text:
from typing import TYPE_CHECKING, Any

from django.contrib.auth.models import BaseUserManager

if TYPE_CHECKING:
    import yak.users.models  # noqa

class UserManager(BaseUserManager):

    def create_user(self, email: str, password: str = None,
                    **extra_fields: Any) -> 'yak.users.models.User':
        email = self.normalize_email(email)
        user = self.model(email=email, **extra_fields)
        user.set_password(password)
        user.save()
        return user

    def create_superuser(self, email: str, password: str,
                         **extra_fields: Any) -> 'yak.users.models.User':
        extra_fields['is_staff'] = True
        return self.create_user(email, password, **extra_fields)

redpajama_set_name: RedPajamaGithub


text:
BanglaCricket Forum > Other Sports > Other Sports
Following Siddiqur Rahman: Bangladesh's golfing star
Other Sports Talk about other Bangladeshi and International sports.
Page 15 of 15 « First < 5 13 14 15 Last »
CricketPagolChele
Location: Texas,USA
Favorite Player: Shakib , Mushfiqur
Originally Posted by Naimul_Hd
Can't wait for that day when BD Cricket team will beat Indian Cricket team in India in front of billions of people
Sweetest !
That day will never come at least in our life time, because India wont let us play there.
Find all posts by CricketPagolChele
সিদ্দিকুর রহমানের ২৯তম জন্মদিন ছিল গতকাল। মেলবোর্নে ভক্ত, শুভানুধ্যায়ীদের ভালোবাসায় সিক্ত হয়েছেন বাংলাদেশের সেরা গলফার। মাত্রই দিল্লি ওপেন জয়ের পর তিনি এখন অস্ট্রেলিয়ায়। নিজের দ্বিতীয় পেশাদার শিরোপা জেতা সিদ্দিকুর আজই নামছেন বড় একটা যুদ্ধে। এই প্রথম বাংলাদেশি হিসেবে খেলছেন গলফের বিশ্বকাপে। বিশ্বকাপের সেরা দশে থাকার আপাতকঠিন লক্ষ্য পূরণের চ্যালেঞ্জ নিয়ে নামছেন রয়্যাল মেলবোর্ন গলফ কোর্সে। যেখানে এই সপ্তাহেই খেলার অভিজ্ঞতা হয়েছে তাঁর।
ফেসবুকের মাধ্যমে সিদ্দিকুর জানালেনও, 'এ ধরনের সবুজ কোর্সে খেলার অভিজ্ঞতা এই প্রথম হয়েছে আমার। গত কদিনে আমি এই কোর্সে অনেক অনুশীলন করেছি। এখন মনে হচ্ছে বিশ্বকাপে মানিয়ে নিতে পারব।'
http://www.prothom-alo.com/sports/ar...A6%86%E0%A6%9C
Good start.. Siddik now in 32nd position out of 60 player..
http://www.golfchannel.com/tours/pga...orld-cup-golf/
Originally Posted by ahnaf
He needs to close in the gaps...if he wants to be in top 10.
Good luck Sid.
December 1, 2013, 01:44 PM
Tiger Manc
BD-SL Test Fantasy Winner 2017
Favorite Player: Suhrawadi Shuvo
Siddikur 4th in Indonesia
Siddikur Rahman produced another excellent performance to finish fourth in the Indonesian Open which concluded at the Pantai Indah Kapuk Course of Damai Indah Golf Club yesterday. The Two-time Asian Tour winner returned with a card of four-under-par (67) on the last day to finish with 12-under-par (272). However, it was not enough for Bangladesh's best golfer as overnight leader Gaganjeet Bhullar raced away with the title with an aggregate score of 16-under-par (268). Nicholas Fung of Malaysia and Chapchai Nirat of Thailand grabbed the joint second position at 13-under-par (271).
http://www.thedailystar.net/beta2/ne...-in-indonesia/
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A good re-bound after the below average performance in the golf Olympic - he is getting better day by day.
Long time no update.
Here's some good news
http://www.prothom-alo.com/sports/ar...A6%A8%E0%A6%BF
http://www.thedailystar.net/sports/s...amburg-1441087
http://www.golfchannel.com/video/no-...-chips-birdie/
http://golfweek.com/2017/07/28/siddi...pean-tour-win/
Jordan Smith Secures Maiden European Tour Title in Germany
By: Derek Clements | Mon 31 Jul 2017 | Comments
JORDAN SMITH crowned his extraordinary maiden season on the European Tour when he defeated defending champion Alex Levy to win the Porsche European Open at Green Eagles Golf Club in Hamburg in a playoff. The Englishman started the final round with a two-shot lead but in a topsy-turvy afternoon, the lead changed hands several times.
Levy led by a single shot as he stood at the 18th tee, but Smith holed a 10-foot birdie putt to get to 13 under and join the Frenchman. At the first extra hole, Levy had a short putt to win but missed, so they headed back to the 18th tee once again.
Smith nearly holed his putt for an eagle and then tapped in to win his first tournament in only his 25th start. It completes an amazing run for Smith - in 2015 he won the EuroPro Tour order of merit and last year won the Road to Oman on the Challenge. He wasted no time securing his card for 2018 with a series of excellent finishes. All that was missing from his CV was a victory, and he has now put that right.
"It's all a bit surreal," Smith said. "Obviously the Challenge Tour last year and the EuroPro the year before that. It's been fun. I had a really good start to the year and the first aim was just to secure my Tour card, which I managed to do. And then the second goal was to win an event, which I've also now managed to do."
Smith extended his advantage to three strokes when he rolled in his six-foot birdie putt at the first to move to 13 under par. But Levy holed his bunker shot at the second to cut Smith's lead back to two shots. And when Smith dropped his first shot of the day, and only his fourth of the week, after sending his approach to the edge of the water hazard at the sixth, Levy cut his lead to a single stroke. A birdie at the eighth took Levy level with Smith and when the Englishman missed a short par putt he suddenly found himself trailing.
They were level again at the 10th after Levy found trouble from the tee and was forced to take a penalty drop, and Smith went back in front with a birdie at the 11th. However, the Englishman left his bunker shot at the 13th well short of the hole and failed to convert his 20-foot par putt.
Levy and Smith both made birdies from inside ten feet at the 15th to get to 12 under before Levy reclaimed the outright lead with another birdie at the 16th. But Smith got up and down from the rough behind the green for a birdie at the last to take it to extra holes.
Smith said: "I've loved it. It's been nerve-wracking at times but I've really enjoyed it. The whole week tee to green is probably the best I have played all year, probably struggled a little bit on the greens, but it came together at the right time."
Levy was disappointed not to have made a successful defence but said he was happy with the way he had played. "I'm really proud of what I did today," he said. "I shoot three under in the conditions that were really tough and I played really well. I think there are a lot of good things coming for me in the next few months. I hit some pure shots. I struggled a bit today on the putting but I think I played really good golf and I improved my game week after week and that's really nice.
"I don't feel a lot of pressure. My dream was to win this tournament for my birthday present in two days but I didn't do that. I'm really pleased with what I did and I'm really happy."
Diminutive Bangladeshi golfer Siddikur Rahman, who finished tied for third with Johan Edfors, was thrilled with his week's work, especially as he is one of the shortest hitters in the game and the course was playing long - really long. He finished with a 68 for an 11 under par total of 277.
"It was a really great day. I couldn't expect this today but I am very fortunate my hitting was good, especially my putting – it was really good all four days – so I am really happy with today's round," Rahman said.
"The second shot unfortunately (on the 18th) went to the left and into the hazard, I talked to myself andsaid no problem as there is still a long way to go, one good shot, maybe one good putt. I was really inspired by the clapping and after that the putt was quite memorable.
"My putting and short game, without out that it's not possible to play like this."
http://www.golfshake.com/news/view/1...n_Germany.html
Favorite Player: Sanga,Mash,Shakib,Fizz
so how good is he really? Don't follow golf so I've never heard of him.
Send a private message to Yankees
Find all posts by Yankees
His rankings improved from 394 reported in the daily star this year to 339 which is still well below his peak ranking of 117 in 2011.

redpajama_set_name: RedPajamaCommonCrawl


Discover the wonder of nature and create a colourful garden kingdom that bugs will love. This Sow n Grow hampa is a great gift to inspire and introduce a child to gardening, they can learn new skills and have lots of fun watching their garden grow.
Seeds from Sow 'n' Sow- This gift of seeds will enchant nature lovers and budding green thumbs young and old. Containing easy to grow old favourites dianthus, cosmos, cornflower and alyssum which will attract bees, butterflies and beneficial bugs as well as producing a delightful display of colour.
Trowel, Transplanter and Weeder- Quality set of 3 gardening tools, perfect size for children to tend to their new garden.


