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PARALLEL BIBLE - 1 Thessalonians 4:16
CHAPTERS: 1 Thessalonians 1, 2, 3, 4, 5
VERSES: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
TEXT: BIB | AUDIO: MISLR - MISC - DAVIS - FOCHT | VIDEO: BIB - COMM
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King James Bible - 1 Thessalonians 4:16
For the Lord himself shall descend from heaven with a shout, with the voice of the archangel, and with the trump of God: and the dead in Christ shall rise first:
For the Lord himself will descend from heaven with a shout, with the voice of the archangel, and with God's trumpet. The dead in Christ will rise first,
Douay-Rheims - 1 Thessalonians 4:16
For the Lord himself shall come down from heaven with commandment, and with the voice of an archangel, and with the trumpet of God: and the dead who are in Christ, shall rise first.
For the Lord himself will descend from heaven with a shout, with the voice of an archangel, and with the trumpet of God: and the dead in Christ shall rise first:
Greek Textus Receptus
οτι 3754 αυτος 846 ο 3588 κυριος 2962 εν 1722 κελευσματι 2752 εν 1722 φωνη 5456 αρχαγγελου 743 και 2532 εν 1722 σαλπιγγι 4536 θεου 2316 καταβησεται 2597 5695 απ 575 ουρανου 3772 και 2532 οι 3588 νεκροι 3498 εν 1722 χριστω 5547 αναστησονται 450 5698 πρωτον 4412
Treasury of Scriptural Knowledge
VERSE (16) - Isa 25:8,9 Mt 16:27; 24:30,31; 25:31; 26:64 Ac 1:11 2Th 1:7
SEV Biblia, Chapter 4:16
Porque el mismo Se�or con aclamaci�n, con voz de Arc�ngel y con trompeta de Dios, descender� del cielo; y los muertos en el Cristo resucitar�n primero;
Clarke's Bible Commentary - 1 Thessalonians 4:16
Verse 16. The Lord himself] That is: Jesus Christ shall descend from heaven; shall descend in like manner as he was seen by his disciples to ascend, i.e. in his human form, but now infinitely more glorious; for thousands of thousands shall minister unto him, and ten thousand times ten thousand shall stand before him; for the Son of man shall come on the throne of his glory: but who may abide the day of his coming, or stand when he appeareth? With a shout] Or order, en keleusmati? and probably in these words.
Arise, ye dead, and come to judgment; which order shall be repeated by the archangel, who shall accompany it with the sound of the trump of God, whose great and terrible blasts, like those on mount Sinai, sounding louder and louder, shall shake both the heavens and the earth! Observe the order of this terribly glorious day: 1. Jesus, in all the dignity and splendour of his eternal majesty, shall descend from heaven to the mid region, what the apostle calls the air, somewhere within the earth's atmosphere. 2. Then the keleusma, shout or order, shall be given for the dead to arise. 3. Next the archangel, as the herald of Christ, shall repeat the order, Arise, ye dead, and come to judgment! 4. When all the dead in Christ are raised, then the trumpet shall sound, as the signal for them all to flock together to the throne of Christ. It was by the sound of the trumpet that the solemn assemblies, under the law, were convoked; and to such convocations there appears to be here an allusion. 5. When the dead in Christ are raised, their vile bodies being made like unto his glorious body, then, 6. Those who are alive shall be changed, and made immortal. 7. These shall be caught up together with them to meet the Lord in the air. 8. We may suppose that the judgment will now be set, and the books opened, and the dead judged out of the things written in those books. 9. The eternal states of quick and dead being thus determined, then all who shall be found to have made a covenant with him by sacrifice, and to have washed their robes, and made them white in the blood of the Lamb, shall be taken to his eternal glory, and be for ever with the Lord. What an inexpressibly terrific glory will then be exhibited! I forbear to call in here the descriptions which men of a poetic turn have made of this terrible scene, because I cannot trust to their correctness; and it is a subject which we should speak of and contemplate as nearly as possible in the words of Scripture.
John Gill's Bible Commentary
Ver. 16. For the Lord himself shall descend from heaven , etc.] Not by proxy, or by representatives; not by the ministry of angels, as on Mount Sinai; nor by the ministers of the word, as under the Gospel dispensation; nor by his spirit, and the discovery of his love and grace, in which sense he descends in a spiritual manner, and visits his people; but in person, in his human nature, in soul and body; in like manner as he went up to heaven will he descend from thence, so as to be visible, to be seen and heard of all: he will come down from the third heaven, whither he was carried up, into which he was received, and where he is retained until the time of the restitution of all things, and from whence the saints expect him: and this descent will be with a shout ; the word here used is observed by many to signify such a noise or shout as is made either by mariners, when they pull and row together; and shout to direct and encourage one another; or to an army with the general at the head of it, when about to undertake some considerable action, to enter on a battle, and make the onset; Christ will now appear as the King of kings, and Lord of lords, as the Judge of the whole earth, attended with the host, or armies of heaven, and the shout of a king will be among them: perhaps the same is intended, as by the voice of a great multitude, as the voice of many waters, and of mighty thunderings upon the coming of Christ, the destruction of antichrist, and the marriage of the Lamb, in ( Revelation 19:1,6,7,14,15). The Vulgate Latin, Syriac, and Ethiopic versions render it, �in�, or �with command�; and the Arabic version, �with his own government�, or �authority�; that is, he shall descend, either by the command of his Father, as man and Mediator, having authority from him, as the son of man, to execute judgment; or with his commanding power and authority over the mighty angels, that shall descend with him: it follows, with the voice of the archangel ; so Michael is called, in ( Jude 1:9) with which compare ( Revelation 12:7) and who perhaps is no other than Christ himself, who is the head of all principality and power; and the sense be, that Christ shall descend from heaven with a voice, or shall then utter such a voice, as will show him to be the archangel; or as the Syriac version renders it, �the head�, or �prince of angels�; and which whether, it will be an articulate voice, such as was expressed at the grave of Lazarus; or a violent clap of thunder, which is the voice of God; or the exertion of the power of Christ, is not certain: it is added, and with the trump of God ; called �the last trump�, ( 1 Corinthians 15:52) because none will be blown after it, and may be the same with the seventh trumpet, ( Revelation 11:15) and here the trump of God, because blown by his order; or by Christ himself, who is God, and so be the same with the voice of the archangel; and these figurative expressions are used, to set forth the grandeur and magnificence in which Christ will come; not in that low, mean, and humble form in which he first came, but with great glory, and marks of honour and respect; with angels shouting, trumpets blowing, and saints rejoicing. This is said in allusion to the trumpet which was heard on Mount Sinai at the giving of the law, and of which the Jews say f28 , that it aytym ajam , �quickened the dead�; for they have a notion, that, when the Israelites first heard the voice of the Lord, they died; but upon hearing it the second time, they returned to life f29 : and they suppose also in the time, to come, at the resurrection of the dead, a trumpet will be blown, which will quicken the dead f30 , and the day of judgment f31 ; and this is reckoned by them as one of the signs of the Messiah's coming f32 : �Michael shall shout with a great shout, and the graves of the dead shall be opened at Jerusalem, and the holy blessed God will restore the dead to life, and Messiah the son of David shall come,� etc. And the dead in Christ shall rise first ; the same with those that are asleep in Jesus, ( 1 Thessalonians 4:14) not only the martyrs that died for the sake of Christ, and his Gospel; nor merely those who die in the lively exercise of faith in Christ; but all that die interested in him, and in union with him: and these shall �rise�, in consequence of their being his; being given to him, made his care and charge, and engaged for by him, and in virtue of their union to him; and shall rise to an entire conformity to his glorious body, and in order to enjoy eternal life and glory with him: and these will rise �first�, before the wicked, which is the first resurrection, ( Revelation 20:5,6) even a thousand years before them; the righteous will rise in the morning of the resurrection, and so will have the dominion in the morning, ( Psalm 49:14) even at the beginning of the thousand years, as soon as Christ will come; but the wicked will not rise till the evening of that day, or till the close of the thousand years: and this agrees with the notions of the Jews, who thought that some will rise before others; �Wheresoever thou findest the dead, take them and bury them, and I will give thee the first place in my resurrection.� (2 Esdras 2:23) Having mentioned those words in ( <19B609> Psalm 116:9) �I will walk before the Lord in the land of the living�, it is asked f33 , �is there no land of the living but Tyre and its neighbours, and Caesarea, and its neighbours, where is cheapness and fulness? says R. Simeon ben Lekish, in the name of Bar Kaphra, the land in which the dead live, hlyjt , �first�, in the days of the Messiah:� and on the same place elsewhere they observe, that �our Rabbins say two things, or give two reasons, why the fathers loved to be buried in the land of Israel, because the dead in the land of Israel hlyjt �yyj , �live�, or �rise first�, in the days of the Messiah, and shall enjoy the years of the Messiah:� and in another place they take notice of what is written in ( Isaiah 26:19) �and the earth shall cast out the dead�: says R. Jochanan, �the dead which are in the land (i.e. of Israel), they shall �live first�; as it is said, �thy dead men shall live, together with my dead body shall they arise�: these are they that are without the land; �awake and sing ye that dwell in the dust�, these are they that die in the wilderness:� and again f36 , �as it is said, Isaiah 40:26 �that bringeth out their host by numbers�, come see, it is said, all that die in the land of Israel atymdqb �wmwqy , �shall rise first�, because the holy blessed God shall awake them, and raise them, according to ( Isaiah 26:19).�
Once more they say f37 , �they that study in the law as they ought, these are they that shall �rise first� to everlasting life, as it is said ( Daniel 12:2), �and many of them that sleep in the dust of the earth shall awake, some to everlasting life�, etc. and these are for everlasting life, because they study in everlasting life, which is the law:� to which may be added the following passage f38 , �they that are worthy to be buried in the land of Israel, shall �be raised first� � and they shall be raised and quickened before the rest of the children of the world, who draw the waters of the law; and they draw, because they study to draw out of the waters of the law; and they are strengthened by the tree of life, and they shall go out �first�, because the tree of life is the cause why they shall �rise first�:� they sometimes endeavour to fix the time, how long they will rise before the rest f39 ; ��many of those that sleep�, etc. these are the righteous that shall �go before� others in life, and how many years shall they go before them? R. Judah says, two hundred and ten years; R. Isaac says, two hundred and fourteen; according to others, the righteous shall go (or be raised) a year before the rest of men; says R. Nachman, it will be according to the computation (of time) that the carcass has been in the dust; R. Jose replies, if so, there will be many resurrections.�
These instances may suffice to show, that the Jews had a notion of some persons rising before others, to which the apostle may have some reference; though his sense is not only this, but also that the dead in Christ shall rise before the living saints are changed, and taken up to be with Christ, and so shall not go before to him; which illustrates and proves what he had before asserted.
Matthew Henry Commentary
Verses 13-18 - Here is comfort for the relations and friends of those who die in the Lord. Grief for the death of friends is lawful; we may weep for our ow loss, though it may be their gain. Christianity does not forbid, an grace does not do away, our natural affections. Yet we must not be excessive in our sorrows; this is too much like those who have no hop of a better life. Death is an unknown thing, and we know little abou the state after death; yet the doctrines of the resurrection and the second coming of Christ, are a remedy against the fear of death, an undue sorrow for the death of our Christian friends; and of thes doctrines we have full assurance. It will be some happiness that all the saints shall meet, and remain together for ever; but the principa happiness of heaven is to be with the Lord, to see him, live with him and enjoy him for ever. We should support one another in times sorrow not deaden one another's spirits, or weaken one another's hands. An this may be done by the many lessons to be learned from the resurrection of the dead, and the second coming of Christ. What comfort a man by telling him he is going to appear before the judgment-seat of God! Who can feel comfort from those words? That ma alone with whose spirit the Spirit of God bears witness that his sin are blotted out, and the thoughts of whose heart are purified by the Holy Spirit, so that he can love God, and worthily magnify his name. We are not in a safe state unless it is thus with us, or we are desirin to be so __________________________________________________________________
Vincent's NT Word Studies
16. The word of the Lord, ver. 15, is apparently not intended to include the specific details which follow. In that word the revelation was to the effect that all believers simultaneously should share the blessings of the advent. The following description of the Lord's descent from heaven is intended to emphasise the fact that the reunion of dead and living believers will be accomplished by the Lord in person (autov). %Oti does not indicate the contents of the word of the Lord (that, as A.V.), but means for or because; and the details are meant to strengthen the more general declaration of ver. 15. In the details themselves there are traces of certain O.T. theophanies, as Exod. xix. 11-18; Micah i. 3. 27 Shall descend from heaven. Used nowhere else of Christ's second coming. Frequently in the Fourth Gospel, of Christ's descent to earth as man. See iii. 13; vi. 33, 38, 41, etc. In Eph. iv. 9, of his descent by the Spirit in order to endow the church.
With a shout (en keleusmati). N.T.o . Once in LXX, Prov. xxiv. 62 (English Bib. xxx. 27). From keleuein to summon. Often in Class. Lit. a shout of command, as of a general to his army, an admiral to his oarsmen, or a charioteer to his horses.
Archangel (arcaggelou). Only here and Jude 9. Not in O.T. The Pauline angelology shows traces of Rabbinical teachings in the idea of orders of angels. See Eph. i. 21; Col. i. 16; Rom. viii. 38. The archangels appear in the apocryphal literature. In the Book of Enoch (see on Jude 14) four are named, Michael, Uriel, Raphael, and Gabriel. Michael is set over the tree which, at the time of the great judgment, will be given over to the righteous and humble, and from the fruit of which life will be given to the elect. In Tob. xii. 15, Raphael appears as one of the seven holy angels. Comp. Apoc. viii. 2. See also on Jude 9, and comp. Dan. xii. 1. 28
With the trump of God (en salpiggi qeou). For the trumpet heralding great manifestations of God, see Exod. xix. 13, 16; Psalm xlvii. 5; Isaiah xxvii. 13; Zech. ix. 14; Zeph. i. 16; Joel ii. 1; Matt. xxiv. 31; 1 Corinthians xv. 52; Apoc. i. 10; iv. 1. Of God does not indicate the size or loudness of the trumpet, but merely that it is used in God's service. Comp. harps of God, Apoc. xv. 2; musical instruments of God, 1 Chronicles xvi. 42. The later Jews believed that God would use a trumpet to raise the dead.
Robertson's NT Word Studies
4:16 {With a shout} (en keleusmati). Note this so-called instrumental use of en. Old word, here only in N.T., from keleuw, to order, command (military command). Christ will come as Conqueror. {With the voice of the archangel} (en fwnei arcaggelou). Further explanation of keleusmati (command). The only archangel mentioned in N.T. is Michael in #Jude 1:9. But note absence of article with both fwnei and arcaggelou. The reference may be thus indefinite. {With the trump of God} (en salpiggi qeou). Trumpet. See same figure in #1Co 15:52. {The dead in Christ shall rise first} (hoi nekroi en Christ"i anast�sontai prwton). {First} here refers plainly to the fact that, so far from the dead in Christ having no share in the parousia, they will rise before those still alive are changed.
CHAPTERS: 1, 2, 3, 4, 5
PARALLEL VERSE BIBLE


