Aziz Esmail's academic qualifications range across the humanities. He originally graduated in literature, followed by studies in philosophy and religion in which he was a lecturer, in the first phase of his academic career, at the University of Nairobi. While there, he helped to establish a curriculum in Islamic Studies in a world-civilisational framework, an area in which he has lectured as well as written. He later lived in the United States where he was associated with the Committee on Social Thought at the University of Chicago followed by a fellowship at the Center for the Study of World Religions at Harvard. During this period he pursued research in psychoanalysis and in the philosophy of medicine, a subject in which he subsequently obtained a diploma in London, where he remains an Associate Member of the Royal Society of Medicine. He is a former Dean of the IIS, and has been a Member of its Board of Governors since 1998.
Naguib Kheraj is a banker by background. He began his career at Salomon Brothers in 1986 and went on to hold a number of senior positions at leading international financial institutions. Over the course of 12 years at Barclays, Naguib served as Group Finance Director and Vice-Chairman and in various business leadership positions. He was Chief Executive Officer of JP Morgan Cazenove, a London-based investment banking business. Naguib is currently Deputy Chairman of Standard Chartered, an international banking group and Chairman of Rothesay Life, a specialist pensions insurer and is a member of the Finance Committee of Oxford University Press.
Mr Kheraj is a former Non-Executive Director of National Health Service England, and has served as a Senior Adviser to Her Majesty's Revenue and Customs Service and to the Financial Services Authority in the UK. He has also served as a member of the Board of the UK-US Fulbright Commission, the Princes Trust Development Board and the Investment Committee of the Wellcome Trust.
Mr Kheraj spends a substantial proportion of his time as a Senior Advisor to the Aga Khan Development Network and serves on the Boards of various entities within the Aga Khan Development Network including the Aga Khan University and the University of Central Asia and is Chairman of the Aga Khan Development Network Endowment Committee.
Naguib was educated at Dulwich College, London and Cambridge University where he graduated with a degree in Economics.
In his capacity as Head of the Department of Jamati Institutions at the Secretariat of His Highness the Aga Khan at Aiglemont in France, Shafik Sachedina is responsible for coordinating the activities and programmes of the Ismaili community's institutions in the 16 principal areas where such institutions exist. Dr Sachedina is also responsible for interfacing with the Aga Khan Development Network programmes and institutions of the Ismaili communities in Central Asia.
Born in Dar-es-Salaam, Tanzania in 1950, Dr Sachedina qualified as a dental surgeon in 1975 from Guy's Hospital Medical and Dental School, University of London, and has practised in England for many years. He has also developed substantial entrepreneurial interests in the health-care sector. His work with the Aga Khan and Jamati Institutions is undertaken in a voluntary capacity, in keeping with the community's long-standing tradition of voluntary service.
In addition to being a member of the Institute's Board of Governors, Dr Sachedina holds several other positions in the work he undertakes for the Ismaili Imamat and community, including Member of the Aga Khan Development Network (AKDN) Committee, Aiglemont; Chairman of FOCUS Humanitarian Assistance International Co-ordinating Committee; and Member of the Ismaili Leaders' International Forum. Previously, Dr Sachedina has served in many other voluntary capacities, including two consecutive terms as President of the Ismaili Council for the United Kingdom.


text:
Clever Geek Handbook
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Maria da Fonte (movement)
"Maria da Fonte" ( port. Movimento Maria da Fonte ) is a Portuguese anti-communist , anti-Marxist movement of the mid-1970s. It operated mainly in the northern regions of the country. Relied on the conservative peasantry under the auspices of the Catholic clergy. Actively participated in the clashes of "Hot Summer" in 1975 , widely used violent methods of political struggle. Played an important role in the counter-offensive of the right forces.
The movement "Maria da Fonte"
port. Movimento Maria da Fonte
Ideology anti-communism , conservatism
Ethnicity Portuguese
Religious affiliation Catholicism
The leaders Waldemar Paradela di Abreu
Headquarters Braga
Active in Portugal
Date of formation 1975
Date of dissolution 1976
Allies Catholic Church ,
Democratic Movement for the Liberation of Portugal ,
Portugal Liberation Army ,
Social Democratic Center
Opponents Portuguese Communist Party ,
left wing of the Armed Forces Movement
Conflict participation "Hot Summer"
Large shares pogroms, arson
Title as historical parallel
Maria da Fonte (according to other data - Maria Louise Balaio or Maria Angelica di Simaesh) is a real historical person, although there is no accurate and reliable information about her. It is known that in 1846 she led a peasant Catholic uprising. Mary was born in the northern region of Fontarkad , belonging to the county and county of Braga . Perhaps it is the collective name of a group of women from the rebel movement [1] .
The uprising of Maria da Fonte - developed in the so-called. "The Minho Revolution" was directed against the government of Costa Cabral , who pursued a policy of state centralism and violated Catholic customs. The image of the militant Catholic Mary, fighting for freedom and justice, was close to the sentiments of the conservative peasants of northern Portugal in 1975. The revolutionary authorities of Lisbon were associated with the anti-people government of Costa Cabral with their communist-socialist policies.
Increasing socio-political tensions
The government of Vaska Gonsalvis , the Communist Party and the ultra-left wing of the ICE carried out a policy of consistent Sovietization after September 1974 . Oppositional sentiments were suppressed by arbitrary arrests [2] . Administrative centralization was intensified, including in the financial sphere (which also reminded of the times of Costa Cabral). Right-wing political activists and Catholic clergy, most notably the archbishop of Braga , called for resistance.
April 25, 1975 in Portugal held elections to the Constituent Assembly. Their results showed the predominance of moderate socialists ( joint venture Mario Soares ), liberals ( NDP Francisco Sa Carneiro ) and conservatives ( CJD Diogo Freitas do Amaral ), who together collected more than 70% of the vote. The Communists received only 12%, but the government of Gonçalves, speaking in alliance with the party Alvaru Cunhal , continued the previous policy. Through the nationalization of industry, transport, banks, the economy was put under the control of the radical radical military and the PCP. The trade union movement was unified under communist control. In the south, collectivization began with direct military pressure. Opposition parties protested, but the ruling Revolutionary Council expressed confidence in Gonçalvesh. The next day, July 13, 1975, a wave of anti-communist protests of a pogrom character swept Braga [3] .
The basis and features of the movement
The initiators of the creation of the movement "Maria da Fonte" were made by the famous journalist Waldemar Paradela de Abreu and the canon of the Cathedral of Braga, Eduardo Mel Peixoto , the leader of the anti-communist forces of the Northern Region of Portugal.
The North was a country of small farmers and small entrepreneurs, independent owners. The working class of the North was engaged not in large enterprises, but in small factories scattered across the fields. And this country faced the South of civil servants, latifundia and large companies - the property of old families, protected by dictatorship and nationalized by revolution. Violent independent communities united around traditional religious authorities. In the summer of 1975, these people took to the streets under the sound of bells to challenge the Lisbon Communism. This happened not for the first time - many then recalled 1846, the popular anti-tax riot of Maria da Fonte against Costa Cabral [4] .
The decision was made in July 1975 at a meeting of right-wing activists chaired by Mel. Financial and organizational assistance was provided by the secret operations expert Jorge Jardin . The consent and support of Archbishop Braga to Francisco Maria da Silva was received.
Paradela di Abreu secretly met with Francisco Maria da Silva. The archbishop agreed to put the church in the service of a large anti-communist struggle project. The Cathedral of canon Mela supported the uprising. In the following weeks, they passed through the north of Portugal with fire and sword [5] .
By this time, radical anti-communist organizations and right-opposition parties were already operating in Portugal. The main base of their influence was precisely the northern regions. However, the movement "Maria da Fonte" was significantly different from them.
The main difference from the terrorist ELP and MDLP was mass and demonstrative. On the other hand, the illegal nature and orientation on the use of force distinguished the "Maria da Fonte" from the NDP and the SAD. In addition, while the Liberation Army and the Democratic Movement consisted mainly of right-oriented military , former PIDE employees and staunch Salazarists , then Maria da Fonte had previously united apolitical citizens, usually peasants, committed to anti-communism , traditional conservatism and Catholicism .
Not the military, not the secret services, but peasants of the northern region of Portugal gathered here. Latifundy was not in the north; in the villages there lived individual peasants, good Catholics, parishioners of the Cathedral of Braga. The pro-communist government set out to cooperate, closing the churches. And the peasants - truly spiritually bound together in Catholic parishes - found the answer to that. In the spirit of revered from generation to generation, Maria da Fonte [6] .
The movement did not have a single ideology and party program. Among its participants were both socialists and future members of the right-wing MIRN crine, Caulza de Arriagi . The unifying idea was anti-communism. The slogans of social Catholicism and national conservatism were also advanced. The most effective organizational centers of the "structure ready for war" were church parishes - which initially corresponded to the idea of Paradela di Abreu and Edouard Melou [7] . The buildings of the churches they considered as "fortresses", the bells as "radio transmitters", the peasant yards as "supply bases", the priests as agitators and organizers of the struggle [8] .
Right offensive - promotions and outcomes
In July-August 1975, the "Maria da Fonte" movement committed up to hundreds of mass power actions against the PCP and pro-communist organizations [9] . Attacks were primarily local party representation. The offices of the Communists were subject to defeat and arson, these actions of the movement were held in close coordination with the MDLP. At the same time, in all major cities of the north and partly of the center - Braga , Bragança , Lamego , Vila Real , Coimbra , Aveiro , Viana do Castelo - Catholic demonstrations were held under anti-communist and anti-government slogans. Clashes occurred in Porto , sunk to Lisbon [10] .
The fighters of "Maria da Fonte" were not only peasants, but also many citizens. A separate power group consisted of three hundred factory workers. Particularly attracted former soldiers of the colonial war in Africa .
"Maria da Fonte" found inspiration even in the red ideologies with which she fought. According to Mao's precepts, the religious army was popularly like fish in water [8] .
Along with the power actions "Maria da Fonte" led an intensive mass agitation. In the parishes spread thousands of leaflets and pamphlets, adapted to the perception of the peasants. Smoothly worked radio station traffic. Not only sermons and political appeals were transmitted, but also songs (sometimes in the traditional Fado style), also of unequivocal content: "Go away, Cunhal! Do not go back scoundrel! Because of you, traitors, Russians drink our wine for nothing! " Among the leaders and agitators there were not only priests and political activists, but also popular fado singers, especially Juan Braga , one of the founders of the movement [8] .
Constantly appeals to Christian images sounded: the arsons of the communist offices were declared " holy fire ", Catholic slogans were pronounced before going to the rally. The atheistic nature of communism as a force hostile to Christianity and the anti-national character of the PCP as the party of Moscow lackeys was emphasized.
The movement acquired such scope that the head of the operational-combat department of the MDLP, Captain Kalvan, proposed to Parade di Abreu to proclaim an alternative state entity, the Free Homeland, in the controlled territories of the north. The commander of the Northern Military District, General Antonio Pires Veloso, spoke positively about the "Maria da Fonte".
On August 10, 1975, Archbishop Francisco Maria da Silva delivered a politicized anti-communist sermon in the Cathedral of Braga (prepared with the participation of canon Melo) [11] . Thousands of parishioners moved to the local PKP headquarters, attacked and burned the premises. This event was the culmination of the Portuguese "Hot Summer".
Paradela was delighted: "There is no such strength that could defeat us!" [8]
Mass anti-communist speeches in combination with party political activity and a series of terrorist acts of the anti-communist underground changed the balance of forces in the country. In the Armed Forces Movement, the right wing was strengthened. The events of November 25, 1975 launched a political process to the right. The prospect of communization has ceased to be relevant. Since 1976, the activity of "Maria da Fonte" has gradually ceased.
Democratic Movement for the Liberation of Portugal
Portugal Liberation Army
CODECO
↑ A região Maria da Fonte
↑ O BISPO, O HOMEM EO CONTRA REVOLUCIONÁRIO ( Unreferenced ) (not available link) . The appeal date is March 25, 2014. Archived March 25, 2014.
↑ Video: Demonstration / Portugal / 1975
↑ O norte da democracia
↑ O Norte a ferro e fogo Archived March 26, 2014.
↑ Hot Braga, or Rifle creates freedom
Hist Doutorando em História (UFRGS) com estágio PDEE-CAPES no Instituto de Ciências Sociais da Universidade de Lisboa (ICS-UL); Mestre em História (UFRGS) / A "Nova" Extrema-Direita: o caráter grupuscular das organizações neofascistas em Portugal e na Argentina. A "nova" extrema-direita em Portugal.
↑ 1 2 3 4 Miguel Carvalho. Quando Portugal Ardeu - Históricas da violência política no pós-25 de Abril / Oficina do livro - Sociedade Editorial, Lda, 2017.
↑ Democrat reluctantly
↑ A CONTRA-REVOLUÇÃO
↑ PORTUGAL: The Anti-Communists Strike Back
Source - https://ru.wikipedia.org/w/index.php?title=Maria_da_Fonte_ ( movement )&oldid = 99603855
Zurabov, Robert Sergeevich
Albania Football Championship 1971/1972
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Church of the Holy Face of Christ the Savior (Sochi)
Teboul Sign Language
Aleksinsky, Mikhail Alexandrovich
Love Story (Lee Freeman's Song)
PSM3
Rudnev, Vladimir Ivanovich (psychiatrist)
Kotovsky (Kemerovo region)
Clever Geek | 2019