As 2010 comes to a close and 2011 lies on the brink of the future, I find myself reflecting a lot on the past 12 months of my life. I've had a tough year in many ways. I've struggled on many fronts. I've learned a lot about myself, about what I need to work on, where I need healing, how I need to proceed in the future in dealing with certain issues, etc. In short, I'm a total work-in-progress. Add to that the following confession: I'm afraid of becoming what I call an "Awkward Academic"! I illustrate this with the picture of The Office's Michael Scott lecturing at "The Temp" Ryan's Business College. In that episode, Scott just bombs because he totally cannot relate to people. His social skills are just terrible, as are his teaching skills!
You know what I'm talking about don't you? The teacher or professor or researcher who thrives in the academic world but struggles in every other area at life. I fear becoming the educator who lights up when standing at the head of classroom but is unable to hold a one-on-one conversation with folks. I fear becoming the instructor who comes to life when sitting on a debate panel or in front of a t.v. camera but lacks the ability to be sensitive to how people perceive me and therefore fails to have any type of fruitful relationship or ministry with/to them. I fear being a widely published author or esteemed presenter only to not be able to "get" jokes or always take myself and others too seriously. I fear these things! I do not want to be the "Awkward Academic"!
But I must admit something else too: The further I get into the academic world, the tougher it is to tame my conversational approaches to people. Let me explain. The academic world is its own culture. In fact, there are many sub-cultures in the academic world. The sub-culture that I am a part of, the culture that I spend the largest amount of my professional time in is in the field of biblical studies. In this field, there is every type of personality imaginable. But in this world, to "make it" you are expected to produce good research, be a good writer, offer good publications, be a sound presenter, have good debate skills, etc.
This spring, I will have presented over 20 academic papers at academic conferences. All of these conferences have similar protocol and expectations. Here's how it typically works: 1) I spend time researching and writing a paper that will make a new contribution to the field of biblical studies; 2) This means that I will have to challenge old ways of thinking and offer new insights; 3) This means that somewhere along the way, I will disagree with people and likewise, they will disagree with me; 4) After I present my paper, I expect that people will challenge me and try to knock the knees out from under my research or to try to poke holes in it; 5) Knowing that I will be challenged by whomever is there, I know that I am going to have to defend myself and debate; 6) I know, however, that this debate will typically not cause strife between myself and whomever challenges me; 7) In fact, I know that when they challenge me (and my work), they are only trying to make me a better researcher, presenter, etc. This is the world I spend the majority of my time in. A world where debating and challenging is the main thrust of discourse.
So, take me out of that culture and put me into another one and there is bound to be some struggle. When you spend your life in a culture of argumentation, which I think is VERY healthy and necessary, and then step into another culture, it is often hard to turn off a lot of the other stuff; there is carry-over. Let me give you an example or two of how this can be problematic. Let's say I spend 10 hours on a Monday studying and researching and writing, etc. I spend my whole day debating with commentators and other researchers, etc. about the Bible or how some passage is supposed to be translated or exegeted. Then, at the end of my day, I stop studying and login to Facebook where someone wrongly, I think, cites the passage (e.g. out of context, etc.) on their status. So, I say something on their wall. More than likely, I am going to say it too aggressively or forcefully or something like that. Probably I will come off as arrogant or intimidating, etc. In short, I become the "Awkward Academic".
Or, here's another scenario. Let's say I spend 80 hours in the week studying the Bible. Then, on Sunday morning I go to church or on Sunday night to a Bible study. In each of those venues, I hear what I perceive to be someone ripping the verse or passage out of context. All week long I have been conditioned to challenge arguments or debate with scholars, etc. So, unwittingly, I do the same thing in one of these settings. Perhaps I confront the preacher afterwards or write an email to him/her. Or in the Bible study I challenge what the leader says.
The fact is, it is quite hard to transition out of one world or culture and into another, where the expectations for discourse are often so different. So, what I hope to work on in the future is how to make that transition more smooth and less rocky. In 2011, I don't want to get closer to being the "Awkward Academic" than I already am. However, I also don't want to downplay education or slack off. Nor do I want to be totally unapproachable by those who might have less education, etc. I just want to be normal, approachable, civil, humble, hospitable and compassionate. That is not to say that I will not speak truth, cave-in to pressures, etc. It just means that I want to find a healthy balance and a good way to transition. That's all.


text:
Koštanjica (; Albanian: Kështenjë) is a village in the municipality of Bar, Montenegro. It is located in the Skadarska Krajina region, by Lake Skadar.

Demographics
According to the 2011 census, its population was 168.

References

Populated places in Bar Municipality
Albanian communities in Montenegro

redpajama_set_name: RedPajamaWikipedia


Information security company Shred-it has been helping GP practices, dental surgeries and other medical service providers in the UK ensure that confidential paperwork, such as medical records, is securely destroyed when no longer needed. This is essential in the medical sector which holds an array of confidential data, from patient and medical records to staff data, all of which is of considerable value to malicious third parties.
Data protection has become a top priority in the medical sector of late, especially since the implementation of the European Union's (EU) General Data Protection Regulation (GDPR). Through its partnership with the LMC Buying Group, a federation with over 6,500 members that negotiates discounts on a wide range of products and services for healthcare providers, Shred-it has helped the industry TO improve data protection measures and therefore provide a timely boost for GDPR compliance plans.
As part of the partnership, practices use Shred-it's services on both an ad hoc basis (when they need to clear out an archive of papers that are no longer needed) as well as on a monthly subscription service basis. This involves practices receiving secure consoles that are regularly collected and shredded by members of Shred-it's team.


text:
Many companies squeezed by crisis and KCCI fields their complaints
The check-in gates at Muan International Airport in South Jeolla are empty after the airport decided Friday to halt all flights until March 28. [YONHAP]
Companies are hurting with the coronavirus outbreak.
Source: KCCI(Korea Chamber of Commerce and Industry)
According to the Korea Chamber of Commerce and Industry (KCCI) Monday, among 357 help requests it received, some 40 percent dealt with plummeting profits; close to 29.7 percent were about difficulties in securing materials and parts; 14.6 percent were about difficulties in exports; 5.3 percent were about shortages in disinfection supplies; and 4.8 percent were issues with employee management.
The industries affected varied by region.
In Seoul, the airlines, travel and education industries were hurt most. Eight local airlines reduced flights to China by 77 percent, which could result in losses of close to 3.7 billion won ($3 million) in operating profit from international flights for the first half of the year. Airline companies are likely to have problems with exports, as travel bans affect transportation of cargo.
Incheon and Gyeonggi, where 36 percent of the country's manufacturing companies are located, are experiencing halts in production and exports. One construction material manufacturer said it's having trouble providing supplies on time to clients in America and Europe after the coronavirus made it difficult to get materials from China.
One semiconductor equipment company in Gyeonggi said it has abandoned its performance goals for this year. The company has lost clients from China.
"The products and services we provide are all made-to-order, so we need to be in China for more than 300 days every year. Since this is now impossible, our operating profit for this year has dropped by more than 15 percent, and we expect the situation to stay grim," said the head of the company. "There's hardly any cash flow in the industry. It's not only us. Close to 30 to 40 partner companies are also struggling," he added.
Ulsan, home to Korea's largest petrochemical complexes, has unique problems. Some companies were unable to do their mandatory air quality measurements due to tight disinfection measures and strict limits on the access of visitors to their facilities.
Jeju Island and Gangwon are likely to sustain losses from a slump in tourism, which the two regions depend on. Operating profits for hotels, tourist sites and golf courses have dropped by 50 percent. For restaurants, profits plummeted by more than 80 percent. One theme park in Gangwon canceled plans to hire new staff this year, after it saw 70 percent of its reservations for the month canceled and its operating profit shrank to a third of the previous year.
Daegu and North Gyeongsang, which have the most number of infected patients, have been pushed to the edge. Factories and businesses have stopped as the two regions experience a serious shortage of masks. The Daegu Chamber of Commerce and Industry said it received numerous requests from factories for masks for workers, so that factories can resume operations.
"We are simply running out of masks to give workers at our plants, which is causing a lot of fear among our staff. It's getting harder to disinfect our facility and run it at the same time," said an official from one car parts manufacturer in Daegu.
Many companies want the government to allow them to extend their working hours. They deserve it after the efforts they've made - allowing remote work and parental leave - to prevent the spread of the virus.
"We are planning to submit a comprehensive proposal to the government. This will include measures to turn around the economy, as well as how to tackle the Covid-19 crisis," said Woo Tae-hee, vice president of the KCCI.
BY LEE SO-A [kang.jaeeun@joongang.co.kr]


text:
\section{Introduction}\label{sec:intro}
\setcounter{equation}{0}

Consider the Laplace operator $\Delta=-\sum_{i=1}^n\partial_{x_i}^2$ on the Euclidean space $\mathbb R^n$

and  the Schr\"odinger equation
 \begin{eqnarray*}\label{e1.00}
\left\{
\begin{array}{ll}
  i{\partial_t u } +\Delta u=0,\\[4pt]
 u|_{t=0}=f
\end{array}
\right.
\end{eqnarray*}
with initial data $f$. Its solution     can be written as
 \begin{eqnarray*}\label{e1.0}
 u(x, t)=e^{it\Delta} f(x)={1\over (2\pi)^n}\int_{{\mathbb R}^n} {\widehat f}(\xi) e^{i(  \langle x, \,  \xi\rangle +t|\xi|^2 )} d\xi,
 \end{eqnarray*}
 where ${\widehat f}$ denotes the Fourier transform of $f$.
It is well-known that the operator $e^{it\Delta}$ acts boundedly on $L^p({\mathbb R}^n)$ if and only if $p=2$;
 see  H\"ormander \cite{H1}. For $p\not= 2, $
it was shown (see for example,  \cite{Br, La, Sj})) that for $s > n|{1/ 2}-{1/p}|$, the operator
 $e^{it\Delta}$ maps the Sobolev space $L^p_{2s}(\RN)$ into $L^p(\RN)$, in other words,
$(I+\Delta)^{-s } e^{it\Delta}$ is bounded on $L^p(\RN)$.  For $s < n|{1/ 2}-{1/p}|$, it is known that the operator
$(I+\Delta)^{-s } e^{it\Delta}$ is unbounded on $L^p(\RN)$.
In \cite{Mi1},  Miyachi obtained the sharp endpoint  estimate for  $e^{it\Delta}$  on Hardy and Lebesgue spaces,
 and showed  that for every $0<p<\infty,$
\begin{eqnarray}\label{e1.1}
 \left\|  (1+\Delta)^{-s} e^{it\Delta} f\right\|_{H^p(\mathbb R^n)} \leq C  (1+|t|)^{s}\|f\|_{H^p (\mathbb R^n)},
  \ \ \ t\in{\mathbb R}, \ \ \ s\geq  n\big|{1\over  2}-{1\over  p}\big|,
\end{eqnarray}
where   $H^p({\mathbb R}^n)$ is  the classical Hardy space (\cite{FS}) on ${\mathbb R}^n$
  and $H^p({\mathbb R}^n)=L^p({\mathbb R}^n)$ if $1<p<\infty$.
  See also Fefferman-Stein's work \cite[Section 6]{FS}.