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When will this madness end? My kid is obsessed!
Mine too! All day every day and she tries to find different versions of the same song.
our babe watched rainbow ruby for 10 minutes on TV so I could shower and now that's all she asks for... "ruby" "ruby" "ruby"
For us it's a wonderful thing during poop diaper changes. Prevents all the wiggles and teying to touch his bum. But I'd go crazy if I had to listen to it all day.
Mine thankfully has moved on to other songs. But, the other day I called her over to her change mat so that I could put on her pyjamas, and she said, "Run away, do do do do do do," then took off down the hall.


text:
ABC7, Disney donates $15K to Alameda Co. Community Food Bank
OAKLAND, Calif. (KGO) -- Our annual ABC7 News Thanksgiving food drive is underway. Our partnership with Bay Area food banks has helped feed local communities for the past 23-years.
On Tuesday, ABC7 weather forecaster Spencer Christian was at the Alameda County Community Food Bank in Oakland. He spoke with executive director Suzan Bateson.
Spencer: "Once again, we do this every year, right?"
Suzan: "It's a great, great way to start out the season."
Spencer: "The need is so staggering, though. 780,000 people in the Bay Area go hungry each month. That translates into 1 in 6 people. And it's an even greater figure here in Alameda County, right?"
Suzan: "Yes, we're serving 1 in 5 in Alameda County. It shouldn't happen in a country as rich as America."
PHOTOS: ABC7 Thanksgiving food drive 2014
This month ABC7 News is highlighting how Bay Area food banks are helping communities in need. You can give where you live with this year's ABC7 Thanksgiving food drive!
Spencer presented Suzan and the food bank a check for $15,000 on behalf of ABC7 and our parent company Disney.
"This will provide 34,000 meals, Spencer," Suzan exclaimed. "It's going to really start things out well!"
This month, ABC7 and our parent company Disney are giving $75,000 to local Feeding America food banks. But they still need more help! Research shows 1 in 6 people in the Bay Area go hungry every month. You can help by making a $10 donation, just by texting the word FEED to 80077.
Click here to find out more about how you can give Where You Live!
societyoaklandalameda countyhomelesswhere you livefood bankhungercommunityabc7 thanksgiving food drive
Give Where You Live: ABC7 Thanksgiving Food Drive
ABC7, Disney donates $15K to Second Harvest Food Bank

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction and Main Results}
Fractional type operators and associated maximal functions are useful tools  in harmonic analysis,
especially in the study of differentiability or smoothness properties of functions.
Recall that for $0<\alpha<n$, the fractional maximal function $M_\alpha$ and the fractional integral operator $I_\alpha$ are defined by
\[
  M_\alpha(f)(x):=\sup_{Q\ni x}\frac{1}{|Q|^{1-\alpha/n}}\int_Q |f(y)|dy
\]
and
\[
  I_\alpha (f)(x):=\int \frac{f(y)}{|x-y|^{n-\alpha}}dy,
\]
respectively, where $f$ is a locally integrable function defined on $\bbR^n$ and $Q$ is a cube in $\bbR^n$.
We refer to \cite{Gra1,Stein} for the basic
properties of these operators.

Let $w$ be a weight, i.e., a non-negative locally integrable function.
Muckenhoupt and Wheeden \cite{MW} showed that for $1<p<n/\alpha$ and $1/q = 1/p-\alpha/n$,
 $I_\alpha$ is bounded from $L^p(w^p)$ to $L^q(w^q)$ if and only if  $w$  belongs to the $A_{p,q}$ class, that is,
\[
  [w]_{A_{p,q}}:=\sup_{Q}\left(\frac{1}{|Q|}\int_Q w(x)^qdx\right)\left(\frac{1}{|Q|}\int_Q w(x)^{-p'}dx\right)^{q/{p'}}<\infty.
\]
Moreover, the fractional maximal function $M_\alpha$  is also  bounded from $L^p(w^p)$ to $L^q(w^q)$.
See also \cite{DL,ISST,LLL,SW,WTL,YY,YYZ,ZC} for more results of fractional integral operators on various function spaces.

In \cite{LMPT}, Lacey, Moen, P\'erez and Torres gave the sharp weighted estimates for both $M_\alpha$ and $I_\alpha$. Specifically, they proved that
\begin{equation}\label{eq:ee1}
\|M_\alpha\|_{L^p(w^p)\rightarrow L^q(w^q)}\le C_{n,p}[w]_{A_{p,q}}^{(1-\frac{\alpha}{n})\frac{p'}{q}}
\end{equation}
and
\begin{equation}\label{eq:ee2}
\|I_\alpha\|_{L^p(w^p)\rightarrow L^q(w^q)}\le C_{n,p}[w]_{A_{p,q}}^{(1-\frac{\alpha}{n})\max\{1,\frac{p'}{q}\}},
\end{equation}
where the exponents of $[w]_{A_{p,q}}$ are sharp.

Multilinear fractional integral operators were studied by many authors, e.g., Grafakos \cite{G}, Kenig and
Stein \cite{KS}, Grafakos and Kalton \cite{GK}, Chen and Xue \cite{CX}, Pradolini \cite{P},
Kuk and Lee \cite{KL},
 and Mei, Xue and Lan \cite{MXL}.
For $\vec f = (f_1,\cdots,f_m)$ and $\quad 0<\alpha<mn$,  the multilinear fractional maximal function $\mathcal{M_\alpha}$ and the multilinear integral operator
$ \mathcal{I}_\alpha $ are defined by
\[
  \mathcal{M}_\alpha (\vec f)(x)=\sup_{Q\ni x}\prod_{i=1}^m\frac{1}{|Q|^{1-\alpha/{mn}}}\int_Q |f_i(y_i)|dy_i
\]
and
\[
  \mathcal{I}_\alpha (\vec{f})(x)=\int_{\bbR^{mn}} \frac{f_1(y_1)\cdots f_m(y_m)}{(|x-y_1|+\cdots+|x-y_m|)^{mn-\alpha}}d\vec{y},
\]
respectively.

To study the weighted estimates for  multilinear fractional integral operators,
Moen \cite{M} introduced the multiple $A_{\vec{P},q}$ weight.
Let $1/p_1 + \cdots + 1/p_m = 1/q + \alpha/n$.
A multiple weight $(w_1,\cdots, w_m)$ is said
to belong to the $A_{\vec{P},q}$ class if and only if
\[
 [\vec{w}]_{A_{\vec{P},q}}:= \sup_{Q}\left(\frac{1}{|Q|}\int_Q u_{\vec{w}}\right)\prod_{i=1}^m \left(\frac{1}{|Q|}\int_Q \sigma_i\right)^{q/{p_i'}}<\infty,
\]
where $u_{\vec{w}}=(\Pi_{i=1}^m w_i)^q$ and $\sigma_i=w_i^{-p_i'}$.
Moen showed that
$\mathcal{I}_\alpha$ is bounded from $L^{p_1}(w_1^{p_1})\times\cdots\times L^{p_m}(w_m^{p_m})$
to $L^q(u_{\vec{w}})$ if and only if $ [\vec{w}]_{A_{\vec{P},q}}<\infty$.

In this paper, we study the sharp estimates for both $\mathcal{M}_\alpha$ and $\mathcal{I}_\alpha$.
For the multilinear fractional maximal function, we give a sharp estimate for $0<\alpha<n$.

\begin{Theorem}\label{thm:m1}
Suppose that $0<\alpha<n$,
$q>0$, $1<p_1,\cdots,p_m<\infty$,
$1/p_1+\cdots +1/p_m=1/q+\alpha/n$,
$p'_{i_0}=\max\{p'_i:\,1\le i\le m\}$ for some $i_0$ and
$p'_{i_0}(1-\alpha/n)\ge \max_{i\ne i_0}\{p_i'\}$. Let
$\vec{w}=(w_1,\cdots,w_m)\in A_{\vec{P},q}$. Then we have
\begin{equation}\label{eq:thm:m1:e1}
  \|\mathcal{M}_{\alpha}(\vec{f})\|_{L^q(u_{\vec{w}})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)
  \frac{\max\{p'_i\}}{q}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})},
\end{equation}
where the exponent of $[\vec{w}]_{A_{\vec{P},q}}$ is sharp.
\end{Theorem}

And for the multilinear fractional integral operators, we also get a sharp estimate in some cases.
\begin{Theorem}\label{thm:m2}
Suppose that $0<\alpha< n$, $q>0$, $1<p_1,\cdots,p_m<\infty$, $1/p_1+\cdots +1/p_m=1/q+\alpha/n$,  $\vec{w}=(w_1,\cdots,w_m)\in A_{\vec{P},q}$ and
\[
  \min\left\{\frac{\max_i{p_i'}}{q},
   \min_j\frac{\max_{i\neq j}\{p_i', q\}}{p_j'}\right\}\le 1-\frac{\alpha}{n}.
\]
Then
\[
  \|\mathcal{I}_\alpha(\vec{f})\|_{L^q(u_{\vec{w}})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)\max_i\{1,\frac{p_i'}{q}\}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})},
\]
where the exponent of $[\vec{w}]_{A_{\vec{P},q}}$ is sharp.
\end{Theorem}

Similar to the estimate for Calder\'on-Zygmund operators,
although the estimate in (\ref{eq:thm:m1:e1}) is sharp, it can be improved whenever mixed estimates are invoked.

\begin{Theorem}\label{thm:m3}
Suppose that $0\le \alpha< mn$,  $q>0$, $1<p_1,\cdots,p_m<\infty$, $1/p_1+\cdots +1/p_m=1/q+\alpha/n$ and  $\vec{w}=(w_1,\cdots,w_m)\in A_{\vec{P},q}$. Then
\[
    \|\mathcal{M}_{\alpha}(\vec{f})\|_{L^q(u_{\vec{w}})}\le
  C_{m,n,\vec{P},q}\left([\vec{w}]_{A_{\vec{P},q}}^{\frac{1}{q}}\prod_{i=1}^m [\sigma_i]_{A_\infty}^{\frac{1}{p_i}(1-\frac{\alpha p}{n})}\right)\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})},
\]
where the exponents are sharp.
\end{Theorem}

Moreover, in the so-called two weight case -- dropping the requirement that $u_{\vec w}=\big(\Pi_i w_i\big)^q$ and taking $u$ to be independent of $\vec w$ -- one can break the homogeneity requirement that $\frac1q=\frac1p-\frac{\alpha}{n}$ and simply require $q\geq p$.  If we are given an $n+1$ tuple of weights $(u,w_1,\ldots,w_n)=(u,\vec w)$ then define
$$[u,\vec w]_{A_{\vec P,q}}=
\sup_Q|Q|^{\frac{\alpha}{n}+\frac1q-\frac1p}\left(\frac{1}{|Q|}\int_Q u\right)^{1/q}\prod_{i=1}^m \left(\frac{1}{|Q|}\int_Q \sigma_i\right)^{1/{p_i'}},$$
where, again, $\sigma_i=w_i^{-p_i'}.$
Then we have the following extension of Theorem \ref{thm:m3}.

\begin{Theorem}\label{thm:m4}
Suppose that $0\le \alpha< mn$, $1<p_1,\cdots,p_m<\infty$, $1/p_1+\cdots +1/p_m=1/p$ and $p\leq q$. If $(u,\vec{w})$ is an $m+1$ tuple of weights satisfying $\sigma_i=w_i^{-p_i'}\in A_\infty$ for all $i=1,\ldots,m$ and $[u,\vec{w}]_{A_{\vec{P},q}}<\infty$, then
\[
    \|\mathcal{M}_{\alpha}(\vec{f})\|_{L^q(u)}\le
  C_{m,n,\vec{P},q}\left([u,\vec{w}]_{A_{\vec{P},q}}\prod_{i=1}^m [\sigma_i]_{A_\infty}^{\frac{p}{q}\frac{1}{p_i}}\right)\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}
\]
\end{Theorem}

Theorem \ref{thm:m4} is the multilinear version of Theorem 2.2 in \cite{CM2}.  To see that Theorem \ref{thm:m4} does indeed extend Theorem \ref{thm:m3} notice that when $\frac{1}{q}=\frac{1}{p}-\frac{\alpha}{n}$ we have $\frac{p}{q}=1-\frac{\alpha p}{n}$ and if $u=u_{\vec w}=\big(\Pi_i w_i\big)^q$ then
$$[u_{\vec{w}},\vec w]_{A_{\vec P,q}}=[\vec w]_{A_{\vec{P},q}}^{1/q}.$$
 Even though Theorem \ref{thm:m3} follows from Theorem \ref{thm:m4} we provide proofs for both Theorems.  We present both proofs because they use different techniques.  For Theorem \ref{thm:m3} we use a sharp reverse H\"older estimate (see Proposition \ref{prop:reverseH} below), while Theorem \ref{thm:m4} will use Carleson sequences (see Lemma \ref{lem:carleson}).