redpajama_set_name: RedPajamaCommonCrawl


Every year, after the apple pies are baked and enjoyed, I make applesauce with all the rest of the apples from my garden. Of course it's usually just in time for Hanukkah which means yummy latkes with applesauce and sour cream. Making applesauce is pretty easy but peeling the apples is very tedious and time consuming, so this year I decided to try leaving the peels on. Okay, I'll admit it, I got lazy, but it worked out for the best. I LOVE this applesauce and it is so easy! It's got a richer and creamier texture AND it's more nutritious since most of the nutients and fiber are in or right under the peel which is usually removed and thrown away. Win! Win! For best results use a combination of both sweet and tart apples. Fuji, Red Delicious, Pink Lady and Golden Delicious apples are sweeter in taste, while Granny Smith and Pippin apples add a nice tartness.
Applesauce is a favorite snack for little kids in particular. The healthiest applesauce is made from organic apples with the skin on and no added sugar. Apples are one fruit in which you can't wash off the pesticides, so buying organic apples is especially important if you don't have your own apple trees. Apples are a great source of natural fiber and vitamin C. The pulp and skin of apples contain flavonoids, which offer numerous health benefits; reduce inflammation, regulate blood pressure. The phytonutrients in apples work as antioxidants to support your heart and helping to lower your bad cholesterol levels. My weight concious readers will like to hear that eating applesauce can also decrease your risk of developing abdominal fat. Studies show that the pectin in apples suppresses your appetite. So what are you waiting for?
Just core the apples, either with a sharp knife or with an apple corer, then cut the apples, peels and all, into slices or chunks. Simmer with a bit of water, lemon juice and a couple of cinnamon sticks until soft. Taste and add sugar if desired. Depending on the apples, it may not need any. Then remove the cinnamon sticks and process the apple mixture until smooth in a blender or food processor. The applesauce can be stored in the refrigerator for a couple of weeks or placed in plastic freezer bags and frozen. It can also be canned while hot. Click HERE for directions from Bell on canning hot applesauce.
Of course, my favorite way to emjoy applesauce is on top of crispy potato pancakes (latkes) with sour cream. Click HERE for my yummy Latke Recipe.
Core apples and slice or cut into chunks. Place in a large stockpot. Pour water and lemon juice over the top.
Add two cinnamon sticks (or ground cinnamon to taste). Cover and simmer for about 30 to 45 minutes. If it starts to stick, add a bit more water but not too much.
When apples are completely soft and falling apart, remove cinnamon sticks and process apples in a food processor or blender until smooth.
For a chunkier applesauce, use a potato masher.
Store in the refrigerator up to 2 weeks.
Core apples and slice or cut into chunks. Place in a large stockpot.
Pour water and lemon juice over the top.


The term "hypocrite" originally referred to an actor who pretended to be something he was not. In its original context, the word "hypocrite" was not a derogatory title; however, it evolved over time so that it came to be associated with dishonesty, a lack of genuineness, and contradictory behavior. It is this latter understanding of hypocrisy that the Bible utilizes.
As Christians, we are consistently instructed to eliminate hypocrisy. We are called to "love…without hypocrisy" (Romans 12:9), we are called to pursue the "wisdom from above" which lacks "hypocrisy" (James 3:17), and we are called to "rid [ourselves] of all…hypocrisy" (1 Peter 2:1). If the Bible is adamant that we eliminate hypocrisy, then we must become acquainted with the Bible's definition of hypocrisy. Consider the following five indicators of a hypocrite according to Scripture.
First, the Bible indicates that you might be a hypocrite if the external exhibition of your faith glorifies yourself rather than God. Jesus called those who draw attention to themselves when they give (Matthew 6:2), pray (Matthew 6:5), and fast (Matthew 6:16) hypocrites. Additionally, He indicated that those who do such things for their own recognition have already been rewarded, insinuating that they will not receive the eternal reward. The expectation of God's people is that their spiritual activities will be conducted for the sole purpose of bringing glory to God. As Paul said in 1 Corinthians 10:31, "whether you eat or drink, or whatever you do, do all to the glory of God." Whenever our spiritual activities are done for personal recognition, then we venture into the realm of hypocrisy.
Second, the Bible indicates that you might be a hypocrite if you hold others to a standard that you do not hold yourself. Jesus said, "Judge not, that you be not judged. For with what judgment you judge, you will be judged; and with the measure you use, it will be measured back to you" (Matthew 7:1-2). He then called those who point out the sins of others without recognizing their own sins hypocrites (Matthew 7:3-5). Jesus' point was that you must examine and correct your sin problem before you can start assisting other people with their sin problem. To ignore your sin and simultaneously point out those of others is hypocritical because you are holding others to a standard to which you refuse to hold yourself.
Third, the Bible indicates that you might be a hypocrite if your behavior is dependent on your environment. Paul said that Peter, Barnabas, and other Jewish Christians "played the hypocrite" in Antioch when they refused to dine with Gentile Christians after "certain men came from James" (Galatians 2:11-13). In other words, they stopped associating with the Gentile group when a Jewish group showed up. Their environment dictated their behavior. Paul would later warn against such conformity (Romans 12:2), and Peter would indicate that we as Christians are expected to be strange to the world (1 Peter 4:3-4). The point of both authors is that Christians are expected to not allow their environment to dictate their behavior. Anytime we adjust who we are, what we believe, or how we act to fit the expectations of our environment we enter into the realm of hypocrisy because we simultaneously stop pursuing the path of holiness.
Fourth, the Bible indicates that you might be a hypocrite if you pick and choose which of God's commands you will obey. In Matthew 23:23-24 Jesus criticized the Scribes and Pharisees saying, "Woe to you, scribes and Pharisees, hypocrites! For you tithe mint and dill and cumin, and have neglected the weightier matters of the law: justice and mercy and faithfulness. These you ought to have done, without neglecting the others." In other words, the Scribes and Pharisees were willing to obey the "lighter" laws, the ones that were predicated on right behavior, but ignored the "weightier" laws, the ones that were predicated on a right heart. They were hypocrites because their obedience to God was selective at best. Whenever we blatantly ignore or rationalize our failure to keep one of God's commands and/or expectations, we become hypocrites because we are practicing selective commitment to God rather than the complete commitment to God.
Fifth, the Bible indicates that you might be a hypocrite if your external actions exist to cover up an internal unrighteousness. This is why Jesus called the scribes and Pharisees "whitewashed tombs." They appeared externally "righteous" but were internally "full of hypocrisy and lawlessness" (Matthew 23:25-28). In other words, their righteousness was just a mask that they wore to cover up their sin. As Christians, we enter the realm of hypocrisy when our faith is simply a mask, and your faith is simply a mask if your lifestyle on Monday through Saturday contradicts your profession on Sunday. We must remember that masks do not fool God because "no creature is hidden from his sight, but all are naked and exposed to the eyes of him to whom we must give account" (Hebrews 4:13). In other words, God will always see past the masks, disguises, and fronts that we use to hide our true identity.
So, are you a Christian or are you a hypocrite? The two are mutually exclusive, and it is important that we know which we are because the hypocrite will face eternal punishment.
In Matthew 24:45-51 Jesus compared a faithful steward to an evil steward, noting that the difference between the two is their preparedness. In this parable, the evil steward failed to be prepared for the master's return and, as a result, was caught behaving unethically and immorally. Jesus stated that the consequence of his actions would be that the Master "will cut him in pieces and put him with the hypocrites. In that place there will be weeping and gnashing of teeth" (Matthew 24:51). The place where the hypocrites are said to be is the same place where the weeds, which represented all sinners and law-breakers, were sent in Matthew 13:42. It's the same place where the garmentless attendee of the wedding feast was sent in Matthew 22:13. It's the same place where the worthless one talent man was sent in Matthew 25:30. In other words, it's the same place that is consistently juxtaposed with heaven. As a result, Scripture indicates that those guilty of hypocrisy will face eternal punishment in hell.
So, are you a Christian or a hypocrite? It's an important question because we must "examine [ourselves], to see whether [we] are in the faith" so that we do not "fail…the test" (2 Corinthians 13:5).


text:
\section{Introduction}

After the Hardy inequality was proved by G. H. Hardy in \cite{Hardy:1920}, a large amount of literature is available on this inequality. The integral inequality of the type
\begin{equation}\label{EQ:Hardy_M}
\left(\int_a^b \left(\int_a^x f(t) dt \right)^q u(x) dx \right)^{1/q} \leq C \left( \int_a^b f^p(x)v(x) dx \right)^{1/p},
\end{equation}
is well-known with $a$ and $b$ real numbers satisfying $-\infty \leq a<b \leq \infty$, and $p, q$ real parameters satisfying $0<q \leq \infty$, $1 \leq p \leq \infty.$ The problem of characterising the weights $u$ and $v$ in this inequality has been investigated by many authors. There are too many references to give an entire overview here, so we refer to only a few: Hardy, Littlewood and Polya \cite{Hardy:1952}, Mitrinovic, Pecaric and Fink \cite{MPF91}, Heinig and Sinnamon \cite{Heinig}, Kokitishvili \cite{VMK},
 Opic and Kufner \cite{OK-book}, Davies \cite{Davies},
Kufner, Persson and Samko \cite{Hardy-weighted-book,KP-Samko},
Edmunds and Evans \cite{DE-book}, Mazya \cite{Mazya-Sobolev-spaces,Mazya-book}, Ghoussoub and Moradifam \cite{GM-book}, Balinsky, Evans and Lewis \cite{BEL-book}, and references therein.
We can also refer to the recent open access book
\cite{RS-book} devoted to Hardy, Rellich and other inequalities in the setting of nilpotent Lie groups.