The Schr\"odinger semigroup $\{e^{it\Delta}\}_{t>0}$ can be defined in terms of the spectral resolution of
the self-adjoint Laplace operator $\Delta.$ A natural question is to determine a sufficient condition so that
\eqref{e1.1} holds when the Laplace $\Delta$ is replaced by a non-negative self-adjoint operator $L.$
  For this purpose we suppose  that $(X,d,\mu) $ is a metric measure space
 with a distance $d$ and a measure  $\mu$, and   $L$ is a non-negative self-adjoint operator on $L^2(X).$
  Such an operator $L$ admits a spectral
resolution
\begin{eqnarray*}
L=\int_0^{\infty} \lambda dE_L(\lambda),
\end{eqnarray*}
where  $E_L(\lambda)$ is the projection-valued measure supported on the spectrum of $L$.
The operator   $   e^{itL}, t\in{\mathbb R},$ is defined by
 \begin{equation}\label{e1.3}
 e^{itL}f =   \int_0^{\infty}    e^{it\lambda}dE_L(\lambda) f
   \end{equation}
 for $f\in L^2(X)$, and forms   the Schr\"odinger group.
   By the spectral theorem (\cite{Mc}),  the operator   $  e^{itL}$  is  continuous on $L^2(X)$.
   Our main interest will be in  the mapping properties
of families of operators derived from the Schr\"odinger group
on Hardy and Lebesgue spaces.

 Depending on the nature of the assumptions regarding the assumption of $e^{-tL}$, there
 are various nuances of  the mapping properties
of the Schr\"odinger group $e^{-tL}$
on $L^p$ spaces  presently available in the literature.
 For example, on Lie groups with polynomial growth and manifolds with non-negative Ricci curvature, similar results as in \eqref{e1.1}
for $s> n\left|{1/ 2}-{1/ p}\right|$ and $1<p<\infty$ have been first announced by Lohou\'e in \cite{Lo},
then Alexopoulos   obtained them in \cite{A}.
   In the abstract setting
of operators on metric measure spaces,   Carron, Coulhon and Ouhabaz \cite{CCO}  showed  that for every $1<p<\infty,$
\begin{eqnarray} \label{e1.555}
 \left\| (I+L)^{-s }e^{itL} f\right\|_{p} \leq C (1+|t|)^{s} \|f\|_{p}, \ \ \ t\in{\mathbb R}, \ \
  \ s> n\big|{1\over  2}-{1\over  p}\big|,
\end{eqnarray}
  provided

the semigroup $e^{-tL}$, generated by $-L$ on $L^2(X)$,  has the kernel  $p_t(x,y)$
which  satisfies
the  Gaussian upper bound, i.e.
\begin{equation*}
 \label{GE}
 \tag{${\rm GE}_m$}
|p_t(x,y)| \leq {C\over V(x,t^{1/m})} \exp\left(-c \, {  \left({d(x,y)^{m}\over    t}\right)^{1\over m-1}}\right)
\end{equation*}
for every $t>0, x, y\in X$, where $c, C$ are   positive constants and $m\geq 2.$
Such estimate  \eqref{GE} is typical for elliptic or sub-elliptic differential operators of order $m$
(see for example, \cite{A,  CCO,   D, DM, DOS,  JN, JN2, O,  Si, Sj}
and the references therein). See also  related results in \cite{BDN, DN, FS, JN, JN2}.

The question whether  estimate \eqref{e1.555} holds with  $s =  n|{1/2}-{1/p}|$ was recently
solved in  \cite{CDLY}. More specifically, if $L$ satisfies the   Gaussian estimate \eqref{GE}, then
 for every  $p\in (1, \infty)$ there exists a  constant $C=C(n,m, p)>0$ independent of $t$ such that
 \begin{eqnarray} \label{e1.5}
 \left\| (I+L)^{-s }e^{itL} f\right\|_{p} \leq C (1+|t|)^{s} \|f\|_{p}, \ \ \ t\in{\mathbb R}, \ \
  \ s= n\big|{1\over  2}-{1\over  p}\big|.
\end{eqnarray}
However, this result does not give any end-point estimate on the Hardy space $H^1_L(X)$ when $p=1.$

This paper   continues a line of study in \cite{CDLY} to  show that under the assumption that
 $L$ satisfies  a Gaussian upper bound  \eqref{GE},
the operator
$(I+L)^{-n/2 }e^{itL}$ is bounded   from  Hardy spaces $H^1_L(X)$ into $L^1(X)$,
  where  $H^1_L(X)$  is a  class of Hardy spaces   associated with  $L$ (\cite{ADM, DY, KU},
see Section 2  below).
  We then apply the duality argument and the   complex interpolation theorem (\cite[Theorem 5.7]{DY}) to obtain
 a new  proof of
 estimate \eqref{e1.5} in \cite{CDLY}    in the case that when $L$
satisfies  a  Gaussian upper bound \eqref{GE}.   Our result can be stated as  follows.

\begin{thm}\label{th1.1}
Suppose  that $(X, d, \mu)$ is  a  space of homogeneous type  with a dimension $n$.  Suppose that $L$
satisfies the property \eqref{GE}.
Then  there exists a  constant $C=C(n, m)>0$ independent of $t$ such that
\begin{eqnarray} \label{e1.6}
  \left\| (I+L)^{-n/2 }e^{itL} f\right\|_{L^1(X)} \leq C (1+|t|)^{n/2} \|f\|_{H_L^1(X)}, \ \ \ t\in{\mathbb R}.
\end{eqnarray}

By a duality and interpolation argument,
estimate  \eqref{e1.5} holds for all $1<p<\infty$.
\end{thm}

 \medskip

We  now apply the result of   Theorem~\ref{th1.1} to  study the property of  the solution to  the Schr\"odinger equation
\begin{eqnarray*}
\left\{
\begin{array}{ll}
  i \partial_t u +Lu=0, \\[4pt]
 u(x, 0)=f(x).
\end{array}
\right.
\end{eqnarray*}
Similarly to the work \cite{Sj} of  Sj\"ostrand
in the case that $L$ is the Laplacian on the Euclidean space ${\mathbb R}^n$,
a possible substitute for this operator
on $L^p$ is its  Riesz means,  defined by
\begin{eqnarray}\label{RieszMean}
I_s(t)(L):= st^{-s}\int_0^t(t-\lambda)^{s-1}e^{-i\lambda L}d\lambda
\end{eqnarray}
  for $t>0$, and $I_s(t)(L)={\overline I}_s(-t)(L)$  for $t<0$. Then we have   the following result.

 \begin{thm}\label{th1.2}
Suppose  that $(X, d, \mu)$ is  a  space of homogeneous type  with a dimension $n$.  Suppose that $L$
satisfies the property \eqref{GE}.
Then  there exists a  constant $C=C(n,m)>0$ independent of $t$ such that
 \begin{eqnarray}\label{nnnw}
 \left\| I_{n/2}(t)(L) f\right\|_{L^1(X)}   \leq    C \|f\|_{H^1_L(X)}, \ \ \ \ t\in {\mathbb R}\backslash\{0\}.
\end{eqnarray}

By a duality and interpolation argument,  for every $1<p<\infty$
 there exists a  constant $C=C(n,m, p)>0$ independent of $t$ such that
 \begin{eqnarray}\label{nnnww}
 \left\| I_{s}(t)(L) f\right\|_{p}   \leq    C \|f\|_{p}, \ \ \ \ t\in {\mathbb R}\backslash\{0\},  \ \
  \ s\geq n\big|{1\over  2}-{1\over  p}\big|.
\end{eqnarray}
\end{thm}

\medskip

We would like to remark that our main result, Theorem~\ref{th1.1},  gives the sharp endpoint  estimate  \eqref{e1.6}
for the  Schr\"odinger group $e^{itL}$ on the Hardy space, namely with the optimal number of derivatives and the optimal time growth for
	 the factor
$(1+|t|)^{s}$ in \eqref{e1.6}. Our endpoint estimate is using the Hardy space $H^1_L(X)$ associated to the operator $L$ instead of
the classical Hardy space but we do not assume certain standard conditions on the operator $L$ such as
 the H\"older continuity on heat kernel and
the conservation property
$
    e^{-t\Delta}1 = 1.
$
We also remark that the results
in \cite{FS, Mi1, Mi} relies on Fourier analysis (e.g., Plancherel's Theorem),
which is not available in the setting of space of homogeneous type in this paper.
 In the proof of Theorem~\ref{th1.1}, the main tool is to use the Phragm\'en-Lindel\"of theorem to show
 \iffalse
   we see that from the Gaussian bounds \eqref{GE},
   there exist constants $C$, $c>0$ such that for any open subsets
$U_1,\,U_2\subset X$,
\begin{equation}\label{e1.16}
|\langle e^{-tL}f_1, f_2\rangle|
\leq C\exp\left(-c \left({{\rm dist}(U_1,U_2)^m\over  t}\right)^{1\over m-1}\right)
\|f_1\|_{L^2(X)}\|f_2\|_{L^2(X)},\quad\forall\,t>0,
\end{equation}
 for every $f_i\in L^2(X)$ with
$\mbox{supp}\,f_i\subset U_i$, $i=1,2$,
where ${\rm dist}(U_1,U_2):=\inf_{x\in U_1, y\in U_2} d(x,y)$ (when $m=2$, we called \eqref{e1.16} the Davies-Gaffney estimate (see for example, \cite{CouS}).
 Using the Phragm\'en-Lindel\"of theorem, we show
 \fi
  that the Gaussian bounds \eqref{GE} implies
  the following off-diagonal estimate of the operator
  $ e^{zL} $ with  $ z=(i\tau-1)R^{-1}, \tau, R>0$:
 \begin{eqnarray} \label{evv}
\|P_{B}   e^{(i\tau-1)R^{-1}L} P_{2^jB\backslash 2^{j-1}B} \|_{2\to 2}\leq
C
  \exp\left(-c\left({ \sqrt[m]{R} 2^j r \over    \sqrt{1+\tau^2}}\right)^{\frac{m}{m-1}}   \right),\ \ \ j=2,3,\ldots
\end{eqnarray}
for all  balls $B\subseteq  X$
 (see Lemma~\ref{le2.2} below).
This new estimate  \eqref{evv} is   crucial in the proof of Theorem~\ref{th1.1}.