\section{Proofs of Theorem~\ref{thm:m1} and Theorem~\ref{thm:m2}}
Recall that the standard dyadic grid in $\bbR^n$ consists of the cubes
\[
  [0,2^{-k})^n+2^{-k}j,\qquad k\in\bbZ, j\in\bbZ^n.
\]
Denote the standard dyadic grid by $\mathcal{D}$.

By a general dyadic grid $\mathscr{D}$ we mean a collection of cubes with the following
properties: (i) for any $Q\in\mathscr{D}$ its sidelength $l_Q$ is of the form $2^k$, $k\in\bbZ$; (ii)
 $Q\cap R \in \{Q,R,\varnothing\}$ for any $Q,R\in\mathscr{D}$; (iii) the cubes of a fixed sidelength $2^k$ form a partition
of $\bbR^n$.

For any $0\le\alpha<n$, we define
\[
  M_{\alpha,w}^{\mathscr{D}}f(x)=\sup_{Q\ni x, Q\in \mathscr{D}}\frac{1}{w(Q)^{1-\frac{\alpha}{n}}}\int_Q |f|w.
\]
When $\alpha=0$, we denote $M_{0,w}^{\mathscr{D}}$ simply by $M_{w}^{\mathscr{D}}$.
We need the following result.
\begin{Proposition}[{\cite[Theorem 2.3]{M1}}]\label{prop:p2}
If $0\le \alpha<n$, $1<p\le \frac n\alpha$ and $1/q=1/p-\alpha/n$, then
\[
  \|M_{\alpha,w}^{\mathscr{D}}f\|_{L^q(w)}\le (1+\frac{p'}{q})^{1-\frac \alpha n}\|f\|_{L^p(w)}.
\]
\end{Proposition}

We say that  $\mathcal{S}:=\{Q_{j,k}\}$ is a sparse family of cubes if:
\begin{enumerate}
\item for each fixed $k$ the cubes $Q_{j,k}$ are pairwise disjoint;
\item if $\Gamma_k=\bigcup_j Q_{j,k}$, then $\Gamma_{k+1}\subset \Gamma_k$;
\item $|\Gamma_{k+1}\bigcap Q_{j,k}|\le \frac{1}{2}|Q_{j,k}|$.
\end{enumerate}
For any $Q_{j,k}\in\mathcal{S}$, we define $E(Q_{j,k})=Q_{j,k}\setminus \Gamma_{k+1}$.
Then the sets $E(Q_{j,k})$ are pairwise disjoint and $|E(Q_{j,k})|\ge \frac{1}{2}|Q_{j,k}|$.

Define
\[
  \mathscr{D}_t:=\{2^{-k}([0,1)^n+m+(-1)^k t): k\in\bbZ, m\in\bbZ^n\},\quad t\in\{0, 1/3\}^n.
\]
The importance of these grids is shown by the following proposition, which
 can be found in \cite[proof of Theorem 1.10]{HP}. See also \cite[Proposition 5.1]{L}.
\begin{Proposition}\label{prop:p1}
There are $2^n$ dyadic grids $\mathscr{D}_t$, $t\in\{0,1/3\}^n$ such that for any cube
$Q\subset \bbR^n$ there exists a cube $Q_t \in \mathscr{D}_t$ satisfying
 $Q \subset Q_t$ and $l(Q_t)\le 6l(Q)$.
\end{Proposition}

Given a dyadic grid $\mathscr{D}$ and  a sparse family $\mathcal{S}$ in $\mathscr{D}$.
Define the dyadic fractional integral operators by
\[
  (\mathcal{I}_\alpha^{\mathcal{S}})(\vec{f})(x)
  =\sum_{Q\in\mathcal{S}}|Q|^{\frac{\alpha}{n}-m}
   \prod_{i=1}^m\int_Q f_i(y_i)dy_i \cdot\chi_Q(x)
\]
and
\[
  (\mathcal{I}_{\alpha,q}^{\mathcal{S}})(\vec{f})(x)
  =\left(\sum_{Q\in\mathcal{S}}\left(|Q|^{\frac{\alpha}{n}-m}
   \prod_{i=1}^m\int_Q f_i(y_i)dy_i \right)^q\cdot\chi_Q(x)\right)^{1/q}.
\]

In \cite{M}, Moen showed that for $q> 1$
\[
  (\mathcal{I}_\alpha(\vec{f}))(x)
   \le C\sum_{Q\in\mathscr{D}}
   \frac{l(Q)^\alpha}{|3Q|^m}\int_{(3Q)^m} f_1(y_1)\cdots f_m(y_m)d\vec{y}\chi^{}_Q(x),
\]
and for $q\le1$,
\[
  (\mathcal{I}_\alpha(\vec{f}))(x)
  \le C\left(\sum_{Q\in\mathscr{D}}\left(\frac{l(Q)^\alpha}{|3Q|^m}
   \int_{(3Q)^m} f_1(y_1)\cdots f_m(y_m)d\vec{y}\right)^{q}\chi^{}_Q(x)\right)^{1/q}.
\]

By Proposition~\ref{prop:p1} and similar arguments as that in \cite{CM}, we get
\begin{equation}\label{eq:q Gt 1}
  \mathcal{I}_\alpha(\vec{f})(x)\lesssim
   \sum_{t\in\{0,1/3\}^n}(\mathcal{I}_\alpha^{\mathcal{S}_t})(\vec{f})(x),\quad q> 1,
\end{equation}
and
\begin{equation}\label{eq:q Lt 1}
  \mathcal{I}_\alpha(\vec{f})(x)\lesssim
   \sum_{t\in\{0,1/3\}^n}(\mathcal{I}_{\alpha,q}^{\mathcal{S}_t})(\vec{f})(x),\quad q\le1.
\end{equation}
We have the following lemmas.
\begin{Lemma} \label{Lm:dual}
Suppose that $\vec{w}=(w_1,\cdots,w_m)\in A_{\vec{P},q}$,
$0\le\alpha<mn$, $1<q,p_1,\cdots,p_m<\infty$ and $1/p_1+\cdots +1/p_m=1/q+\alpha/n$.
Then
\[
  \vec{w}^i:=(w_1,\cdots,w_{i-1},(\Pi_{i=1}^m w_i)^{-1},w_{i+1},\cdots,w_m)\in A_{\vec{P}^i,p_i'},
\]
where $\vec{P}^i=(p_1,\cdots,p_{i-1},q',p_{i+1},\cdots,p_m)$ and $[\vec{w}^i]_{A_{\vec{P}^i,p_i'}}=[\vec{w}]_{A_{\vec{P},q}}^{p_i'/q}$.
\end{Lemma}
\begin{proof}
By the definition we have
\begin{eqnarray*}
[\vec{w}^i]_{A_{\vec{P}^i,p_i'}}&=&\sup_Q \left(\frac{1}{|Q|}\int_Q w_i^{-p_i'}\right)\left(\frac{1}{|Q|}\int_Q u_{\vec{w}}\right)^{p_i'/q}
\prod_{j\neq i}\left(\frac{1}{|Q|}\int_Q w_j^{-p_j'}\right)^{p_i'/{p_j'}}\\
&=&[\vec{w}]_{A_{\vec{P},q}}^{p_i'/q}.
\end{eqnarray*}
\end{proof}

\begin{Lemma}\label{lm:l1}
Suppose that $0<\alpha< n$,  $1<p_1,\cdots,p_m<\infty$,
$1/p_1+\cdots +1/p_m=1/q+\alpha/n$, $q(1-\alpha/n)\ge max\{p_i'\}$
 and
that $\vec{w}=(w_1,\cdots,w_m)\in A_{\vec{P},q}$. Then
\[
  \|\mathcal{I}_\alpha^{\mathcal{S}}(|f_1|,\cdots,|f_m|)\|_{L^q(u_{\vec{w}})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{1-\alpha/n}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\]
\end{Lemma}
\begin{proof}
It is equivalent to prove the following
\[
  \|\mathcal{I}_\alpha^{\mathcal{S}}(|f_1|\sigma_1,\cdots,|f_m|\sigma_m)\|_{L^q(u_{\vec w})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{1-\alpha/n}\prod_{i=1}^m \|f_i\|_{L^{p_i}(\sigma_i)}.
\]
Let $1/p=1/q+\alpha/n$. We have
\begin{eqnarray}
  \label{eq:Lm:L1:e1}
&&\|\mathcal{I}_\alpha^{\mathcal{S}}(|f_1|\sigma_1,\cdots,|f_m|\sigma_m)\|_{L^q(u_{\vec w})}
   \\
&=&\sup_{\|h\|_{L^{q'}(u_{\vec w})}=1}\int_{\bbR^n}\mathcal{I}_\alpha^{\mathcal{S}}
  (|f_1|\sigma_1,\cdots,|f_m|\sigma_m)h u_{\vec w}\nonumber \\
&=&\sup_{\|h\|_{L^{q'}(u_{\vec w})}=1}\sum_{Q\in\mathcal{S}}|Q|^{\frac{\alpha}{n}-m}
  \prod_{i=1}^m\int_Q |f_i|\sigma_i dy_i \cdot\int_Q h u_{\vec w}\nonumber \\
&=&\sup_{\|h\|_{L^{q'}(u_{\vec w})}=1}\sum_{Q\in\mathcal{S}}
   \left(\frac{u_{\vec w}(Q)}{|Q|}\right)^{1-\frac{\alpha}{n}}
\prod_{i=1}^m\left(\frac{\sigma_i(Q)}{|Q|}\right)^{\frac{q(1-\alpha/n)}{p_i'}}
\prod_{i=1}^m \sigma_i(Q)^{1-\frac{q}{p_i'}(1-\frac{\alpha}{n})} \nonumber \\
&&\times|Q|^{(m-\frac{\alpha}{n})((1-\frac{\alpha}{n})q-1)}
  \cdot\frac{1}{u_{\vec w}(Q)^{1-\alpha/n}}\int_Q h u_{\vec w}\cdot
\prod_{i=1}^m\frac{1}{\sigma_i(Q)}\int_Q |f_i|\sigma_i dy_i\nonumber \\
&\le&[\vec{w}]_{A_{\vec{P},q}}^{1-\alpha/n}\sup_{\|h\|_{L^{q'}(u_{\vec w})}=1}
  \sum_{Q\in\mathcal{S}}
|Q|^{(m-\frac{\alpha}{n})\frac{q}{p'}}
\prod_{i=1}^m \sigma_i(Q)^{1-\frac{q}{p_i'}(1-\frac{\alpha}{n})}\nonumber  \\
&&\times\frac{1}{u_{\vec w}(Q)^{1-\alpha/n}}\int_Q h u_{\vec w}\cdot
  \prod_{i=1}^m\frac{1}{\sigma_i(Q)}\int_Q |f_i|\sigma_i dy_i. \nonumber
\end{eqnarray}
By H\"older's inequality, we have
\begin{eqnarray}
\label{eq:e4}
|E(Q)|&=&\int_{E(Q)} u_{\vec w}^{\frac{1}{(m-\frac{\alpha}{n})q}}\prod_{i=1}^m \sigma_i^{\frac{1}{(m-\frac{\alpha}{n})p_i'}}dx\\
&\le&u_{\vec w}(E(Q))^{\frac{1}{(m-\frac{\alpha}{n})q}}\prod_{i=1}^m \sigma_i(E(Q))^{\frac{1}{(m-\frac{\alpha}{n})p_i'}}.\nonumber
\end{eqnarray}
Since $|Q|\le 2|E(Q)|$ and $q(1-\alpha/n)\ge \max\{p_i'\}$, we have
\begin{eqnarray*}
&&|Q|^{(m-\frac{\alpha}{n})\frac{q}{p'}}
\prod_{i=1}^m \sigma_i(Q)^{1-\frac{q}{p_i'}(1-\frac{\alpha}{n})}\\
&\lesssim& u_{\vec w}(E(Q))^{\frac{1}{p'}}\prod_{i=1}^m
  \sigma_i(E(Q))^{\frac{q}{p'p_i'}}\prod_{i=1}^m
     \sigma_i(Q)^{1-\frac{q}{p_i'}(1-\frac{\alpha}{n})}\\
&\le&u_{\vec w}(E(Q))^{\frac{1}{p'}}\prod_{i=1}^m \sigma_i(E(Q))^{1/{p_i}}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
&&\sum_{Q\in\mathcal{S}}|Q|^{(m-\frac{\alpha}{n})\frac{q}{p'}}
\prod_{i=1}^m \sigma_i(Q)^{1-\frac{q}{p_i'}(1-\frac{\alpha}{n})}
  \frac{1}{u_{\vec w}(Q)^{1-\alpha/n}}\int_Q h u_{\vec w}\\
&&\times\prod_{i=1}^m\frac{1}{\sigma_i(Q)}\int_Q |f_i|\sigma_i dy_i\\
&\lesssim& \sum_{Q\in\mathcal{S}}
\frac{1}{u_{\vec w}(Q)^{1-\alpha/n}}\int_Q h u_{\vec w}
  \cdot u_{\vec w}(E(Q))^{\frac{1}{p'}}\cdot\prod_{i=1}^m
    \frac{1}{\sigma_i(Q)}\int_Q |f_i|\sigma_i dy_i\\
&&\times\prod_{i=1}^m \sigma_i(E(Q))^{1/{p_i}}\\
&\le&\left(\sum_{Q\in\mathcal{S}} \left(\frac{1}{u_{\vec w}(Q)^{1-\alpha/n}}
  \int_Q h u_{\vec w}\right)^{p'}u_{\vec w}(E(Q))\right)^{1/{p'}}\\
&&\times \prod_{i=1}^m \left(\sum_{Q\in\mathcal{S}} \left(\frac{1}{\sigma_i(Q)}
  \int_Q |f_i|\sigma_i dy_i\right)^{p_i}\sigma_i(E(Q))\right)^{1/{p_i}}\\
&\le&\|M_{\alpha,u_{\vec w}}^{\mathscr{D}}h\|_{L^{p'}(u_{\vec w})}\prod_{i=1}^m
  \|M_{\sigma_i}^{\mathscr{D}}f_i\|_{L^{p_i}(\sigma_i)}\\
&\lesssim& \|h\|_{L^{q'}(u_{\vec w})}\prod_{i=1}^m \|f_i\|_{L^{p_i}(\sigma_i)}.
\end{eqnarray*}
It follows from (\ref{eq:Lm:L1:e1}) that
\[
  \|\mathcal{I}_\alpha^{\mathcal{S}}(|f_1|\sigma_1,\cdots,|f_m|\sigma_m)\|_{L^q(u_{\vec w})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{1-\alpha/n}\prod_{i=1}^m \|f_i\|_{L^{p_i}(\sigma_i)}.
\]
\end{proof}