In our previous paper  \cite{RV-18}, for the case $1< p \leq q < \infty$, we characterised the weights $u$ and $v$ for the Hardy inequalities \eqref{EQ:Hardy_M} to hold on general metric measure spaces with polar decompositions.
In this paper, complementary to \cite{RV-18}, we consider the weight characterisations for the case $$0<q<p, 1<p<\infty.$$

  The setting of these papers is rather general, and we consider {\em polarisable metric measure spaces}. These are
  metric spaces $(\mathbb X,d)$ with a Borel measure $dx$ allowing for the following {\em polar decomposition} at $a\in{\mathbb X}$: we assume that there is a locally integrable function $\lambda \in L^1_{loc}$  such that for all $f\in L^1(\mathbb X)$ we have
   \begin{equation}\label{EQ:polar}
   \int_{\mathbb X}f(x)dx= \int_0^{\infty}\int_{\Sigma_r} f(r,\omega) \lambda(r,\omega) d{\omega_r} dr,
   \end{equation}
   where $(r,\omega)\rightarrow a $ as $r\rightarrow0$.
    Here the sets $\Sigma_r=\{x\in G:d(x,a)=r \}\subset \mathbb X$ are equipped with measures, which we denote by $d\omega_r$.

The polar decomposition \eqref{EQ:polar} is a rather general condition in the sense that we allow the function $\lambda$ to be dependent on the full variable $x=(r,\omega)$.
In the examples described below, in the presence of the differential structure, the function $\lambda(r,\omega)$ can appear naturally as the Jacobian of the polar change of variables.
However, since we do not assume that $\mathbb X$ must have a differentiable structure, we impose
\eqref{EQ:polar} as a condition on metric and measure.

In our previous paper \cite{RV-18} we gave several important examples of polarisable metric spaces, let us briefly recapture them here:
\begin{itemize}
\item[(I)] On the Euclidean space ${\mathbb R}^n$, we can take $ \lambda (r,\omega)= {r}^{n-1}$, and more generally, we have \eqref{EQ:polar} on all
homogeneous groups with $ \lambda (r,\omega)= {r}^{Q-1}$, where $Q$ is the homogeneous dimension of the group. We can also refer to Folland and Stein \cite{FS-Hardy} and to \cite{FR} for details of such groups.
\item[(II)] Hyperbolic spaces $\mathbb H^n$ with $\lambda(r,\omega)=(\sinh {r})^{n-1}$, or more general symmetric spaces of noncompact type.
\item[(III)] Cartan-Hadamard manifolds, that is, complete, simply connected Riemannian manifolds with non-positive sectional curvature. In this case $\lambda(\rho,\omega)$ depends on both variables $\rho$ and $\omega$.
We refer to Section \ref{SEC:CH} for this example, and to \cite{RV-18} for more details on  $\lambda(\rho,\omega)$ in this case.
 \item[(IV)] Arbitrary complete Riemannian manifolds $M$: let $C(p)$ denote the cut locus of a point $p\in M$, which we may fix.
 Let us denote by $M_{p}$ the tangent space to $M$ at $p$, and by $|\cdot|$ the Riemannian length.
 We also denote $D_{p}:=M\backslash C(p)$ and $S(p;r):=\{x\in M_{p}:|x|=r\}$. Then for any integrable function $f$ on $M$ we have the polar decomposition
        \begin{equation}\label{pol_decom_manif}
        \int_{M}f dV=\int_{0}^{+\infty}dr\int_{r^{-1}S(p;r)\cap D_{p}}f(\exp r\xi)\sqrt{g}(r;\xi)d\mu_{p}(\xi)
        \end{equation}
        for some function $\sqrt{g}$ on $D_{p}$, where $r^{-1}S(p,r)\cap D_{p}$ is the subset of $S_{p}$ obtained by dividing each of the elements of $S(p,r)\cap D_{p}$ by $r$, and $S_{p}:=S(p;1)$. The measure $d\mu_{p}(\xi)$ is the Riemannian measure on $S_{p}$ induced by the Euclidean Lebesgue measure on $M_{p}$. We refer to \cite[Formula III.3.5, P.123]{Cha06}, \cite[Chapter 4]{Li12} and \cite[Chapter 1, Paragraph 12]{CLN06} for more information on this decomposition.
   \end{itemize}

In this paper, as usual, we will write $ A\approx B$ to indicate  that the expressions $A$ and $B$ are equivalent.

The authors would like to thank Aidyn Kassymov and Bolys Sabitbek for checking some calculations in this paper.

\section{Main results}

Let $d$ is the metric on $\mathbb X$.
We denote by $B(a,r)$ the corresponding balls with respect to $d$, centred at $a\in\mathbb X$ and having radius  $r$, namely,
$$B(a,r):= \{x\in\mathbb X : d(x,a)<r\}.$$
To simplify the notation, for all arguments, we fix some point $a\in {\mathbb X}$, and then we will  denote  $${\vert x \vert}_a := d(a,x).$$

The main result of this paper is to characterise the weights $u$ and $v$ for which the corresponding Hardy inequality holds on $\mathbb X$. For $\mathbb X$=$\mathbb {R}$, such result has been proved by Sinnamon and Stepanov \cite{SS-96}.
For an alternative approach to these estimates in the case $1<q<p<\infty$ we refer to \cite[Theorem 1.13]{RY18b}.
Now, we formulate one of our main results:
\begin{thm}\label{Theorem1}
Suppose $0<q<p$, $1<p<\infty$ and $1/r=1/q-1/p$. Let $\mathbb X $ be a metric measure space with a polar decomposition at $a$. Let $u,v> 0$ be measurable and positive a.e in $\mathbb X$ such that $u\in L^1(\mathbb X\backslash \{a\})$ and $v^{1-p'}\in L^1_{loc}(\mathbb X)$.

Then the inequality
\begin{equation}\label{EQ:Hardy1}
\bigg(\int_\mathbb X\bigg(\int_{B(a,\vert x \vert_a)}\vert f(y) \vert dy\bigg)^q u(x)dx\bigg)^\frac{1}{q}\le C\bigg\{\int_{\mathbb X} {\vert f(x) \vert}^pv(x)dx\bigg\}^{\frac1p}
\end{equation}

holds for all measurable functions $f:\mathbb X\to{\mathbb C}$ if and only if
$$
\mathcal A_2:=  \bigg(\int_\mathbb X \bigg(\int_{\mathbb X\backslash{B(a,|x|_a )}} u(y) dy\bigg)^{r/p}  \bigg( \int_{B(a,|x|_a  )}v^{1-p'}(y)dy\bigg)^{r/p'}u(x)dx\bigg)^{1/r}<\infty.
$$
Moreover, the smallest constant $C$ for which \eqref{EQ:Hardy1} holds satisfies
$$
(p')^{1/p'} q^{1/p} (1-q/p) \mathcal A_2 \le C \le (r/q)^{1/r} p^{1/p} p'^{1/p'} \mathcal A_2.
$$
\end{thm}

Before proving the above theorem, we will need to prove several auxiliary facts.
Throughout this paper, we will use the following notations:
\begin{align}\label{EQ:U}
U(x)= { \int_{\mathbb X\backslash{B(a,|x|_a )}} u(y) dy},
\end{align}
\begin{align} \label{EQ:V}
V(x)= \int_{B(a,|x|_a  )}v^{1-p'}(y)dy,
\end{align}
 \begin{align}\label{EQ:U_tilde}
  \tilde{U}(t)= \int_t^\infty\int_{\Sigma_\rho} \lambda(\rho,\sigma)u(\rho,\sigma) d\sigma_{\rho} d\rho,
 \end{align}
 \begin{align}\label{EQ:V_tilde}
   \tilde{V}(t)= \int_0^t \int_{\Sigma_\rho} \lambda(\rho,\sigma)v^{1-p'}(\rho,\sigma) d\sigma_{\rho} d\rho,
 \end{align}
 \begin{align}\label{EQ:U_1}
 U_1(\rho)=\int_{\Sigma_\rho} \lambda(\rho,\sigma)u(\rho,\sigma) d\sigma_{\rho},
 \end{align}
 \begin{align}\label{EQ:V_1}
  V_1(\rho)=\int_{\Sigma_\rho} \lambda(\rho,\sigma)v^{1-p'}(\rho,\sigma) d\sigma_\rho.
 \end{align}
 \begin{lemma}
 \label{lemma1}
 Let us denote
$$
\mathcal A_{1} := \bigg\{\int_{\mathbb X} U^{r/q}(x) V^{r/q'}(x)v^{1-p'}(x)dx \bigg\}^{1/r}.
$$
Then
 \begin{align}\label{EQ:Equi1}
 \mathcal A_2^r =(q/p')  \mathcal A_1^r.
  \end{align}
  \end{lemma}
  \begin{proof} Using integration by parts, we have
   \begin{align}
&  \mathcal A_2^r =\int_{\mathbb X} U^{r/p}(x)V^{r/p'}(x)u(x)dx \nonumber
\\
&= \int_0^\infty \tilde{U}^{r/p}(t) \tilde{V}^{r/p'}(t) U_1(t)dt\nonumber
\\
&=\int_0^\infty \bigg(\int_t^\infty U_1(\rho)d\rho \bigg)^{r/p} \tilde{V}^{r/p'}(t) U_1(t)dt\nonumber
\\
&=-(q/r)U^{r/q}(\infty)V^{r/p'}(\infty)+(q/r)U^{r/q}(0)V^{r/p'}(0)+({q/r})(r/p')\nonumber
\\
&\quad\times \int_{\mathbb X} U^{r/q}(x)V^{r/q'}(x)v^{1-p'}(x)dx\nonumber
\\
& = ({q/p'}) \int_{\mathbb X} U^{r/q}(x)V^{r/q'}(x)v^{1-p'}(x)dx \nonumber
\\
&=({q/p'}) \mathcal A_1 ^r, \nonumber
  \end{align}
  completing the proof.
\end{proof}

\begin{lemma}
\label{lemma2}
Suppose that $\alpha, \beta$ and $\gamma$ are non-negative functions and $\gamma $ is a radial non-decreasing function of $\vert \cdot \vert_a$.
 If $\int_{\mathbb X \backslash{B(a,|x|_a )}} \alpha(y)dy\le \int_{\mathbb X \backslash{B(a,|x|_a )}} \beta(y)dy $ for all $x$, then $\int_ \mathbb X\gamma \alpha \le \int_\mathbb X\gamma \beta$.
 \end{lemma}
 \begin{proof}
  Let us denote
 $$\alpha_1(r)=\int_{\Sigma_r} \lambda(r,\sigma)\alpha(r,\sigma) d\sigma_r,$$
$$\beta_1(r)=\int_{\Sigma_r} \lambda(r,\sigma) \beta(r,\sigma) d\sigma_r,$$
$$\tilde{\gamma}(r)=\gamma(x),$$ \rm for $ \vert x \vert_a=r.$
 Given that,
$ \int_{\mathbb X \backslash{B(a,|x|_a )}} \alpha(y)dy\le \int_{\mathbb X \backslash{B(a,|x|_a )}} \beta(y)dy$,
changing to polar coordinates, we get
$$ \int_{\vert x \vert_a}^\infty {\alpha_1}(r)dr \le  \int_{\vert x \vert_a}^\infty {\beta_1}(r)dr.$$
\\
Using \cite[Lemma 2.1]{SS-96} which says if $\alpha, \beta$, $\gamma $ are non-negative functions and $\gamma $ is non-decreasing, and if $\int_x^\infty \alpha(y)dy\le \int_x^\infty \beta(y)dy $ for all $x$, then $\int_0^\infty\gamma\alpha\le\int_0^\infty\gamma\beta$.
Therefore,
\begin{align}
&&
\int_\mathbb X \gamma(x) \alpha(x)dx= \int_0^\infty \tilde{\gamma}(r) {\alpha_1}(r)dr\nonumber
& \le  \int_0^\infty \tilde{\gamma}(r) {\beta_1}(r)dr\nonumber=\int_\mathbb X \gamma(x) \beta(x)dx,
\end{align}
completing the proof.
\end{proof}

\begin{prop}
\label{proposition1}
 Suppose that $u,b$ and $F$ are non-negative functions with $F$ non-decreasing such that $\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y)dy<\infty$ for all $x\neq a$ and $\int_{\mathbb X} b(x)dx=\infty$.
If $0<q<p<\infty$, $F$ is radial in $\vert x \vert_a$, and $ 1/r=1/q-1/p$, then
\begin{align}
&\bigg(\int_{\mathbb X}  F^q(x) u(x)dx \bigg)^{1/q}\nonumber
\\
&  \le(r/p)^{1/r} \bigg(\int_{\mathbb X} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} u(y)dy \bigg)^{r/q} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y) dy\bigg)^{-r/q} b(x) dx\bigg)^{1/r}\nonumber
\\
& \quad\times \bigg(\int_{\mathbb X} F^p(x) b(x)dx \bigg)^{1/p}.\nonumber
\end{align}
\end{prop}
\begin{proof}

