# MIMO Antenna Elements Effect on Chassis Modes

Asim Ghalib and Mohammad S. Sharawi  Electrical Engineering Department
King Fahd University of Petroleum and Minerals (KFUPM)
<EMAIL_ADDRESS>

###### Abstract

In this paper, a 4-element printed multiple-input-multiple-output (MIMO) loop
antenna having a bandwidth of 400 MHz is proposed. The effect of the MIMO
antenna on the chassis modes system is analyzed via the theory of
characteristic modes (TCM), and it is shown that chassis CM are not enough for
the full analysis. A defected-ground-structure (DGS) is also proposed to
enhance the isolation between the antenna elements by blocking the coupling
modes.

††This work is supported by DSR-KFUPM under Project no. KAUST-002.

##  I Introduction

Theory of characteristic modes (TCM) was initially developed by Garbacz [1]
but it gained importance after it was revisited in [2]. Characteristic modes
(CM) are given [3, 4]

$[X]J_{n}=\lambda_{n}[R]J_{n}$ (1)

where, $X$ and $R$ represents the imaginary and real part of the impedance
matrix $Z$. $\lambda_{n}$ is the eigenvalue corresponding to eigenfunction
i.e. current density ($J_{n}$). TCM is widely used for the design and analysis
of various type of antennas and designs made with TCM are more robust [5, 6,
7, 8, 9].

One of the performance metrics of multiple-input-multiple-output (MIMO) is
port isolation [10]. Alot of empirical methods such as decoupling networks,
parasitic element, neutralization technique and defected ground structure
(DGS) are proposed in literature to enhance the isolation. All the method
relies on empirical approach. TCM was also used to improve the isolation
between MIMO antennas [11, 12, 13]. In [11], it was observed that if we
restrict our study to less than 1 GHz, only one chassis mode will be present.
So, if two antennas are placed in a such way that one of them excite the
chassis while the other antenna does not excite the chassis, this can yield
better isolation. So, one antenna was placed at the electric field maxima
(coupling to chassis) and the other one at electric field minima (not coupling
to chassis), as a result 5dB isolation improvement was achieved.

Several LTE bands use frequencies greater than 1.5 GHz, thus it means that we
have to deal with more than one chassis mode. Let us assume the case when we
have two chassis modes (for a frequency greater than 1.83 GHz and a chassis
size of 120x60 mm2), we can see that there are no locations on the chassis
where the modes have electric field minima or maxima at the same time.
Normally when one mode has maxima the other mode has minima at that location.
This means that the method proposed in [11] has limitations. Secondly, only
the chassis modes are discussed and analysis are made based on it. The effect
of the single antenna and MIMO antennas on the chassis modes are totally
ignored. In this work, we try to answer these effects. Besides we present a
4-element MIMO loop antenna with enhanced isolation.

##  II Analysis Process

To analyze the effect of multiple antennas, we followed a systematic
procedure. We considered a normal mobile chassis of 120x60 mm2 dimensions. The
effect of 1-element, 2 and 4-element loop antennas on the chassis modes was
investigated. For brevity reasons, the 1-element and 2-element printed loop
antenna cases are not shown separately. The top view and bottom view of the
proposed design is shown in the Fig. 1(a) and 1(b), respectively, while the
reflection and isolation curves are shown in the Fig. 1(c). The antenna is
designed on an FR4 substrate having a dielectric constant of 4.0 and substrate
thickness of 0.8 mm.

Figure 1: 4 element MIMO loop antenna, where (a) top view, (b) bottom view and
(c) Reflection and isolation curves without the DGS. SP and FP refers to the
shorting and feeding point respectively. All dimensions are in mm.

The CM and the modal significance (MS) curve of the ground plan are shown in
the Figs. 2(1a-1f) and 3(a) respectively. The MS curves of the chassis modes
are significantly affected after the introduction of the antenna as shown in
the Fig. 3(b). For brevity the current distributions are not shown. Mode 1
radiating BW is severely affected. The other modes i.e. mode 2 to 6 are
slightly affected. This proves that the presence of the antenna affects the
CM. In the presence of the antenna all the current maxima shift to the antenna
element.

The MS curves of a 2-element MIMO antenna placed at the shorter edge of the
chassis are shown in the Figure 3(c). We can observe that as compared to the
case of chassis and single antenna, the CM of the chassis are affected more.
We can observe that the MS curves of the modes 1, 4 and 6 are severely
affected. Mode 4 seems to be effected because the antennas are placed at the
corners of the shorter edge and mode 4 has current maxima at the shorter edge
of the chassis. Mode 6 has started contributing to the radiating BW. The
amount of affect on mode 1 seems to be same for the single and 2-element
cases.

The CM and the MS curves for the 4-element MIMO case are shown in Fig.
2(2a-2f) and 3(d) respectively. As expected the 4-element MIMO case has highly
affected the chassis modes as compared to 1 and 2-element MIMO cases. The
current maxima lies on the antenna element and the radiating BW of almost all
the modes are affected.This validates that the affect of the antennas
especially multiple antennas cannot be ignored and we need to take them into
account while analyzing them.

Figure 2: CM current distribution, where (1) ground plane, (2) 4-element MIMO
loop antenna.(a)-(f) represents mode 1 to 6.

Figure 3: MS curve where, (a) chassis (120x60 mm2), (b) 1 element printed loop
antenna on a chassis, (c) 2-element printed MIMO loop antenna installed on
shorter edge of the chassis and (d) 4-element printed MIMO loop antenna.

An isolation of 10 dB (poor) can be observed between Ant-1 and 3 as shown in
the Fig. 1(c). If we carefully observe the CM current distribution of the
chassis and 4-element MIMO in Figs. 2(1a-2f), we can observe that modes 1, 5
and 6 are contributing to the coupling while modes 2 and 3 are not
contributing because they have a current null (observe the blue spot in the
middle of the chassis). This means to improve the isolation we need to block
the coupling modes while the non-coupling modes will remain unaffected.
Remember that in the impedance BW of interest, only the first four modes are
present in the radiating BW and all the modes have current maxima across the
antenna element. So, placing a defected ground structure (DGS) at the middle
of the chassis will not affect the non-coupling mode but will block the
coupling mode. The proposed DGS is shown in Fig. 1(b). It enhanced the
isolation by 11 dB i.e. from -11 dB to -22 dB as shown in the Fig. 4(a). It
slightly shifted the resonance of Ant-1 and 3 but the effective BW is still
from 2.2 GHz to 2.6 GHz i.e. 400 MHz on a VSWR<2 criteria. We can observe in
Fig. 4(b) and 4(c) that the current coupling has been significantly reduced.

Figure 4: 4-element MIMO loop antenna, where (a) Reflection and isolation
curves, (b) current distribution when antenna 1 is excited in the absence of
the DGS and (c) presence of DGS.

##  III Conclusion

The effect of MIMO antennas on mobile chassis cannot be ignored because they
severely affect the chassis modes. In the presence of the antenna the current
distribution shifts to the antenna elements. A DGS was proposed to stop the
coupling modes that enhanced the isolation of a 4-element printed loop based
MIMO design by 11 dB.

## References

  * [1] R. Garbacz, “A generalized expansion for radiated and scattered fields,” Ph.D. dissertation, Ohio State University, Columbus, 1968.
  * [2] M. Cabedo Fabres, “Systematic design of antennas using the theory of characteristic modes,” Ph.D. dissertation, University of Politecnica de Valencia, 2007. [Online]. Available: https://riunet.upv.es/handle/10251/1883
  * [3] A. Ghalib, R. Hussain, and M. S. Sharawi, “Analysis of slot-based radiators using TCM and its application in MIMO antennas,” _International Journal of RF and Microwave Computer-Aided Engineering_ , p. e21544, oct 2018. [Online]. Available: http://doi.wiley.com/10.1002/mmce.21544
  * [4] A. Ghalib and M. S. Sharawi, “Analyzing antenna effects on mobile chassis currents using theory of characteristic modes,” _Microwave and Optical Technology Letters_ , vol. 60, no. 8, pp. 1898–1905, aug 2018. [Online]. Available: http://doi.wiley.com/10.1002/mop.31257
  * [5] I. Vasilev and B. K. Lau, “On User Effects in MIMO Handset Antennas Designed Using Characteristic Modes,” _IEEE Antennas and Wireless Propagation Letters_ , vol. 15, pp. 758–761, 2016. [Online]. Available: http://ieeexplore.ieee.org/document/7219365/
  * [6] A. Ghalib, R. Hussain, and M. S. Sharawi, “Low profile frequency agile MIMO slot antenna with TCM characterization,” in _2017 11th European Conference on Antennas and Propagation (EUCAP)_. IEEE, mar 2017, pp. 2652–2655. [Online]. Available: http://ieeexplore.ieee.org/document/7928198/
  * [7] M. Ikram, R. Hussain, A. Ghalib, and M. S. Sharawi, “Compact 4-element MIMO antenna with isolation enhancement for 4G LTE terminals,” in _2016 IEEE International Symposium on Antennas and Propagation (APSURSI)_. IEEE, jun 2016, pp. 535–536. [Online]. Available: http://ieeexplore.ieee.org/document/7695976/
  * [8] A. Ghalib and M. S. Sharawi, “Effects of actual antenna excitation on natural radiation modes,” in _2017 11th European Conference on Antennas and Propagation (EUCAP)_. IEEE, mar 2017, pp. 3467–3470. [Online]. Available: http://ieeexplore.ieee.org/document/7928168/
  * [9] ——, “Analyzing DGS behavior for a MIMO antenna system using theory of characteristic modes,” in _2016 IEEE Middle East Conference on Antennas and Propagation (MECAP)_. IEEE, sep 2016, pp. 1–4. [Online]. Available: http://ieeexplore.ieee.org/document/7790087/
  * [10] M. S. Sharawi, “Printed Multi-Band MIMO Antenna Systems and Their Performance Metrics [Wireless Corner],” _IEEE Antennas and Propagation Magazine_ , vol. 55, no. 5, pp. 218–232, oct 2013. [Online]. Available: http://ieeexplore.ieee.org/document/6735522/
  * [11] H. Li, Y. Tan, and B. K. Lau, “Characteristic Mode Based Tradeoff Analysis of Antenna-Chassis Interactions for Multiple Antenna Terminals,” _IEEE Transactions on Antennas and Propagation_ , vol. 60, no. 2, pp. 490–502, feb 2012\. [Online]. Available: http://ieeexplore.ieee.org/document/6060882/
  * [12] A. Ghalib and M. S. Sharawi, “{TCM} Analysis of Defected Ground Structures for MIMO Antenna Designs in Mobile Terminals,” _IEEE Access_ , vol. 5, pp. 19 680–19 692, 2017. [Online]. Available: http://ieeexplore.ieee.org/document/8010273/
  * [13] H. Li, B. K. Lau, Z. Ying, and S. He, “Decoupling of Multiple Antennas in Terminals With Chassis Excitation Using Polarization Diversity, Angle Diversity and Current Control,” _IEEE Transactions on Antennas and Propagation_ , vol. 60, no. 12, pp. 5947–5957, dec 2012. [Online]. Available: http://ieeexplore.ieee.org/document/6266702/


# Sharp upper bounds for integral quantities related to the Bellman function
of three variables of the dyadic maximal operator

Eleftherios N. Nikolidakis

###### Abstract

We obtain sharp upper bounds for integral quantities related to the Bellman
function of three integral variables of the dyadic maximal operator.