The paper is organized as follows.
In Section 2 we provide some preliminary results on Hardy and BMO spaces
and   spectral multipliers. In Section 3 we   apply the Phragm\'en-Lindel\"of theorem
to
give off-diagonal bounds for \eqref{evv} and the operator $F(L)$
for some  compactly supported function $F,$ which play a crucial role in the proof of   Theorem~\ref{th1.1}.
The  proof of Theorem~\ref{th1.1}  will be given in Section 4. In Section 5 we   apply Theorem~\ref{th1.1}
to obtain $L^p$-boundedness of the Riesz means of the solution to the Schr\"odinger equation.
In Section 6 we discuss some extensions
of Theorem~\ref{th1.1}
for operators satisfying weaker assumption than   the Gaussian bound \eqref{GE}.

\medskip

\section{Notations and preliminaries on Hardy spaces}
\setcounter{equation}{0}

We start by introducing  some notation and assumptions.  Throughout this paper,
unless we mention the contrary, $(X,d,\mu)$ is a metric measure  space where $\mu$
is a Borel measure with respect to the topology defined by the metric $d$.
Next, let
$B(x,r)=\{y\in X,\, {d}(x,y)< r\}$ be  the open ball
with centre $x\in X$ and radius $r>0$. To simplify notation we often just use $B$ instead of $B(x, r)$ and
given $\lambda>0$, we write $\lambda B$ for the $\lambda$-dilated ball
which is the ball with the same centre as $B$ and radius $\lambda r$. Let $B^c$ be the set $X\backslash B$.
We set $V(x,r)=\mu(B(x,r))$ the volume of $B(x,r)$ and we say that $(X, d, \mu)$ satisfies
 the doubling property (see Chapter 3, \cite{CW})
if there  exists a constant $C>0$ such that
\begin{eqnarray}
V(x,2r)\leq C V(x, r)\quad \forall\,r>0,\,x\in X. \label{eq2.1}
\end{eqnarray}
If this is the case, there exist  $C, n$ such that for all $\lambda\geq 1$ and $x\in X$
\begin{equation}
V(x, \lambda r)\leq C\lambda^n V(x,r). \label{eq2.2}
\end{equation}
In the Euclidean space with Lebesgue measure, $n$ corresponds to
the dimension of the space.

  For $1\le p\le+\infty$, we denote the
norm of a function $f\in L^p(X,{\rm d}\mu)$ by $\|f\|_p$, by $\langle \cdot,\cdot \rangle$
the scalar product of $L^2(X, {\rm d}\mu)$, and if $T$ is a bounded linear operator from $
L^p(X, {\rm d}\mu)$ to $L^q(X, {\rm d}\mu)$, $1\le p, \, q\le+\infty$, we write $\|T\|_{p\to q} $ for
the  operator norm of $T$.
Given a  subset $E\subseteq X$, we  denote by  $\chi_E$   the characteristic
function of   $E$ and  by $P_E$ the projection
$
P_Ef(x):=\chi_E(x) f(x).
$
We denote the dilation of a function $F$ by $\delta_r F(\cdot):=F(r\cdot)$ and
$\widehat{f}\,$ denotes the Fourier  transform, i.e.
of $f$,
$$
\widehat{f}(\xi)={1\over (2\pi)^{n/2}}\int_{\mathbb R^n} f(x)e^{-ix\xi} dx, \ \ \ \ \xi\in \mathbb R^n.
$$
Sometimes we also use  $\widehat{f}$ for ${\mathcal F} f$.

\medskip
\subsection{Hardy and BMO spaces associated with operators}
 A theory of Hardy spaces associated with certain operators was introduced and developed in \cite{ADM, DL, DY2, DY3, HLMMY, KU}
 and the references therein, similar to
 the way that classical Hardy spaces are adapted to the Laplacian. We present some main features of this theory
 in this section for reader's convenience.

  Suppose that $L$ is a non-negative
 self-adjoint operator on $L^2(X)$ which satisfies the Gaussian upper bounds  \eqref{GE} with $m\geq 2$.
We  consider the following quadratic operators associated
to $L$
\begin{eqnarray}
S_{L}f(x)=\left(\int_0^{\infty}\!\!\!\!\int_{\substack{  d(x,y)<t}}
|(t^mL) e^{-t^mL} f(y)|^2 {d\mu(y)\over V(x,t)}{dt\over t}\right)^{1/2},
\quad x\in X
\label{e2.1}
\end{eqnarray}
where $f\in L^2(X)$.
Under the assumption  of   the Gaussian upper bounds  \eqref{GE} of an operator $L$,
it is known that  the null space $N(L)= \{0\}$ (see \cite[Section 2.6]{HLMMY}) and
 the function   $S_{L}$  is   bounded on $L^p(X), 1<p<\infty$
and
$$
  \|S_Lf\|_{L^p(X)} \simeq \|f\|_{L^p(X)}.
$$
The Hardy space $H^1_{L}(X)$ associated with
$L$  is the completion of the space
$$
\{ f\in L^2(X):\  S_{L}f\in L^1(X) \}
$$
in the norm
$$
 \|f\|_{H_{L}^1(X)}=  \|S_{L}f\|_{L^1(X)}.
$$

 Let us describe the notion of a
{\it $(1,2,M,\varepsilon)$-molecule}
associated to an operator $L$  on spaces $(X,d,\mu)$.  Denote by ${\mathcal D}(T)$
the domain of an operator $T$. For every ball $B$, we set
\begin{equation}
U_0(B)=B,  \ \  {\rm and}\ \ \ U_j(B)=2^j B\backslash 2^{j-1}B
\,\,\mbox{ for }\,\,j=1,2, \dots.
\label{e2.3}
\end{equation}

\begin{definition} \label{def2.7} Let   $ \varepsilon>0$ and
$M\in{\mathbb N}$.
A function $m(x)\in L^2(X)$ is called a
$(1,2,M,\varepsilon)$-molecule associated with $L$ if there exist a function
$b\in {\mathcal D}(L^M)$ and a ball $B$ such that

\medskip

{ (i)}\ $m=L^M b$;

\medskip

{ (ii)}\  For every $k=0,1,2,\dots,M$ and $j=0,1,2,\dots$, there holds
$$
\|(r_B^mL)^{k}b\|_{L^2(U_j(B))}\leq 2^{-j\varepsilon} r_B^{mM}
V(2^jB)^{-{1/2}},
$$
\noindent where the annuli $U_j(B)$ are defined in (\ref{e2.3}).
\end{definition}

Next we give the definition of the molecular Hardy spaces associated with $L$.

\begin{definition}\label{def 2.4} We fix $\varepsilon > 0$ and
$M\in{\mathbb N}$.
The Hardy space $\HML$ is defined as follows.
We say that
$f= \sum\lambda_j m_j$ is a molecular
$(1,2,M,\varepsilon)$-representation (of $f$) if $ \{\lambda_j\}_{j=0}^{\infty}\in {\ell}^1$,
each $m_j$ is a $(1,2,M,\varepsilon)$-molecule, and the sum converges in $L^2(X).$
Set
\begin{equation*}
\HHML= \Big\{f:
\mbox{$f$ has a molecular $(1,2,M,\varepsilon)$-representation} \Big\},
\end{equation*}
with the norm given by
$$
||f||_{\HML}= {\rm inf}\Big\{\sum_{j=0}^{\infty}|\lambda_j|: f=\sum\limits_{j=0}^{\infty}\lambda_jm_j\,
\mbox{ is a molecular $(1,2,M,\varepsilon)$-representation} \Big\}.
$$
The space $\HML$ is then defined as the completion of $\HHML$ with respect to this norm.
\end{definition}

As a direct consequence of the definition, we note that $H^1_{L, {mol}, M_2, \varepsilon}(X)\subset
H^1_{L, {mol}, M_1, \varepsilon}(X)$ for $\varepsilon>0$ and $M_1, M_2\in{\mathbb N}$ with $M_1\leq M_2$.
We  have the following characterization. For its  proof, see Section 3, \cite{DL}.

\begin{lemma}\label{prop2.10} Suppose   $M\geq 1$.
  Then we have $\HML = \HSL$.  Moreover,
$$
\|f\|_{\HML} \approx \|f\|_{H^1_{L}(X)},
$$
\noindent
where the implicit constants depend only on $M, m$ and  $n$   in \eqref{eq2.2}  only.
\end{lemma}

Next we turn to the  space  ${{\rm BMO}_L(X)}$ associated with operators.
\iffalse

Recall
a function $f\in L^2_{\rm loc}(X)$ of type $(x_0,\beta)$, where $x_0\in X$
and $\beta>0$, if it satisfies

\begin{equation}\label{e7.5}
\|f\|_{{\mathcal M}_{(x_0,\beta)}}
:=\left(\int_X{|f(x)|^2\over (1+d(x_0, x))^{\beta}
V(x_0,1+d(x_0, x))}d\mu(x)\right)^{1/2}<\infty.
\end{equation}

\noindent We denote by ${\mathcal M}_{(x_0,\beta)}$ the collection of all
functions of type $(x_0,\beta)$. For a fixed $x_0\in X$, it is easy to see that
${\mathcal M}_{(x_0, \beta)}$ is a Banach space under the norm (\ref{e7.5}).
Also, for any $x_1\in X$, we have
${\mathcal M}_{(x_1, \beta)}={\mathcal M}_{(x_0, \beta)}$ with equivalent
norms. Set

\begin{eqnarray*}
{\mathcal M}:=\bigcup\limits_{x_0\in X}\
\bigcup\limits_{\beta:\,0<\beta<\infty}
{\mathcal M}_{(x_0,\beta)}
\end{eqnarray*}