\begin{Lemma}\label{lm:lq}
Let the hypotheses be as in Theorem~\ref{thm:m1}.
Moreover, let $\mathcal S$ be a sparse family of cubes.
Then we have
\[
  \|\mathcal{I}_{\alpha,q}^{\mathcal{S}}(|f_1|,\cdots,|f_m|)\|_{L^q(u_{\vec w})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)\max_i\{\frac{p_i'}{q}\}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\]
\end{Lemma}
\begin{proof}
Without loss of generality, assume that $p_1=\min\{p_1,\cdots,p_m\}$.
As in Lemma~\ref{lm:l1},
it is equivalent to prove the following
\[
  \|\mathcal{I}_{\alpha,q}^{\mathcal{S}}
   (|f_1|\sigma_1,\cdots,|f_m|\sigma_m)\|_{L^q(u_{\vec w})}
   \!\le\!
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\alpha/n)\max\{p'_i/q\}}
   \prod_{i=1}^m \|f_i\|_{L^{p_i}(\sigma_i)}.
\]
We have
\begin{eqnarray*}
&&\|\mathcal{I}_{\alpha,q}^{\mathcal{S}}(|f_1|\sigma_1,\cdots,|f_m|\sigma_m)
   \|_{L^q(u_{\vec w})}^q\\
&=&\sum_{Q\in\mathcal{S}}\left(|Q|^{\frac{\alpha}{n}-m}\prod_{i=1}^m\int_Q
  |f_i|\sigma_i dy_i \right)^q\cdot u_{\vec w}(Q)\\
&=&\sum_{Q\in\mathcal{S}}\left(\frac{u_{\vec w}(Q)}{|Q|}\right)^{(1-\frac{\alpha}{n})p_1'}
\prod_{i=1}^m\left(\frac{\sigma_i(Q)}{|Q|}\right)^{\frac{(1-\alpha/n)qp_1'}{p_i'}}
  |Q|^{q(m-\frac{\alpha}{n})((1-\frac{\alpha}{n})p_1'-1)}\\
&&\times\prod_{i=2}^m \sigma_i(Q)^{q-\frac{qp_1'}{p_i'}(1-\frac{\alpha}{n})}
    u_{\vec w}(Q)^{1-(1-\alpha/n)p_1'}\left(\frac{\sigma_1(Q)^{\frac{\alpha}{n}}}{\sigma_1(Q)}
       \int_Q |f_1|\sigma_1 dy_1 \right)^{q}\\
&&\times\prod_{i=2}^m \left(\frac{1}{\sigma_i(Q)}\int_Q |f_i|\sigma_i dy_i \right)^{q}\\
&\lesssim&[\vec{w}]_{A_{\vec{P},q}}^{(1-\alpha/n)p_1'}\sum_{Q\in\mathcal{S}}
  \left(\frac{\sigma_1(Q)^{\frac{\alpha}{n}}}{\sigma_1(Q)}\int_Q |f_1|\sigma_1 dy_1
    \right)^{q}\prod_{i=2}^m \left(\frac{1}{\sigma_i(Q)}\int_Q |f_i|\sigma_i dy_i
       \right)^{q}\\
&&\times|E(Q)|^{q(m-\frac{\alpha}{n})((1-\frac{\alpha}{n})p_1'-1)}
   u_{\vec{w}}(Q)^{1-(1-\alpha/n)p_1'}
\prod_{i=2}^m \sigma_i(Q)^{q-\frac{qp_1'}{p_i'}(1-\frac{\alpha}{n})}\\
&\le& [\vec{w}]_{A_{\vec{P},q}}^{(1-\alpha/n)p_1'}\sum_{Q\in\mathcal{S}}
  \left(\frac{\sigma_1(Q)^{\frac{\alpha}{n}}}{\sigma_1(Q)}\int_Q |f_1|
    \sigma_1 dy_1 \right)^{q}\prod_{i=2}^m \left(\frac{1}{\sigma_i(Q)}
      \int_Q |f_i|\sigma_i dy_i \right)^{q}\\
&&\times\sigma_1(E(Q))^{q((1-\frac{\alpha}{n})-\frac{1}{p_1'})}\prod_{i=2}^m
  \sigma_i(E(Q))^{q/{p_i}},
\end{eqnarray*}
where we use (\ref{eq:e4}) and the fact that $p_1'(1-\alpha/n) \ge \max_{i\neq 1}\{p_i'\}$
in the last step.

Let $1/q_1={1}/{p_1}-{\alpha}/{n}$.
By H\"older's inequality, we have
\begin{eqnarray*}
&&\|\mathcal{I}_{\alpha,q}^{\mathcal{S}}(|f_1|\sigma_1,\cdots,|f_m|\sigma_m)
   \|_{L^q(u_{\vec w})}^q\\
&\le& [\vec{w}]_{A_{\vec{P},q}}^{(1-\alpha/n)p_1'}
  \left(\sum_{Q\in\mathcal{S}}\left(\frac{\sigma_1(Q)^{\alpha/n}}{\sigma_1(Q)}
     \int_Q |f_1|\sigma_1 dy_1 \right)^{q_1}\sigma_1(E(Q))\right)^{q/{q_1}}\\
&&\times\prod_{i=2}^m\left(\sum_{Q\in\mathcal{S}}\left(\frac{1}{\sigma_i(Q)}
  \int_Q |f_i|\sigma_i dy_i \right)^{p_i}\sigma_i(E(Q))\right)^{q/{p_i}}\\
&\le&[\vec{w}]_{A_{\vec{P},q}}^{(1-\alpha/n)p_1'}\|M_{\alpha,\sigma_1}^{\mathscr{D}}
   f_1\|_{L^{q_1}(\sigma)}^q \prod_{i=2}^m
     \|M_{\sigma_i}^{\mathscr{D}}f_i\|_{L^{p_i}(\sigma_i)}^q
 \\
&\lesssim&[\vec{w}]_{A_{\vec{P},q}}^{(1-\alpha/n)p_1'}\prod_{i=1}^m
   \|f_i\|_{L^{p_i}(\sigma_i)}^{q},
\end{eqnarray*}
where $p_1'(1-\frac \alpha n)>1$ ensure that $p_1<\frac n \alpha$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:m1}]
By Proposition~\ref{prop:p1}, we have
\[
  \mathcal{M}_{\alpha}(f_1,\cdots, f_m)(x)
  \le C_{m,n}\sum_{t\in\{0,1/3\}^n}\mathcal{M}_{\alpha}^{\mathscr{D}_t}
    (f_1,\cdots,f_m)(x).
\]
So it suffices to prove the desired conclusion for $\mathcal{M}_{\alpha}^{\mathscr{D}}$.
Let $a=2^{(m-\alpha/n)(n+1)}$  and $\Omega_k=\{x\in\bbR^n:
 \mathcal{M}_{\alpha}^{\mathscr{D}}(f_1,\cdots,f_m)(x)>a^k\}$. Suppose that
$\Omega_k=\bigcup_j Q_j^k$, where $Q_j^k$ are pairwise disjoint maximal dyadic cubes
 in $\Omega_k$. Then $\mathcal S:=\{Q_j^k\}$ is a sparse family.
To see this, it suffices to prove that
\[
  | Q_j^k\bigcap \Omega_{k+1}| \le \frac{1}{2} |Q_j^k|.
\]
In fact, by the definition of $Q_j^k$, we have
\[
  a^k < \prod_{i=1}^m \frac{1}{|Q_j^k|^{1-\alpha/mn}} \int_{Q_j^k} |f_i|
    \le 2^{mn-\alpha} a^k.
\]
It follows that
\begin{eqnarray*}
  | Q_j^k\bigcap \Omega_{k+1}|
&=& \sum_{Q_l^{k+1}\subset Q_j^k} |Q_l^{k+1}| \\
&\le& \sum_{Q_l^{k+1}\subset Q_j^k}  a^{-(k+1)/(m-\alpha/n)} \prod_{i=1}^m
   \Big(\int_{Q_l^{k+1}} f_i\Big)^{1/(m-\alpha/n)} \\
&\le& a^{-(k+1)/(m-\alpha/n)}\prod_{i=1}^m
\!\left(\sum_{Q_l^{k+1}\subset Q_j^k}
   \!\!\!\Big(\int_{Q_l^{k+1}} f_i\Big)^{1/(1-\alpha/nm)}\!\right)^{1/m}\! \\
&\le& a^{-(k+1)/(m-\alpha/n)}\prod_{i=1}^m
\left(  \int_{Q_j^k} f_i\right)^{1/(m-\alpha/n)} \\
&\le&
  a^{-1/(m-\alpha/n)} 2^n  |Q_j^k|
   \\
&=& \frac{1}{2} |Q_j^k|.
\end{eqnarray*}
Hence $\mathcal S:=\{Q_j^k\}$ is a sparse family.
Therefore,
\begin{eqnarray}
\label{eq:e1}&&\int_{\bbR^n}\mathcal{M}_{\alpha}^{\mathscr{D}}(f_1,\cdots,f_m)^q
  u_{\vec w}dx\\
&=& \sum_k \int_{\Omega_k\setminus \Omega_{k+1}}\mathcal{M}_{\alpha}^{\mathscr{D}}
  (f_1,\cdots,f_m)^q u_{\vec w}dx\nonumber\\
&\le& a^q\sum_{k,j}\left(\prod_{i=1}^m\frac{1}{|Q_j^k|^{1-\alpha/{mn}}}\int_{Q_j^k}
  |f_i(y_i)| dy_i\right)^q u_{\vec w}(E(Q_j^k))\nonumber\\
&\lesssim& \int_{\bbR^n}(\mathcal{I}_{\alpha,q}^{\mathcal{S}})(f_1,\cdots,
    f_m)^q u_{\vec w}dx\nonumber
\end{eqnarray}
Now the desired conclusion follows from Lemma~\ref{lm:lq}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:m2}]
There are two cases.

