proof. For a[12,1), we consider (μ,p)=(a,2a) and (ν,p)=(a,p0) satisfying (3.6), which yields (4.1) upon substitution into (3.5).

    To establish that a and p0 are the best possible constants, we observe that the Hölder mean is monotonically increasing with respect to p. Consequently, it suffices to analyze the case of 2a<p<p0.

    According to Theorem 3.2, the inequality

    π2H2(1a)p(1,r;1σ0)Eaπ2H2(1a)p(1,r;b) (4.2)

    holds for all r(0,1), where 1σ0 and b are sharp, with b given as in Theorem 3.2. From Theorem 3.1, together with the monotonicity of ωHp(1,r;ω), we have 1σ0<a<b for p(2a,p0), implying

    π2H2(1a)p(1,r;1σ0)π2H2(1a)p(1,r;a)π2H2(1a)p(1,r;b).

    Therefore, considering the sharpness of 1σ0 and b in inequality (4.2), we conclude that there exist two numbers r1,r2(0,1) such that

    π2H2(1a)p(1,r1;a)>Ea(r1),π2H2(1a)p(1,r2;a)<Ea(r2).

    Thus, the proof is completed.

    Figure 3 demonstrates that the sharp bounds of Ea are attained at p1=2a and p2=p0 with a=0.7 as an example.

    Figure 3.  Best constants for (4.1) with a=0.7 as an example.

    Furthermore, it is observed that computing the lower bound in (3.6) for the case μ(a,p)=1σ0 is challenging, while the case ν(a,p)=1 is trivial. Thus, we propose using μ(a,p)=b for p[2,) and ν(a,p)=b for p(0,p0) to establish new bounds. The specific inequality is as follows.

    Corollary 4.2. Inequality

    π2{(sin(πa)(1a)π)11a+[1(sin(πa)(1a)π)11a]r2}1a<Ea<π2{(sin(πa)(1a)π)p02(1a)+[1(sin(πa)(1a)π)p02(1a)]rp0}2(1a)p0 (4.3)

    holds for r(0,1).

    Lemma 4.3. Let a[12,1),

    Δ(p,r)=H2(1a)p(1,r;b)={(sin(πa)(1a)π)p2(1a)+[1(sin(πa)(1a)π)p2(1a)]rp}2(1a)p.

    Then, the function Δ(p,r) with respect to p is strictly decreasing on (0,) for r(0,1).

    Proof. By differentiating logΔ(p,r):

    1Δ(p,r)Δ(p,r)p=˜Δ(p,rp)p2ψ(p,rp), (4.4)

    where

    ψ(p,x)=(sin(πa)(1a)π)p2(1a)+[1(sin(πa)(1a)π)p2(1a)]x,

    and

    ˜Δ(p,x)=2(1a)ψ(p,x)log(ψ(p,x))p(1x)(sin(πa)(1a)π)p2(1a)log(sin(πa)(1a)π)2(1a)[1(sin(πa)(1a)π)p2(1a)]xlogx.

    Differentiating ˜Δ(p,x) with respect to x yields

    ˜Δ(p,x)x=2(1a)[1(sin(πa)(1a)π)p2(1a)]logψ(p,x)x+p(sin(πa)(1a)π)p2(1a)log(sin(πa)(1a)π)

    Give the observation that \Delta_0(p, x) is strictly decreasing for x\in(0, 1) . In fact,

    \begin{equation} \begin{aligned} \frac{\partial\Delta_0(p,x)}{\partial x} = -2(1-a)\frac{\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\right]\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}}{x\psi(p,x)} < 0. \end{aligned} \notag \end{equation}

    And

    \begin{equation} \Delta_0(p,0^+) = \infty,\; \; \; \; \; \; \Delta_0(p,1^-) = p\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\log\left(\frac{\sin(\pi a)}{(1-a)\pi}\right) < 0 \notag \end{equation}

    indicate that \tilde{\Delta}(p, x) first strictly increases on (0, x_0) and then strictly decreases on (x_0, 1) for some x_0\in (0, 1) . Note that for p > 0 , it is observed that

    \begin{equation} \tilde{\Delta}(p,0^+) = \tilde{\Delta}(p,1^-) = 0. \end{equation} (4.5)

    Hence, \tilde{\Delta}(p, x) > 0 for x \in (0, 1) .

    Consequently, monotonicity of \Delta(p, r) with respect to p follows from (4.4).

    Remark 4.1. Following Lemma 4.3 and inequality (3.5), we observe that

    \begin{equation} \begin{cases} \mathcal{E}_a > \frac{\pi}{2}H^{2(1-a)}_2(1,r';b^{{\frac{1}{1-a}}}_1)\geq \frac{\pi}{2}H^{2(1-a)}_p(1,r';b^{{\frac{p}{2(1-a)}}}_1), &if \text{ $p\in\lbrack2,\infty)$},\\ \mathcal{E}_a < \frac{\pi}{2}H^{2(1-a)}_{p_0}(1,r';b^{{\frac{p_0}{2(1-a)}}}_1)\leq \frac{\pi}{2}H^{2(1-a)}_p(1,r';b^{{\frac{p}{2(1-a)}}}_1), &if \text{ $p\in(0,p_0\rbrack$}{,} \end{cases} \end{equation} (4.6)

    where

    b_1 = \frac{sin(\pi a)}{(1-a)\pi}.

    According to the proof of (3.2), if p\in(p_0, 2) , it follows that

    \begin{equation} 1-\sigma_0 < b < a. \notag \end{equation}

    Therefore, it results in

    \begin{equation*} \frac{\pi}{2}H^{2(1-a)}_{p}(1,r';1-\sigma_0) < \frac{\pi}{2}H^{2(1-a)}_{p}(1,r';b) < \frac{\pi}{2}H^{2(1-a)}_{p}(1,r';a) \end{equation*}

    by the monotonicity of H^{2(1-a)}_p(1, r';\zeta) with respect to \zeta .Theorem 3.2 presents that, for p\in(p_0, 2) , 1-\sigma_0 is sharp weight of H^{2(1-a)}_p(1, r';\zeta) as the lower bound of \mathcal{E}_a , while a is sharp weight as the upper bound of \mathcal{E}_a .

    Hence, as a bound of \mathcal{E}_a , H^{2(1-a)}_p(1, r';b) can attain the best upper bound at p = p_0 and the best lower bound at p = 2 by (4.6).

    In this article, we have proved the monotonicity of \mathcal{F}_{a, p}(r) , where \mathcal{F}_{a, p}(r) is given as in Theorem 3.1. Moreover, we find the sharp weighted Hölder mean approximating \mathcal{E}_a :

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,r';\mu)\leq\mathcal{E}_a\leq\frac{\pi}{2}H^{2(1-a)}_p(1,r';\nu) \notag \end{equation}

    holds for all r\in (0, 1) if and only if \mu \leq \mu(a, p) and \nu \geq \nu(a, p) , where \mu(a, p) and \nu(a, p) are given as in (3.6). Besides, we derive several bounds of \mathcal{E}_a in terms of weights and power, which are given by Corollary 4.1, Corollary 4.2, and Remark 4.1. These conclusions provide an extension of the work of [15].

    Zixuan Wang: Investigation, Writing – original draft. Chuanlong Sun: Validation. Tiren Huang: Writing – review & editing. All authors have read and approved the final version of the manuscript for publication.

    The authors declare that they have not used artificial intelligence (AI) tools in the creation of this article.

    The authors declare no conflicts of interest regarding the publication for the paper.



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