Let us denote
\\
\begin{align*}
B(x) &= \int_{\mathbb X \backslash{B(a,|x|_a )}} b(y)dy,\\

  \tilde{B}(t) &=\int_{t}^ {\infty}\int_{\Sigma_\rho} \lambda(\rho,\omega) b(\rho,\omega)d\omega_{\rho} d\rho,\\
  B_1(\rho) &=\int_{\Sigma_\rho} \lambda(\rho,\omega) b(\rho,\omega)d\omega_{\rho},\\
   \tilde{F}(t) &=F(x) , \ \textrm {for} \ \vert x \vert_a=t.
   \end{align*}
Applying H\"older's inequality with indices $q/r$ and $q/p$, we get

\begin {align}
&
\bigg(\int_{\mathbb X}  F^q(x) u(x)dx \bigg)^{1/q}\nonumber
\\
&=\bigg(\int_{\mathbb X}  \bigg(\int_{B(a,|x|_a )} U^{r/p}(y) B^{-r/q}(y) b(y)dy \bigg)^{q/r} F^q(x)\nonumber
\\
& \quad\times \bigg(\int_{B(a,|x|_a )} U^{r/p}(y) B^{-r/q}(y) b(y)dy \bigg)^{-q/r} u(x)dx \bigg)^{1/q} \nonumber
\\
& =\bigg(\int_0^\infty  \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{q/r} \tilde{F}^q(t)  \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-q/r} U_1(t) dt\bigg)^{1/q}\nonumber
\\
& =\bigg(\int_0^\infty  \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{q/r} \tilde{F}^q(t)  \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-q/r}\nonumber
\\
 &\quad\times U_1^{q/r+q/p}(t) dt\bigg)^{1/q}\nonumber
\\
&\le  \bigg(\int_0^\infty \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg) U_1(t)dt \bigg)^{1/r} \bigg(\int_0^\infty \tilde {F}^p(t) \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-p/r}\nonumber
\\
&\quad\times U_1(t)dt \bigg)^{1/p} .\nonumber
\end{align}

On interchanging the order of integration and using $r/p+1=r/q$, the first factor becomes
\begin{align}
 & \bigg(\int_0^\infty  \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho) \bigg(\int_\rho^{\infty} U_1(t)dt\bigg) d\rho \bigg)^{1/r}\nonumber
\\
&= \bigg(\int_0^\infty  \tilde {U}^{r/q}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho) d\rho \bigg)^{1/r}\nonumber
\\
 &=\bigg(\int_{\mathbb X} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} u(y)dy \bigg)^{r/q} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y) dy\bigg)^{-r/q} b(x) dx\bigg)^{1/r}.\nonumber
 \end{align}
 \\
  To complete the proof we apply Lemma \ref{lemma2} to the second factor. We take $\alpha(x)=\bigg(\int_{B(a,|x|_a )} U^{r/p}(y) B^{-r/q}(y)b(y)dy \bigg)^{-p/r} u(x)$, $\beta(x)=(r/p)^{p/r}b(x)$, and $\gamma(x)=F^p(x)$ in Lemma  \ref{lemma2}. As $\gamma$ is  non-decreasing by assumption, it remains to check that
  $$\int_{\mathbb X \backslash{B(a,|x|_a )}} \alpha(y)dy \le \int_{\mathbb X \backslash{B(a,|x|_a )}} \beta(y)dy,$$ for all $x$. Since $\bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-p/r}$ and $\tilde{U}$ are non-increasing,
\begin{align}
\int_{\mathbb X \backslash{B(a,|x|_a )}} \alpha(y)dy\nonumber
&= \int_{\vert x \vert_a}^\infty \bigg(\int_0^t \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q} (\rho)B_1(\rho)d\rho \bigg)^{-p/r} U_1(t)dt\nonumber
\\
& \le \bigg(\int_0^{ \vert x \vert_a}  \tilde {U}^{r/p}(\rho) \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-p/r} \int_{\vert x \vert_a}^\infty U_1(t)dt\nonumber
\\
& \le \bigg(\int_0^{ \vert x \vert_a}   \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-p/r} \tilde U^{-1}(\vert x \vert_a) \int_{\vert x \vert_a}^\infty U_1(t)dt\nonumber
\\
& =\bigg(\int_0^{ \vert x \vert_a}  \tilde {B}^{-r/q}(\rho) B_1(\rho)d\rho \bigg)^{-p/r}\nonumber
\\
&=\bigg(\int_0^{ \vert x \vert_a}  \tilde {B}^{-r/q}(\rho) d(-\tilde{B}(\rho)) \bigg)^{-p/r}\nonumber
\\
&=\bigg((p/r)  B^{-r/p}(x)\bigg)^{-p/r} =\int_{\mathbb X \backslash{B(a,|x|_a )}} \beta(y)dy. \nonumber
\end{align}
Finally, by using Lemma \ref{lemma2} we get
\begin {align}
&
\bigg(\int_{\mathbb X}  F^q(x) u(x)dx \bigg)^{1/q}\nonumber
\\
&\le  \bigg(\int_{\mathbb X} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} u(y)dy \bigg)^{r/q} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y) dy\bigg)^{-r/q} b(x) dx\bigg)^{1/r}  \bigg(\int_{\mathbb X} (r/p)^{p/r} F^p(x) b(x)dx \bigg)^{1/p}\nonumber
\\
&= (r/p)^{1/r} \bigg(\int_{\mathbb X} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} u(y)dy \bigg)^{r/q} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y) dy\bigg)^{-r/q} b(x) dx\bigg)^{1/r}  \bigg(\int_{\mathbb X} F^p(x) b(x)dx \bigg)^{1/p}\nonumber,
\end{align}
which completes the proof.
\end{proof}

\begin{prop}
\label{proposition2}
Suppose $1<p<\infty$ and $w$ is a non-negative function satisfying
\\
  \begin{align}\label{EQ:Condn}
   0<\displaystyle \int_{B(a,|x|_a )} w(y)dy <\infty , \forall x\neq a, \displaystyle \int_{\mathbb X} w(x)dx =\infty.
   \end{align}
\\
Then
\\
\begin{align}\label{EQ:Prepo}
\bigg(\displaystyle \int_{\mathbb X}\bigg( \int_{B(a,|x|_a )} f(y)dy \bigg)^p \bigg( \int_{B(a,|x|_a )} w(y)dy \bigg)^{-p} w(x)dx\bigg)^{1/p}\nonumber
\\
\le p' \bigg(\int_{\mathbb X} f^p(x) w^{1-p}(x)dx\bigg)^{1/p},
\end{align}
for all measurable functions $f\ge0.$
\end{prop}
\begin{proof} Let us denote:
${f_1}(\rho)=\int_{\Sigma_\rho}  \lambda(\rho,\sigma) f(\rho,\sigma) d\sigma_\rho $,
${w_1}(\rho)=\int_{\Sigma_\rho}  \lambda(\rho,\sigma) w(\rho,\sigma) d\sigma_\rho .$

Consider the left hand side of \eqref{EQ:Prepo} and change it into polar coordinates, to get

\begin{align}
& \bigg(\displaystyle \int_{\mathbb X}\bigg( \int_{B(a,|x|_a )} f(y)dy \bigg)^p \bigg( \int_{B(a,|x|_a )} w(y)dy \bigg)^{-p} w(x)dx\bigg)^{1/p}\nonumber
\\
&=\bigg(\displaystyle \int_0^\infty \bigg( \int_0^t {f_1}(\rho)d\rho \bigg)^p \bigg( \int_0^t {w_1}(\rho)d\rho \bigg)^{-p} {w_1}(t)dt\bigg)^{1/p}\nonumber.
\end{align}
Now, let us use \cite[Proposition 2.3]{SS-96} which says that if $1<p<\infty$ and $w$ is a non-negative function satisfying
\\
  \begin{align}
   & 0<\displaystyle \int_0^x w(y)dy <\infty , \forall x> 0, \displaystyle \int_0^\infty w(x)dx =\infty,\nonumber
   \end{align}
   then
   \begin{align}
\bigg(\displaystyle \int_0^\infty \bigg( \int_0^t f(\rho)d\rho \bigg)^p \bigg( \int_0^t w(\rho)d\rho \bigg)^{-p} w(t)dt\bigg)^{1/p}\nonumber
\\
\le p' \bigg(\int_0^\infty f^p(t) w^{1-p}(t)dt\bigg)^{1/p}.\nonumber
\end{align}

 By using  H\"older's inequality to the indices $1/p$ and $1/p'$, the LHS of \eqref{EQ:Prepo} can be estimated by
\begin{align}
&
 \le  p' \bigg(\int_0^\infty {f_1}^p(t) {w_1}^{1-p}(t)dt\bigg)^{1/p}\nonumber
 \\
 &= p' \bigg(\int_0^\infty \bigg(\int_{\Sigma_t}  \lambda(t,\sigma) f(t,\sigma) d\sigma_t \bigg)^p \bigg(\int_{\Sigma_t}  \lambda(t,\sigma) w(t,\sigma) d\sigma_t \bigg)^{1-p} dt\bigg)^{1/p}\nonumber
 \\
 &= p' \bigg(\int_0^\infty \bigg(\int_{\Sigma_t}  \lambda(t,\sigma) f(t,\sigma) w^{(1-p)/p+(p-1)/p}d\sigma_t \bigg)^p \bigg(\int_{\Sigma_t}  \lambda(t,\sigma) w(t,\sigma) d\sigma_t \bigg)^{1-p} dt \bigg)^{1/p}\nonumber
 \\
&\le p' \bigg(\int_0^\infty \bigg(\int_{\Sigma_t}  \lambda(t,\sigma) f^p(t,\sigma) w^{1-p}d\sigma_t \bigg)\bigg(\int_{\Sigma_t}  \lambda(t,\sigma) w^{p'(p-1)/p}(t,\sigma) d\sigma_t \bigg)^{p-1}\nonumber
\\
& \quad\times \bigg(\int_{\Sigma_t}  \lambda(t,\sigma) w(t,\sigma) d\sigma_t \bigg)^{1-p} dt \bigg)^{1/p}\nonumber
\\
&
= p' \bigg(\int_0^\infty \int_{\Sigma_t}  \lambda(t,\sigma) f^p(t,\sigma) w^{1-p}(t,\sigma) d\sigma_t dt \bigg)^{1/p}\nonumber
\\
&= p' \bigg(\displaystyle\int_{\mathbb X} f^p(x) w^{1-p}(x) dx \bigg)^{1/p},\nonumber
\end{align}
completing the proof.
\end{proof}