(i).\,\, $q> 1$.  By (\ref{eq:q Gt 1}),
it suffices to prove that
\[
  \|\mathcal{I}_\alpha^{\mathcal{S}}(\vec{f})\|_{L^q(u_{\vec w})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)\max_i\{1,\frac{p_i'}{q}\}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\]
If $q(1-\alpha/n)\ge max\{p_i'\}$, the desired conclusion follows from Lemma~\ref{lm:l1}. If $p_j'(1-\frac{\alpha}{n})\ge \max_{i\neq j}\{p_i', q\}$,
without loss of generality, assume that $p_1'(1-\frac{\alpha}{n})\ge \max_{i\neq 1}\{p_i', q\}$.
By duality, we have
\begin{eqnarray*}
&& \|\mathcal{I}_\alpha^{\mathcal{S}}\|_{L^{p_1}(w_1^{p_1})\times\cdots\times L^{p_m}(w_m^{p_m})\rightarrow L^q(u_{\vec w})}\\
  &=&\|\mathcal{I}_\alpha^{\mathcal{S}}\|_{L^{q'}(\Pi_{i=1}^m w_i^{-q'})\times L^{p_2}(w_2^{p_2})\times\cdots
  \times L^{p_m}(w_m^{p_m})\rightarrow L^{p_1'}( w_1^{-p_1'})}\\
  &\lesssim&[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac{\alpha}{n})\frac{p_1'}{q}},
\end{eqnarray*}
where we use Lemma~\ref{Lm:dual} in the last step.

(ii).\, $q\le1$. In this case,
\[
   \min_j \{\frac{\max_{i\neq j}\{p_i'\}}{p_j'}\}
 = \min\left\{\frac{\max_i{p_i'}}{q},
   \min_j\frac{\max_{i\neq j}\{p_i', q\}}{p_j'}\right\}
  \le 1-\frac{\alpha}{n}.
\]
By (\ref{eq:q Lt 1}),
it suffices to prove that
\[
  \|\mathcal{I}_{\alpha,q}^{\mathcal{S}}(\vec{f})\|_{L^q(u_{\vec w})}\le
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)\max_i\{1,\frac{p_i'}{q}\}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\]
By Lemma~\ref{lm:lq}, we get
\begin{eqnarray*}
\|\mathcal{I}_{\alpha,q}^{\mathcal{S}}(\vec{f})\|_{L^q(u_{\vec w})}&\le&
  C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)\max_i\{\frac{p_i'}{q}\}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}\\
  &=&C_{m,n,\vec{P},q}[\vec{w}]_{A_{\vec{P},q}}^{(1-\frac\alpha n)\max_i\{1,\frac{p_i'}{q}\}}\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\end{eqnarray*}
This completes the proof.
\end{proof}

\section{Proof of Theorem~\ref{thm:m3}}
First, we introduce the sharp reverse H\"older's property of $A_\infty$ weights which was proved in \cite{HP} and \cite[Theorem 2.3]{HPR}. Recall that
\[
  [w]_{A_\infty}:=\sup_Q \frac{1}{w(Q)}\int_Q M(w\chi_Q).
\]
\begin{Proposition}[{\cite[Theorem 2.3]{HP}}]\label{prop:reverseH}
Let $w\in A_\infty$. Then
\begin{equation}\label{eq:e2}
\left(\frac{1}{|Q|}\int_Q w^{r(w)}\right)^{1/{r(w)}}\le 2\frac{1}{|Q|}\int_Q w,
\end{equation}
where $r(w)=1+\frac{1}{\tau_n [w]_{A_\infty}}$ and $\tau_n=2^{11+n}$. Notice that the conjugate $r(w)'\simeq [w]_{A_\infty}$.
\end{Proposition}
We also need the following characterization of $A_{\vec{P},q}$ weights.
\begin{Proposition}[{\cite[Theorem 3.4]{M}}]
Suppose that $1<p_1,\cdots,p_m<\infty$, and $\vec{w}\in A_{\vec{P},q}$. Then
\[
  u_{\vec w}\in A_{mq}\quad\mbox{and}\quad \sigma_i\in A_{mp_i'}.
\]
\end{Proposition}
\begin{proof}[Proof of Theorem~\ref{thm:m3}]
 Set $\alpha_i=(p_i'r_i)'$,
where $r_i$ is the exponent in the sharp reverse H\"older's inequality (\ref{eq:e2})
for the weights $\sigma_i$ which are in $A_\infty$ for $i = 1$, $\cdots$, $m$.
By (\ref{eq:e1}), we have
\begin{eqnarray*}
&&\int_{\bbR^n}\mathcal{M}_{\alpha}^{\mathscr{D}}(f_1,\cdots,f_m)^q
  u_{\vec w}dx\\
&\le&\sum_{k,j}\left(\prod_{i=1}^m\frac{1}{|Q_j^k|^{1-\alpha/{mn}}}
  \int_{Q_j^k} |f_i(y_i)| dy_i\right)^q u_{\vec w}(E(Q_j^k))\\
&\le&\sum_{k,j}\frac{u_{\vec w}(E(Q_j^k))}{|Q_j^k|^{-\alpha q/n}}
  \prod_{i=1}^m \left(\frac{1}{|Q_j^k|}\int_{Q_j^k}|f_i|^{\alpha_i}
     w_i^{\alpha_i}\right)^{\frac{q}{\alpha_i}}
\cdot\left(\frac{1}{|Q_j^k|}\int_{Q_j^k}w_i^{-\alpha_i'}
  \right)^{\frac{q}{\alpha_i'}}\\
&\le&\sum_{k,j}\frac{u_{\vec w}(E(Q_j^k))}{|Q_j^k|^{-\alpha q/n}}
   \prod_{i=1}^m \left(\frac{1}{|Q_j^k|}\int_{Q_j^k}|f_i|^{\alpha_i}
      w_i^{\alpha_i}\right)^{\frac{q}{\alpha_i}}
   \cdot \left(\frac{2}{|Q_j^k|}\int_{Q_j^k}\sigma_i\right)^{\frac{q}{p_i'}}\\
&&\hskip 70mm \mbox{(by Proposition~\ref{prop:reverseH})}    \\
&\le&C[\vec{w}]_{A_{\vec{P},q}}\sum_{k,j}|E(Q_j^k)|\prod_{i=1}^m
  \left(\frac{|Q_j^k|^{\frac{\alpha p \alpha_i}{n p_i}}}{|Q_j^k|}
    \int_{Q_j^k}|f_i|^{\alpha_i}w_i^{\alpha_i}\right)^{\frac{q}{\alpha_i}}\\
&\le&C[\vec{w}]_{A_{\vec{P},q}}\int_{\bbR^n}\prod_{i=1}^m
   M_{\frac{\alpha p \alpha_i}{p_i}}^{\mathscr{D}}
     (|f_i|^{\alpha_i} w_i^{\alpha_i})^{q/{\alpha_i}}dx\\
&\le&C[\vec{w}]_{A_{\vec{P},q}}\prod_{i=1}^m\left(\int_{\bbR^n}
   M_{\frac{\alpha p \alpha_i}{p_i}}^{\mathscr{D}}(|f_i|^{\alpha_i}
      w_i^{\alpha_i})^{q_i/{\alpha_i}}dx\right)^{q/{q_i}},
\end{eqnarray*}
where
\[
  \frac{1}{q_i}=\frac{1}{p_i}-\frac{\alpha p}{n p_i},\quad i=1,\cdots,m.
\]
Substituting $ \alpha p \alpha_i/p_i$ for $\alpha$ and letting $w=1$ in
Proposition~\ref{prop:p2}, we get
\begin{eqnarray*}
&&\int_{\bbR^n}\mathcal{M}_{\alpha}^{\mathscr{D}}(f_1,\cdots,f_m)^q
  u_{\vec{w}}dx\\
&\le& C[\vec{w}]_{A_{\vec{P},q}}\prod_{i=1}^m
  \left(1+\frac{(p_i/{\alpha_i})'}{(q_i/{\alpha_i})}
  \right)^{(1-\frac{\alpha p\alpha_i}{n p_i})
    \frac{q}{\alpha_i}}\bigg\||f_i|^{\alpha_i} w_i^{\alpha_i}
       \bigg\|_{L^{\frac{p_i}{\alpha_i}}(\bbR^n)}^{\frac{q}{\alpha_i}}\\
&=& C[\vec{w}]_{A_{\vec{P},q}}\prod_{i=1}^m \left(1+\frac{(p_i/{\alpha_i})'}
  {(q_i/{\alpha_i})}\right)^{(1-\frac{\alpha p\alpha_i}{n p_i})
    \frac{q}{\alpha_i}}\|f_i\|_{L^{p_i}(w_i^{p_i})}^q.
\end{eqnarray*}
It is easy to check that $\frac{p_i}{\alpha_i}-1\simeq [\sigma_i]_{A_\infty}^{-1}$.
Therefore
\[
  1+\frac{(p_i/{\alpha_i})'}{(q_i/{\alpha_i})}\lesssim 1+\frac{p_i}{q_i}[\sigma_i]_{A_\infty}\lesssim [\sigma_i]_{A_\infty}
\]
and
\begin{eqnarray*}
  (1-\frac{\alpha p\alpha_i}{n p_i})\frac{q}{\alpha_i}
&=&\frac{q}{p_i}-\frac{\alpha pq}{np_i}+\frac{q}{p_i}(\frac{p_i}{\alpha_i}-1) \\
&\le&
  \frac{q}{p_i} \left(1-\frac{\alpha p}{n}\right)
  +  \frac{q}{p_i}\cdot C [\sigma_i]_{A_\infty}^{-1}.
\end{eqnarray*}
Consequently,
\[
  \left(1+\frac{(p_i/{\alpha_i})'}{(q_i/{\alpha_i})}\right)^{(1-\frac{\alpha p\alpha_i}{n p_i})\frac{q}{\alpha_i}}
  \lesssim [\sigma_i]_{A_\infty}^{\frac{q}{p_i}(1-\frac{\alpha p}{n})}.
\]
Now we get
\[
  \|\mathcal{M}_{\alpha}(\vec{f})\|_{L^q(u_{\vec{w}})}\le
  C_{m,n,\vec{P},q}\left([\vec{w}]_{A_{\vec{P},q}}^{\frac{1}{q}}
  \prod_{i=1}^m [\sigma_i]_{A_\infty}^{\frac{1}{p_i}(1-\frac{\alpha p}{n})}\right)
  \prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\]
\end{proof}

\section{Proof of Theorem \ref{thm:m4}}
To prove Theorem we will need the following Lemma about Carleson sequences (see \cite[Theorem 4.5]{HP} and \cite[Lemma 5.3]{CM2}).

\begin{Lemma} \label{lem:carleson}Suppose $a=\{a_Q\}_{Q\in \mathscr D}$ and $c=\{c_Q\}_{Q\in \mathscr D}$ are sequences with $c_Q$ nonnegative, and $\mu$ is a positive Borel measure.  Set $M^{\mathscr D} a(x)=\sup_{Q\ni x} |a_Q|$ and
\begin{equation}\label{carlconst}\mathscr C(c)=\sup_{R\in \mathscr D} \frac{1}{\mu(R)}\sum_{Q\subseteq R} c_Q.\end{equation}
If $r>0$ and $\mathscr C(c)<\infty$, then
$$\sum_{Q\in \mathscr D} |a_Q|^rc_Q\leq \mathscr C(c) \int_{\mathbb R^n} (M^{\mathscr D} a)^r\,d\mu.$$
\end{Lemma}

If $\mathscr C(c)<\infty$ in \eqref{carlconst} we say that $c=\{c_Q\}$ is a Carleson sequence with respect to $\mu$.   The constant $\mathscr C(c)$ is called the Carleson constant.  We are now ready to prove Theorem \ref{thm:m4}.