##  1 Introduction

The dyadic maximal operator on $\mathbb{R}^{n}$ is a useful tool in analysis
and is defined by

$\mathcal{M}_{d}\varphi(x)=\sup\left\\{\frac{1}{|S|}\int_{S}|\varphi(u)|\,\mathrm{d}u:x\in
S,\ S\subseteq\mathbb{R}^{n}\ \text{is a dyadic cube}\right\\},$ (1.1)

for every $\varphi\in L^{1}_{\text{loc}}(\mathbb{R}^{n})$, where $|\cdot|$
denotes the Lebesgue measure on $\mathbb{R}^{n}$, and the dyadic cubes are
those formed by the grids $2^{-N}\mathbb{Z}^{n}$, for $N=0,1,2,\ldots$.
It is well known that it satisfies the following weak type (1,1) inequality

$\left|\left\\{x\in\mathbb{R}^{n}:\mathcal{M}_{d}\varphi(x)>\lambda\right\\}\right|\leq\frac{1}{\lambda}\int_{\left\\{\mathcal{M}_{d}\varphi>\lambda\right\\}}|\varphi(u)|\,\mathrm{d}u,$
(1.2)

for every $\varphi\in L^{1}(\mathbb{R}^{n})$, and every $\lambda>0$, from
which it is easy to get the following $L^{p}$-inequality

$\|\mathcal{M}_{d}\varphi\|_{p}\leq\frac{p}{p-1}\|\varphi\|_{p},$ (1.3)

for every $p>1$, and every $\varphi\in L^{p}(\mathbb{R}^{n})$. It is easy to
see that the weak type inequality (1.2) is the best possible. For refinements
of this inequality one can consult [6].

It has also been proved that (1.3) is best possible (see [1] and [2] for
general martingales and [18] for dyadic ones). An approach for studying the
behaviour of this maximal operator in more depth is the introduction of the
so-called Bellman functions which play the role of generalized norms of
$\mathcal{M}_{d}$. Such functions related to the $L^{p}$-inequality (1.3) have
been precisely identified in [4], [5] and [11]. For the study of the Bellman
functions of $\mathcal{M}_{d}$, we use the notation
$\operatorname{Av}_{E}(\psi)=\frac{1}{|E|}\int_{E}\psi$, whenever $E$ is a
Lebesgue measurable subset of $\mathbb{R}^{n}$ of positive measure and $\psi$
is a real valued measurable function defined on $E$. We fix a dyadic cube $Q$
and define the localized maximal operator $\mathcal{M}^{\prime}_{d}\varphi$ as
in (1.1) but with the dyadic cubes $S$ being assumed to be contained in $Q$.
Then for every $p>1$ we let

$B_{p}(f,F)=\sup\left\\{\frac{1}{|Q|}\int_{Q}(\mathcal{M}^{\prime}_{d}\varphi)^{p}:\operatorname{Av}_{Q}(\varphi)=f,\
\operatorname{Av}_{Q}(\varphi^{p})=F\right\\},$ (1.4)

where $\varphi$ is nonnegative in $L^{p}(Q)$ and the variables $f,F$ satisfy
$0<f^{p}\leq F$. By a scaling argument it is easy to see that (1.4) is
independent of the choice of $Q$ (so we may choose $Q$ to be the unit cube
$[0,1]^{n}$). In [5], the function (1.4) has been precisely identified for the
first time. The proof has been given in a much more general setting of tree-
like structures on probability spaces.

More precisely we consider a non-atomic probability space $(X,\mu)$ and let
$\mathcal{T}$ be a family of measurable subsets of $X$, that has a tree-like
structure similar to the one in the dyadic case (the exact definition will be
given in Section 3). Then we define the dyadic maximal operator associated to
$\mathcal{T}$, by

$\mathcal{M}_{\mathcal{T}}\varphi(x)=\sup\left\\{\frac{1}{\mu(I)}\int_{I}|\varphi|\,\mathrm{\;}d\mu:x\in
I\in\mathcal{T}\right\\},$ (1.5)

for every $\varphi\in L^{1}(X,\mu)$, $x\in X$.

This operator is related to the theory of martingales and satisfies
essentially the same inequalities as $\mathcal{M}_{d}$ does. Now we define the
corresponding Bellman function of four variables of
$\mathcal{M}_{\mathcal{T}}$, by

$B_{p}^{\mathcal{T}}(f,F,L,k)=\sup\left\\{\int_{K}\left[\max(\mathcal{M}_{\mathcal{T}}\varphi,L)\right]^{p}\mathrm{\;}d\mu:\varphi\geq
0,\int_{X}\varphi\,\mathrm{\;}d\mu=f,\right.\\\
\left.\int_{X}\varphi^{p}\,\mathrm{\;}d\mu=F,\ K\subseteq X\ \text{measurable
with}\ \mu(K)=k\right\\},$ (1.6)

the variables $f,F,L,k$ satisfying $0<f^{p}\leq F$, $L\geq f$, $k\in(0,1]$.
The exact evaluation of (1.6) is given in [5], for the cases where $k=1$ or
$L=f$. In the first case the author (in [5]) precisely identifies the function
$B_{p}^{\mathcal{T}}(f,F,L,1)$ by evaluating it in a first stage for the case
where $L=f$. That is he precisely identifies $B_{p}^{\mathcal{T}}(f,F,f,1)$
(in fact $B_{p}^{\mathcal{T}}(f,F,f,1)=F\omega_{p}(\frac{f^{p}}{F})^{p}$,
where $\omega_{p}:[0,1]\to[1,\frac{p}{p-1}]$ is the inverse function
$H^{-1}_{p}$, of $H_{p}(z)=-(p-1)z^{p}+pz^{p-1}$).

The proof of the above mentioned evaluation relies on a one-parameter integral
inequality which is proved by arguments based on a linearization of the dyadic
maximal operator. More precisely the author in [5] proves that the inequality

$F\geq\frac{1}{(\beta+1)^{p-1}}f^{p}+\frac{(p-1)\beta}{(\beta+1)^{p}}\int_{X}(M_{\mathcal{T}}\varphi)^{p}\;d\mu,$
(1.7)

is true for every non-negative value of the parameter $\beta$ and sharp for
one that depends on $f$, $F$ and $p$, namely for
$\beta=\omega_{p}(\frac{f^{p}}{F})-1$. This gives as a consequence an upper
bound for $B_{p}^{\mathcal{T}}(f,F,f,1)$, which after several technical
considerations is proved to be best possible.Then by using several calculus
arguments the author in [5] provides the evaluation of
$B_{p}^{\mathcal{T}}(f,F,L,1)$ for every $L\geq f$.

Now in [11] the authors give a direct proof of the evaluation of
$B_{p}^{\mathcal{T}}(f,F,L,1)$ by using alternative methods. Moreover in the
second case, where $L=f$, the author (in [5]) uses the evaluation of
$B_{p}^{\mathcal{T}}(f,F,f,1)$ and provides the evaluation of the more general
$B_{p}^{\mathcal{T}}(f,F,f,k)$, $k\in(0,1]$.

Our aim (in the future) is to use the results of this article in order to
approach the following Bellman function problem (of three integral variables)

$B_{p,q}^{\mathcal{T}}(f,A,F)=\sup\left\\{\int_{X}\left(\mathcal{M}_{\mathcal{T}}\varphi\right)^{p}\mathrm{\;}d\mu:\varphi\geq
0,\int_{X}\varphi\,\mathrm{\;}d\mu=f,\right.\\\
\left.\int_{X}\varphi^{q}\,\mathrm{\;}d\mu=A,\
\int_{X}\varphi^{p}\,\mathrm{\;}d\mu=F\right\\},$ (1.8)

where $1<q<p$, and the variables $f,A,F$ lie in the domain of definition of
the above problem.

In this article we prove that whenever $0<\frac{x^{q}}{\kappa^{q-1}}<y\leq
x^{\frac{p-q}{p-1}}\cdot
z^{\frac{q-1}{p-1}}\;\Leftrightarrow\;0<s_{1}^{\frac{q-1}{p-1}}\leq s_{2}<1$,
(where $s_{1},s_{2})$ are defined below), we find a constant
$t=t(s_{1},s_{2})$ for which if $h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$
satisfies $\int_{0}^{\kappa}h=x$ , $\int_{0}^{\kappa}h^{q}=y$ and
$\int_{0}^{\kappa}h^{p}=z$ then

$\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt\leq
t^{p}(s_{1},s_{2})\cdot\int_{0}^{\kappa}h^{p},$

where $t(s_{1},s_{2})=t$ is the greatest element of $\big{[}{1,t(0)}\big{]}$
for which $F_{s_{1},s_{2}}(t)\leq 0$ and $F_{s_{1},s_{2}}$ is defined in
Section 4. Moreover for each such fixed $s_{1},s_{2}$

$t=t(s_{1},s_{2})=\min\Big{\\{}{t(\beta)\;:\;\beta\in\big{[}0,\tfrac{1}{p-1}\big{]}}\Big{\\}}$

where $t(\beta)=t(\beta,s_{1},s_{2})$ is defined by equation (4). That is we
find a constant $t=t(s_{1},s_{2})$ for which the above inequality is satisfied
for all $h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$ as mentioned above. Note
that $s_{1},s_{2}$ depend by a certain way on $x,y,z$, namely
$s_{1}=\frac{x^{p}}{\kappa^{p-1}z}$, $s_{2}=\frac{x^{q}}{\kappa^{q-1}y}$ and
$F_{s_{1},s_{2}}(\cdot)$ is given in terms of $s_{1},s_{2}$.

We need to mention that the extremizers for the standard Bellman function
$B_{p}^{\mathcal{T}}(f,F,f,1)$ have been studied in [7], and in [9] for the
case $0<p<1$. Also in [8] the extremal sequences of functions for the
respective Hardy operator problem have been studied. Additionally further
study of the dyadic maximal operator can be seen in [10, 11] where
symmetrization principles for this operator are presented, while other
approaches for the determination of certain Bellman functions are given in
[13, 14, 15, 16, 17]. Moreover results related to applications of dyadic
maximal operators can be seen in [12].

##  2 Preliminaries

The following theorem has been proven in [3]

######  Theorem 2.1.

Let $q\in(1,p)$, $f>0$ and $\varphi\in L^{p}(X,\mu)$, with $\int_{X}\varphi=f$
. Then the inequality

$\displaystyle\int_{X}(\mathcal{M}_{\mathcal{T}}\varphi)^{p}\,d\mu$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\int_{X}(\mathcal{M}_{\mathcal{T}}\varphi)^{p-q}\varphi^{q}\,d\mu+\frac{(p-q)(\beta+1)}{G}\,f^{p}\,+$
$\displaystyle\quad+\frac{p(q-1)\beta}{G}\,f^{p-q}\int_{X}(\mathcal{M}_{\mathcal{T}}\varphi)^{q}\,d\mu-\frac{p(\beta+1)^{q}}{G}\,f^{p-q}\int_{X}\varphi^{q}\,d\mu\,,$

is true for every value of $\beta>0$, where $G$ is defined by
$G=G(p,q,\beta)=$
$q(p-1)(\beta+1)-p(q-1)$.

By using the proof of the result in [11], one can now see that the following
inequality is an immediate consequence:

If $h:(0,1]\longrightarrow\mathbb{R}^{+}$ is non-increasing, $h\in L^{p}$ with
$\int_{0}^{1}h=f$, $\int_{0}^{1}h^{q}=A$, then for every $\beta>0$ the
following inequality is true:

$\displaystyle\int_{0}^{1}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\int_{0}^{1}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p-q}h^{q}(t)\,dt+\frac{(p-q)(\beta+1)}{G}f^{p}\,+$
$\displaystyle+\frac{p(q-1)\beta}{G}f^{p-q}\int_{0}^{1}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{q}dt-\frac{p(\beta+1)^{q}}{G}f^{p-q}\\!\cdot\\!A$
(2.1)

where $G$ is as in the statement of Theorem (2.1).