\noindent We will say that $f\in{\mathcal M} $ is in \fi
Following \cite{DY}, we  say that $f  $ is in
${{\rm BMO}_L(X)}$, the space of
{\it functions of bounded mean oscillation} associated with $\{e^{-tL}\}_{t>0}$,
if there exists a positive constant $C$ such that

\begin{eqnarray}\label{e7.6b}
\hskip 0.30in
\|f\|_{ {\rm BMO}_L(X)}:=\sup_B {1\over V(B)}
\int_B|f(x)-e^{-r_B^m L}f(x)|d\mu(x)\leq C <\infty,
\end{eqnarray}
 where the supremum is taken over all balls $B$ in $X$.

\begin{lemma}\label{th7.5}
Assume that the operator $L$ satisfies  the  Gaussian upper bounds \eqref{GE}. Then, we have
$$
(H^1_{L}(X))^{\ast}={{\rm BMO}_L(X)}.
$$
\end{lemma}

\begin{proof} This result can be proved in a similar, but slightly simpler,
fashion to the duality results in \cite[Section 6]{HLMMY}. We omit the proof.
\end{proof}

 Similar to the classical Hardy spaces, Hardy spaces associated with operators
form a complex interpolation scale.
Let $[\cdot, \, \cdot]_{\theta}$ stand  for the complex interpolation bracket. Then we have the following result.

\begin{lemma}\label{th7.55}
Assume that the operator $L$ satisfies the property \eqref{GE}.   Then  for every  $0<\theta<1$ and $1<p_0<\infty$,  we have
$$
[ H^1_L,   L^{p_0} ]_{\theta} =L^{p}, \ \ \ \  \  {1\over p}=(1-\theta) +{\theta\over p_0},
$$
and
$$
[ L^{p_0},   {\rm BMO}_L ]_{\theta} =L^{p} , \ \  \ \ \ \  {1\over p}={ 1-\theta  \over p_0}.
$$
\end{lemma}

\begin{proof}
The proof can be verified that by viewing these spaces via the framework of tent spaces
and by using the interpolation properties  of tent spaces (see for example, Lemma 4.20, \cite{KU}).
\end{proof}

 \medskip
\subsection{Spectral multipliers  on the Hardy space.}
 The following result is a standard known result in the theory of spectral multipliers of non-negative self-adjoint
operators.

\begin{prop}\label{prop2.7}
Let $m\geq 2$. Suppose that $(X, d, \mu)$ is a space of homogeneous type with a dimension $n$.
Assume that the operator $L$ satisfies the property \eqref{GE}.
Assume in addition that $F$ is  an even bounded Borel function  such that
 $
\sup_{R>0}\|\eta\delta_RF\|_{C^\alpha}<\infty
 $
for some integer $\alpha> n/2  $ and some non-trivial function $\eta\in C_c^{\infty}(0, \infty)$.
Then the operator  $F(L)$ is bounded on $H^{1}_L(X)$,
\begin{eqnarray}\label{n}
\|F({L})\|_{H^1_L(X)\to  H^1_L(X)}\leq C \left(\sup_{R>0}\|\eta\delta_RF\|_{C^\alpha}+F(0)\right).
\end{eqnarray}
\end{prop}

\begin{proof}
For the proof,  see for example, \cite[Theorem 1.4]{KU}) and \cite[Theorem 1.1]{DY3}.
\end{proof}

\bigskip

\section{Off-diagonal bounds for compactly  supported spectral multipliers  }
\setcounter{equation}{0}

Let us start with stating the Phragm\'en-Lindel\"of Theorem
 for sectors in the complex plane $\mathbb C$. For its proof, we refer to \cite[Lemma 4.2]{SW}.

\begin{thm}\label{th2.1} Let $S$ be the open region in ${\mathbb C}$ bounded by two rays meeting at an angle $\pi/a$ for some $a>1/2$.
Suppose that $F$ is analytic on $S$, continuous on ${\bar S}$ and satisfies $|F(z)|\leq C\exp (c|z|^{b})$ for some $b\in [0, a)$
and for all $z\in S$. Then the condition $|F(z)|\leq B$
 on the two bounding rays implies that $|F(z)|\leq B$ for all $z\in S$.
 \end{thm}

  The following result is a   consequence of
 Theorem~\ref{th2.1}.

 \begin{lemma}\label{le2.22}
\label{tw0} Suppose that $F$ is an analytic function on $\mathbb{C}_+=\{z\in{\mathbb C}:  {{\rm Re }z}>0 \}$, the open
right half-plane.
Assume that, for given numbers $M_1, M_2, \gamma>0$, $0<\alpha\leq 1$,
\begin{equation}  \label{a1}
|F(z)|\le M_1, \quad \forall \, z \in \mathbb{C}_+
\end{equation}
and
\begin{equation}  \label{a2}
|F(t)|\le M_2 \exp\Big({-\frac{\gamma}{t^\alpha}}\Big), \quad \forall \, t \in \mathbb{R}_+.
\end{equation}
Then for every $ z \in \mathbb{C}_+$,
\begin{equation}  \label{bb2}
|F(z)| \le \max\big\{ M_1 ,M_2 \big\}\exp \left(-\alpha \gamma \, { {{\rm Re }z} \over |z|^{\alpha+1} }\right).
\end{equation}
\end{lemma}

\begin{proof}  Lemma~\ref{le2.22} was proved in  \cite[Lemma 9]{D2}. See also  \cite[Proposition 2.2]{CouS}
 and \cite[Lemma 6.18]{O}.
   We give a brief argument
of this proof for completeness and convenience for the reader.

Consider the function
\begin{equation}  \label{gg1}
u_+(\zeta):=F\left(\frac{1}{\zeta}\right)\exp\left(\gamma e^{i(\pi/2-\pi \alpha/2)}{\zeta^\alpha}\right),
\end{equation}
which is also defined on $\mathbb{C}_+$.
By (\ref{a1}),
\begin{equation*}
|u_+(\zeta)|\le M_1 \exp\left(\gamma|\zeta|^\alpha\right), \quad \forall \, \zeta \in \mathbb{C}_+ .
\end{equation*}
Again by (\ref{a1}) we have, for any $\varepsilon > 0$ and $\zeta=\varepsilon+iy=:C_\varepsilon e^{i\theta_\varepsilon}$,
\begin{eqnarray*}
|u_+(\zeta)|&=&\left|F\left(\frac{1}{\zeta}\right)\exp\left(\gamma e^{i(\pi/2-\pi \alpha/2)}{\zeta^\alpha}\right)\right|
 \leq  M_1\exp\left( C_\varepsilon^\alpha \gamma\sin\Big(\frac{\pi \alpha}{2}-\alpha\theta_\varepsilon\Big) \right).
\end{eqnarray*}
For $y\geq 0$, it follows from $0<\alpha\leq 1$ that
\begin{eqnarray*}
|u_+(\zeta)|
&\leq& M_1\exp\left( C_\varepsilon^\alpha \gamma\sin\Big(\frac{\pi \alpha}{2}-\alpha\theta_\varepsilon\Big) \right)
 \leq  M_1\exp\left(  \gamma\varepsilon^\alpha \right),
\end{eqnarray*}
which implies that
\begin{equation}  \label{22a}
\qquad \sup_{{\mbox{\Small{\rm Re}}}\zeta=\varepsilon, {\mbox\Small{\rm Im}}\zeta\geq 0}|u_+(\zeta)|\le
M_1 e^{\gamma \varepsilon^\alpha}.
\end{equation}
By (\ref{a2}),
\begin{equation}  \label{22b}
\sup_{ \zeta\in [\varepsilon,\infty)}|u_+(\zeta)|\le M_2.
\end{equation}
Hence, by Phragm\'en-Lindel\"of theorem~\ref{th2.1} with angle $\pi/2$ and $b=\alpha$,
applied to
\begin{equation*}
S^+_{\varepsilon}= \{z\in \mathbb{C} \colon \, {\mbox{\Small{\rm Re}}} z >
\varepsilon \quad \mbox{and} \quad {\mbox{\small{\rm Im}}} z > 0\},
\end{equation*}
we  obtain
\begin{equation*}
\sup_{{\mbox{\Small{\rm Re}}}\zeta\geq\varepsilon, {\mbox\Small{\rm Im}}\zeta\geq 0}|u_+(\zeta)|\le
\max\{ M_2 , M_1 e^{\gamma \varepsilon^a} \}, \quad \forall \,
\varepsilon > 0 .
\end{equation*}

Next  we consider the function
\begin{equation}  \label{ggg2}
u_-(\zeta):=F\left(\frac{1}{\zeta}\right)\exp\left(\gamma e^{i(-\pi/2+\pi \alpha/2)}{\zeta^\alpha}\right).
\end{equation}
A similar argument  shows that
\begin{equation*}
\sup_{{\mbox{\Small{\rm Re}}}\zeta\geq\varepsilon,\,  {\mbox\Small{\rm Im}}\zeta\leq 0}|u_-(\zeta)|\le
\max\{M_2 , M_1e^{\gamma \varepsilon^a} \}, \quad \forall \,
\varepsilon > 0 .
\end{equation*}
Letting $\varepsilon \to 0$ we obtain
\begin{equation*}
\sup_{{\mbox{\Small{\rm Re}}}\zeta>0, \, {\mbox\Small{\rm Im}}\zeta\geq 0}|u_+(\zeta)|\le
\max\{ M_1 ,M_2 \}
\end{equation*}
and
\begin{equation*}
\sup_{{\mbox{\Small{\rm Re}}}\zeta>0, \,  {\mbox\Small{\rm Im}}\zeta\leq 0}|u_-(\zeta)|\le
\max\{ M_1, M_2\}.
\end{equation*}
Putting $\zeta =\frac{1}{z}$, we obtain for all $\mbox{Re}\,z>0$
\begin{equation*}
|F(z)|\le
\max\{ M_1, M_2\}\exp\left(- \alpha\sin\left(\pi/2+|\theta_z|\right) {\gamma \over |z|^{\alpha}}\right),
\end{equation*}
where $\theta_z=\arg z$. From this, (\ref{bb2}) follows readily.
 \end{proof}