\begin{proof} Using the same decomposition as in the proof of Theorem \ref{thm:m3} we arrive at
\begin{align*}
\lefteqn{\left(\int_{\bbR^n}\mathcal{M}_{\alpha}^{\mathscr{D}}(f_1,\cdots,f_m)^q
  u\,dx\right)^{1/q}}\\
&\le\left(\sum_{k,j}\left(|Q_{j^k}|^{\frac{\alpha}{n}}\prod_{i=1}^m\frac{1}{|Q_j^k|}
  \int_{Q_j^k} |f_i(y_i)| dy_i\right)^q u(E(Q_j^k))\right)^{1/q}\\
  &\le [u,\vec w]_{A_{\vec{P},q}}\left(\sum_{k,j}\left(\prod_{i=1}^m
  \int_{Q_j^k} |f_i(y_i)| dy_i\,\sigma_i(Q_j^k)^{-\frac1{p_i'}}\right)^q\right)^{1/q}.
\end{align*}
Let $q_i=\frac{qp_i}{p}\geq p_i$ and note that $\frac{1}{q}=\frac{1}{q_1}+\cdots+\frac{1}{q_m}$ so we may use a discrete H\"older's inequality to obtain:
\begin{multline*}
\left(\sum_{k,j}\left(\prod_{i=1}^m
  \int_{Q_j^k} |f_i(y_i)| dy_i\,\sigma_i(Q_j^k)^{-\frac1{p_i'}}\right)^q\right)^{1/q}\\
 \leq \prod_{i=1}^m \left(\sum_{k,j}
  \left(\int_{Q_j^k} |f_i(y_i)| dy_i\,\sigma_i(Q_j^k)^{-\frac1{p_i'}}\right)^{q_i}\right)^{1/q_i}.
\end{multline*}
Next define $\beta_i$ by
$$\frac{\beta_i}{n}=\frac{1}{p_i}-\frac{1}{q_i}$$
and note that
$$
  \left(\int_{Q_j^k} |f_i(y_i)| dy_i\,\sigma_i(Q_j^k)^{-\frac1{p_i'}}\right)^{q_i}= \left(\frac{1}{\sigma_i(Q_i^k)^{1-\frac{\beta_i}{n}}}\int_{Q_j^k} |f_i(y_i)| dy_i\,\right)^{q_i}\sigma_i(Q_j^k).$$
We let
$$a_Q=\frac{1}{\sigma_i(Q)^{1-\frac{\beta_i}{n}}}\int_{Q} |f_i(y_i)| dy_i$$
if $Q=Q_j^k$ for some $j,k$ and $a_Q=0$ otherwise.  Likewise let $c_Q=\sigma_i(Q)$ if $Q=Q_j^k$ and $c_Q=0$ otherwise.  Further, notice that $\{c_Q\}$ is a Carleson sequence with respect to $\sigma_i$ with constant $\mathscr C(c)\leq [\sigma_i]_{A_\infty}$.  Indeed,
\begin{align*}
\sum_{Q\subset R} c_Q= \sum_{j,k: Q^k_j\subset R} \sigma_i(Q_j^k)&=\sum_{j,k: Q^k_j\subset R} \sigma_i(Q_j^k)\\
&\lesssim\sum_{j,k: Q^k_j\subset R} \frac{\sigma_i(Q_j^k)}{|Q_j^k|}|E(Q_j^k)|\\
&\leq\int_R M(\chi_R\sigma_i)\,dx\\
&\leq [\sigma_i]_{A_\infty} \sigma_i(R).
\end{align*}
Moreover
$$M^{\mathscr D} a(x)\leq M^{\mathscr D}_{\beta_i,\sigma_i}(f\sigma_i^{-1})(x)$$
and by Proposition \ref{prop:p2} we have
$$\|M^{\mathscr D}_{\beta_i,\sigma_i}(f\sigma_i^{-1})\|_{L^{q_i}(\sigma_i)}\lesssim \|f\sigma_i^{-1}\|_{L^{p_i}(\sigma_i)}=\|f\|_{L^{p_i}(w_i^{p_i})}.$$
Combining everything, we have
\begin{multline*}
{\left(\int_{\bbR^n}\mathcal{M}_{\alpha}^{\mathscr{D}}(f_1,\cdots,f_m)^q
  u\,dx\right)^{1/q}}\lesssim\\
 [u,\vec{w}]_{A_{\vec P,q}}\prod_{i=1}^m [\sigma_i]_{A_\infty}^{1/q_i}  \|f_i\|_{L^{p_i}(w_i^{p_i})}= \Big([u,\vec{w}]_{A_{\vec P,q}}\prod_{i=1}^m [\sigma_i]_{A_\infty}^{\frac{p}{qp_i}} \Big) \prod_{i=1}^m\|f_i\|_{L^{p_i}(w_i^{p_i})}.
\end{multline*}
\end{proof}

\section{Examples}
Finally, we end with some examples to show that our bounds are sharp.  First we show that Theorem~\ref{thm:m1} is sharp.  Consider the case $m=2$ (we leave it to the reader to modify the example for $m>2$) and suppose that
\begin{equation}\label{sharp}
\|\mathcal{M}_\alpha\|_{L^{p_1}(w_1^{p_1})\times L^{p_2}(w_2^{p_2}) \rightarrow L^q(w_1^q w_2^q)}\lesssim [\vec{w}]_{A_{\vec{P},q}}^{r(1-\frac{\alpha}{n})\max(\frac{p_1'}{q},\frac{p_2'}{q})}
\end{equation}
for some $r<1$.  Further suppose that $p_1'\geq p_2'$.  For $0<\ep<1$, let $f_1(x)=|x|^{\ep-n}\chi^{}_{B(0,1)}(x)$, $f_2(x)=|x|^{\frac{\ep-n}{p_2}}\chi_{B(0,1)}(x)$, $w_1(x)=|x|^{(n-\ep)/{p_1'}}$ and $w_2(x)=1$.  Calculations show that
$\|f_1\|_{L^{p_1}(w_1^{p_1})}\simeq\ep^{-{1}/{p_1}}, \|f_2\|_{L^{p_2}(w_2^{p_2})}\simeq\ep^{-1/p_2},$
and
$[\vec{w}]_{A_{\vec{P},q}}\simeq\ep^{-{q}/{p_1'}}.$
For $x\in B(0,1)$ we have
\begin{eqnarray*}
\mathcal{M}_\alpha(f_1,f_2)(x)&\gtrsim&\frac{1}{|x|^{n-\frac{\alpha}{2}}}\int_{B(0,|x|)} |y_1|^{\ep-n}\,dy_1 \cdot \frac{1}{|x|^{n-\frac{\alpha}{2}}}\int_{B(0,|x|)} |y_2|^{\frac{\ep-n}{p_2}}\,dy_2\\
&\gtrsim&\frac{1}{\ep}|x|^{\ep-n+\frac{\ep-n}{p_2}+\alpha}.
\end{eqnarray*}
Hence,
\begin{eqnarray}\label{eq:e3}
\|\mathcal{M}_\alpha(f_1,f_2)\|_{L^q(w_1^{q}w_2^q)}&\gtrsim &\frac{1}{\ep}\Big(\int_{B(0,1)}|x|^{(\ep-n)(q+\frac{q}{p_2}-\frac{q}{p_1'})+\alpha q}\,dx\Big)^{1/q}\\
&\simeq&\frac1\ep\Big(\int_0^1x^{(1+\frac{\alpha q}{n})\ep-1}\,dx\Big)^{1/q}\nonumber\\
&\simeq&\frac{1}{\ep}\Big(\frac{1}{\ep}\Big)^{1/q}.\nonumber
\end{eqnarray}
Combining this with inequality (\ref{sharp}) we see for some $r<1$,
$\Big(\frac1\ep\Big)^{1+\frac1q}\lesssim \Big(\frac1\ep\Big)^{r(1-\frac\alpha n)+\frac1p},$
which is impossible as $\ep\rightarrow 0$.

Next we show that Theorem~\ref{thm:m2} is sharp.
It is easy to notice that $\mathcal{M}_\alpha\le C_{m,n,\alpha} \mathcal{I}_\alpha$. If
  $\max_i p_i'\ge q$, then using the same $f_i$ and $w_i$ as above we get (\ref{eq:e3}) with $\mathcal{M}_\alpha$
replaced by $\mathcal{I}_\alpha$, showing the sharpness. On the other hand, if $\max_i p_i'< q$, the sharpness
follows from the standard duality argument used in the proof of Theorem~\ref{thm:m2}.

Finally we show that Theorem~\ref{thm:m3} is sharp.
For $0<\ep<1$, let
\[
  w_i(x)=|x|^{(n-\ep)/{p_i'}}\quad\mbox{and}\quad f_i(x)=|x|^{\ep-n}\chi^{}_{B(0,1)}(x), \quad i=1,\cdots,m.
\]
Then $u_{\vec{w}}=|x|^{(n-\ep)(m-\frac 1 p)q}$ and it is easy to check that
\[
  [\vec{w}]_{A_{\vec{p},q}}\simeq \left(\frac{1}{\ep}\right)^{q(m-\frac{1}{p})},\quad [\sigma_i]_{A_\infty}\lesssim \frac{1}{\ep}
  \quad\mbox{and}\quad \prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}\simeq \left(\frac{1}{\ep}\right)^{1/p}.
\]
For $x\in B(0,1)$, we have
\begin{eqnarray*}
\mathcal{M}_\alpha(\vec{f})(x)&\gtrsim&\prod_{i=1}^m\frac{1}{|x|^{n-\frac{\alpha}{m}}}\int_{B(0,|x|)} |y_i|^{\ep-n}\,dy_i\\
&\gtrsim&\left(\frac{1}{\ep}\right)^m |x|^{m(\ep-n)+\alpha}.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
\|\mathcal{M}_\alpha(\vec{f})\|_{L^q(u_{\vec{w}})}&\gtrsim& \left(\frac{1}{\ep}\right)^m \left(\int_{B(0,1)}|x|^{mq(\ep-n)+\alpha q +(n-\ep)(m-\frac{1}{p})q}dx\right)^{\frac{1}{q}}\\
&\simeq & \left(\frac{1}{\ep}\right)^m \left(\int_0^1 t^{(\alpha+1)\ep -1}dt\right)^{\frac{1}{q}}\\
&\simeq& \left(\frac{1}{\ep}\right)^{m+\frac{1}{q}}\\
&\gtrsim&\left([\vec{w}]_{A_{\vec{P},q}}^{\frac{1}{q}}\prod_{i=1}^m [\sigma_i]_{A_\infty}^{\frac{1}{p_i}(1-\frac{\alpha p}{n})}\right)\prod_{i=1}^m \|f_i\|_{L^{p_i}(w_i^{p_i})}.
\end{eqnarray*}

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text:
James Washburn
8 June 1939 - 21 December 2020
Under: Obituary
James "Jim" Washburn passed away on December 21, 2020, after a valiant 7 year battle with melanoma.
Jim was born on June 8, 1939, in Longmont, Colorado, but spent the majority of his first 8 years in Iowa, including a year and a half at a Catholic boarding school. When his mother remarried, the family moved west to California and then to Portland, Oregon, where the family settled. Jim attended 3 elementary schools in the Grant district and graduated from Grant High in 1957.
The first in his family to attend college, Jim worked at Boeing and surveying to put himself through college. He attended PSU for 2 years, before transferring to the University of Oregon for another 2 years, where he became a member of Beta Theta Phi.
Jim was a lifelong adventurer and, after 4 years of college, worked for a year on Grand Bahama Island life-guarding and instructing scuba. He then traveled to Europe and then spent another year working and traveling through Europe by Vespa.
Returning to the US, Jim worked in Montana on the Minuteman Missile silos before returning to Oregon to reestablish residence and complete his Bachelor's degree at OSU. He then continued his education with graduate work in oceanography. While working at OSU in 1968, Jim met Kathy Steinbach, who happened to live in the same apartment complex. They married in 1971 and shared life's adventures for next 49 years. In 1980, their lives were enriched by the birth of their son, Doug.
Jim spent the majority of his career working for OSU's Department of Ocean Engineering as a research associate and as the university's first diving officer. Jim participated in many long cruises on the OSU research vessel, the Yaquina, being responsible for the hydrographic data. He made numerous dives on the TOTEM project to moor instruments designed to monitor oceanographic and meteorological conditions. He participated in oceanographic experiments in Tahiti, New Caledonia, and the Bahamas. He was in charge of hydrographic data on cruises to Costa Rica and Peru, which provided him with the opportunity to dive in the Galapagos before the tourist boom! He also set up experiments in the OSU wave tank to study the effects of storm and tsunamis.
As a diving officer, Jim worked with as many as 500 divers on university related projects and estimated he made around 150-200 dives a year. Jim had the opportunity to dive around the world-from the Arctic to South America and from the Persian Gulf to Polynesia. One of the nicknames he was given as a diver was "Iceman," for calmly playing Tic-Tac-Toe with a fellow diver while being circled by a great white shark.
On leave from the university, Jim worked as a diving supervisor on the Exxon Valdez oil spill, spent two summers instructing scuba at ARAMCO and worked as a consultant to the EPA on a sewage dumping project which allowed him the opportunity to spend a week living in an underwater habitat.
Jim enjoyed ski trips and traveling with his family to Hawaii, Mexico, Europe, the Caribbean, Canary Islands, and French Polynesia. Jim was an avid reader, enjoyed coffee chats with friends and cheering on the Beavers He was thrilled sharing his diving passion with his son, coaching his son's soccer and basketball teams, and visiting in Sweden to spend time with his son and other family. He appreciated the many wonderful sailing adventures he shared with good friends in Hawaii, Canada, Tahiti, the Bahamas, and Croatia.
Jim was definitely a people person and keenly interested in others. He had an engaging smile, punctuated by dimples, which drew people to him. Jim had an intellectual curiosity, a good sense of humor, was an excellent listener, and loved a good conversation. He felt fortunate to work in a stimulating university environment, live in an enlightened community, and have talented, intelligent, and caring friends and family.
Jim is survived by his wife, Kathy; his son, Doug (Paula); and his two beloved grandchildren: Olivia and William. He is also survived by his three sisters: Judy Brennan, Pat Minor, and Lisa Trickle.
In line with Jim's wishes, a celebration of life will be held once the pandemic has passed. For those interested, donations in Jim's memory can be made to the Union of Concerned Scientists, with hope that science can help our planet be safe and healthy for future generations,
McHenry Funeral Home is handling the arrangements.