Now, we prove our Theorem \ref{Theorem1} :
\begin{proof}
Set $w=v^{1-p'}$. Suppose that inequality \eqref{EQ:Hardy1}  holds for all $f\ge0$ and let $u_0$ and $w_0$ be $L^1$ functions  such that $0 < u_0 \le u$ and $0 < w_0 \le w$.
We denote
 $$ \tilde{u_0}(\rho)=\int_{\Sigma_{\rho}} \lambda(\rho,\omega)u_0(\rho, \omega)d\omega_{\rho},$$
 $$ \tilde{w_0}(\rho)=\int_{\Sigma_{\rho}} \lambda(\rho,\omega)w_0(\rho, \omega)d\omega_{\rho}.$$

Let us apply inequality \eqref{EQ:Hardy1} to the function
$$f(x)=\displaystyle\bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/pq} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/pq'} w_0(x).$$

After changing to polar coordinates and using $$ \frac{r}{(pq')}+1=r(\frac{1}{pq'}+\frac{1}{r})\\
=r\bigg(\frac{1}{p}(1-\frac{1}{q})+\frac{1}{q}-\frac{1}{p}\bigg)=\frac{r}{p'q},$$
we have
\begin{align*}
&
\int_{B(a,|x|_a )} f(y)dy = \int_0^{\vert x \vert_a} \int_{\Sigma_{t}} \lambda(t,\sigma) \bigg(\int_{t}^\infty \int_{\Sigma_{\rho}} \lambda(\rho,\omega)u_0(\rho,\omega)d\rho d\omega_{\rho} \bigg)^{r/pq}\nonumber
\\
& \quad\times \bigg(\int_0^t \int_{\Sigma_{\rho}} \lambda(\rho,\omega) w_0 (\rho, \omega)d\rho d\omega_{\rho} \bigg)^{r/pq'}w_0(t,\sigma)dt d\sigma_t\nonumber
\\
&=\int_0^{\vert x \vert_a} \bigg(\int_{t}^\infty  \tilde u_0(\rho)d\rho \bigg)^{r/pq} \bigg(\int_0^t \tilde w_0 (\rho)d\rho \bigg)^{r/pq'} \tilde w_0(t)dt\nonumber
\\
& \ge  \bigg(\int_{{\vert x \vert_a}}^\infty  \tilde u_0(\rho)d\rho \bigg)^{r/pq} \int_0^{\vert x \vert_a} \bigg(\int_0^t \tilde w_0 (\rho)d\rho \bigg)^{r/pq'} \tilde w_0(t)dt\nonumber
\\
& =(p'q/r) \bigg(\int_{\vert x \vert_a}^\infty  \tilde u_0(\rho)d\rho \bigg)^{r/pq} \bigg(\int_0^{\vert x \vert_a} \tilde w_0 (\rho)d\rho \bigg)^{r/p'q}\nonumber
\\
&=(p'q/r) \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/pq} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/p'q}. \nonumber
\end{align*}

Observe that $w=v^{1-p'}$ implies that $v=w^{1/(1-p')}=w^{1-p}$ since $$1-1/(1-p')=-p'/(1-p')=-1/(1/p'-1)=p.$$
We then have
\begin{align}
 &\bigg(\int_{\mathbb X} (p'q/r)^q  \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/p} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/p'} u_0(x)dx \bigg)^{1/q}\nonumber
 \\
 & \le \bigg(\int_\mathbb X\bigg(\int_{B(a,\vert x \vert_a)} f(y)  dy\bigg)^q u(x)dx\bigg)^{1/q}\nonumber
 \\
 & \le C \bigg(\int_{\mathbb X} f^p(x) w^{1-p}(x)dx \bigg)^{1/p}\nonumber
 \\
 &=C \bigg(\int_{\mathbb X}  \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/q} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/q'} w_0^p(x) w^{1-p}(x)dx \bigg)^{1/p}\nonumber
 \\
 & \le C \bigg(\int_{\mathbb X}  \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/q} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/q'} w_0^p(x) w_0^{1-p}(x)dx \bigg)^{1/p}\nonumber
 \\
 &= C  \bigg(\int_{\mathbb X}  \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/q} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/q'} w_0(x)dx \bigg)^{1/p}\nonumber
 \\
 &=C   \bigg(\int_{\mathbb X} (p'/r)(r/q) \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/p} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/p'} u_0(x)dx \bigg)^{1/p}\nonumber
 \\
 &= C (p'/q)^{1/p}  \bigg(\int_{\mathbb X}  \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u_0(y)dy \bigg)^{r/p} \bigg(\int_{B(a,|x|_a )} w_0(y)dy\bigg)^{r/p'} u_0(x)dx \bigg)^{1/p},\nonumber
 \end{align}
where the second last equality is integration by parts. Since   $u_0$ and $w_0$ are in $L^1$ and are positive, the integral on the right hand side is finite. Therefore, we have

$$
 (p'q/r)(q/p')^{1/p}  \bigg(\int_\mathbb X \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} u_0(y)dy \bigg)^{r/p}  \bigg( \int_{B(a,|x|_a )}w_0(y)dy \bigg)^{r/p'} u_0(x)dx\bigg)^{1/r} \le C.
$$
Approximating $u$ and $w$ by increasing sequence of $L^1$ functions, using $$ (p'q/r)(q/p')^{1/p} =(p' )(p')^{-(1/p)}q^{1/p}(q/r)=(p')^{1/p'}q^{1/p}(q(1/q-1/p))=(p')^{1/p'} q^{1/p} (1-q/p) $$  and applying the Monotone Convergence Theorem, we conclude that
$$
(p')^{1/p'} q^{1/p} (1-q/p) \mathcal A_2 \le C.
$$

Suppose now that $\mathcal A_2<\infty$ and, for the moment, that \eqref{EQ:Condn} holds for $w$.
Set $V(x)=\displaystyle \int_{B(a,|x|_a )} w(y)dy$ and apply Proposition \ref{proposition1} with $b=V^{-p}w$ and $F(x)=\int_{B(a,|x|_a )} f(y)dy $.
Let us denote $$ \tilde{V}(t)= \int_0^t \int_{\Sigma_\rho}\lambda(\rho,\sigma) w(\rho,\sigma) d\rho d\sigma_\rho ,$$ $$ V_1(\rho)=\int_{\Sigma_\rho} \lambda(\rho,\sigma)w(\rho,\sigma) d\sigma_\rho, $$ $$ U_1(\rho)= \int_{\Sigma_\rho} \lambda(\rho,\sigma)u(\rho,\sigma) d\sigma_\rho. $$

Also,
$$\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y)dy <\infty,$$
since $$\int_{\mathbb X \backslash{B(a,|x|_a )}} b(y)dy=\int_{\mathbb X \backslash{B(a,|x|_a )}} V^{-p}(y)w(y)dy =\int_{\vert x \vert_a}^\infty \tilde V^{-p}(\rho) V_1(\rho)d\rho=(p'/p) \tilde V^{1-p}(\vert x \vert_a) .$$

The conclusion of Proposition \ref{proposition1} becomes

 \begin{align}
 &
 \bigg(\int_\mathbb X \bigg(\int_{B(a,|x|_a )} f(y)dy \bigg)^q u(x)dx \bigg)^{1/q}\nonumber
 \\
 &\le (r/p)^{1/r}  \bigg(\int_\mathbb X  \bigg( \int_{\mathbb X \backslash{B(a,|x|_a )}}u(y)dy \bigg)^{r/q} \bigg(\int_{\mathbb X \backslash{B(a,|x|_a )}} V^{-p}(y)  w(y)dy\bigg)^{-r/q} V^{-p}(x)  w(x)dx \bigg)^{1/r}\nonumber
 \\
  & \quad\times  \bigg(\int_\mathbb X  \bigg(\int_{B(a,|x|_a )} f(y)dy \bigg)^p V^{-p}(x) w(x)dx \bigg)^{1/p}\nonumber
 \\
 & = (r/p)^{1/r} \bigg(\int_0^\infty \bigg( \int_\rho^\infty U_1(t)dt \bigg)^{r/q} \bigg( \int_\rho^\infty \tilde V^{-p}(t) V_1(t)dt \bigg)^{-r/q} \tilde V^{-p}(\rho) V_1(\rho)d\rho\bigg)^{1/r}\nonumber
 \\
 &\quad\times  \bigg(\int_\mathbb X  \bigg(\int_{B(a,|x|_a )} f(y)dy \bigg)^p V^{-p}(x) w(x)dx \bigg)^{1/p}.\nonumber
  \end{align}
Using $ \int_s^\infty \tilde V^{-p}(t) V_1(t)dt = (p'/p) \tilde V^{1-p}(s) $  in the first factor and applying Proposition \ref{proposition2} to the second factor, we reach the inequality
{\small \begin{align}
 &
  \bigg(\int_\mathbb X \bigg(\int_{B(a,|x|_a )} f(y)dy \bigg)^q u(x)dx \bigg)^{1/q}\nonumber
  \\
  & \le (r/p)^{1/r} (p/p')^{1/q} p' \bigg(\int_0^\infty \bigg( \int_\rho^\infty U_1(t) dt\bigg)^{r/q}  \tilde{V}(\rho)^{(p-1)r/q} \tilde V^{-p}(\rho)V_1(\rho)d\rho \bigg)^{1/r} \bigg(\int_\mathbb X f^p(x) v(x)dx \bigg)^{1/p}\nonumber
  \\
  & = (r/p)^{1/r} (p/p')^{1/q} p' \bigg(\int_0^\infty \bigg( \int_\rho^\infty U_1(t) dt \bigg)^{r/q} \bigg(\int_0^\rho V_1(t) dt \bigg)^{r/q'} V_1(\rho)d\rho \bigg)^{1/r} \bigg(\int_\mathbb X f^p(x) v(x)dx \bigg)^{1/p}\nonumber
    \\
    & = (r/p)^{1/r} (p/p')^{1/q} p'
    \mathcal A_1  \bigg(\int_\mathbb X f^p(x) v(x)dx \bigg)^{1/p}\nonumber
    \\
  & = (r/p)^{1/r} (p/p')^{1/q} p' (p'/q)^{1/r} \mathcal A_2  \bigg(\int_\mathbb X f^p(x) v(x)dx \bigg)^{1/p}\nonumber
  \\
  & =(r/q)^{1/r} (p')^{1/p'} p^{1/p}  \mathcal A_2  \bigg(\int_\mathbb X f^p(x) v(x)dx \bigg)^{1/p}.\nonumber
     \end{align}
    }
     In the fourth last equality, we used $\frac{r(p-1)}{q}-p=r\bigg(\frac{p-1}{q}-\frac{p}{r}\bigg)=r\bigg(\frac{p-1}{q}-\frac{p(p-q)}{pq}\bigg)=\frac{r}{q'}$ and in the second last equality, we used Lemma \ref{lemma1}.\\