By using (2) and a simple linear change of variables (in the variable $t$),
one easily shows that if $\kappa\in(0,1]$ and
$h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$ is $L^{p}$-integrable and non-
increasing, then the following inequality is true:

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p-q}h^{q}(t)\,dt\,+$
$\displaystyle\quad+\frac{(p-q)(\beta+1)}{G}\cdot\frac{1}{\kappa^{p-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p}\,+$
$\displaystyle\qquad+\frac{p(q-1)\beta}{G}\bigg{(}\frac{1}{\kappa}\int_{0}^{\kappa}h\bigg{)}^{p-q}\cdot\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{q}dt\,-$
$\displaystyle\hskip
25.6073pt-\frac{p(\beta+1)^{q}}{G}\bigg{(}\frac{1}{\kappa}\int_{0}^{\kappa}h\bigg{)}^{p-q}\cdot\int_{0}^{\kappa}h^{q}\,.$
(2.2)

Suppose now that $(f,A,F)$ are given such that there exists
$\kappa_{0}\in(0,1]$ with

$\displaystyle\omega_{q}\Big{(}{\frac{f^{q}}{\kappa_{0}^{q-1}A}}\Big{)}=\omega_{p}\Big{(}{\frac{f^{p}}{\kappa_{0}^{p-1}F}}\Big{)}:=\varepsilon_{0}\,.$
(2.3)

Define $g:(0,\kappa_{0}]\longrightarrow\mathbb{R}^{+}$ by
$g(t)=\vartheta\,t^{-1+\frac{1}{\varepsilon_{0}}}$, $t\in(0,\kappa_{0}]$,
where $\varepsilon_{0}$ is defined by (2.3) and $\vartheta$ is such that

$\displaystyle\int_{0}^{\kappa_{0}}g=f\quad\Leftrightarrow\quad\vartheta=\frac{f}{\kappa_{0}^{1/\varepsilon_{0}}\varepsilon_{0}}\,.$
(2.4)

We notice now the following

  1. A)

By using (2.3), (2.4) and the form that $g$ has, we easily see that the
following relations are satisfied:

$\int_{0}^{\kappa_{0}}g^{q}=A\,,\qquad\int_{0}^{\kappa_{0}}g^{p}=F\,.$

Moreover, after some standard simple calculations, we see that, for every
$\kappa\in(0,\kappa_{0}]$, the following identity holds true:

$\frac{\big{(}{\textstyle\int_{0}^{\kappa}g}\big{)}^{p}}{\kappa^{p-1}\textstyle\int_{0}^{\kappa}g^{p}}=\frac{f^{p}}{\kappa_{0}^{p-1}F}=\frac{\big{(}{\textstyle\int_{0}^{\kappa_{0}}g}\big{)}^{p}}{\kappa_{0}^{p-1}\textstyle\int_{0}^{\kappa_{0}}g^{p}}\,.$

  2. B)

Moreover for the function that is constructed in comment (A) above, the
following equality is true:

$\displaystyle\frac{1}{t}\int_{0}^{t}g(u)\,du=\varepsilon_{0}\cdot
g(t)\,,\quad\forall\,t\in(0,\kappa_{0}]\,.$ (2.5)

By using (2.5), it is straight forward to see that we obtain equality in (2),
if we replace $h$ by $g$ and $\kappa$ by $\kappa_{0}$.

##  3 A first approach

As we have already noticed in in the previous section, if
$h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$ is $L^{p}$-integrable and non-
increasing, where $0<\kappa\leq 1$, the following inequality is true:

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p-q}h^{q}(t)\,dt\,+$
$\displaystyle\quad+\frac{(p-q)(\beta+1)}{G}\cdot\frac{1}{\kappa^{p-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p}\,+$
$\displaystyle\qquad+\frac{p(q-1)\beta}{G}\cdot\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\cdot\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{q}dt\,-$
$\displaystyle\hskip
25.6073pt-\frac{p(\beta+1)^{q}}{G}\cdot\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\cdot\int_{0}^{\kappa}h^{q}\,,$
(3.1)

where $G=G(p,q,\beta)=q(p-1)(\beta+1)-p(q-1)$.

It is now well known, that the inequalities

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{q}dt$
$\displaystyle\leq\int_{0}^{\kappa}h^{q}\,\omega_{q}\bigg{(}{\frac{\big{(}{\textstyle\int_{0}^{\kappa}h}\big{)}^{q}}{\kappa^{q-1}\textstyle\int_{0}^{\kappa}h^{q}}}\bigg{)}^{q}\leq$
$\displaystyle\leq\frac{\beta+1}{\beta}\cdot\frac{(\beta+1)^{q-1}{\textstyle\int_{0}^{\kappa}h^{q}}-\frac{1}{\kappa^{q-1}}\big{(}{\textstyle\int_{0}^{\kappa}h}\big{)}^{q}}{q-1}$
(3.2)

are true, for every $\beta>0$, where the right-hand side of (3) attains its
minimum value for
$\beta=\omega_{q}\Big{(}{\frac{\big{(}{\textstyle\int_{0}^{\kappa}h}\big{)}^{q}}{\kappa^{q-1}\textstyle\int_{0}^{\kappa}h^{q}}}\Big{)}-1$
and this value is the middle term in the above chain of inequalities. We
substitute the estimate described in (3) in (3) and we get:

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt\leq\frac{p(\beta+1)^{q}}{G}\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p-q}h^{q}(t)\,dt\,+$
$\displaystyle\qquad+\frac{(p-q)(\beta+1)}{G}\cdot\frac{1}{\kappa^{p-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p}-\frac{p(\beta+1)^{q}}{G}\int_{0}^{\kappa}h^{q}\cdot\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\,+$
$\displaystyle+\frac{p(q-1)\beta}{G}\,\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\frac{\beta+1}{\beta(q-1)}\bigg{[}{(\beta+1)^{q-1}\int_{0}^{\kappa}h^{q}-\frac{1}{\kappa^{q-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{q}}\bigg{]}$
(3.3)

We apply now Hölder’s inequality on the first term on the right side of (3)
and, denoting
$I:=\int_{0}^{\kappa}\big{(}\frac{1}{t}\int_{0}^{t}h\big{)}^{p}dt$, we get

$\displaystyle
I\leq\frac{p(\beta+1)^{q}}{G}\,I^{\frac{p-q}{p}}\bigg{(}{\int_{0}^{\kappa}h^{p}}\bigg{)}^{\frac{q}{p}}+\frac{(p-q)(\beta+1)}{G}\cdot\frac{1}{\kappa^{p-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p}\,+$
$\displaystyle+\frac{p(q-1)\beta}{G}\,\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\frac{\beta+1}{\beta(q-1)}\bigg{[}{(\beta+1)^{q-1}\int_{0}^{\kappa}h^{q}-\frac{1}{\kappa^{q-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{q}}\bigg{]}\,-$
$\displaystyle\hskip
122.34692pt-\frac{p(\beta+1)^{q}}{G}\int_{0}^{\kappa}h^{q}\cdot\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\,.$
(3.4)

We denote now $\int_{0}^{\kappa}h=x$ , $\int_{0}^{\kappa}h^{q}=y$ and
$\int_{0}^{\kappa}h^{p}=z$ . Then (3) becomes:

$\displaystyle I$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\,I^{\frac{p-q}{p}}\cdot
z^{\frac{q}{p}}+\frac{(p-q)(\beta+1)}{G}\cdot\frac{1}{\kappa^{p-1}}\cdot
x^{p}\,+$
$\displaystyle\;\;+\frac{p(q-1)\beta}{G}\,\frac{1}{\kappa^{p-q}}\,x^{p-q}\frac{(\beta+1)^{q}}{\beta(q-1)}\,y-\frac{p(q-1)\beta}{G}\,\frac{1}{\kappa^{p-q}}\,x^{p-q}\frac{\beta+1}{\beta(q-1)}\,\frac{1}{\kappa^{q-1}}\,x^{q}\,-$
$\displaystyle\hskip
156.49014pt-\frac{p(\beta+1)^{q}}{G}\,y\cdot\frac{1}{\kappa^{p-q}}\,x^{p-q}\,.$
(3.5)

Then (3), gives after simple cancellations of terms, the following inequality

$\displaystyle I\leq\frac{p(\beta+1)^{q}}{G}\,I^{\frac{p-q}{p}}\cdot
z^{\frac{q}{p}}-\frac{q}{G}\,(\beta+1)\,\frac{x^{p}}{\kappa^{p-1}}\,.$ (3.6)

Set now
$J:=\frac{\int_{0}^{\kappa}\big{(}\frac{1}{t}\int_{0}^{t}h\big{)}^{p}dt}{\int_{0}^{\kappa}h^{p}}=\frac{I}{z}$
, and (3.6) gives:

$\displaystyle
J\leq\frac{p(\beta+1)^{q}}{G}\,J^{1-\frac{q}{p}}-\frac{q}{G}\,(\beta+1)\,\frac{x^{p}}{\kappa^{p-1}z}\,.$
(3.7)

Denoting by $s_{1}=s_{1}(x,z)=\frac{x^{p}}{\kappa^{p-1}z}$ , (3.7) becomes:

$\displaystyle
J^{1-\frac{q}{p}}-A_{\beta}J\geq\frac{q}{p}\,\frac{1}{(\beta+1)^{q-1}}\,s_{1}\,,$
(3.8)

where

$\displaystyle A_{\beta}$
$\displaystyle:=\frac{G}{p(\beta+1)^{q}}=\frac{p(q-1)\beta+(p-q)(\beta+1)}{p(\beta+1)^{q}}=$
$\displaystyle=\frac{p-q}{p}\,\frac{1}{(\beta+1)^{q-1}}+\frac{(q-1)\beta}{(\beta+1)^{q}}\,.$
(3.9)

Write $h_{\beta}(y)=y^{p-q}-A_{\beta}y^{p}$ , for $y\geq 1$, thus by setting
$J^{\prime}=J^{\frac{1}{p}}$, (3.8) gives

$h_{\beta}(J^{\prime})\geq\frac{q}{p}\,\frac{1}{(\beta+1)^{q-1}}\,s_{1}\,.$

But the function $h_{\beta}:[1,+\infty)\longrightarrow\mathbb{R}$ is strictly
decreasing as we shall see now. Indeed

$\displaystyle h_{\beta}^{\prime}(y)$
$\displaystyle=(p-q)y^{p-q-1}-A_{\beta}\cdot py^{p-1}=$
$\displaystyle=y^{p-1}\big{(}{y^{-q}(p-q)-p\cdot A_{\beta}}\big{)}\cong
y^{-q}(p-q)-p\cdot A_{\beta}\,,$ (3.10)

where the symbol $\cong$ means that the respective quantities involved are of
the same sign. Now (3), since $y\geq 1$ gives

$\displaystyle h^{\prime}_{\beta}(y)\underset{y\geq
1}{\underset{\raisebox{1.0pt}{$\sim$}}{\leq}}(p-q)-p\cdot A_{\beta}\,,$ (3.11)

with the obvious meaning for the symbol
$\underset{\raisebox{1.0pt}{$\sim$}}{\leq}$ .

We now prove that
$A_{\beta}>\frac{p-q}{p},\;\forall\,\beta\in\big{(}0,\frac{1}{p-1}\big{)}$ .
We write

$A_{\beta}=\frac{p-q}{p}\,\frac{1}{(\beta+1)^{q-1}}+\,\frac{(q-1)\beta}{(\beta+1)^{q}}\,,$

using (3) and then by setting $\mathcal{K}(\beta)$ the right side of the above
equality we get $\mathcal{K}(0)=\frac{p-q}{p}$, and
$\mathcal{K}^{\prime}(\beta)>0,\;\forall\,\beta\in\big{(}0,\frac{1}{p-1}\big{)}$
. Indeed:

$\displaystyle\mathcal{K}^{\prime}(\beta)$
$\displaystyle=\bigg{(}{\frac{G(\beta)}{p(\beta+1)^{q}}}\bigg{)}^{\prime}_{\beta}\cong
q(p-1)(\beta+1)^{q}\cdot p-G\cdot p\cdot q\cdot(\beta+1)^{q-1}\cong$
$\displaystyle\cong(p-1)(\beta+1)-G=(p-1)(\beta+1)-q(p-1)(\beta+1)+p(q-1)=$
$\displaystyle=(\beta+1)\big{(}{(p-1)-qp+q}\big{)}+p(q-1)=$
$\displaystyle=(\beta+1)\big{(}{(p-1)-q(p-1)}\big{)}+p(q-1)=$
$\displaystyle=-(p-1)(q-1)(\beta+1)+p(q-1)\stackrel{{\scriptstyle
0\leq\beta<\frac{1}{p-1}}}{{>}}$
$\displaystyle>-(p-1)(q-1)\,\frac{p}{p-1}+p(q-1)=0\,.$

Then

$\mathcal{K}(\beta)=A_{\beta}>\mathcal{K}(0)=\frac{p-q}{p}\,,\quad\forall\,\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}\,.$

Thus by (3.11) above, we set $h^{\prime}_{\beta}(y)<0\,,\;\;\forall\,y\geq 1$,
thus
$h_{\beta}:[1,+\infty)\longrightarrow\big{(}{-\infty,h_{\beta}(1)}\big{]}$ is
strictly decreasing.