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text:
Tata in Trade Secret Trouble, US Jury Slaps $940 Million Fine
A US grand jury has slapped two companies of India's Tata group – Tata Consultancy Services and Tata America International Corporation – with a $940 million fine for stealing software.
The federal grand jury in the US State of Wisconsin ruled that the two companies must pay $240 million in a trade secret lawsuit filed against them by Epic Systems. The company has also been asked to pay another $ 700 million in punitive damages.
The lawsuit was filed in October 2014 in US District Court in Madison.
The lawsuit was amended in January and December 2015, of "brazenly stealing the trade secrets, confidential information, documents and data" belonging to Epic.
In its lawsuit, Epic had said that TCS took that data during consultation for a customer but fraudulently used the software beyond the consulting contract – to improve their own competing product.
One TCS employee's account, which was used in India and several US locations, downloaded 6,477 documents, according to Epic. The lawsuit said:
Rather than compete lawfully with Epic, TCS has engaged in an apparently elaborate campaign of deception to steal documents, confidential information, trade secrets, and other information and data from Epic, for the purpose of realising technical expertise developed by Epic over years of hard work and investment.
Mark Zuckerberg Pays Kochi-Based College Student $700 For Website
Intel May Cut Thousands of Jobs by End of 2016


text:
So Tell Me About the Last Time You Had Sex: Laying Bare and Learning to Repair Our Love Lives
By Ian Kerner PhD
Ian Kerner Grand Central Publishing 9781538734858
04/20/2021 • 1
Format : CD (In Stock)
1 Format: CD
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Renowned sex therapist and New York Times bestselling author Ian Kerner shares the unique and indispensable program he uses to help thousands of couples achieve more intimacy and better sex. Dr. Ian Kerner is a Sherlock Holmes of the bedroom—a sexual detective helping individuals and couples solve the mystery of their sexual distress. His secret weapon? Informed curiosity. Kerner has perfected the art of the "sex script analysis," a method of inquiry to examine your sex life in action moment by moment. In those details—the what, where, when, and why of your last sexual experience—all the clues of what went wrong are laid bare and the mystery of how to create mutual pleasure can be solved. When our sex scripts work, we let go into arousal and lose ourselves in pleasure. But when the sex script fails, it's all we can do not to ruminate over the details. Sometimes you can be lying in bed right next to someone and feel a million miles apart. In those moments of silent desperation, the sex script is our guide to everything happening between the sheets and beneath the surface. With wit and warmth, the nationally recognized sex therapist and author of the smash hit She Comes First uses his tried and true techniques and tools to show readers how to tap into their erotic personalities and realize their sexual potential. He'll help you figure out what's working and what's not in your sex script and discuss many common sexual problems, from low desire and mismatched libidos to male and female sexual function issues, that may be interfering with your sex life. With the help of decades of clinical insight, the latest sexual science and research, valuable homework assignments, and more, this insightful and original book strips away discomfort and gives readers the ability to not just talk about sex openly and knowingly, but to actually do something about it.
Renowned sex therapist and New York Times bestselling author Ian Kerner shares the unique and indispensable program he uses to help thousands of couples achieve more intimacy and better sex.
Dr. Ian Kerner is a Sherlock Holmes of the bedroom—a sexual detective helping individuals and couples solve the mystery of their sexual distress. His secret weapon? Informed curiosity. Kerner has perfected the art of the "sex script analysis," a method of inquiry to examine your sex life in action moment by moment. In those details—the what, where, when, and why of your last sexual experience—all the clues of what went wrong are laid bare and the mystery of how to create mutual pleasure can be solved. When our sex scripts work, we let go into arousal and lose ourselves in pleasure. But when the sex script fails, it's all we can do not to ruminate over the details. Sometimes you can be lying in bed right next to someone and feel a million miles apart. In those moments of silent desperation, the sex script is our guide to everything happening between the sheets and beneath the surface.
With wit and warmth, the nationally recognized sex therapist and author of the smash hit She Comes First uses his tried and true techniques and tools to show readers how to tap into their erotic personalities and realize their sexual potential. He'll help you figure out what's working and what's not in your sex script and discuss many common sexual problems, from low desire and mismatched libidos to male and female sexual function issues, that may be interfering with your sex life. With the help of decades of clinical insight, the latest sexual science and research, valuable homework assignments, and more, this insightful and original book strips away discomfort and gives readers the ability to not just talk about sex openly and knowingly, but to actually do something about it.
Author Bio: Ian Kerner PhD
Ian Kerner is a nationally recognized sex therapist and New York Times bestselling author. He is a regular contributor to NBC's Today show and the CBS Early Show, a contributing editor to Cosmopolitan, and CNN's sex expert. He is also the founder of GoodinBed.com, a new web destination for sex / relationship advice that brings together some of the world's most renowned sex experts. He lives in New York City with his wife and two sons.
Available Formats : CD
Category: Nonfiction/Health & Fitness
CDs: 6
Purchased:
0 copies
AE Catalog ID: N/A


text:
Ever thought of running to Paris?
If you can run a half marathon - you can Run2Paris - achieve the incredible as part of a team. Plus win a space on the 2022 event!
Run2Paris
We see, hear and read incredible stories of unbelievable endurance challenges every day. Events are getting longer, harder, tougher and faster but what about the rest of us? We long for adventures but don't have the time to train, money to spend or knees to face it. Run2Paris is new, it's different and is soon to be top of every runner's bucket list.
This epic 4-day relay event will be the highlight of your year as you travel from London 2 Paris on foot with your best (or soon to be best) friends. Each runner is part of a team of 5 and runs 1 leg, roughly a half marathon distance each day, before passing the baton onto their teammate. The final mile of the event is run together, right up to the Eiffel Tower.
The event exists to raise money to support Parkinson's UK and SSAFA the Armed Forces charity and everyone taking part has a fundraising target of £400. The registration fee includes a return ferry ticket for your team car, camping each night with tents provided, breakfast and dinner each day, incredible event support, brilliant t-shirts, medals and events bling but most importantly a party in Paris to celebrate RUNNING THERE!
A giant road trip, with a running problem, with your best friends. Achieve something you never thought possible. Run2Paris.
Website - www.run2paris.com
Enter the competition here
Win 2 Free Places on the Run2Paris
Answer one simple question to win two free places on the Run2Paris 2022. T&C's below.
What is the name of the famous landmark in Paris where teams finish?
ENTER HERE -->>http://eepurl.com/gEKMv1
Competition closes at midnight on 14th February. A winner will be selected at random from all correct entries on 15th February 2022 and notified within 5 days. They will have two free places for the event and must supply all details by 1st March 2022. Remember to take part you will be part of a team of 5. You can get 3 friends to join you and your pal or we have a team ready and waiting for you to join.
There is a fundraising target of £400pp which is split between Parkinson's UK and SSAFA the Armed Force Charity, £300 must be raised by 26th April 2022. Places must be taken on the 2022 edition of Run2Paris between 26-29 May 2022. You can unsubscribe at any time by clicking the link in the footer of our emails.
This competition is organised by Run2Paris. For information about their privacy practices, please visit our website www.run2paris.com
Outside & Active is the home for those who love the outdoors. Our mission is to inspire, inform and educate people about being active outdoors in a fun, safe and sustainable way. We provide inspiration, kit, tech and advice on adventure, camping, climbing, cycling, hiking, running water and winter.
More articles from Outside & Active
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Páramo extends its instore offering with Elliot Brown Watches
Ethical outdoor clothing brand Páramo has opened its doors to British watch brand Elliot Brown with the introduction of a special edition Mountain Rescue charity watch now sold in all three of its stores covering London, Keswick and Páramo's home town of Wadhurst.
Jamie Ramsay - Save Our Outdoors (Pt.2)
Join Dom on the latest Outside & Active Podcast where he is joined by award-winning Endurance Adventure Athlete, Jamie Ramsay. This is the second of two parts discussing Cotswold Outdoor's #SaveOurOutdoors campaign.
How can I have achieved so much yet feel so empty?!
Previous Outside & Active Podcast guest, Men & Mountains founder, and 'Rethinking Masculinity' author, Dan Stanley, has made the first chapter of his book freely available for Outside & Active readers.
Most recent articles in
A Recipe for Love: Outside and Active Style
Valentine's Day Instructions: Find the ingredients, give them a good mix and sprinkle in your own magic. Contemplate for 24 hours, then wait for the unique feeling of "huggyness" to take hold. Recipe should be taken with a large pinch of salt and a dose of sense of humour.


Kenpo Karate / Keith Curts / Video Library > Thundering Hammers	 Video Length: 1min. 31sec.
Used to defend against a straight or rounded punch. First you step off the line and block followed by a reverse hammer-fist to the groin and then a strike to the kidneys and back of the head.


Typical price (TYP) is technical analysis indicator which averages price for each period. Some technicians use it instead of closing price. Money flow index indicator uses typical price.
Typical price is used by some technicians instead of closing price.
Open chart for google with typical price indicator.


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Sanskriti, CRPF school may reserve 60% nursery seats under public servant quota – ( the high court said it is like apartheid and here the AAP champion of the underclass is promoting this elitist segregation amazing )
Posted on Jan 6 2017 - 6:49pm by admin
The Delhi government has proposed a special exemption to let two private schools continue reserving 60% of their nursery seats for children of public servants, suggesting they be kept out of a new rule mandating distance as the only yardstick for admission.
According to sources, the exemption proposal — sent to lieutenant governor Anil Baijjal for approval — has been mooted for Sanskriti School and CRPF Public School, which reserve seats for children of central government employees and CRPF personnel respectively.
"This matter has been discussed but we are yet to get a formal word from the government", said Abha Sehgal, principal, Sanskriti School, which is run by a society formed by the wives of the civil servants.
Sanskriti and CRPF Public schools are among 298 institutions running on Delhi Development Authority land. They will need to prioritise nursery admissions on the basis of how close a child lives.
Guidelines for the process are being drawn up, and admissions to these schools have been embargoed till the process is complete.
According to the proposal, however, Sanskriti and CRPF Public schools will need to follow the distance criteria for admission to the 15 general category seats.
The matter was first raised in a meeting between the government and school heads on December 5 and a proposal was sent to the then L-G, Najeeb Jung, on December 20.
"Schools set up for specific category establishments such as defence forces/ paramilitary forces etc shall have the right to reserve seats for admissions for the wards of specific government servants. The extent of seats reserved shall be notified in the notice boards and website," the minutes of the meeting said. "The remaining seats, if any, shall be filled on the basis of neighbourhood limit fixed by the school".
It was discussed again on Wednesday when the directorate of education met representatives of eight schools to discuss their concerns about the new policy.
"The file is with the new L-G now. We have also received a query on this from the L-G office," said a government official.
In June 2015, the Delhi High Court quashed Sanskriti school's 60% reservation for children of civil servants. The school moved the Supreme Court and as an interim measure, the court allowed the reservation to continue.
"The entire purpose of establishing a school will be defeated if we have to conduct admission under the neighbourhood/distance criteria. The school was established for CRPF cadets and if the government enforces this rule, they will lose out," Meenu Tiwari, principal, CRPF Public School, said
http://www.hindustantimes.com/delhi/sanskriti-crpf-school-may-reserve-60-nursery-seats-under-public-servant-quota/story-uFd4gLqeAbgl7XfpJk1YKN.html


Nobody Counts The Number Of Ads You Run They Just Remember The Impression You Make.
There is no denying it- Facebook plays a big part of billions of our lives, and the channel is continuously improving… Click here to read on.
Facebook Advertising has quickly become a mainstream marketing method businesses use to effectively reach their target audience and brand goals. Looking to grow your customer base through Facebook Ads? Click here for 6 proven Facebook advertising offers you can use to attract more business.
Facebook Advertising, How Does It Work?
For your Facebook ads to succeed, first you need to grab your audiences attention- but how do you create a hook which works? Read this article.
On the 30th of November, it was confirmed that Facebook had entirely stopped the advertising tool giving advertisers the ability to refine their targeting through "multicultural affinity segments". Click here to read on.
Facebook's Advertising campaigns go way beyond the simple "boosting" of a post. This strategy is widely used by marketing genius' to help businesses achieve and exceed their goals and get their message across to the right target audience. Click here to learn how to make money with Facebook Ads.
Facebook is starting to take a more firm approach with its ad policies and principles, as it aims to make users clear on how the platform functions in the marketplace. Click here to read on.
Ever heard of Messenger Broadcast? Well, it'll be everywhere in the marketing world very soon… Messenger Broadcast, Facebook's new marketing interface, will allow businesses to send custom advertisement messages to users! Click here to read on.