  To establish sufficiency for general $w,$ we fix positive functions $u$ and $w$. If $w=0$ almost everywhere on some ball $B(a,\vert x \vert_a)$ then translating $u$, $w$ on the left will reduce the problem to the one in which this does not occur. (If $w=0$  almost everywhere on $ \mathbb X$, sufficiency holds trivially). We therefore assume that $0<\displaystyle \int_{B(a,|x|_a )} w(y)dy,$ for all $x \neq a $. For each $n>0,$ set $u_n=u \chi_{(B(a,n))}$ and $w_n=\min(w,n)+\chi_{( \mathbb X \backslash{B(a,n )})}$. Then $w_n$ clearly satisfies \eqref{EQ:Condn}, so from previous arguments we have
  \begin{align}
  & \bigg(\int_\mathbb X \bigg(\int_{B(a,|x|_a )} f(y)dy \bigg)^q u_n(x)dx \bigg)^{1/q}\nonumber
  \\
  & \le c \bigg(\int_0^\infty \bigg( \int_\rho^\infty \tilde u_n(t)dt \bigg)^{r/p}  \bigg(\int_0^\rho \tilde w_n(t)dt \bigg)^{r/p'} \tilde u_n(\rho)d\rho \bigg)^{1/r} \bigg( \int_{\mathbb X}  f^p(y)  w_n^{1-p}(y)dy \bigg)^{1/p},\nonumber
    \end{align}
    for all $f\ge0$. Here $c=(r/q)^{1/r} (p')^{1/p'} p^{1/p}$. If we take $f = g \min(w,n)^{1/p'} \chi_{(B(a,n) )}$ and use the definitions of $u_n$ and $w_n,$ the inequality becomes
\begin{align}
&\bigg(\int_{B(a,n)}   \bigg(\int_{B(a,\vert x \vert_a)}  {g}(y) \min( w,n)^{1/p'}dy \bigg)^q  u (x)dx \bigg)^{1/q}\le c \bigg(\int_0^n \bigg(\int_\rho^n \tilde u(t) dt\bigg)^{r/p}\nonumber
\\
& \quad\times  \bigg(\int_0^\rho \min(\tilde w,n)\bigg)^{r/p'} \tilde u(\rho)d\rho \bigg)^{1/r} \bigg(\int_ {\mathbb X}  g^p(y)dy \bigg)^{1/p},\nonumber
\end{align}
for all non-negative $g$. We let $n \rightarrow \infty$, apply the Monotone Convergence Theorem and substitute $fw^{-1/p'}$ for $g$ to get the desired inequality and complete the proof.
\end{proof}

  \section{Applications and examples}

 In this section we present several examples of applications of our results to characterise the weights $u$ and $v$ in several settings: homogeneous groups, hyperbolic spaces, and more general Cartan-Hadamard manifolds.

 \subsection{Homogeneous groups}  Let $\mathbb X=\mathbb G$ be a homogeneous group in the sense of Folland and Stein \cite{FS-Hardy}, see also an up-to-date exposition in the open access books \cite{FR} and \cite{RS-book}.
Here condition \eqref{EQ:polar} is always satisfied with function $ \lambda (r,\omega)= {r}^{Q-1}$, with $Q$ being the homogeneous dimension of the group.

Without loss of generality, let us fix $a=0$ to be the identity element of the group $\mathbb G$. To simplify the notation further, we denote $\vert x \vert_a$ by $\vert x \vert$. We note that this is consistent with the notation for the quasi-norm $|\cdot|$ on a homogeneous group $\mathbb G$.

Let us consider an example  of the power weights
$$u(x) =
\left\{
	\begin{array}{ll}
		{\vert x \vert}^{\alpha_1}  & \mbox{if }  \vert x \vert < 1, \\
		{\vert x \vert}^{\alpha_2} & \mbox{if }  \vert x \vert \geq 1,
	\end{array}
\right.
\qquad v(x)= {\vert x \vert }^\beta.$$
Then by Theorem \ref{Theorem1} the inequality
$$
\bigg(\int_\mathbb G \bigg( \int_{B(a,{\vert x \vert}_a)} \vert f(y) \vert dy \bigg)^q u(x) dx\bigg)^\frac{1}{q} \le C\bigg( \int_\mathbb G {\vert f(y) \vert}^p v(x) dx \bigg)^{\frac{1}{p}}
$$
 holds for $0<q<p,$ $1<p <\infty$, if and only if
 \begin{align}
 &
 \mathcal A_2^r =\bigg(\int_\mathbb G \bigg(\int_{\mathbb G\backslash{B(a,|x|_a )}} u(y) dy\bigg)^{r/p}  \bigg( \int_{B(a,|x|_a  )}v^{1-p'}(y)dy\bigg)^{r/p'}u(x)dx\bigg)\nonumber
 \\
 & \approx \int_0^1 \bigg(\int_t^1 \rho^{\alpha_1+Q-1} d\rho+ \int_1^\infty \rho^{\alpha_2+Q-1} d\rho \bigg)^{r/p} \bigg(\int_0^t \rho^{\beta(1-p')+Q-1} d\rho \bigg)^{r/p'} t^{\alpha_1+Q-1}dt\nonumber
 \\
& + \int_1^\infty \bigg(\int_t^\infty \rho^{\alpha_2+Q-1} d\rho \bigg)^{r/p} \bigg(\int_0^t \rho^{\beta(1-p')+Q-1} d\rho \bigg)^{r/p'} t^{\alpha_2+Q-1}dt <\infty. \nonumber
 \end{align}

Let us consider
\begin{align}
&
\int_0^1 \bigg(\int_t^1 \rho^{\alpha_1+Q-1} d\rho+ \int_1^\infty \rho^{\alpha_2+Q-1} d\rho \bigg)^{r/p} \bigg(\int_0^t \rho^{\beta(1-p')+Q-1} d\rho \bigg)^{r/p'} t^{\alpha_1+Q-1}dt\nonumber
\\
&=\int_0^1 \bigg(1/(\alpha_1+Q)- {t}^{\alpha_1+Q}/(\alpha_1+Q)-1/(\alpha_2+Q) \bigg)^{r/p} \bigg(t^{(\beta(1-p')+Q)}/(\beta(1-p')+Q) \bigg)^{r/p'}\nonumber
\\
& \quad\times t^{\alpha_1+Q-1}dt,\nonumber
\end{align}
which is finite for
$$\alpha_2+Q<0, \beta(1-p')+Q>0, (\alpha_1+Q){r/p}+(\beta(1-p')+Q){r/p'}+\alpha_1+Q>0,$$
which means
$$\alpha_2+Q<0, \beta(1-p')+Q>0, (\alpha_1+Q){r/q}+(\beta(1-p')+Q){r/p'}>0,$$
since we have  $r/p+1=r(1/p+1/r)=r(1/p+1/q-1/p)=r/q$.
\\

Now, consider the other part
\begin{align}
&
\int_1^\infty \bigg(\int_t^\infty \rho^{\alpha_2+Q-1} d\rho \bigg)^{r/p} \bigg(\int_0^t \rho^{\beta(1-p')+Q-1} d\rho \bigg)^{r/p'} t^{\alpha_2+Q-1}dt\nonumber
\\
&= \int_1^\infty  \{(-t)^{(\alpha_2+Q)}/(\alpha_2+Q)\}^{r/p} \{ t^{(\beta(1-p')+Q)}/(\beta(1-p')+Q) \}^{r/p'} t^{\alpha_2+Q-1} dt, \nonumber
\end{align}
which is finite for
$$\alpha_2+Q<0, \beta(1-p')+Q>0, (\alpha_2+Q){r/p}+(\beta(1-p')+Q){r/p'}+\alpha_2+Q<0,$$
or for
$$\alpha_2+Q<0,\beta(1-p')+Q>0, (\alpha_2+Q){r/q}+(\beta(1-p')+Q){r/p'}<0.$$

Summarising that we get

  \begin{cor}\label{COR:hom}
Let $\mathbb G$ be a homogeneous group of homogeneous dimension $Q$, an we equip it with a quasi-norm $|\cdot|$. Let  $0<q<p,$ $1<p <\infty,$ $1/r=1/q-1/p,$ and let $\alpha_1, \alpha_2, \beta \in \mathbb R$. Assume that $\alpha_1+Q\not=0$.
Let
\begin{equation}\label{EQ:uv}
u(x) =
\left\{
	\begin{array}{ll}
		{\vert x \vert}^{\alpha_1}  & \mbox{if } \vert x \vert < 1, \\
		{\vert x \vert}^{\alpha_2} & \mbox{if }  \vert x \vert \geq 1,
	\end{array}
\right.
\qquad
 v(x)= {\vert x \vert }^\beta.
 \end{equation}
 Then the inequality
\begin{equation}\label{EQ:Hardy1hg}
\bigg(\int_\mathbb G\bigg(\int_{B(a,\vert x \vert_a)}\vert f(y) \vert dy\bigg)^q u(x) dx\bigg)^\frac{1}{q}\le C\bigg\{\int_{\mathbb G} {\vert f(x) \vert}^p v(x) dx\bigg\}^{\frac1p}
\end{equation}
holds for all measurable functions $f:\mathbb G\to{\mathbb C}$  if and only if
the parameters satisfy the following conditions:
 $\alpha_2+Q<0$, $\beta(1-p')+Q>0$, $(\alpha_1+Q){r/q}+(\beta(1-p')+Q){r/p'}>0,$ $(\alpha_2+Q){r/q}+(\beta(1-p')+Q){r/p'}<0.$
 \end{cor}

It is interesting to note that in view of the last two conditions, it is not possible to have Hardy inequality \eqref{EQ:Hardy1hg} with weights $u$ and $v$ in \eqref{EQ:uv} with $\alpha_1=\alpha_2$. This is why we consider different powers $\alpha_1, \alpha_2$ in this example. This is different from the case $p\leq q$ which was considered as an application in \cite{RV-18}.

The case $\alpha_1+Q=0$ can be treated in a similar way.