Note that as we have seen above:

$h_{\beta}(1)\geq
h_{\beta}(J^{\prime})\geq\frac{q}{p}\,\frac{1}{(\beta+1)^{q-1}}\,s_{1}\,,$

(since obviously $J^{\prime}\geq 1$), thus if we denote by
$L:=\frac{q}{p}\,\frac{1}{(\beta+1)^{q-1}}\,s_{1}$ we have that:

$\displaystyle h_{\beta}(J^{\prime})\geq L\quad$
$\displaystyle\Rightarrow\quad J^{\prime}\leq h^{-1}_{\beta}(L)$
$\displaystyle\Rightarrow\quad\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt\leq
t_{1}^{p}\int_{0}^{\kappa}h^{p}\,,$

where $t_{1}=t_{1}(\beta)$, is the value of $h^{-1}_{\beta}(L)$, that is

$\displaystyle t_{1}=h^{-1}_{\beta}(L)\quad$
$\displaystyle\Leftrightarrow\quad h_{\beta}(t_{1})=L$
$\displaystyle\Leftrightarrow\quad
t_{1}^{p-q}-A_{\beta}t_{1}^{p}=L:=\frac{q}{p}\,\frac{1}{(\beta+1)^{q-1}}\,s_{1}\,.$
(3.12)

Until now we have proved that for every
$\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}$ there exists $t_{1}(\beta)\geq 1$
such that for every $h$ as above the following inequality is true

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt\leq
t_{1}^{p}(\beta)\cdot\int_{0}^{\kappa}h^{p}\,,$ (3.13)

where $t_{1}=t_{1}(\beta)$ is given by the equation (3). We wish now to find
the minimum of
$\big{\\{}{t_{1}(\beta)\;:\;\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}}\\}$, in
order to find the exact estimate that (3.13) gives for the integrals that are
involved.
First of all if we set $\beta=\beta_{0}:=\omega_{p}(s_{1})-1$, we notice that
$t_{1}(\beta_{0})=\omega_{p}(s_{1})$. For the proof of this fact we move in
the following manner: We prove the equality:

$\displaystyle(t^{\prime})^{p-q}-\frac{G(\beta_{0})}{p(\beta_{0}+1)^{q}}(t^{\prime})^{p}=\frac{q}{p}\,\frac{1}{(\beta_{0}+1)^{q-1}}\,s_{1}$
(3.14)

where $t^{\prime}=\beta_{0}+1=\omega_{p}(s_{1})$ . Indeed (3.14) is equivalent
to:

$\displaystyle(\beta_{0}+1)^{p-q}-\frac{G(\beta_{0})}{p(\beta_{0}+1)^{q}}(\beta_{0}+1)^{p}=\frac{q}{p}\,\frac{1}{(\beta_{0}+1)^{q-1}}\,s_{1}$
$\displaystyle\Leftrightarrow$
$\displaystyle-\frac{G(\beta_{0})}{p}(\beta_{0}+1)^{p-q}+(\beta_{0}+1)^{p-q}=\frac{q}{p}\,\frac{1}{(\beta_{0}+1)^{q-1}}\,H_{p}(\beta_{0}+1)$
$\displaystyle\Leftrightarrow$
$\displaystyle(\beta_{0}+1)^{p-q}(\beta_{0}+1)^{q-1}\bigg{[}{1-\frac{G(\beta_{0})}{p}}\bigg{]}=\frac{q}{p}\,\Big{(}{p(\beta_{0}+1)^{p-1}-(p-1)(\beta_{0}+1)^{p}}\Big{)}$
$\displaystyle\Leftrightarrow$ $\displaystyle
1-\frac{G(\beta_{0})}{p}=q-\frac{q}{p}(p-1)(\beta_{0}+1)$
$\displaystyle\Leftrightarrow$ $\displaystyle
p-G(\beta_{0})=pq-q(p-1)(\beta_{0}+1)$ $\displaystyle\Leftrightarrow$
$\displaystyle G(\beta_{0})=q(p-1)(\beta_{0}+1)-p(q-1)\,,$

which is true by the definition of $G(\beta_{0})$ . Thus by (3.14) and the
definition of $t_{1}(\beta_{0})$ we get
$t_{1}(\beta_{0})=t^{\prime}=\beta_{0}+1=\omega_{p}(s_{1})$ .
We now show that
$\min\big{\\{}{t_{1}(\beta)\;:\;\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}}\big{\\}}=t_{1}(\beta_{0})=\omega_{p}(s_{1})$
(obviously by definition $\beta_{0}\in\big{(}0,\tfrac{1}{p-1}\big{)}$) .
More precisely we prove that:

$\displaystyle
t_{1}(\beta)>\omega_{p}(s_{1})\,,\quad\forall\,\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}\setminus\\{{\beta_{0}}\\}$
(3.15)

Let us suppose that (3.15) is not true. Then there exists
$\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}\setminus\\{{\beta_{0}}\\}$ for which

$\displaystyle t_{1}(\beta)$
$\displaystyle\leq\omega_{p}(s_{1})\quad\Rightarrow$ $\displaystyle
t_{1}(\beta)^{p-q}-\frac{G}{p(\beta+1)^{q}}\,t_{1}(\beta)^{p}$
$\displaystyle\geq\big{(}{\omega_{p}(s_{1})}\big{)}^{p-q}-\frac{G}{p(\beta+1)^{q}}\,\big{(}{\omega_{p}(s_{1})}\big{)}^{p}\,,$
(3.16)

because of the type of the monotonicity of $h_{\beta}(y),\,y\geq 1$. Then

$(\ref{eq:2p16})\quad\Rightarrow\quad\frac{q}{p}\,\frac{1}{(\beta+1)^{q-1}}\,s_{1}\geq\big{(}{\omega_{p}(s_{1})}\big{)}^{p-q}-\frac{G}{p(\beta+1)^{q}}\,\big{(}{\omega_{p}(s_{1})}\big{)}^{p}\,.$

At this point we prove the following:

######  Lemma 3.1.

The inequality

$\displaystyle\frac{q}{p}\,\frac{s_{1}}{(\beta+1)^{q-1}}\leq\big{(}{\omega_{p}(s_{1})}\big{)}^{p-q}-\frac{G}{p(\beta+1)^{q}}\,\big{(}{\omega_{p}(s_{1})}\big{)}^{p}$
(3.17)

is true, for every $\beta\in\big{(}0,\tfrac{1}{p-1}\big{)}$, with the only
case of equality being: $\beta=\omega_{p}(s_{1})-1=:\beta_{0}$ .

###### Proof.

$\displaystyle(\ref{eq:2p17})\quad\Leftrightarrow\quad\frac{\big{(}{\omega_{p}(s_{1})}\big{)}^{p-q}}{s_{1}}(\beta+1)^{q-1}-\frac{G(\beta)}{p\,s_{1}(\beta+1)}\,\big{(}{\omega_{p}(s_{1})}\big{)}^{p}\geq\frac{q}{p}\,.$
(3.18)

Set $\omega_{p}(s_{1})=\lambda\;\;\Leftrightarrow\;\;s_{1}=H_{p}(\lambda)$,
where since $s_{1}\in(0,1]$, we get that
$\lambda\in\big{[}{1,\frac{p}{p-1}}\big{)}$. Thus (3.18) becomes

$\displaystyle\Delta(\lambda):=\frac{\lambda^{p-q}}{H_{p}(\lambda)}(\beta+1)^{q-1}-\frac{q(p-1)(\beta+1)-p(q-1)}{p(\beta+1)}\,\frac{\lambda^{p}}{H_{p}(\lambda)}\geq\frac{q}{p}\,.$
(3.19)

We check first the case of equality: We just need to prove that:
$\Delta(\beta+1)=\frac{q}{p}$ . We calculate

$\displaystyle\Delta(\beta+1)$
$\displaystyle=\frac{(\beta+1)^{p-1}}{p(\beta+1)^{p-1}-(p-1)(\beta+1)^{p}}\,-$
$\displaystyle\qquad-\Big{(}{\frac{q(p-1)}{p}-\frac{q-1}{\beta+1}}\Big{)}\frac{(\beta+1)^{p}}{p(\beta+1)^{p-1}-(p-1)(\beta+1)^{p}}=$
$\displaystyle=\frac{1}{p-(p-1)(\beta+1)}\biggl{\\{}{1-\Big{(}{\frac{q(p-1)}{p}-\frac{q-1}{\beta+1}}\Big{)}(\beta+1)}\biggr{\\}}=$
$\displaystyle=\frac{1}{p-(p-1)(\beta+1)}\Bigl{\\{}{q-\frac{q}{p}(p-1)(\beta+1)}\Bigr{\\}}=\frac{q}{p}\,.$

We now prove that in (3.19), we have strict inequality when
$\lambda\neq\beta+1$ .Thus we have to prove that for every
$\lambda\in\big{[}{1,\frac{p}{p-1}}\big{)}\setminus\\{{\beta+1}\\}$, the
following chain of equivalent inequalities are true:

$\displaystyle\lambda^{p-q}(\beta+1)^{q-1}-\Big{(}{\frac{q(p-1)}{p}-\frac{q-1}{\beta+1}}\Big{)}\lambda^{p}>\frac{q}{p}\big{(}{p\lambda^{p-1}-(p-1)\lambda^{p}}\big{)}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\beta+1)^{q-1}-\lambda^{q}\Big{(}{\frac{q(p-1)}{p}-\frac{q-1}{\beta+1}}\Big{)}>\frac{q}{p}\,\lambda^{q-1}\big{(}{p-(p-1)\lambda}\big{)}$
$\displaystyle\Leftrightarrow$
$\displaystyle(\beta+1)^{q-1}+\lambda^{q}\frac{q-1}{\beta+1}>q\,\lambda^{q-1}$
$\displaystyle\Leftrightarrow$ $\displaystyle
q\,\lambda^{q-1}-(q-1)\frac{\lambda^{q}}{\beta+1}<(\beta+1)^{q-1}$
$\displaystyle\Leftrightarrow$ $\displaystyle
q\,\Big{(}{\frac{\lambda}{\beta+1}}\Big{)}^{q-1}-(q-1)\Big{(}{\frac{\lambda}{\beta+1}}\Big{)}^{q}<1\,,\;\forall\,\lambda\in\big{[}{1,\tfrac{p}{p-1}}\big{)}\setminus\\{{\beta+1}\\}\,.$

But this last inequality is obviously true since the function
$H_{q}(s)=q\,s^{q-1}-(q-1)s^{q}$ attains its maximum value on the interval
$(0,+\infty)$, at the unique point $s_{0}=1$, and this value is $H_{q}(1)=1$ .
∎

The results of this section are thus described as follows:
By using the inequalities (3) and (3) we reached to the bound:

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt\leq\omega_{p}\Big{(}{\frac{\big{(}{\textstyle\int_{0}^{\kappa}h}\big{)}^{p}}{\kappa^{p-1}\textstyle\int_{0}^{\kappa}h^{p}}}\Big{)}^{p}\cdot\int_{0}^{\kappa}h^{p}$
(3.20)

which holds for every $h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$
$L^{p}$-integrable and non-increasing, where $0<\kappa\leq 1$ . This bound is,
of course, well known (see [5], [11]), but the approach that we gave in this
section, will give us a direction of the methods that we shall apply in order
to find estimates as in (3.20), by considering also the variable
$y=\int_{0}^{\kappa}h^{q}$, as given. This will be done in the next section.