\begin{lemma}\label{le2.2}
Suppose that  $L$ satisfies the  Gaussian upper bounds \eqref{GE}.
There exist  two positive constants  $C$ and $c$ such that
for every $j=2,3,\ldots$
 \begin{eqnarray*}
\|P_{B}   e^{(i\tau-1)R^{-1}L} P_{U_j(B)} \|_{2\to 2}&\leq&
C
  \exp\left(-c\left({ \sqrt[m]{R} 2^j r \over    \sqrt{1+\tau^2}}\right)^{\frac{m}{m-1}}   \right)
\end{eqnarray*}
for all  balls $B\subseteq  X$.
 \end{lemma}

\begin{proof}   For any open sets $U$ and $V$, and ${\rm Re}z>0$,

 we define a function
 $$
 F(z):=\langle e^{-zL}f_1,f_2\rangle,
 $$
 where $\supp f_1\subset U$ and $\supp f_2\subset V$.
 Then $F(z)$ is an analytic function  on the complex half plain ${\rm Re} \, z>0$.  It is seen that
 $$
 |F(z)|\leq \|e^{-zL}f_1\|_{2} \|f_2\|_2\leq \|e^{-z\lambda}\|_{L^\infty}\|f_1\|_2\|f_2\|_2\leq
  \|f_1\|_2\|f_2\|_2, \quad \forall \, z \in \mathbb{C}_+
 $$
 and it follows from the  Gaussian upper bounds \eqref{GE} that
 $$
 |F(t)|\le C \exp\left(-c{ \frac{d(U,V)^{\frac{m}{m-1}}}{ t^{\frac{1}{m-1}}}}\right)\|f_1\|_2\|f_2\|_2, \quad \forall \, t \in \mathbb{R}_+.
 $$
  Let $M_1=\|f_1\|_2\|f_2\|_2$, $M_2=C\|f_1\|_2\|f_2\|_2$,
 $\gamma= cd(U,V)^{m/(m-1)} $ and $\alpha=1/(m-1)$.
We apply  Lemma~\ref{le2.22} to get
 $$
 |\langle e^{-zL}f_1,f_2\rangle|\leq C\exp\left( {-  c\, {{\rm Re} z}\, \frac{d(U,V)^{\frac{m}{m-1}} } { |z|^{\frac{1}{m-1}+1}} }\right)\|f_1\|_{2} \|f_2\|_2.
 $$
From it, we have that
 \begin{eqnarray*}
\|P_{B}   e^{(i\tau-1)R^{-1}L} P_{U_j(B)} \|_{2\to 2}&\leq&
C
  \exp\left(-c \left({ \sqrt[m]{R} 2^j r_B \over    \sqrt{1+\tau^2}}\right)^{\frac{m}{m-1}}   \right).
\end{eqnarray*}
This ends the proof of Lemma~\ref{le2.2}.
\end{proof}

Next we define a Besov type norm of $F$ by
$$
\|F\|_{{{\bf B}^{s}}}:=\int_{-\infty}^{\infty} |\widehat{F}(\tau)|(1+|\tau|)^{s}d\tau,
$$
 where ${\widehat f}$ denotes the Fourier transform of $f$.
Since for every  functions $F$ and $G$,  it can be checked that
 \begin{eqnarray*}
 \|FG\|_{{{\bf B}^{s}}}&=&\int_{-\infty}^{\infty} |\widehat{(FG)}(\tau)|(1+|\tau|)^{s}d\tau\\

 &\leq & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\big|  (\widehat{F} (\tau-\eta)   \widehat{G}(\eta) \big|
 |(1+|\tau-\eta|)^{s} |(1+|\eta|)^{s}d\eta  d\tau\\

\end{eqnarray*}
 and so by the Fubini theorem,
$$
\|FG\|_{{{\bf B}^{s}}}\leq \|F\|_{{{\bf B}^{s}}}\|G\|_{{{\bf B}^{s}}}.
$$

Finally, we can show the following result.

\begin{prop}\label{le2.3}
Suppose that  $L$ satisfies  the  Gaussian upper bounds \eqref{GE}.
Then for every  $s\geq 0$, there exists  a constant $C>0$ such that
for every $j=2,3,\ldots$
\begin{eqnarray}\label{e3.8} \hspace{1cm}
 \big\|P_{B}    F({L}) P_{U_j(B)}\big\|_{2\to 2}\leq
 C  \big(\sqrt[m]{R}2^j r)^{-s}   \|F(R\cdot)\|_{{{\bf B}^{s}}}
\end{eqnarray}
for all  balls $B\subseteq  X$, and all Borel functions $F$  such that supp $F\subseteq [-R, R]$.
 \end{prop}

\begin{proof}  Let $G(\lambda)=F({R\lambda}) e^{\lambda}.$ In virtue of the Fourier inversion formula
$$
F(L)=G(L/R)e^{-L/R}={1\over 2\pi} \int_{\mathbb R} e^{(i\tau-1)R^{-1}L} {\hat G}(\tau)d\tau
$$
we have that
$$
 \|P_{B}    F({L}) P_{U_j(B)}\|_{2\to 2} \leq
{1\over 2\pi} \int_{\R} |{\hat G}  (\tau)| \,   \|P_{B}   e^{(i\tau-1)R^{-1}L} P_{U_j(B)} \|_{2\to 2}
  d\tau.
$$
   Then it follows from Lemma~\ref{le2.2} for every $s\geq 0$,
 \begin{eqnarray*}
\|P_{B}   e^{(i\tau-1)R^{-1}L} P_{U_j(B)} \|_{2\to 2}&\leq&
C
  \exp\left(-c\left({ \sqrt[m]{R} 2^j r \over    \sqrt{1+\tau^2}}\right)^{\frac{m}{m-1}}   \right)\\
  &\leq& C_s     \left( {  \sqrt[m]{R} 2^jr \over \sqrt{1+\tau^2}    } \right)^{-s}.

\end{eqnarray*}
Therefore (compare \cite[(4.4)]{DOS})
  \begin{eqnarray*}
   \|P_{B}    F({L}) P_{U_j(B)}\|_{2\to 2}
   &\leq&
 C \big(\sqrt[m]{R} 2^j r)^{-s}
   \int_{\R} |{\hat G}  (\tau)|  \big(1+|\tau|)^{ s}  d\tau\nonumber\\
   &\leq&    C   \big(\sqrt[m]{R} 2^j r)^{-s}   \|G\|_{{{\bf B}^{s}}}.
\end{eqnarray*}
 Note that supp $F\subseteq [-R, R]$ and so supp $ F(R\cdot)\subseteq  [-1, 1]$. Thus taking a function
 $\psi\in C_c^\infty$ such that supp $\psi\subset [-2,2]$ and $\psi(\lambda)=1$ for $\lambda\in [-1,1]$, we have
 $$
 G(\lambda)=F({R\lambda}) e^{\lambda}=F({R\lambda}) \psi(\lambda)e^{\lambda}
 $$
 and so
 $$
  \|G\|_{{{\bf B}^{s}}} \leq C \|F(R\cdot)\|_{{{\bf B}^{s}}} \|\psi(\lambda)e^{\lambda}\|_{{{\bf B}^{s}}}
  \leq C \|F(R\cdot)\|_{{{\bf B}^{s}}} .
 $$
This ends the proof of Proposition~\ref{le2.3}.
\end{proof}

\begin{rem}\label{re3.4}\
In \cite[Proposition 4.1]{CCO}, Carron, Coulhon and Ouhabaz used some techniques introduced by Davies (\cite{D2})  to show that the upper
Gaussian estimate \eqref{GE}  on $e^{-tL}, t>0, $ extends to a similar estimate on $e^{-zL}$ where
$z$ belongs to  the whole complex right half-plane  and all $x, y\in X$,
\begin{eqnarray*}
|p_z(x,y)|\leq {C\over \left(V(x, ({|z|\over (\cos \theta)^{m-1} })^{1/m} )\, V(y, ({|z|\over (\cos \theta)^{m-1} })^{1/m} )\right)^{1/2}}
 \exp\left(-c \, {  \left({d(x,y)^{m}\over    |z|}\right)^{1\over m-1}}\cos\theta \right) {1\over (\cos\theta)^{n}}
 \end{eqnarray*}
where $\theta={\rm Arg} \, z$. It follows that
  for every $j=2,3,\ldots$
 \begin{eqnarray} \label{ecc}
\|P_{B}   e^{(i\tau-1)R^{-1}L} P_{U_j(B)} \|_{2\to 2}&\leq&
C 2^{jn}
  \exp\left(-c\left({ \sqrt[m]{R} 2^j r \over    \sqrt{1+\tau^2}}\right)^{\frac{m}{m-1}}   \right)
\end{eqnarray}
for all  balls $B\subseteq  X$. In our Lemma~\ref{le2.2}, we made an important improvement in obtaining
the upper bound on the right hand side  of \eqref{ecc} without the factor ``$2^{jn}$".
This plays an essential role in
 estimate \eqref{e3.8} of Proposition~\ref{le2.3} and
in the proof of Theorem~\ref{th1.1} in Section 4.
\end{rem}

\iffalse
\begin{rem}\label{re3.4}\
 When $L$  is a self-adjoint non-negative operator in $L^2(X)$   satisfying the generalized Gaussian estimate
   \eqref{GGE}  for some
  $1\leq p_0< 2$, it was shown in  \cite[Lemma 2.5]{KU} (see also \cite[Theorem 2.1]{Bl} and \cite[Lemma 2.1]{CDLY})
  that generalized Gaussian estimates can be extended from
real times $t>0$ to complex times   $z\in {\mathbb C}$ with ${\rm Re} z>0$ such that for every $j=2,3,\ldots$
 \begin{eqnarray} \label{ecc}
\|P_{B}   e^{(i\tau-1)R^{-1}L} P_{U_j(B)} \|_{2\to 2}&\leq&
C 2^{jn}
  \exp\left(-c\left({ \sqrt[m]{R} 2^j r \over    \sqrt{1+\tau^2}}\right)^{\frac{m}{m-1}}   \right)
\end{eqnarray}
for all  balls $B\subseteq  X$. In our Lemma~\ref{le2.2}, we can drop the factor ``$2^{jn}$" in \eqref{ecc}
in the case that when $L$
satisfies  a  Gaussian upper bound \eqref{GE}. From it,
we obtain  estimate \eqref{e3.8} in Proposition~\ref{le2.3}, which plays an  essential role
in the proof of Theorem~\ref{th1.1} in Section 4.
\end{rem}
\fi