\subsection{Hyperbolic spaces}

 Let $\mathbb H^n$ denote the hyperbolic space of dimension $n$. In this case  condition \eqref{EQ:polar} is always satisfied with $\lambda(r,\omega)=(\sinh {r})^{n-1}$. Let $a\in \mathbb H^n$, and
 let us fix the weights
 $$u(x) =
 \left\{
	\begin{array}{ll}
		{(\sinh {\vert x \vert_a})}^{\alpha_1}  & \mbox{if }  \vert x \vert < 1, \\
		{(\sinh {\vert x \vert_a})}^{\alpha_2} & \mbox{if }  \vert x \vert \geq 1,
	\end{array}
\right.,\quad v(x)= (\sinh {\vert x \vert_a})^\beta.$$
 We note that $\mathcal A_2$ is equivalent to
 \begin{align}
& \mathcal A_2^r \approx \int_0^1  \bigg(\int_{t}^1 (\sinh \rho)^{\alpha_1+n-1}d\rho + \int_{1}^\infty (\sinh \rho)^{\alpha_2+n-1}d\rho \bigg)^{r/p} \bigg( \int_0^{t} (\sinh \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}\nonumber
\\
& \quad\times( \sinh {t})^{\alpha_1+n-1} dt + \int_1^\infty\bigg( \int_{t}^\infty (\sinh \rho)^{\alpha_2+n-1}d\rho \bigg)^{r/p}  \bigg( \int_0^{t} (\sinh \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}  \nonumber
\\
 &\quad\times( \sinh {t})^{\alpha_2+n-1} dt. \nonumber
\end{align}

In the first part, for ${\alpha_2+n-1}<0$ and $ {\beta(1-p')+n}>0,$
 \begin{align}
&\int_0^1  \bigg(\int_{t}^1 (\sinh \rho)^{\alpha_1+n-1}d\rho + \int_{1}^\infty (\sinh \rho)^{\alpha_2+n-1}d\rho \bigg)^{r/p} \bigg( \int_0^{t} (\sinh \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}\nonumber
\\
& \quad\times( \sinh {t})^{\alpha_1+n-1} dt\nonumber
\\
&\approx \int_0^1  \bigg(\int_{t}^1 ( \rho)^{\alpha_1+n-1}d\rho + \int_{1}^\infty (\exp \rho)^{\alpha_2+n-1}d\rho \bigg)^{r/p}\bigg( \int_0^{t} ( \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}\nonumber
\\
& \quad\times( {t})^{\alpha_1+n-1} dt\nonumber
\\
&=\int_0^1 \bigg(1/(\alpha_1+n)- {t}^{\alpha_1+n}/(\alpha_1+n)-(\exp 1)^{\alpha_2+n-1}/({\alpha_2+n-1}) \bigg)^{r/p}\nonumber
\\
& \quad\times \bigg( t^{\beta(1-p')+n}/({\beta(1-p')+n}) \bigg)^{r/p'} t^{\alpha_1+n-1} dt,\nonumber
\end{align}
which is finite for,

(a) $\alpha_1+n \geq 0$, $({\beta(1-p')+n})r/p' +{\alpha_1+n} >0,  $
\\

(b) $\alpha_1+n < 0$, $(\alpha_1+n){r/p}+{(\beta(1-p')+n})r/p' +{\alpha_1+n} >0.  $
\\

However, we can note that in (a), if $\alpha_1+n \geq 0$, then the second condition is automatically satisfied under our assumption $ {\beta(1-p')+n}>0.$

In the second part, for  ${\alpha_2+n-1}<0,$
 \begin{align}
 &\int_1^\infty \bigg( \int_{t}^\infty (\sinh \rho)^{\alpha_2+n-1}d\rho \bigg)^{r/p}  \bigg( \int_0^{t} (\sinh \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}
( \sinh {t})^{\alpha_2+n-1} dt \nonumber
\\
& \approx  \int_1^\infty \bigg( \int_{t}^\infty (\exp \rho)^{\alpha_2+n-1}d\rho \bigg)^{r/p} \bigg( \int_0^{t} (\exp \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}  (\exp {t})^{\alpha_2+n-1} dt \nonumber
\\
&= \int_1^\infty  \bigg( -(\exp t)^{\alpha_2+n-1}/(\alpha_2+n-1) \bigg)^{r/p}\bigg( (\exp t)^{\beta(1-p')+n-1}/(\beta(1-p')+n-1) \bigg)^{r/p'}  \nonumber
\\
& \quad\times ( (\exp {t})^{\alpha_2+n-1} dt, \nonumber
 \end{align}
 which is finite for
 $$
 ({\alpha_2+n-1})r/p+ (\beta(1-p')+n-1)r/p' + \alpha_2+n-1<0,
 $$
which is the same as
$$
 ({\alpha_2+n-1})r/q+ (\beta(1-p')+n-1)r/p' <0.
$$
\begin{cor}\label{COR:hyp}
Let $\mathbb H^n$ be the hyperbolic space, $a\in \mathbb H^n$, and let $|x|_a$ denote the hyperbolic distance from $x$ to $a$. Let $0<q<p,$ $1<p <\infty,$ $1/r=1/q-1/p,$ and let $\alpha_1, \alpha_2, \beta\in\mathbb R$. Assume that $\alpha_1+n\not=0$.
Let
$$u(x) =
 \left\{
	\begin{array}{ll}
		{(\sinh {\vert x \vert_a})}^{\alpha_1}  & \mbox{if }  \vert x \vert < 1, \\
		{(\sinh {\vert x \vert_a})}^{\alpha_2} & \mbox{if }  \vert x \vert \geq 1,
	\end{array}
\right.
\qquad
 v(x)= (\sinh {\vert x \vert_a})^\beta.$$
 Then the inequality
\begin{equation*}\label{EQ:Hardy1hyp}
\bigg(\int_{\mathbb H^n}\bigg(\int_{B(a,\vert x \vert_a)}\vert f(y) \vert dy\bigg)^q u(x) dx\bigg)^\frac{1}{q}\le C\bigg\{\int_{\mathbb H^n} {\vert f(x) \vert}^p v(x) dx\bigg\}^{\frac1p}
\end{equation*}
holds for all measurable functions $f:\mathbb H^n\to{\mathbb C}$  if and only if
the parameters satisfy the following conditions:
${\alpha_2+n-1}<0,$ $ {\beta(1-p')+n}>0,$ $ (\alpha_1+n){r/q}+(\beta(1-p')+n){r/p'}>0,$ $({\alpha_2+n-1})r/q+ (\beta(1-p')+n-1)r/p' <0.$
\end{cor}

\subsection{Cartan-Hadamard manifolds} \label{SEC:CH}
Let $(M,g)$ be a Cartan-Hadamard manifold. This means that $M$ is
a complete and simply connected Riemannian manifold with non-positive sectional curvature, that is, the sectional curvature of $M$ satisfies $K_M\le 0$ along each (plane) section at each point of $M$. Then condition \eqref{EQ:polar} is automatically satisfied by taking $\lambda(\rho,\omega)= J(\rho,\omega) \rho^{n-1}$, where $J(\rho,\omega)$ is the density function on $M$, see e.g. \cite{DV1}, Helgason \cite{DV3}.

Let us fix a point $a\in M$ and denote by
$\rho(x)=d(x,a)$ the geodesic distance from $x$ to $a$ on $M$. The exponential map ${\rm exp}_a :T_a M \to  M$ is a diffeomorphism, see e.g. Helgason \cite{DV3}.  Let us assume that the sectional curvature $K_M$ is constant, in which case it is known that the function $J(t,\omega)$ depends only on $t$. More precisely, let us denote $K_M=-b$ for $b\ge0$. Then we have
$J(t,\omega)= 1$ if $b=0$, and $J(t,\omega)=(\frac{\sinh \sqrt{b}t}{\sqrt{b}t})^{n-1}$ for $b>0$,
see e.g. \cite {DV5}.
 In the case $b=0$, then let us take the weights
 $$u(x) =
 \left\{
	\begin{array}{ll}
		{(\sinh {\vert x \vert_a})}^{\alpha_1}  & \mbox{if }  \vert x \vert < 1, \\
		{(\sinh {\vert x \vert_a})}^{\alpha_2} & \mbox{if }  \vert x \vert \geq 1,
	\end{array}
\right. \qquad
v(x)= (\sinh {\vert x \vert_a})^\beta,$$
then   inequality \eqref{EQ:Hardy1} holds for $0<q<p$, $1<p <\infty,$ $1/r=1/q-1/p,$ if and only if\\
$ \mathcal A_2 \approx \bigg(\displaystyle\int_0^1  \bigg( \int_t^1 \rho^{\alpha_1+n-1} d\rho +\int_1^\infty \rho^{\alpha_2+n-1} d\rho  \bigg)^{r/p} \bigg( \int_0^t \rho^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}  t^{\alpha_1+n-1}dt + \displaystyle+ \int_1^\infty  \bigg( \int_t^\infty \rho^{\alpha_2+n-1} d\rho   \bigg)^{r/p} \bigg( \int_0^t \rho^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}  t^{\alpha_2+n-1}dt\bigg)^{1/r}<\infty,\nonumber
$\\
which is finite if and only if conditions of Corollary \ref{COR:hom} hold with $Q=n$ (which is natural since the curvature is zero).
  \\
When $b>0$, let
$$u(x) =
 \left\{
	\begin{array}{ll}
		{(\sinh \sqrt{b}{\vert x \vert_a})}^{\alpha_1}  & \mbox{if }  \vert x \vert < 1, \\
		{(\sinh \sqrt{b}{\vert x \vert_a})}^{\alpha_2} & \mbox{if }  \vert x \vert \geq 1,
	\end{array}
\right.
\qquad
v(x)=(\sinh \sqrt b{\vert x \vert_a})^\beta.$$
Then   inequality \eqref{EQ:Hardy1} holds for  $0<q<p$, $1<p <\infty,$ $1/r=1/q-1/p$, if and only if
  $\mathcal A_2$ is finite. We have
 {\small
 \begin{align}
& \mathcal A_2 \approx \bigg(\int_0^1  \bigg(\int_t^1 (\sinh \sqrt b \rho)^{\alpha_1} {(\frac{\sinh \sqrt b \rho}{{\sqrt b} \rho})}^{n-1} {\rho}^{n-1}d\rho + \int_1^\infty (\sinh \sqrt b \rho)^{\alpha_2} {(\frac{\sinh \sqrt b \rho}{{\sqrt b} \rho})}^{n-1} {\rho}^{n-1}d\rho \bigg)^{r/p}\nonumber
\\
&\quad\times  \bigg(\displaystyle \int_0^{t} (\sinh \sqrt b \rho)^{\beta(1-p')} {(\frac{\sinh \sqrt b \rho}{{\sqrt b} \rho})}^{n-1} {\rho}^{n-1}d\rho \bigg)^{r/p'}\nonumber
\\
& \quad\times ( \sinh \sqrt b {t})^{\alpha_1} {(\frac{\sinh \sqrt b t}{{\sqrt b} t})}^{n-1} {t}^{n-1}dt +\int_1^\infty  \bigg(\int_t^\infty (\sinh \sqrt b \rho)^{\alpha_2} {(\frac{\sinh \sqrt b \rho}{{\sqrt b} \rho})}^{n-1} {\rho}^{n-1}d\rho \bigg)^{r/p}\nonumber
\\
&\quad\times\bigg(\displaystyle \int_0^{t} (\sinh \sqrt b \rho)^{\beta(1-p')} {(\frac{\sinh \sqrt b \rho}{{\sqrt b} \rho})}^{n-1} {\rho}^{n-1}d\rho \bigg)^{r/p'}\nonumber
\\
& \quad\times ( \sinh \sqrt b {t})^{\alpha_2} {(\frac{\sinh \sqrt b t}{{\sqrt b} t})}^{n-1} {t}^{n-1}dt  \bigg)^{1/r}\nonumber
\\
&  \approx \bigg(\int_0^1  \bigg(\int_{t}^1 (\sinh \sqrt b \rho)^{\alpha_1+n-1} d\rho+\int_1^\infty (\sinh \sqrt b \rho)^{\alpha_2+n-1} d\rho \bigg)^{r/p}\nonumber
\\
&\quad\times  \bigg(\displaystyle \int_0^{t} (\sinh \sqrt b \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}( \sinh \sqrt b {t})^{\alpha_1+n-1} dt + \int_1^\infty  \bigg(\int_{t}^\infty (\sinh \sqrt b \rho)^{\alpha_2+n-1} d\rho \bigg)^{r/p
}\nonumber
\\
&\quad\times  \bigg(\displaystyle \int_0^{t} (\sinh \sqrt b \rho)^{\beta(1-p')+n-1} d\rho \bigg)^{r/p'}( \sinh \sqrt b {t})^{\alpha_2+n-1} dt \bigg)^{1/r},\nonumber
\end{align} }%
which has the same conditions for finiteness as the case of the hyperbolic space in Corollary \ref{COR:hyp}  (which is also natural since it is the negative constant curvature case).

redpajama_set_name: RedPajamaArXiv


WATCH: Can the PIC inquiry unravel the Ayo riddle?
The commission of inquiry into impropriety at the Public Investment Corporation (PIC) on Monday looked at whether Ayo Technology benefited unduly from PIC investments.
The testimony from Independent Media chair Iqbal Survé, who owns a large portion of Ayo via Sekunjalo Group, has resulted in the JSE being summoned to the inquiry on account of the JSE's decision not to list his IT business, Sagarmatha.
Business Day journalist Warren Thompson, who has been attending the inquiry, joined Business Day TV to discuss Ayo Technology, its executives and the company's role in what is essentially the capture of the PIC.