##  4 Second approach

We now determine a bound for the quantity:
$I=\int_{0}^{\kappa}\big{(}\frac{1}{t}\int_{0}^{t}h\big{)}^{p}dt$, where
$h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$ is $L^{p}$-integrable, non-
increasing, with $\int_{0}^{\kappa}h=x$ , $\int_{0}^{\kappa}h^{q}=y$ and
$\int_{0}^{\kappa}h^{p}=z$, for $1<q<p$, without ignoring the variable $y$. We
move in the following manner: Remember that the following inequality is true
(see the Preliminaries):

$\displaystyle
I=\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p-q}h^{q}(t)\,dt\,+$
$\displaystyle\quad+\frac{(p-q)(\beta+1)}{G}\cdot\frac{1}{\kappa^{p-1}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p}\,+$
$\displaystyle\qquad+\frac{p(q-1)\beta}{G}\cdot\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\cdot\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{q}dt\,-$
$\displaystyle\hskip
25.6073pt-\frac{p(\beta+1)^{q}}{G}\cdot\frac{1}{\kappa^{p-q}}\bigg{(}\int_{0}^{\kappa}h\bigg{)}^{p-q}\cdot\int_{0}^{\kappa}h^{q}\qquad\Rightarrow$

(by applying Hölder’s inequality on the first term on the right)

$\displaystyle I$
$\displaystyle\leq\frac{p(\beta+1)^{q}}{G}\,I^{\frac{p-q}{p}}\cdot
z^{\frac{q}{p}}+\frac{(p-q)}{G}(\beta+1)\cdot\frac{1}{\kappa^{p-1}}\cdot
x^{p}\,-$
$\displaystyle\quad-\frac{p(\beta+1)^{q}}{G}\cdot\frac{1}{\kappa^{p-q}}x^{p-q}y+\frac{p(q-1)\beta}{G}\,\frac{1}{\kappa^{p-q}}\,x^{p-q}y\cdot\omega_{q}\Big{(}{\frac{x^{q}}{\kappa^{q-1}y}}\Big{)}^{q}\,,$
(4.1)

where we have used the first of the inequalities (3).
Set
$\delta=\Big{(}{\frac{\int_{0}^{\kappa}\big{(}\frac{1}{t}\int_{0}^{t}h\big{)}^{p}dt}{\int_{0}^{\kappa}h^{p}}}\Big{)}^{1/p}=\big{(}{\frac{I}{z}}\big{)}^{1/p}$
. Thus we get:

$\displaystyle\delta^{p}\leq\frac{p(\beta+1)^{q}}{G}\,\delta^{p-q}\,+$
$\displaystyle+\Bigg{[}{\frac{(p-q)}{G}(\beta+1)\,\frac{x^{p}}{\kappa^{p-1}z}-\frac{p(\beta+1)^{q}}{G}\,\frac{\frac{x^{p}}{\kappa^{p-1}z}}{\frac{x^{q}}{\kappa^{q-1}y}}+\frac{p(q-1)\beta}{G}\,\frac{\frac{x^{p}}{\kappa^{p-1}z}}{\frac{x^{q}}{\kappa^{q-1}y}}\cdot\omega_{q}\Big{(}{\frac{x^{q}}{\kappa^{q-1}y}}\Big{)}^{q}}\Bigg{]}$
$\displaystyle=\frac{p(\beta+1)^{q}}{G}\,\delta^{p-q}+T_{1}(\beta)\,,$ (4.2)

where $T_{1}(\beta)$ is the quantity in brackets in (4). Denote also
$s_{1}=\frac{x^{p}}{\kappa^{p-1}z}$ and $s_{2}=\frac{x^{q}}{\kappa^{q-1}y}$ .
Note that, since $\int_{0}^{\kappa}h=x$ , $\int_{0}^{\kappa}h^{q}=y$ ,
$\int_{0}^{\kappa}h^{p}=z\Rightarrow\frac{x^{q}}{\kappa^{q-1}}\leq y\leq
x^{\frac{p-q}{p-1}}\cdot z^{\frac{q-a}{p-1}}$ .
By the second inequality above we immediately get $s_{1}^{q-1}\leq
s_{2}^{p-1}$. Further $0<s_{1}$ and $s_{2}\leq 1$.
Define also (as we have done in Section 3) :
$h_{\beta}(y)=y^{p-q}-A_{\beta}y^{p}$ , $y\geq 1$, where
$A_{\beta}:=\frac{G}{p(\beta+1)^{q}}$ . Thus (4) describes the following
inequality:

$\displaystyle
h_{\beta}(\delta)\geq-\frac{G}{p(\beta+1)^{q}}\big{(}{T_{1}(\beta)}\big{)}\,.$
(4.3)

As it is proved in Section 3, the function $h_{\beta}$ is strictly decreasing
on $[1,+\infty)$, and $A_{\beta}$ satisfies:

$A_{\beta}>\frac{p-q}{p}\,,\quad\forall\,\beta\in\big{(}{0,\tfrac{1}{p-1}}\big{)}\,.$

Since $\delta\geq 1$ and thus $h_{\beta}(\delta)\leq h_{\beta}(1)$, and also
$\lim_{y\to+\infty}h_{\beta}(y)=-\infty$, we have that for every
$\beta\in\big{(}{0,\tfrac{1}{p-1}}\big{)}$ there exists unique
$t(\beta)=t(\beta,s_{1},s_{2})\geq 1$, for which:

$\displaystyle h_{\beta}(t(\beta))$
$\displaystyle=-\frac{G}{p(\beta+1)^{q}}\cdot T_{1}(\beta)\quad$
$\displaystyle\Leftrightarrow$
$\displaystyle\frac{G}{p(\beta+1)^{q}}\,t^{p}(\beta)-t^{p-q}(\beta)$
$\displaystyle=\frac{G}{p(\beta+1)^{q}}\cdot T_{1}(\beta)\quad$
$\displaystyle\Leftrightarrow$ $\displaystyle
t^{p}(\beta)-\frac{p(\beta+1)^{q}}{G}\,t^{p-q}(\beta)$
$\displaystyle=T_{1}(\beta)=$ $\displaystyle=\frac{p-q}{G}(\beta+1)\,s_{1}+\,$
$\displaystyle\frac{s_{1}}{s_{2}}\Big{(}{\frac{p(q-1)\beta}{G}\,\omega_{q}(s_{2})^{q}-\frac{p(\beta+1)^{q}}{G}}\Big{)}$
(4.4)

Note that for this choice of $t(\beta)$ we obviously have $\delta\leq
t(\beta)$, while all the inequalities stated above, remain to hold, even in
the case $\beta=0$, (by letting $\beta\to 0^{+}$ in (4) ). Moreover if we
consider $t=\min_{0\leq\beta<\frac{1}{p-1}}t(\beta)$ we should have that
$t\leq\omega_{p}(s_{1})$ since we have used the strong part of inequality (3)
in order to obtain the bound $t(\beta)$, and all the remaining arguments are
exactly the same, as in Section 3, where we have obtain the bound
$t^{\prime}_{1}=\min_{0<\beta<\frac{1}{p-1}}t_{1}(\beta)=\omega_{p}(s_{1})$.
We remind also that the quantity $G=G(p,q,\beta)$ is given by
$G=q(p-1)(\beta+1)-p(q-1)\,,\;\forall\,\beta$.

Our aim is to differentiate (4) in terms of $\beta$, so for this reason we
calculate the following derivatives (with respect to $\beta)$ :

  1. i)

$\Big{(}{\dfrac{\beta+1}{G}}\Big{)}^{\prime}=\dfrac{G-(\beta+1)q(p-1)}{G^{2}}=\dfrac{-p(q-1)}{G^{2}}$
.

  2. ii)

$\Big{(}{\dfrac{\beta}{G}}\Big{)}^{\prime}=\Big{(}{\dfrac{\beta+1}{G}}\Big{)}^{\prime}-\Big{(}{\dfrac{1}{G}}\Big{)}^{\prime}=\dfrac{-p(q-1)}{G^{2}}+\dfrac{G^{\prime}}{G^{2}}=$
${}\hskip 25.6073pt=\dfrac{-p(q-1)+q(p-1)}{G^{2}}=\dfrac{p-q}{G^{2}}$ , and

  3. iii)

$\displaystyle\Big{(}{\dfrac{(\beta+1)^{q}}{G}}\Big{)}^{\prime}$
$\displaystyle=\dfrac{q(\beta+1)^{q-1}G-(\beta+1)^{q}q(p-1)}{G^{2}}=$
$\displaystyle=\dfrac{q(\beta+1)^{q-1}}{G^{2}}\big{(}{G-(p-1)(\beta+1)}\big{)}=$
$\displaystyle=\dfrac{q(\beta+1)^{q-1}}{G^{2}}\big{(}{q(p-1)(\beta+1)-p(q-1)-(p-1)(\beta+1)}\big{)}=$
$\displaystyle=\dfrac{q(\beta+1)^{q-1}}{G^{2}}\big{(}{(\beta+1)(p-1)(q-1)-p(q-1)}\big{)}=$
$\displaystyle=\dfrac{q(q-1)(\beta+1)^{q-1}}{G^{2}}\big{(}{(\beta+1)(p-1)-p}\big{)}\,.$

We restrict ourselves, in the following range of values for $\beta$ :
$\beta\in\big{[}{0,\frac{1}{p-1}}\big{)}$ . We now proceed to differentiate
(4), with respect to $\beta$ in the range
$\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ .
The relations that will come after, remain to hold even for $\beta=0$ or
$\beta=\frac{1}{p-1}$, by defining $t^{\prime}(0)=\lim_{\beta\to
0^{+}}t^{\prime}(\beta)$ and
$t^{\prime}\big{(}{\frac{1}{p-1}}\big{)}=\lim_{\beta\to{\frac{1}{p-1}}^{-}}t^{\prime}(\beta)$
.
We get by (4) that:

$\displaystyle
p\,t^{p-1}(\beta)\,t^{\prime}(\beta)-p(p-q)\,t^{p-q-1}(\beta)\,\frac{(\beta+1)^{q}}{G}\,t^{\prime}(\beta)-p\,t^{p-q}(\beta)\,q(q-1)\,\frac{\vartheta(\beta)}{G^{2}}=$
$\displaystyle=(p-q)s_{1}\frac{\big{(}{-p(q-1)}\big{)}}{G^{2}}+\frac{s_{1}}{s_{2}}\bigg{[}{p(q-1)\frac{p-q}{G^{2}}\,\omega_{q}(s_{2})^{q}-\frac{p\,q(q-1)\vartheta(\beta)}{G^{2}}}\bigg{]},$
(4.5)

where $\vartheta(\beta)=(p-1)(\beta+1)^{q}-p\,(\beta+1)^{q-1}$ .
We consider the following case: There exists
$\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ for which $t^{\prime}(\beta)=0$ .
Then (4) implies

$\displaystyle-q\,t^{p-q}(\beta)\,\vartheta(\beta)$
$\displaystyle=-(p-q)s_{1}+\frac{s_{1}}{s_{2}}\Big{[}{(p-q)\,\omega_{q}(s_{2})^{q}-q\,\vartheta(\beta)}\Big{]}$
$\displaystyle\Rightarrow$
$\displaystyle(p-q)s_{1}\Big{(}{\frac{\omega_{q}(s_{2})^{q}}{s_{2}}-1}\Big{)}$
$\displaystyle=q\,\vartheta(\beta)\Big{(}{\frac{s_{1}}{s_{2}}-t^{p-q}(\beta)}\Big{)}$
$\displaystyle\Rightarrow$ $\displaystyle t^{p-q}(\beta)$
$\displaystyle=\frac{s_{1}}{s_{2}}-(p-q)\,\frac{s_{1}\cdot\alpha(s_{2})}{q\,\vartheta(\beta)}\,,$
(4.6)

where $\alpha(s_{2})$ is defined by:

$\alpha(s_{2})=\frac{\omega_{q}(s_{2})^{q}}{s_{2}}-1\,,\quad s_{2}\in(0,1]\,.$

Also, since $\vartheta(\beta)=(p-1)(\beta+1)^{q}-p\,(\beta+1)^{q-1}$ we have

$\displaystyle\vartheta^{\prime}(\beta)$
$\displaystyle=q\,(p-1)(\beta+1)^{q-1}-p(q-1)\,(\beta+1)^{q-2}\cong$
$\displaystyle\cong q(p-1)(\beta+1)-p(q-1)=$
$\displaystyle=G>q(p-1)-p(q-1)=p-q>0\,.$

Thus $\vartheta:\big{(}{0,\frac{1}{p-1}}\big{)}\longrightarrow\mathbb{R}^{-}$
is increasing with $\vartheta(0^{+})=-1$ ,
$\vartheta\big{(}{{\frac{1}{p-1}^{-}}}\big{)}=0$ .