\bigskip

\section{Proof of main Theorem}
\setcounter{equation}{0}

To prove   \eqref{e1.6}, it follows by Lemma~\ref{prop2.10}  and a standard argument (see for example, \cite{DY3, HLMMY, HM, KU}) that
it suffices to show  that for every $(1,2,M,\varepsilon)$-molecule $a$ associated to a
ball $B$,
\begin{eqnarray}\label{e3.3}
\|(1+L)^{-n/2}e^{itL}a\|_{L^1}\leq C(1+|t|)^{n/2}, \ \ \  t\in{\mathbb R}
\end{eqnarray}
where $M\in{\mathbb N}$  is large enough so that $M>n/2.$

Suppose that $a$ is a $(1,2,M,\varepsilon)$-molecule
supported in a ball $B=B(x_B,r)$ and $a=L^Mb$ such that  for every $k=0,1,2,\dots,M$ and $j=0,1,2,\dots$, there holds
\begin{eqnarray}\label{ee3.0}
\|(r^mL)^{k}b\|_{L^2(U_j(B))}\leq 2^{-j\varepsilon} r^{mM}
V(2^jB)^{-{1/2}},
\end{eqnarray}
where the annuli $U_j(B)$ were defined in (\ref{e2.3}).
  Following  \cite{HM},
we write
\begin{eqnarray}
&&\hspace{-1.2cm}I=m\, \Big( r^{-m}\int_{r}^{\sqrt[m]{2}r}s^{m-1}ds\Big)   \cdot I\nonumber\\
&=&mr^{-m} \int_{r}^{\sqrt[m]{2}r}s^{m-1}(I-e^{-s^mL})^Mds
 +  \sum_{\nu=1}^M C_{\nu,M}  r^{-m}
\int_{r}^{\sqrt[m]{2}r}
s^{m-1} e^{- \nu s^mL} ds,
\label{e3.5}
\end{eqnarray}
\noindent
where $C_{\nu, M}$ are some constants depending   on $\nu$ and $M$ only. However,
 $\partial_se^{-\nu s^mL}=-m\nu s^{m-1}Le^{-\nu s^mL} $ and therefore,
\begin{eqnarray}
m\nu L\int_{r}^{\sqrt[m]{2}r} s^{m-1} e^{-\nu s^mL} ds
&=&e^{-\nu r^mL}-e^{-2\nu r^mL}
=
e^{-\nu r^mL} (I-e^{-\nu r^mL})     \nonumber\\ [4pt]
&=&  e^{-\nu r^mL}
(I-e^{-r^mL})    \sum_{\mu=0}^{\nu -1}e^{-ir^mL}.
\label{e3.6}
\end{eqnarray}

In the following, we set $F(\lambda):=(1+\lambda)^{-n/2}e^{it\lambda}, t>0$.
Applying the procedure outline in (\ref{e3.5})-(\ref{e3.6})  $M$ times,
we have for every $x\in X,$
\begin{eqnarray*}
 F(L)a(x) &=&(1+L)^{-n/2}e^{itL}a(x)\\
&=&m^M\left(r^{-m}\int_r^{\sqrt[m] 2 r} s^{m-1}(1-e^{-s^mL})^M ds\right)^M F(L)a(x) \\
 &+&\sum_{k=1}^M (1-e^{-r^mL})^k \left(r^{-m}\int_r^{\sqrt[m] 2 r} s^{m-1}(1-e^{-s^mL})^{M}ds\right)^{M-k}\times\\
 &&\ \ \ \times  \sum_{\nu =1}^{(2M-1)k} C(\nu ,k, M) e^{-\nu  r^mL}F(L) (r^{-mk}L^{M-k} b)(x)\\
&=&\sum_{k=0}^{M-1} r^{-m}  \int_r^{\sqrt[m] 2 r} s^{m-1}(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L) (r^{-mk}L^{M-k} b)ds\\
 & +& \sum_{\nu =1}^{(2M-1)M} C(\nu, k, M) e^{-\nu  r^mL}F(L)(1-e^{-r^mL})^M(r^{-mM} b)(x)\\
&=:& \sum_{k=0}^{M-1} E_k(x)+ E_M(x),
\end{eqnarray*}
where for $k=1,2,\cdots,M-1$
$$
G_{k,r,M}(\lambda):=(1-e^{-r^m\lambda})^k \left(r^{-m}\int_r^{\sqrt[m] 2 r} s^{m-1}(1-e^{-s^m\lambda})^{M}ds\right)^{M-k-1}
\sum_{\nu =1}^{(2M-1)k} C(\nu,k, M) e^{-\nu r^m\lambda}
$$
and for $k=0$
$$
G_{0,r,M}(\lambda):=m^M\left(r^{-m}\int_r^{\sqrt[m] 2 r} s^{m-1}(1-e^{-s^m\lambda})^{M}ds\right)^{M-1}.
$$
We will establish an adequate bound on each $E_k, k=0, 1, \cdots, M$, by considering  two cases $k=0, 1, \cdots, M-1$
and $k=M.$

\smallskip

\noindent
{\bf Case 1.} \ $k=0, 1, \cdots, M-1$. \ In this case, we see that
  \begin{eqnarray}\label{e1}
\|E_k\|_{L^1(X)}&\leq& C\sup_{s\in [r,\sqrt[m] 2 r]}\left\|(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L) (r^{-mk}L^{M-k} b)\right\|_{L^1(X)}\nonumber\\
&\leq& C\sum_{j\geq 0}\sup_{s\in [r,\sqrt[m] 2 r]}\|(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L)P_{U_{j}(B)} (r^{-mk}L^{M-k} b)\|_{L^1(X)}\nonumber\\
&=:& C\sum_{j\geq 0}\sup_{s\in [r,\sqrt[m] 2 r]} \|E(k, j, s)\|_{L^1(X)},
\end{eqnarray}
where
$$
E(k, j, s)=(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L)P_{U_{j}(B)} (r^{-mk}L^{M-k} b).
$$

Let us estimate the term $\|E(k, j, s)\|_{L^1(X)}.$ Note that  $\|G_{k,r,M}\|_{L^\infty}+ \|F\|_{L^\infty}\leq C$.
We apply estimate  \eqref{ee3.0}  and   the doubling condition \eqref{eq2.2},
$$
\left({\mu((1+t)2^j B) \over  \mu(2^j B)} \right)\leq C (1+t)^n, \ \ \ \ j\geq 0,
$$
to get
\begin{eqnarray} \label{eddd}
\|E(k, j, s)\|_{L^1(4(1+t)2^j B)}&=&\left\|(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L) P_{U_{j}(B)} (r^{-mk}L^{M-k} b)\right\|_{L^1(4(1+t)2^j B)}\nonumber\\
&\leq& \left\|(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L) P_{U_{j}(B)} (r^{-mk}L^{M-k} b)\right\|_{L^2(4(1+t)2^j B)} \mu(4(1+t)2^j B)^{1/2}\nonumber\\
&\leq& C r^{-mk}\left\|P_{U_{j}(B)} L^{M-k}b\right\|_{2}\mu((1+t)2^j B)^{1/2}\nonumber\\
&\leq& C 2^{-j \varepsilon} \mu(2^j B)^{-1/2}\mu((1+t)2^j B)^{1/2}\nonumber\\
&\leq& C 2^{-j \varepsilon} (1+t)^{n/2}.
\end{eqnarray}
Next we show that for some $\varepsilon'>0$,
\begin{eqnarray} \label{eddde}
\|E(k, j, s)\|_{L^1((4(1+t)2^j B)^c)}   \leq  C2^{-j \varepsilon'}(1+t)^{n/2},
\end{eqnarray}
 and this is the major one.

Let
$\phi$ be a non-negative $C_c^{\infty}$ function on $\mathbb R$ such
that ${\rm supp}\  \phi \subseteq ({1/4}, 1)$ and
$$
\sum_{\ell\in {\Bbb Z}}\phi(2^{-\ell}\lambda)=1, \ \ \ \ \ \forall   \lambda>0,
$$
and let
$\phi_\ell(\lambda)$ denote the function $\phi(2^{-\ell}\lambda)$. Then
\begin{eqnarray} \label{eddde2}
(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L)
&=&
\sum_{\ell\in{\mathbb Z}} F_{\ell,s}(L),
\end{eqnarray}
where
\begin{eqnarray} \label{eddde1}
F_{\ell,s}(\lambda):=\phi_\ell(\lambda)(1-e^{-s^m\lambda})^{M}G_{k,r,M}(\lambda)F(\lambda).
\end{eqnarray}
Let $\nu_0\in {\mathbb Z}_+$ be a  positive integer such that
\begin{eqnarray}\label{bbb}
 1\leq 2^{\nu_0+j-\ell(m-1)/m}r\leq 2 \ \ \ \ {\rm if} \ \ \ 2^{j-\ell(m-1)/m}r<1;\nonumber\\[1pt]
 \nu_0=0 \ \ \ \ {\rm if} \ \ \ 2^{j-\ell(m-1)/m}r<1.
 \end{eqnarray}
 Then
by \eqref{eddde2},
\begin{eqnarray*}
\|E(k, j, s)\|_{L^1((4(1+t)2^j B)^c)}&=&\|(1-e^{-s^mL})^{M}G_{k,r,M}(L)F(L) P_{U_{j}(B)}(r^{-mk}L^{M-k} b)\|_{L^1((4(1+t)2^j B)^c)}\\
&\leq& \sum_{\ell\in {\Bbb Z}}\sum_{\nu\geq \nu_0} \|F_{\ell,s}(L)P_{U_{j}(B)}(r^{-mk}L^{M-k} b)\|_{L^1(P_{U_{\nu+j}((1+t)B)} )}\\
&+&\sum_{\ell\in {\Bbb Z}} \|F_{\ell,s}(L)P_{U_{j}(B)}(r^{-mk}L^{M-k} b)\|_{L^1(B(x_B,(1+t)2^{\ell(m-1)/m}))}\\
&=:&I(k, j, s)+II(k, j, s).
\end{eqnarray*}
Note that there is no term $II(k, j, s)$ if $2^{j-\ell(m-1)/m}r\geq 1$. So if $2^{j-\ell(m-1)/m}r\leq 1$,
then for the   term $II(k, j, s)$, we use estimate  \eqref{ee3.0} to get
\begin{eqnarray*}
II(k, j, s)&\leq& \sum_{\ell\in {\Bbb Z}}\|F_{\ell,s}(L)P_{U_{j}(B)}(r^{-mk}L^{M-k} b)\|_{L^2(B(x_B,(1+t)2^{\ell(m-1)/m}))} V\big(x_B,(1+t)2^{\ell(m-1)/m}\big)^{1/2}\\
&\leq& \sum_{\ell\in {\Bbb Z}}\|F_{\ell,s}\|_{L^\infty} \|r^{-mk}L^{M-k} b\|_{L^2(U_{j}(B))} V\big(x_B,(1+t)2^{\ell(m-1)/m}\big)^{1/2}\\
&\leq& \sum_{\ell\in {\Bbb Z}}C\min\{1, (2^{\ell/m} r)^M\} 2^{-n\ell/2} 2^{-j \varepsilon} V(x_B,2^j r)^{-1/2}V(x_B,(1+t)2^{\ell(m-1)/m})^{1/2}\\
&\leq& C2^{-j \varepsilon}\sum_{\ell\in {\Bbb Z}}\min\{1, (2^{\ell/m} r)^M\} (2^{\ell/m}r)^{-n/2}(1+t)^{n/2}\\
&\leq& C2^{-j \varepsilon}(1+t)^{n/2}.
\end{eqnarray*}