If I wanted to have flow control and I have 3 16 Zone controllers can I install 3 flow meters in series, each connected to its own controller. If so what are the installation restrictions.
You will want to make sure you have enough linear space to accomplish this. I have done several installations with 2 WFM that are back to back. I found it is important to have @8inches between them for best results.
This requires you to have a full 6 inches prior to the first one, 1st flow meter, 8 inches, 2nd flow meter, 8 inches, 3rd flow meter and then 14 inches of clean pipe (no bends) after the 3rd flow meter. That comes out to about a 57 inch long area to accomplish this.


text:
Quick Answer: How The School Bus Go Into Gated Community In Ga?
1 What is a crossing gate on a school bus?
2 What is the danger zone around a school bus in Georgia?
3 Is the bus stop considered school property?
4 What is the purpose of a school bus crossing arm?
5 How do school bus stop signs work?
6 What is the danger zone around a school bus?
7 Can you pass a school bus with flashing yellow lights in Georgia?
8 Do schools have to provide transportation in Georgia?
9 How many miles can a school bus go on a full tank?
10 When a school bus is loading or unloading students you do not need to stop?
11 When you approach a school bus with red signal lights flashing that is stopped on the opposite side of a median you must?
12 When was the yellow school bus invented?
What is a crossing gate on a school bus?
A school bus crossing arm is a safety device intended to protect children from being struck while crossing in front of a school bus. This ensures that the bus driver can see them as they cross, avoiding a common blind spot immediately in front of the bus, closest to the bumper.
What is the danger zone around a school bus in Georgia?
The Danger Zone is the 12 foot area that surrounds a stopped school bus where the bus driver might not see you or where cars might be moving when you are getting on and off the school bus. Stay away from the dangerous areas where it is difficult for the bus driver to see you with their direct vision.
Is the bus stop considered school property?
Part of you private property is public property if they need if for roads or whatever. Seems to follow that, while a school bus is picking up/dropping off kids, that portion of property could be considered school property.
You might be interested: Readers ask: What Elisode Does The Magic School Bus Go Inside The Belley?
What is the purpose of a school bus crossing arm?
A school bus crossing arm is a safety device intended to protect children from being struck while crossing in front of a school bus.
How do school bus stop signs work?
When a school bus has the stop sign out and the red lights flashing, drivers are required to stop: on two-lane, undivided roads, in both directions. on roads that are divided to separate the directions of travel, in the direction that the bus is traveling.
What is the danger zone around a school bus?
The "Danger Zone" is the area on all sides of the bus where children are in the most danger of not being seen by the driver (ten feet in front of the bus where the driver may be too high to see a child, ten feet on either side of the bus where a child may be in the driver's blind spot, and the area behind the school
Can you pass a school bus with flashing yellow lights in Georgia?
You should slow down when approaching a school bus that is flashing its yellow warning lights (a telltale sign the bus is preparing to load or unload passengers). It's illegal for a vehicle to pass a bus with its red lights flashing.
Do schools have to provide transportation in Georgia?
Local school systems have the responsibility of operating student transportation programs within their systems or contracting for the service. Local school systems are required to have all school buses inspected annually by the Georgia Department of Public Safety.
You might be interested: Quick Answer: Is It The Law To Stop For A School Bus When There Is A Divided Highway?
How many miles can a school bus go on a full tank?
On average, a clean-diesel school bus can travel about 510 miles on a tank of diesel vs. only 270 miles on gasoline, based on the same standard-sized fuel tanks.
When a school bus is loading or unloading students you do not need to stop?
You must stop at least thirty feet from a stopped school bus that is loading or unloading children. This is required by law whether you are meeting the bus or traveling behind it.
When you approach a school bus with red signal lights flashing that is stopped on the opposite side of a median you must?
You must stop your vehicle at least 20 feet before reaching a school bus that is stopped with its red lights flashing whether it is on your side of the road, the opposite side of the road, or at an intersection you are approaching. You must remain stopped until the flashing red lights are no longer operating.
When was the yellow school bus invented?
Origin. In April 1939, Dr. Frank W. Cyr, a professor at Teachers College, Columbia University in New York organized a conference that established 44 uniform national design, construction, and safety standards for school buses in America, including the exterior body color.


text:
Roche May Defend Rituximab, Bevacizumab Sales With New Drug Combinations
Roche has announced positive data on 2 of its innovator oncology biologics in combination with other therapies. The combinations using rituximab (MabThera, Rituxan) and bevacizumab (Avastin) could help the Swiss drug maker to protect sales of these monoclonal antibodies in the face of oncoming biosimilar competition.
Venetoclax Plus Rituximab
On December 12, Roche announced the first results from the phase 3 MURANO study, an open-lable, international, multi-center, randomized study evaluating the efficacy and safety of venetoclax (Venclexta, Venclyxto) in combination with Roche's rituximab compared with bendamustine in combination with rituximab.
The study included 389 patients with relapsed or refractory chronic lymphocytic leukemia (CLL). Patients receiving venetoclax plus rituximab had significantly improved progression-free survival (PFS) compared with those who received bendamustine plus rituximab; at 2 years, 84.9% of patients in the venetoclax plus rituximab arm had not experienced disease progression versus 36.3% of patients in the bendamustine plus rituximab arm. Venetoclax plus rituximab reduced the risk of disease progression or death by 81% compared to bendamustine plus rituximab (HR, 0.19; 95% CI, 0.13, 0.28; P <.0001).
No new safety signals were observed with the treatment combination. Roche plans to submit data from the MURANO study to the FDA, which has granted venetoclax plus rituximab a Breakthrough Therapy Designation for the treatment of refractory CLL.
Use of Roche's branded rituximab in this new combination could help the Swiss drug maker to defend against the eventual introduction of a rituximab biosimilar in the US market; while no biosimilar of the top-selling drug has yet been approved in the United States, Sandoz submitted a Biologics License Application (BLA) for a rituximab biosimilar to the FDA in September 2017. In Europe, Roche already faces competition from approved biosimilars, including Truxima, Riximyo, and Rixathon.
Atezolizumab Plus Bevacizumab
On December 11, Roche announced that the phase 3 IMmotion151 study met its co-primary endpoint of investigator-assessed PFS, and demonstrated that the combination of atezolizumab (Tecentriq) and bevacizumab provided a statistically significant, clinically meaningful improvement in PFS versus sunitinib (Sutent) in patients with advanced kidney cancer whose disease expressed programmed death-ligand 1.
The study was conducted in 915 patients with inoperable, locally advanced or metastatic renal cell carcinoma who had not received prior systemic active or experimental therapy. The results of the multi-center, randomized, open-label study will be presented in 2018.
Roche's innovator bevacizumab will face competition from Amgen and Allergan's biosimilar, Mvasi, approved by the FDA in September 2017, once the drug is launched in the US marketplace. Mvasi is also under review by European regulators; on November 9, 2017, the biosimilar was issued a positive opinion by the European Medicines Agency's Committee for Medicinal Products for Human Use. While it is not yet clear how deep a discount Amgen and Allergan will provide on their product, Roche could use a new indication for its originator product to insulate itself from the market impact of a less expensive bevacizumab treatment.
OncologyBusinessPracticeRegulatory


text:
THE NATION PRESS NETWORK LLC
Publisher - Editor - in- Chief: Abdulnasser Mugali
Arab American news network separate comprehensive, legally registered and commercially in the state of Michigan, United States. Began its work on the electronic network in the on the second of June 2006 and was re-activated under the name of (the nation Press news) in the ninth From May 2008 for the expansion at all levels, technical and absorptive capacity, flexibility and ease of navigation, as well as to add more than a universal language alive and important and required the services of the network of media space for more openness and broader and global.
Issued on the company network (Arab American Media, Culture, Heritage and Arts Center)
: ♦Press Television Radio♦ information electronically and print publication), USA, and supervised by selected media Arabs and the Americans and the young staff is edited by experienced journalist known for their efficient and distributed some of the most correspondents in Arab and foreign capitals, Interested in Arab affairs and issues in various areas, as well as American Muslims and world news for several languages.
*The first Arab-American news site in North America
Copyright ownership of the business name of the company reserved the network (The Nation Press Network LLC),
according to the Federal Copyright Act in the United States


This is a very special organic nut roast (almost impossible to come by outside Devon). Lucky ducks that we are we've managed to get our hands on it. Fit for a proper Sunday lunch and handmade in Devon, this delicious, crumbly nut roast is free from gluten and vegan friendly. It's jam cram full of nuts, seeds and veg. Fantastically moist, yet with a satisfying crunch. It's bowled over veggies and meat eaters alike.
Chilled: Store below 5°C Store in packaging. See 'use by' date label on box. Freeze as soon as possible after purchase and before Use By date and consume within six months. Store below -18°C Do not refreeze once defrosted. Once defrosted use within 48 hours.
Remove from box / plastic and place pie in the centre of a pre-heated oven at 180°C / Gas mark 4 or 160°C for a fan oven for 10-15 minutes. Enjoy!


When I was researching Negative Zero online, I found a few bands with this name, which isn't so promising when you're trying to find out everything you can about a specific group. This particular Negative Zero — based out of Oklahoma — show a lot of promise, but rely heavily on the bands of yesteryear like Rufio and latter-era MxPx.
The album starts off well enough with "Let's Bust Outta This Joint," but from there, the songs become a little indistinguishable before picking up again midway with the powerpop jam, "Wondergirl". There's also an ill-placed ballad, "In the Shadows," that should be shuffled towards the end of the record, if only for consistency.
All in all, it's not a terrible record, but it's not necessarily a burner, either. The music are written well enough and comparable to that of any other band that is heavily influenced by '90s Tooth And Nail bands, but the fact that many of songs are based on relationships with girls and the Lord make the album edge on the redundant side. There are some solid tracks and Negative Zero can do well in the future, but Call It A Day definitely would've made a much better EP than a full-length.


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Derby County midfielder Jacob Butterfield has joined League One side Bradford City on loan for the rest of the season.
The 28-year-old looks to have played his last game for the Rams as his contract is up at the end of the season and he is unlikely to be offered a new deal.
Butterfield has not figured for Derby this season under manager Frank Lampard.
He made 89 appearances for the Rams, scoring nine goals, but his last appearance was in the League Cup win at Grimsby Town in August 2017.
He had a loan spell at Sheffield Wednesday last season - and has previously attracted interest from another Championship side, Hull City.
Bradford are in a relegation battle in League One.
They are fourth from bottom of the table, just inside the drop zone on goal difference.
David Hopkin's men are away to Wycombe Wanderers on Saturday.