Claim 1. $t=t(\beta)$ satisfies:

$\displaystyle
t^{p}-\dfrac{p(\beta+1)^{q}}{G}\,t^{p-q}=s_{1}-\dfrac{p(\beta+1)^{q}}{G}\,\frac{s_{1}}{s_{2}}+p(q-1)\,\frac{\beta}{G}\,s_{1}\,\alpha(s_{2})\,.$
(4.7)

Indeed from (4) it is enough to prove the following:

$\displaystyle\frac{p-q}{G}(\beta+1)s_{1}+\frac{s_{1}}{s_{2}}\bigg{(}{\frac{p(q-1)\beta}{G}\,\omega_{q}(s_{2})^{q}-\frac{p(\beta+1)^{q}}{G}}\bigg{)}=$
$\displaystyle{}\hskip
56.9055pt=s_{1}-\frac{p(\beta+1)^{q}}{G}\,\frac{s_{1}}{s_{2}}+p(q-1)\frac{\beta}{G}\,s_{1}\bigg{(}{\frac{\omega_{q}(s_{2})^{q}}{s_{2}}-1}\bigg{)}$
$\displaystyle\Leftrightarrow$
$\displaystyle\frac{p-q}{G}(\beta+1)s_{1}+\frac{s_{1}}{s_{2}}\,\frac{p(q-1)\beta}{G}\,\omega_{q}(s_{2})^{q}=$
$\displaystyle{}\hskip
56.9055pt=s_{1}+p(q-1)\frac{\beta}{G}\,\frac{s_{1}}{s_{2}}\,\omega_{q}(s_{2})^{q}-p(q-1)\frac{\beta}{G}\,s_{1}$
$\displaystyle\Leftrightarrow$ $\displaystyle(p-q)(\beta+1)s_{1}=s_{1}\cdot
G-p(q-1)\beta\,s_{1}$ $\displaystyle\Leftrightarrow$
$\displaystyle(p-q)(\beta+1)=q(p-1)(\beta+1)-p(q-1)-p(q-1)\beta$
$\displaystyle\Leftrightarrow$
$\displaystyle(p-q)(\beta+1)+p(q-1)(\beta+1)-p(q-1)=q(p-1)(\beta+1)-p(q-1)$
$\displaystyle\Leftrightarrow$
$\displaystyle(\beta+1)(pq-p+p-q)=q(p-1)(\beta+1)\,,$

which is obviously true. Thus (4.7) is also true.
Now by (4) and (4.7) we deduce

$\displaystyle
t^{p}=\frac{p(\beta+1)^{q}}{G}\bigg{(}{\frac{s_{1}}{s_{2}}-\frac{p-q}{q}\,\frac{s_{1}\,\alpha(s_{2})}{\vartheta(\beta)}}\bigg{)}+s_{1}-\frac{p(\beta+1)^{q}}{G}\,\frac{s_{1}}{s_{2}}\,+$
$\displaystyle{}\hskip
239.00298pt+p(q-1)\,\frac{\beta}{G}\,s_{1}\,\alpha(s_{2})=$
$\displaystyle=s_{1}+p(q-1)\,\frac{\beta}{G}\,s_{1}\,\alpha(s_{2})-\frac{p(\beta+1)^{q}}{G}\,\frac{p-q}{q}\,\frac{s_{1}\,\alpha(s_{2})}{\vartheta(\beta)}=$
$\displaystyle=s_{1}+p(q-1)\,\frac{\beta}{G}\,s_{1}\,\alpha(s_{2})-\frac{p(\beta+1)^{q}}{G}\,\frac{p-q}{q}\,\frac{s_{1}\,\alpha(s_{2})}{(p-1)(\beta+1)^{q}-p\,(\beta+1)^{q-1}}=$
$\displaystyle=s_{1}+p(q-1)\,\frac{\beta}{G}\,s_{1}\cdot\alpha(s_{2})-\frac{p(\beta+1)}{G}\,\frac{p-q}{q}\,\frac{s_{1}\,\alpha(s_{2})}{(p-1)(\beta+1)-p}\qquad\Rightarrow$
$\displaystyle
t^{p}=s_{1}+p\,\frac{s_{1}}{G}\bigg{[}{(q-1)\beta\,\alpha(s_{2})-\frac{p-q}{q}\,\frac{(\beta+1)\,\alpha(s_{2})}{(p-1)(\beta+1)-p}}\bigg{]}\qquad\Rightarrow$
$\displaystyle
t^{p}=s_{1}+\frac{p\,s_{1}}{G}\,\alpha(s_{2})\bigg{[}{(q-1)\beta-\frac{p-q}{q}\,\frac{(\beta+1)}{(p-1)(\beta+1)-p}}\bigg{]}=$
$\displaystyle\;\;\;=s_{1}+\frac{p\,s_{1}\,\alpha(s_{2})}{G}\,\bigg{[}{(q-1)\beta-\frac{(p-q)(\beta+1)}{q(p-1)(\beta+1)-p(q-1)-p}}\bigg{]}\qquad\Rightarrow$
$\displaystyle
t^{p}=s_{1}+\frac{p\,s_{1}\,\alpha(s_{2})}{G}\,\bigg{[}{(q-1)\beta-\frac{(p-q)(\beta+1)}{G-p}}\bigg{]}\,.$
(4.8)

Note that $(G-p)^{-1}$ is well defined since
$\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$. Moreover

$\displaystyle q\vartheta(\beta)$
$\displaystyle=q(\beta+1)^{q-1}\big{(}{(p-1)(\beta+1)-p}\big{)}=$
$\displaystyle=(\beta+1)^{q-1}\big{(}{q(p-1)(\beta+1)-p(q-1)-p}\big{)}=$
$\displaystyle=(\beta+1)^{q-1}(G-p)\,.$

Thus (4) becomes:

$\displaystyle
t^{p-q}=\frac{s_{1}}{s_{2}}-(p-q)\,\frac{s_{1}\,\alpha(s_{2})}{(\beta+1)^{q-1}(G-p)}\,.$
(4.9)

Then we simplify the form that (4) has, by considering the term
$\Big{[}{(q-1)\beta-\frac{(p-q)(\beta+1)}{G-p}}\Big{]}\Big{/}G$, which equals
to:

$\displaystyle\frac{\Big{[}{(q-1)\beta-\frac{(p-q)(\beta+1)}{\frac{q\,\vartheta(\beta)}{(\beta+1)^{q-1}}}}\Big{]}}{G}$
$\displaystyle=\frac{(q-1)\beta-\frac{p-q}{q}\,\frac{(\beta+1)^{q}}{\vartheta(\beta)}}{G}\quad\Rightarrow$
$\displaystyle\frac{\Big{[}{(q-1)\beta-\frac{(p-q)(\beta+1)}{G-p}}\Big{]}}{G}$
$\displaystyle=\frac{(q-1)\beta-\frac{p-q}{q}\,\frac{(\beta+1)^{q}}{\vartheta(\beta)}}{G}\,.$
(4.10)

Claim 2. Either of the quantities in (4) equals to
$-\frac{H_{p}(\beta+1)}{q\,\vartheta(\beta)}$ .
Indeed, we just need to prove that

$\displaystyle\frac{(q-1)\beta-\frac{p-q}{q}\,\frac{(\beta+1)^{q}}{\vartheta(\beta)}}{G}=-\frac{H_{p}(\beta+1)}{q\,\vartheta(\beta)}$
$\displaystyle\Leftrightarrow$
$\displaystyle\frac{q(q-1)\beta\,\vartheta(\beta)-(p-q)(\beta+1)^{q}}{q\,\vartheta(\beta)}=-\frac{H_{p}(\beta+1)\cdot
G}{q\,\vartheta(\beta)}$ $\displaystyle\Leftrightarrow$ $\displaystyle
q(q-1)\big{(}{(p-1)(\beta+1)^{q}-p(\beta+1)^{q-1}}\big{)}\,\beta-(p-q)(\beta+1)^{q}=$
$\displaystyle{}\hskip
28.45274pt=\big{(}{q(p-1)(\beta+1)-p(q-1)}\big{)}(-1)\big{(}{q(\beta+1)^{q-1}-(q-1)(\beta+1)^{q}}\big{)}$
$\displaystyle\Leftrightarrow$ $\displaystyle
q(q-1)\big{(}{(p-1)(\beta+1)-p}\big{)}\beta-(p-q)(\beta+1)=$
$\displaystyle{}\hskip
91.04872pt=-\big{(}{q(p-1)(\beta+1)-p(q-1)}\big{)}\big{[}{q-(q-1)(\beta+1)}\big{]}$
$\displaystyle\Leftrightarrow$ $\displaystyle
q(q-1)(p-1)(\beta^{2}+\beta)-pq(q-1)\beta-(p-q)(\beta+1)=$
$\displaystyle{}\hskip
14.22636pt=-q^{2}(p-1)(\beta+1)+pq(q-1)+q(p-1)(q-1)(\beta^{2}+2\beta+1)\,-$
$\displaystyle{}\hskip 241.84842pt-p(q-1)^{2}(\beta+1)$
$\displaystyle\Leftrightarrow$
$\displaystyle\big{[}{q(q-1)(p-1)-pq(q-1)-(p-q)}\big{]}\beta-(p-q)=$
$\displaystyle{}\hskip
56.9055pt=\big{[}{-q^{2}(p-1)+2q(p-1)(q-1)-p(q-1)^{2}}\big{]}\beta\,+$
$\displaystyle{}\hskip
76.82234pt+\big{(}{-q^{2}(p-1)+pq(q-1)+q(p-1)(q-1)-p(q-1)^{2}}\big{)}$
$\displaystyle\Leftrightarrow$ $\displaystyle\big{[}{q(pq-
q-p+1)-p(q^{2}-q)-(p-q)}\big{]}\beta-(p-q)=$ $\displaystyle{}\hskip
28.45274pt=\big{[}{-pq^{2}+q^{2}+2q(pq-p-q+1)-p(q^{2}-2q+1)}\big{]}\beta\,+$
$\displaystyle{}\hskip 51.21504pt+\big{(}{-q^{2}p+q^{2}+pq^{2}-pq+q(pq-
p-q+1)-p(q^{2}-2q+1)}\big{)}$ $\displaystyle\Leftrightarrow$
$\displaystyle\big{[}{-q^{2}+2q-p}\big{]}\beta-(p-q)=\big{[}{-q^{2}+2q-p}\big{]}\beta+(-pq+q-pq+2pq-p)\,,$

which is obviously true. Thus Claim 2 is proved.

By using Claim 2 and (4), we obtain the following:

$\displaystyle
t^{p}=s_{1}+\big{(}{p\,s_{1}\,\alpha(s_{2})}\big{)}\cdot\bigg{(}{-\frac{H_{p}(\beta+1)}{q\,\vartheta(\beta)}}\bigg{)}\,.$
(4.11)

But by (4) we have that:

$\displaystyle t^{p-q}-\frac{s_{1}}{s_{2}}$
$\displaystyle=-(p-q)\,\frac{s_{1}\cdot\alpha(s_{2})}{q\,\vartheta(\beta)}\qquad\Rightarrow$
$\displaystyle\frac{s_{1}\cdot\alpha(s_{2})}{q\,\vartheta(\beta)}$
$\displaystyle=\frac{1}{p-q}\,\Big{(}{\frac{s_{1}}{s_{2}}-t^{p-q}}\Big{)}\,.$
(4.12)

Thus (4.11) becomes:

$\displaystyle t^{p}$
$\displaystyle=s_{1}-\frac{p}{p-q}\Big{(}{\frac{s_{1}}{s_{2}}-t^{p-q}}\Big{)}\cdot
H_{q}(\beta+1)\qquad\Rightarrow$
$\displaystyle\frac{p-q}{p}\,\frac{t^{p}-s_{1}}{t^{p-q}-\frac{s_{1}}{s_{2}}}$
$\displaystyle=H_{q}(\beta+1)\,.$ (4.13)

Define now
$\tau(t)=\dfrac{p-q}{p}\,\dfrac{t^{p}-s_{1}}{t^{p-q}-\frac{s_{1}}{s_{2}}}\,,\;t\geq
1\,.$
Now by (4), for $\beta=0$, we have:

$\displaystyle t^{p}(0)-\frac{p}{G(0)}\,t^{p-q}(0)$
$\displaystyle=\frac{p-q}{G(0)}\,s_{1}-\frac{p}{G(0)}\,\frac{s_{1}}{s_{2}}\qquad\Rightarrow$
$\displaystyle t^{p}(0)-\frac{p}{p-q}\,t^{p-q}(0)$
$\displaystyle=s_{1}-\frac{p}{p-q}\,\frac{s_{1}}{s_{2}}\,.$ (4.14)

Define the following function: $g(y)=y^{p}-\frac{p}{p-q}\,y^{p-q}\,,\;\;y\geq
1$ . Then $g$ is strictly increasing on $[1,+\infty)$, with
$g(1)=\frac{-q}{p-q}$ and $\lim_{y\to+\infty}g(y)=+\infty$ .