Barbie and Ken were two of the "actors" in the third computer animated film Toy Story (2010). Accompanying to the film Mattel issued some Toy Story dolls. At the New York Toy Fair Mattel presented a Barbie/Ken Gift set, titled "made for each other". The Set included a modern version of the Great Shape Barbie and the Animal Lovin Ken. Ken had got a brand new facemold.
But later Mattel had the idea, that Ken should look more alike the computer animated doll in the film. So a second Toy Story Ken was issued.
The following picture shows the original Animal Lovin Ken (left), the first Toy Story Ken, who was presented on the Toy Fair (middle) and the second Tory Story Ken with the look alike face.
The second face has like the original Aninmal loving Ken (Crystal Facemold) a little dimple in the chin. The right mouth corner is slightly pulled upward, which makes his grin look a little bit silly/naïve. The light points in Ken eyes seems to reflect on photos, which makes them diluted. It looks like as Ken would start crying in the next second. His right eyebrow is also slightly raised.
Both headmolds were only used for the Toy Story Ken.


Bring to a boil. Simmer for 15 minutes. Keep warm.
Strain and serve in a mug with a lemon slice.
Cut the orange peel into strips. Combine orange juice, cranberry juice, and brown sugar in crockpot.
Combine orange peel, cinnamon stick, and cloves in a cheesecloth pouch tied with a string. Place in crockpot. Cover; cook on low for 5-7 hours or on high for 2-3 hours. Remove spice bag.
Buy half-gallon container of Ziegler's spiced apple cider.
Pour spiced cider into microwave-safe mug.
Heat to your preferred temperature for a hot beverage.
Simplicity is one of the highest virtues, Jeff! Thanks for the tip (and the chuckle).
Pour wine in the pitcher and squeeze the juice wedges from the orange,lemon and lime into the wine. Toss in the fruit wedges, and add sugar. Chill overnight. Add ginger ale or club soda just before serving.
DO NOT DRIVE AFTER DRINKING THIS!
This sounds delicious, and less dangerous for the carpet at a party. I am laughing at the idea of white "sangria" though. Is it made only of white blood platelets?
Probably, LOL. Actually in spanish it is called Limoná, but the right translation is white sangria, even though it doesn't make much sense.
Quarter cup of eggnog ..
Mix and enjoy!! Simple as that, but pure heaven, imo!
One of my favorites as well. Bailey's also goes well with hot chocolate.


text:
Hornets boss wants all three points
23rd April 2019, 02:14 by Michael Holland
Watford boss Javi Gracia insists his side need the points just as much as lowly visitors Southampton when the two sides clash tonight.
With an FA Cup final meeting with Manchester City on the horizon and the potential for a seventh-placed finish in the Premier League, Gracia has enjoyed a fine first full season at Vicarage Road.
Southampton, meanwhile, are still not mathematically safe from relegation and are currently five points clear of Cardiff in 18th - despite winning three of their last five matches.
The trip to Hertfordshire is their game in hand and a chance for Ralph Hasenhuttl's side to all-but secure another season in the top-flight, but Gracia will be sending his players out with a clear message.
"I think before playing the game, my players know perfectly that we need the points more than them," he said. "It is the best mentality to get the points available. If we don't play with that feeling it is difficult to get results.
"They really need the points as well. We try to compete as well as possible because I'm sure we will have to if we want the points. All of the teams are honest and always they play very well and it is always difficult to get points in the different circumstances.
"We have to be concentrated until the end, if we don't do it in the next games, I'm sure we won't be able to achieve the results we want."
More Watford
Bachmann keen to fight for Hornets spot
Daniel Bachmann knows he is unlikely to be first-choice goalkeeper at Watford but wants to stay and fight for his place.
Coughlan keen on fresh Jakubiak deal
Watford starlet Alex Jakubiak is wanted by Bristol Rovers boss Graham Coughlan on a fresh season-long loan deal.
Hornets confirm Dalby arrival
Leeds United striker Sam Dalby has switched Yorkshire for Hertfordshire after joining Watford for an undisclosed fee.
Sarr hints at Watford move
Watford target Ismaila Sarr has informed his Senegal team-mates that he wants to move to Vicarage Road, reports claim.
Watford defender set for loan move
Watford defender Marc Navarro is on the verge of joining Spanish outfit Leganes on loan.
Milan may let Watford in to land Uruguayan
AC Milan have been warned to firm up their interest in PSV Eindhoven's Gaston Pereiro as Watford are reportedly set to offer £13m.


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Dacey
Animated Views Staff
Joined: February 8th, 2005
Location: The US of A
Re: The Lego Movie
Post by Dacey » April 6th, 2014, 1:16 pm
Hmmmm. Ben might know everything. But that doesn't mean he's always right.
I thought the ending was brilliant. I was already enjoying the movie up until that point, but that's what officially made it "great" for me. Not only did it fully explain the motives behind the "villain," but it clarified why all of the Lego worlds were existing in "different dimensions."
click to reveal content
The camera may have been focused on Lego City, but you could see that he had various sets set up in "different areas" in the dad's basement. Which is why Emmett and the others weren't allowed to go to different universes, because the dad was just setting them up like store displays, and not letting his son play with them.
That also explained why the son had made Lord Business the "villain" in the world, because he's modeling him more or less after his father. Also, the villain's name, "Business," can be seen as the name the son would give his bad guy, as his father, while clearly well off, is probably at work a lot, and distant from his son as a result. When they were playing together at the end, I took that as the two of them finally connecting with each other.
What I also loved is that Emmett (and the son, likewise) doesn't defeat the "villain" by having to fight him or by telling him that he's wrong. No, he tells the dad (and Lord Business) that what he has done is amazing, and is beautiful in its own right, and he has created something great...even if he was just following the directions. The dad, meanwhile, sees the wonder of what the son has been doing, and is able to join him in his creativity then.
But I love that, telling a "Toy Story" of sorts from the perspective of the toy who's the star of a child's giant, wide-eyed imagination. I love "the great hand" being like a God, and the "chosen one" ultimately being whichever toy figure the child happens to decide he loves playing with. That explains how the Millennium Falcon was just able to show up then. It didn't have to make sense! In a child's mind, he needed the spaceship for his heroes' then, so that's when it showed up in his story.
To me, The Lego Movie captured what playtime as a kid really meant. The fun, the wonder, and, yes, the sheer insanity. So while I can understand while some may not have liked the "twist," this 26 year old male found himself getting teary.
"Yesterday is history, tomorrow is a mystery, but today is a gift--that is why it's called the present."
LotsoA113
AV Forum Member
Joined: March 19th, 2010
Location: Probably Cinemark
Post by LotsoA113 » April 6th, 2014, 3:30 pm
All of that beautiful, thought-provoking symbolism...and it was one of the funniest movies I've seen in ages!
"I know that sounds like something from a cat poster...but it's true"
I love all things cinema, from silent movies to world cinema to animated cinema to big blockbusters to documentaries and everything in between!
droosan
Joined: December 16th, 2004
Location: Burbank, Calif.
Post by droosan » April 6th, 2014, 6:22 pm
The only problem I'd had with the third-act 'twist' was its pacing; the movie had built toward something of a climax just prior .. and suddenly, the story had to take a back seat to long lingering shots in which not much was happening.
I mean, I get that Warner Bros probably spent a truckload of money to have Will Ferrell in their movie, so they made darn certain he was 'on-camera' for as long as possible to get their money's worth .. but that workbench scene (in which Emmet tries to move unseen) was freakin' interminable.
I got the symbolism, and such .. I just felt it slowed things down a bit much. Not enough to keep me from enjoying the heck out of the rest of the movie, though..!
Animated Views Admin
Joined: October 23rd, 2004
Location: SaskaTOON, Canada
Post by Randall » April 6th, 2014, 9:42 pm
Whenever I watch this one, I'll always be bracing myself for when it goes off the rails in the end. But the rest is too good to ignore.
Buzz Bumble
Joined: August 16th, 2013
Post by Buzz Bumble » April 6th, 2014, 9:44 pm
Ben wrote: Well, I always make sure to just use the singular! I think "Legos" plural is more an American thing...?
"Legos" is a word made up and used by Americans, but it's meaningless nonsense. "Lego" is the company, so it is only ever singular (if you work at Lego and use "Legos" you'll get your hand smacked, technically it should also always be in capital letters). There is no plural version - you don't go around calling all the action figures "Habros", for example. The sets and plastic pieces are exactly that, "Lego sets" and "Lego pieces". I found this on the Interweb-thingy a few weeks ago:
There is no such thing as "Legos". The progression is:
- "A bit of Lego",
- "two bits of Lego",
- "several bits of Lego",
- "lots of Lego",
and - "leave me alone I'm making a tower."
Joined: October 22nd, 2004
Post by Ben » April 7th, 2014, 4:44 am
Dacey wrote: Hmmmm. Ben might know everything. But that doesn't mean he's always right.
And I do get all that...but the movie set up the rules (that this was a Lego universe), then broke those rules (ok, so it wasn't), but then conflicted with itself by...having the Lego figure 'come to life' in our real world.
These a things. They don't a make much a sense!
Post by Dan » November 15th, 2015, 3:29 pm
A bump to give some interesting heads up about a Target exclusive out on blu ray that just came out.
The Special Special Edition
It's essentially the original blu ray release plus a bonus second blu ray disc with around two hours of bonus material with the key feature being a new commentary track from the directors, animation co-director Chris McKay, and film critic FILM CRIT HULK. There's also more deleted scenes, two of the How It Should Have Ended shorts (Alternate Ending and Super Cafe), the film in 90 seconds as told by a kid, the Trapped in the Ice short that was used to help create the look of the film, the "Wizard Team Go!" short (to explain the "origin" of the Vitruvius character), and a feature where Batman points out some of the hidden Easter Eggs in the film. They also moved the Dream Job: Meet the LEGO Builders features onto this disc for some reason, but at least it's still there.
Post by Dacey » November 16th, 2015, 11:00 am
The Dream Job: Meet the Lego Builders feature was not included on the original 2D set, which is unfortunately what my family bought before I was able to stop them.
Post by Dacey » February 12th, 2019, 11:18 pm
If you liked the first Lego Movie, you're probably going to like The Lego Movie 2.
And if you didn't like the first Lego Movie...you're probably still going to like The Lego Movie 2. It's less "manic" than the original while still being loaded with jokes, and the songs are actually pretty great (similar to Trolls, which Mike Mitchell also directed, there are at least four showstoppers here).
For me the original Lego Movie probably has the slight edge. But both are terrific at what they do, and it's a shame this film has been getting a fairly lukewarm reception box office wise.
Joined: October 16th, 2004
Post by James » February 13th, 2019, 12:29 am
Speaking of which... http://animatedviews.com/2019/the-lego- ... part-film/
Post by Ben » February 13th, 2019, 7:24 pm
Trolls was a fun surprise. I'll catch Lego 2 down the line sometime.
KathrynBerck
Joined: July 17th, 2019
Post by KathrynBerck » July 18th, 2019, 5:41 am
Seems it might actually be a great film
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Jones Isandlwana Durnford james George 2022 Russell williams horse Prince drift smith harford martini rifles john brown Mounted taylor 24th Roll Franklin Natal Police anstey henry
WWW.1879ZULUWAR.COM :: 1879 ZULU WAR CAMPAIGN MEDAL
Zulu War 1879, Colony of Natal, Exempted from Native Law Medal
Subject: Zulu War 1879, Colony of Natal, Exempted from Native Law Medal Wed Mar 31, 2010 6:46 am
I have come across this medal and does anyone have any further information on it?
This medal was instituted by the Colony of Natal after the passing of Law 28 in 1865 whereby certain natives could be exempted from native law. Those awarded in 1879 largely rewarded natives who had remained loyal to Britain during the Zulu War.
Subject: Native Law Medal Wed Mar 31, 2010 11:17 am
hi 1879graves.
Very interesting , I haven"t heard of or seen one before , they may also be an interesting item ( along with the pen )
to be put up by an Auction House , I wouldnt think there would be many that have survived to this day. Mons14 or maybe
one of our Sth African Members may be able to shed more light on this award . Any information would be appreciated .
cheers 90th.
Subject: Re: Zulu War 1879, Colony of Natal, Exempted from Native Law Medal Wed Mar 31, 2010 5:36 pm
Hi 90th
This Medal was sold at an Auction House on the 2nd September 2002 for £420.00
Subject: Native exemption medal Wed Mar 31, 2010 5:52 pm
I would bet it would go for a lot more Today. .
» COLONY OF NATAL, EXEMPTED FROM NATIVE LAW MEDAL
» The natal native Contingent in the Anglo-Zulu War 1879
» The Natal Native Contingent in the Anglo-Zulu War, 1879
» Memorial to men of the Natal Carbineers, Natal Mounted Police, Buffalo Border Guard & Natal Native Horse killed in action at Isandlwana
» ZULU WAR medal 1879 100% Genuine