Claim 3. For every $(s_{1},s_{2})$ as above we have that:

$s_{1}-\frac{p}{p-q}\,\frac{s_{1}}{s_{2}}>-\frac{q}{p-q}\,.$

We prove the above claim as follows: Remember that $(s_{1},s_{2})$ satisfies:

$s_{1}^{q-1}\leq s_{2}^{p-1}\quad\Rightarrow\quad s_{2}\geq
s_{1}^{\frac{q-1}{p-1}}\,.$

Thus for the proof of the claim it is enough to prove

$s_{1}-\frac{p}{p-q}\,s_{1}^{\frac{p-q}{p-1}}>-\frac{q}{p-q}\,,\quad\forall\,s_{1}\in(0,1]\,.$

We define
$\lambda(s_{1}):=s_{1}-\frac{p}{p-q}\,s_{1}^{\frac{p-q}{p-1}}\,,\;s_{1}\in(0,1)$
. Then it is easily seen that:

$\displaystyle\lambda^{\prime}(s_{1})=1-\frac{p}{p-1}\,s_{1}^{-\frac{q-1}{p-1}}<1-\frac{p}{p-1}=-\frac{1}{p-1}<0\,.$

Thus $\lambda(s_{1})$ is strictly decreasing on $(0,1)$, therefore
$\lambda(s_{1})>\lambda(1)$, or
$s_{1}-\frac{p}{p-q}\,s_{1}^{\frac{p-q}{p-1}}>-\frac{q}{p-q}$, which is the
desired result.
Now we observe that the function $\tau(t)\,,\,t\geq 1$, satisfies the
following properties:

  1. i)

$\tau\big{(}{t(0)}\big{)}=1$ .
For this claim it is enough to show that
$\dfrac{p-q}{p}\,\dfrac{t^{p}(0)-s_{1}}{t^{p-q}(0)-\frac{s_{1}}{s_{2}}}=1$ ,
which is immediate consequence of (4).

  2. ii)

$\tau(t)$ is strictly increasing function of $t$ on $[1,+\infty)$.
Indeed we calculate

$\displaystyle\tau^{\prime}(t)$ $\displaystyle\cong
p\,t^{p-1}\Big{(}{t^{p-q}-\frac{s_{1}}{s_{2}}}\Big{)}-(t^{p}-s_{1})(p-q)\,t^{p-q-1}\cong$
$\displaystyle\cong
p\,t^{q}\Big{(}{t^{p-q}-\frac{s_{1}}{s_{2}}}\Big{)}-(p-q)(t^{p}-s_{1})=$
$\displaystyle=q\,t^{p}-p\,\frac{s_{1}}{s_{2}}\,t^{q}+(p-q)\,s_{1}=:\mu(t)\,,$
$\displaystyle\mu^{\prime}(t)$
$\displaystyle=qp\,t^{p-1}-pq\,\frac{s_{1}}{s_{2}}\,t^{q-1}\cong
t^{p-q}-\frac{s_{1}}{s_{2}}\,.$

But for $t>1$ and since $s_{1}^{q-1}\leq
s_{2}^{p-1}\;\Rightarrow\;\big{(}{\frac{s_{1}}{s_{2}}}\big{)}^{q-1}\leq
s_{2}^{p-q}\leq 1$ , we have that $\mu(t)$ is strictly increasing, thus
$\mu(t)>\mu(1)$, for every $t>1$, which gives:

$\displaystyle\mu(t)$
$\displaystyle>q-p\,\frac{s_{1}}{s_{2}}+(p-q)s_{1}>q-p\,\frac{s_{1}}{s_{1}^{\frac{q-1}{p-1}}}+(p-q)s_{1}=$
$\displaystyle=(p-q)\,s_{1}+q-p\,s_{1}^{\frac{p-q}{p-1}}=\Big{(}{s_{1}-\frac{p}{p-q}\,s_{1}^{\frac{p-q}{p-1}}}\Big{)}(p-q)+q\,>$
$\displaystyle\hskip 147.95424pt>\Big{(}{-\frac{q}{p-q}}\Big{)}(p-q)+q=0\,.$

Thus $\mu(t)>0\,,\;\forall\,t>1$ .
From the above we conclude that $\tau^{\prime}(t)>0\,,\;\forall\,t>1$, thus
the desired result.

By (4) we have $\tau(t)=H_{q}(\beta+1)<1$, for every
$\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ for which $t^{\prime}(\beta)=0$,
where $t=t(\beta)$. Thus $\beta+1=\omega_{q}(\tau(t))$, and since
$\tau(t(0))=1$ and $\tau$ strictly increasing, we have from $\tau(t)<1$ that
$t<t(0)$.
Moreover

$\displaystyle\vartheta(\beta)$
$\displaystyle=(p-1)(\beta+1)^{q}-p\,(\beta+1)^{q-1}=$
$\displaystyle=(p-1)\,\omega_{q}(\tau)^{q}-p\,\omega_{q}(\tau)^{q-1}\,.$

Thus by (4) we obtain

$\displaystyle t^{p-q}-\frac{s_{1}}{s_{2}}$
$\displaystyle=\frac{p-q}{q}\,\frac{s_{1}\,\alpha(s_{2})}{p\,\omega_{q}(\tau)^{q-1}-(p-1)\,\omega_{q}(\tau)^{q}}\qquad\Rightarrow$
$\displaystyle F_{s_{1},s_{2}}(t)$
$\displaystyle:=q\,\big{(}{p\,\omega_{q}(\tau)^{q-1}-(p-1)\,\omega_{q}(\tau)^{q}}\big{)}\Big{(}{t^{p-q}-\frac{s_{1}}{s_{2}}}\Big{)}-(p-q)\,s_{1}\,\alpha(s_{2})=0\,.$

We have proved the following:

######  Theorem 4.1.

Let $s_{1},s_{2}$ fixed such that $0<s_{1}^{q-1}\leq s_{2}^{p-1}\leq 1$. If
there exists $\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ such that
$t^{\prime}(\beta)=0$, then $t=t(\beta)$ should satisfy
$F_{s_{1},s_{2}}(t)=0$, where $F_{s_{1},s_{2}}(t)$ as given right above, and
$\tau(t)$ is described previously. Moreover in this case $t(0)>1$ and
$t=t(\beta)\in\big{[}{1,t(0)}\big{)}$ .

Note that
$A_{\beta}=\frac{G(\beta)}{p(\beta+1)^{q}}=\mathcal{K}(\beta)>\mathcal{K}(0)=\frac{p-q}{p}\,,\;\forall\,\beta\in\big{(}0,\tfrac{1}{p-1}\big{]}$
. Thus $A_{\frac{1}{p-1}}>\frac{p-q}{p}$ , thus the function
$h_{\frac{1}{p-1}}:[1,+\infty)\longrightarrow\big{(}{-\infty,h_{\frac{1}{p-1}}(1)}\big{]}$
defined by $h_{\frac{1}{p-1}}(y)=y^{p-q}-A_{\frac{1}{p-1}}y^{p}$ is strictly
decreasing on $[1,+\infty)$.
Thus the quantity $t\big{(}{\frac{1}{p-1}}\big{)}$ is well defined, from (4)
for $\beta=\frac{1}{p-1}$. Now define
$t^{\prime}\big{(}{\frac{1}{p-1}}\big{)}=\lim_{\begin{smallmatrix}\beta\to\frac{1}{p-1}\\\
0<\beta<\frac{1}{p-1}\end{smallmatrix}}t^{\prime}(\beta)$. This limit is well
defined for the following reasons: Remember that $t^{\prime}(\beta)$ satisfies
(4), for every $\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$, that is

$\displaystyle
p\,t^{p-q-1}(\beta)\,\Big{[}{t^{q}(\beta)-(p-q)\,\frac{(\beta+1)^{q}}{G(\beta)}}\Big{]}\,t^{\prime}(\beta)-pt^{p-q}(\beta)q(q-1)\frac{\vartheta(\beta)}{G^{2}(\beta)}=$
$\displaystyle=(p-q)s_{1}\frac{-p(q-1)}{G^{2}(\beta)}+\frac{s_{1}}{s_{2}}\bigg{[}{p(q-1)\frac{p-q}{G^{2}(\beta)}\,\omega_{q}(s_{2})^{q}-\frac{p\,q(q-1)\vartheta(\beta)}{G^{2}(\beta)}}\bigg{]},$
(4.15)

where $G=G(\beta)=q(p-1)(\beta+1)-p(q-1)$ .
Letting $\beta\to\tfrac{1}{p-1}$ in (4), since $G(\beta)\to qp-pq+p=p$ we get
that the right side tends to

$(p-q)s_{1}\frac{-p(q-1)}{p^{2}}+\frac{s_{1}}{s_{2}}\,p(q-1)\frac{p-q}{p^{2}}\,\omega_{q}(s_{2})^{q}\in\mathbb{R}\,.$

As we have noted above $t\big{(}{\frac{1}{p-1}}\big{)}$ is well defined, and
by its definition equals to $\lim_{\beta\to{\frac{1}{p-1}}^{-}}t(\beta)$. Thus
by (4) we have as a consequence that the limit
$\lim_{\beta\to{\frac{1}{p-1}}^{-}}t^{\prime}(\beta)$ exists if we prove that
$t^{q}\,(\frac{1}{p-1})-(p-q)\frac{\big{(}{\frac{p}{p-1}}\big{)}^{q}}{p}\neq
0$ . Indeed this is the case, since

$\displaystyle\frac{p-q}{p}\,\Big{(}{\frac{p}{p-1}}\Big{)}^{q}<1\leq
t\,\Big{(}{\frac{1}{p-1}}\Big{)}^{q}\,.$ (4.16)

The second inequality in (4.16) is obviously true, while for the first we
argue as follows:
For $1<q<p$ consider the $L^{p}$-integrable function on $(0,1]$, defined by
$\delta(t)=t^{-1/p}$ . Then since $\delta(t)$ is non-constant on $(0,1]$ we
set (by Hölder’s inequality):

$\displaystyle\bigg{(}{\int_{0}^{1}t^{-1/p}\,dt}\bigg{)}^{q}<\int_{0}^{1}t^{-q/p}\,dt\quad$
$\displaystyle\Rightarrow\quad\Big{(}{\frac{p}{p-1}}\Big{)}^{q}<\frac{1}{1-\frac{q}{p}}=\frac{p}{p-q}$
$\displaystyle\Rightarrow\quad\Big{(}{\frac{p}{p-1}}\Big{)}^{q}\,\frac{p-q}{p}<1\,.$

Thus taking limits in (4) as $\beta\to{\frac{1}{p-1}}^{-}$ we have:

$\displaystyle
p\,t^{p-1}\big{(}{\tfrac{1}{p-1}}\big{)}\,t^{\prime}\big{(}{\tfrac{1}{p-1}}\big{)}-p(p-q)\,t^{p-q-1}\big{(}{\tfrac{1}{p-1}}\big{)}\,\frac{\big{(}{\tfrac{p}{p-1}}\big{)}^{q}}{p}\,t^{\prime}\big{(}{\tfrac{1}{p-1}}\big{)}=$
$\displaystyle\hskip
56.9055pt=\frac{-p(p-q)(q-1)}{p^{2}}\,s_{1}+\frac{s_{1}}{s_{2}}\,\frac{p(q-1)(p-q)}{p^{2}}\,\omega_{q}(s_{2})^{q}$
$\displaystyle\Rightarrow$ $\displaystyle
t^{\prime}\big{(}{\tfrac{1}{p-1}}\big{)}\,t^{p-q-1}\big{(}{\tfrac{1}{p-1}}\big{)}\,\bigg{[}{t^{q}\big{(}{\tfrac{1}{p-1}}\big{)}-\frac{p-q}{p}\,\Big{(}{\frac{p}{p-1}}\Big{)}^{q}}\bigg{]}=$
$\displaystyle\hskip
28.45274pt=\frac{-(p-q)(q-1)}{p^{2}}\,s_{1}+\frac{s_{1}}{s_{2}}\,\frac{(q-1)(p-q)}{p^{2}}\,\omega_{q}(s_{2})^{q}=$
$\displaystyle\hskip
42.67912pt=\frac{(p-q)(q-1)}{p^{2}}\,s_{1}\Big{(}{\frac{\omega_{q}(s_{2})^{q}}{s_{2}}-1}\Big{)}=\frac{(p-q)(q-1)}{p^{2}}\,s_{1}\alpha(s_{2})\,.$
(4.17)

At this point we note that we restrict to ourselves (see the beginning of
Section 4) to the study of those functions $h$ which are not constant on
$(0,\kappa]$. Thus $x,y$ should satisfy:
$\frac{x^{q}}{\kappa^{q-1}}<y\;\Rightarrow\;s_{2}=\frac{x^{q}}{\kappa^{q-1}y}<1$
.
It is obvious that the function
$\alpha(s_{2})=\big{(}{\frac{\omega_{q}(s_{2})^{q}}{s_{2}}-1}\big{)}$ is
strictly decreasing on $s_{2}\in(0,1]$ so, $\alpha(s_{2})>\alpha(1)=0$. Thus
by (4) and (4.16) we get that $t^{\prime}\big{(}{\tfrac{1}{p-1}}\big{)}>0$ .
Again now by (4), or (4) we get that $t^{\prime}(0):=\lim_{\beta\to
0^{+}}t^{\prime}(\beta)$ exists if $t^{q}(0)\neq(p-q)\,\frac{1}{G(0)}=1$ . But
$t(0)$ satisfies (by its definition) :
$\delta^{p}=\frac{\int_{0}^{\kappa}\big{(}\frac{1}{t}\int_{0}^{t}h\big{)}^{p}dt}{\int_{0}^{\kappa}h^{p}}\leq
t^{p}(0)$, and since $h$ is not constant we easily see that $\delta>1$, which
gives $t(0)>1$. Thus $t^{\prime}(0):=\lim_{\beta\to 0^{+}}t^{\prime}(\beta)$
is well defined.
Now we consider the following general cases:

  1. I)

$t^{\prime}(0)<0$. Then since
$t^{\prime}\big{(}{\tfrac{1}{p-1}}\big{)}>0\;\Rightarrow\;\exists\,\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}:\;t^{\prime}(\beta)=0$,
for which if we set $t=t(\beta)$, $t$ satisfies $F_{s_{1},s_{2}}(t)=0$. But as
can be easily seen the function $F_{s_{1},s_{2}}(\cdot)$ is strictly
increasing in $[1,+\infty)$, that is $F$ is one to one. Thus $t(\beta)$ is
uniquely determined when $t^{\prime}(\beta)=0$.
This easily gives in this case that for a
$\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ such that $t^{\prime}(\beta)=0$ we
have that

$t=t(\beta)=\min\Big{\\{}{t(\gamma)\;:\;\gamma\in\big{[}0,\tfrac{1}{p-1}\big{]}}\Big{\\}}$

and of course $F_{s_{1},s_{2}}(t)=0$. This case is then completely studied. We
turn now to the other cases.

  2. II)

$t^{\prime}(0)>0$. Then we must have that $t^{\prime}(\beta)>0$,
$\forall\,\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$, otherwise there would
exist $\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ for which $t(\beta)>t(0)$ with
$t^{\prime}(\beta)=0$. But the condition $t^{\prime}(\beta)=0$ implies
$t(\beta)<t(0)$ as we have seen in the proof of Theorem 4.1. Thus in this case

$\min\Big{\\{}{t(\gamma)\;:\;\gamma\in\big{[}0,\tfrac{1}{p-1}\big{]}}\Big{\\}}=t(0)\,.$

  3. III)

$t^{\prime}(0)=0$. There does not exist
$\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ for which $t^{\prime}(\beta)=0$ and
$t(\beta)<t(0)$, by using the same arguments that are mentioned above. Thus
$t(\beta)\geq t(0)$, $\forall\,\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$, since
otherwise would exist $\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ with
$t(\beta)<t(0)$ that is $t(\gamma)$ would be strictly decreasing on a
subinterval of $\big{(}{0,\frac{1}{p-1}}\big{)}$. This fact, and the
observation that $t^{\prime}\big{(}{\frac{1}{p-1}}\big{)}>0$ would then give
the existence of an $\beta_{1}\in\big{(}{0,\frac{1}{p-1}}\big{)}$ for which
$t^{\prime}(\beta_{1})=0$ and $t(\beta_{1})<t(0)$ which is a case that is
excluded. Thus $t(\beta)\geq t(0)$,
$\forall\,\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ and we get in this case
also that

$\min\Big{\\{}{t(\gamma)\;:\;\gamma\in\big{[}0,\tfrac{1}{p-1}\big{]}}\Big{\\}}=t(0)\,.$

Moreover we can argue alternatively for the third case as follows: If
$t^{\prime}(0)=0$, then $F_{s_{1},s_{2}}(t(0))=0$. Thus we must have that
$t^{\prime}(\beta)>0$, $\forall\,\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$.
Indeed if there exists $\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ with
$t^{\prime}(\beta)\leq 0$, then since
$t^{\prime}\big{(}{\tfrac{1}{p-1}}\big{)}>0\;\Rightarrow\;\exists\,\beta_{1}\in\big{[}{\beta,\frac{1}{p-1}}\big{)}:\;t^{\prime}(\beta_{1})=0\;\Rightarrow\;F_{s_{1},s_{2}}(t(\beta_{1}))=0$
and $t(\beta_{1})<t(0)$, as already proved. Thus
$F_{s_{1},s_{2}}(t(0))=0=F_{s_{1},s_{2}}(t(\beta_{1}))$ with
$t(\beta_{1})<t(0)$ which can not be true. Thus $t^{\prime}(\beta)>0$,
$\forall\,\beta\in\big{(}{0,\frac{1}{p-1}}\big{)}$ and as consequence

$\min\Big{\\{}{t(\gamma)\;:\;\gamma\in\big{[}0,\tfrac{1}{p-1}\big{]}}\Big{\\}}=t(0)\,,$

in case where $t^{\prime}(0)=0$.

Note that in case where $t^{\prime}(0)=0$ we get by following the lines of the
proof of Theorem 4.1, that:

$\tau(t(0))=H_{q}(1)=1\,.$

Thus $F_{s_{1},s_{2}}(t(0))=0$ (see right above the statement of Theorem 4.1).
Note also that in case where
$t^{\prime}(0)>0\;\Rightarrow\;F_{s_{1},s_{2}}(t(0))<0$.
The above claim is true and for this we argue as follows:
If $t^{\prime}(0)>0$, then by (4) or (4) we have

$\displaystyle p\,t^{p-q-1}(0)\,t^{\prime}(0)\,\big{[}{t^{q}(0)-1}\big{]}=$
$\displaystyle\qquad=p\,t^{p-q}(0)\,q(q-1)\frac{-1}{(p-q)^{2}}+(p-q)\,s_{1}\frac{-p(q-1)}{(p-q)^{2}}\,+$
$\displaystyle\hskip
85.35826pt+\frac{s_{1}}{s_{2}}\bigg{[}{p(q-1)\frac{p-q}{(p-q)^{2}}\,\omega_{q}(s_{2})^{q}-\frac{p\,q(q-1)(-1)}{(p-q)^{2}}}\bigg{]}\,.$
(4.18)

But since $t(0)$ bounds the quantity
$\delta=\big{(}{\frac{I}{z}}\big{)}^{1/p}=\Big{(}{\frac{\int_{0}^{\kappa}\big{(}\frac{1}{t}\int_{0}^{t}h\big{)}^{p}dt}{\int_{0}^{\kappa}h^{p}}}\Big{)}^{1/p}$
and $h$ is considered to be non-constant we have $t(0)>1$, and since also
$t^{\prime}(0)>0$, we obtain by (4)that its right hand side is positive, that
is

$\displaystyle-q\,t^{p-q}(0)-(p-q)\,s_{1}+\frac{s_{1}}{s_{2}}\big{[}{(p-q)\,\omega_{q}(s_{2})^{q}+q}\big{]}>0$
$\displaystyle\Rightarrow$ $\displaystyle
q\,t^{p-q}(0)<(p-q)\,\frac{s_{1}}{s_{2}}\,\omega_{q}(s_{2})^{q}-(p-q)\,s_{1}+q\,\frac{s_{1}}{s_{2}}$
$\displaystyle\Rightarrow$ $\displaystyle
q\,\Big{(}{t^{p-q}(0)-\frac{s_{1}}{s_{2}}}\Big{)}<(p-q)\,s_{1}\cdot\alpha(s_{2})\,.$
(4.19)

But then by definition of $F_{s_{1},s_{2}}(t(0))=$
$q\,\big{(}{p\,\omega_{q}(\tau(t(0)))^{q-1}-(p-1)\,\omega_{q}(\tau(t(0)))^{q}}\big{)}\Big{(}{t(0)^{p-q}-\frac{s_{1}}{s_{2}}}\Big{)}-(p-q)\,s_{1}\,\alpha(s_{2})$
and since $\tau(t(0))=1$, and (4) is true, we get $F_{s_{1},s_{2}}(t(0))<0$.
[In the same way we prove that: if $t^{\prime}(0)=0$, then
$F_{s_{1},s_{2}}(t(0))=0$].

At this point we mention that accordingly to the results of this section, we
have that whenever $0<\frac{x^{q}}{\kappa^{q-1}}<y\leq
x^{\frac{p-q}{p-1}}\cdot
z^{\frac{q-1}{p-1}}\;\Leftrightarrow\;0<s_{1}^{\frac{q-1}{p-1}}\leq s_{2}<1$,
there is defined a constant $t=t(s_{1},s_{2})$ for which if
$h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$ satisfies $\int_{0}^{\kappa}h=x$ ,
$\int_{0}^{\kappa}h^{q}=y$ and $\int_{0}^{\kappa}h^{p}=z$ then

$\displaystyle\int_{0}^{\kappa}\bigg{(}\frac{1}{t}\int_{0}^{t}h\bigg{)}^{p}dt\leq
t^{p}(s_{1},s_{2})\cdot\int_{0}^{\kappa}h^{p}$ (4.20)

where $t(s_{1},s_{2})=t$ is the greatest element of $\big{[}{1,t(0)}\big{]}$
for which $F_{s_{1},s_{2}}(t)\leq 0$. Moreover for each such fixed
$s_{1},s_{2}$

$t=t(s_{1},s_{2})=\min\Big{\\{}{t(\beta)\;:\;\beta\in\big{[}0,\tfrac{1}{p-1}\big{]}}\Big{\\}}$

where $t(\beta)=t(\beta,s_{1},s_{2})$ is defined by equation (4). That is we
have found a constant $t=t(s_{1},s_{2})$ for which inequality (4.20) is
satisfied for all $h:(0,\kappa]\longrightarrow\mathbb{R}^{+}$ as mentioned
above. Note that $s_{1},s_{2}$ depend by a certain way on $x,y,z$, namely
$s_{1}=\frac{x^{p}}{\kappa^{p-1}z}$, $s_{2}=\frac{x^{q}}{\kappa^{q-1}y}$ and
$F_{s_{1},s_{2}}(\cdot)$ is given in terms of $s_{1},s_{2}$.

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Nikolidakis Eleftherios
Assistant Professor
Department of Mathematics
Panepistimioupolis, University of Ioannina, 45110
Greece
E-mail address<EMAIL_ADDRESS>