To estimate  term $I(k, j, s)$,  we apply   \eqref{ee3.0} and  the doubling condition \eqref{eq2.2} to get
\begin{eqnarray}\label{kkkk}
 I(k, j, s) &\leq&  \sum_{\ell\in {\Bbb Z}}\sum_{\nu\geq \nu_0} \|P_{U_{\nu+j}((1+t)B)}F_{\ell,s}(L)P_{U_{j}(B)}\|_{2\to 2}
\|r^{-mk}L^{M-k} b\|_{L^2(U_{j}(B))}V(x_B,2^{\nu+j}(1+t)r)^{1/2}\nonumber\\
&\leq&  \sum_{\ell\in {\Bbb Z}}\sum_{\nu\geq \nu_0} \|P_{U_{\nu+j}((1+t)B)}F_{\ell,s}(L)P_{U_{j}(B)}\|_{2\to 2}
2^{-j \varepsilon}
\left({V(x_B,2^{\nu+j}(1+t)r) \over V(x_B, 2^jr)}\right)^{1/2}\nonumber\\
  &\leq& C2^{-j \varepsilon}(1+t)^{n/2}\sum_{\ell\in {\Bbb Z}}\sum_{\nu\geq \nu_0} 2^{n\nu/2}\|P_{U_{\nu+j}((1+t)B)}F_{\ell,s}(L)P_{U_{j}(B)}\|_{2\to 2}.
\end{eqnarray}
By Proposition~\ref{le2.3},
\begin{eqnarray}\label{2to2offinproof}
\|P_{U_{\nu+j}((1+t)B)} F_{\ell,s}(L)P_{U_{j}(B)}\|_{2\to 2}\leq C \left(2^\nu 2^{\ell/m} (1+t)r\right)^{-\alpha}
\|\delta_{2^\ell} F_{\ell,s}\|_{{{\bf B}}^{\alpha } }
\end{eqnarray}
 for every $\alpha> 0$.
To go on, we claim  that
\begin{eqnarray}\label{claimBw}
\|\delta_{2^\ell} F_{\ell,s}\|_{{\bf B}^{\alpha } }
&=& \int_{-\infty}^{\infty} |\widehat{\delta_{2^\ell} F_{\ell,s}}(\tau)|(1+|\tau|)^{\alpha}d\tau\nonumber\\
&\leq&
C \max\{1,2^{(\alpha -n/2)\ell}\}(1+t)^\alpha \min\{1, (2^{\ell/m} r)^M\}.
\end{eqnarray}
Let us show the claim \eqref{claimBw}. In the sequel we let
 $\psi\in C_c^\infty$   supported in  $\psi\subset [1/8, 2]$ and $\psi(\lambda)=1$ for $\lambda\in [1/4,1]$.
We have that  for ${\ell\in {\Bbb Z}}$,
\begin{eqnarray*}
\|\delta_{2^\ell} F_{\ell,s}\|_{{\bf B}^{\alpha } }
&=& \| \phi(\lambda)(1-e^{-s^m2^\ell\lambda})^{M}G_{k,r,M}(2^\ell\lambda)F(2^\ell\lambda)\|_{{\bf B}^{\alpha } }\\
&\leq& \| \psi(\lambda)(1-e^{-s^m2^\ell\lambda})^{M}\|_{{\bf B}^{\alpha } }
\| \psi(\lambda)G_{k,r,M}(2^\ell\lambda)\|_{{\bf B}^{\alpha } }
\| \phi(\lambda)F(2^\ell\lambda)\|_{{\bf B}^{\alpha } }\\
&\leq& C\| \psi(\lambda)(1-e^{-s^m2^\ell\lambda})^{M}\|_{C^{\alpha+2}}
\| \psi(\lambda)G_{k,r,M}(2^\ell\lambda)\|_{C^{\alpha+2}}
\| \phi(\lambda)F(2^\ell\lambda)\|_{{\bf B}^{\alpha } }.
\end{eqnarray*}
Note that for every $s\in [r, \sqrt[m]{2}r]$,
$$
\| \psi(\lambda)(1-e^{-s^m2^\ell\lambda})^{M}\|_{C^{\alpha+2}}\leq C\min\{1,(2^{\ell/m}r)^M\}
$$
and
$$
\| \psi(\lambda)G_{k,r,M}(2^\ell\lambda)\|_{C^{\alpha+2}}\leq C
$$
with $C$ independent of $k,\ell$ and $r$. Let us  estimate $\| \phi(\lambda)F(2^\ell\lambda)\|_{{\bf B}^{\alpha } }$.
It follows from the Fourier transform ${\mathcal F} \big(\phi F(2^\ell\cdot)\big)$ of $\phi F(2^\ell\cdot)$ that
\begin{eqnarray*}
{\mathcal F} \big(\phi F(2^\ell\cdot)\big)(\tau)=\int_{\mathbb{R}} \phi(\lambda)\frac{e^{i( 2^\ell t-\tau)\lambda}}{(1+2^\ell\lambda)^{n/2}}d\lambda.
\end{eqnarray*}
Integration by parts gives for every $N\in \mathbb{N}$,
\begin{eqnarray*}
\left|{\mathcal F} \big(\phi F(2^\ell\cdot)\big)(\tau)\right|\leq
C_N\min\{1,2^{-\ell n/2}\}(1+|2^\ell t-\tau|)^{-N},
\end{eqnarray*}
which yields
\begin{eqnarray*}
\| \phi(\lambda)F(2^\ell\lambda)\|_{{\bf B}^{\alpha } }&\leq& C\min\{1,2^{-\ell n/2}\}\int_{\mathbb{R}} (1+|2^\ell t-\tau|)^{-N}
(1+|\tau|)^\alpha d\tau\\
&\leq& C\max\{1,2^{(\alpha-n/2) \ell}\}\left(1+ t\right)^\alpha.
\end{eqnarray*}
Hence,  for ${\ell\in {\Bbb Z}}$,
\begin{eqnarray*}
\|\delta_{2^\ell} F_{\ell,s}\|_{{\bf B}^{\alpha } }
&\leq&
C \max\{1,2^{(\alpha-n/2) \ell}\}(1+t)^\alpha \min\{1, (2^{\ell/m} r)^M\}.
\end{eqnarray*}
\iffalse
When $\ell=0$, it is seen that
\begin{eqnarray*}
\|\delta_{1} F_{0,s}\|_{{\bf B}^{\alpha } }
&\leq& C\| \psi(\lambda)(1-e^{-s^m\lambda})^{M}\|_{C^{\alpha+2}}
\| \psi(\lambda)G_{k,r,M}(\lambda)\|_{C^{\alpha+2}}
\| \phi_0(\lambda)F(\lambda)\|_{{\bf B}^{\alpha } }
\end{eqnarray*}
Integration by parts gives that for every $N\in \mathbb{N}$,
$$
\left|{\mathcal F} \big(\phi_0(\lambda)F(\lambda)\big)(\tau)\right|\leq C_N(1+| t-\tau|)^{-N}.
$$
From this, it is easy to see that $
\| \phi_0(\lambda)F(\lambda)\|_{{\bf B}^{\alpha } }
\leq C(1+t)^\alpha,
$
so
$$\|\delta_{1} F_{0,s}\|_{{\bf B}^{\alpha } }\leq C  (1+t)^\alpha \min\{1,  r ^M\}.
$$\fi
This proves our claim  \eqref{claimBw}.

Now, letting $\alpha$ be a fixed number such that $0<\alpha-n/2<\varepsilon$,
we substitute  \eqref{claimBw} and \eqref{2to2offinproof} back into   \eqref{kkkk} to
see that the term $ I(k, j, s)$ is controlled by
\begin{eqnarray*}
  C2^{-j \varepsilon}(1+t)^{n/2}
 \sum_{\ell\in {\Bbb Z}}\sum_{\nu\geq \nu_0} 2^{(-\alpha+n/2)\nu} ( 2^{\ell/m} r )^{-\alpha} \max\{1,2^{(\alpha-n/2) \ell}\}\min\{1, (2^{\ell/m} r)^M\}.
\end{eqnarray*}
From the  definition  of $\nu_0\geq 0$ in \eqref{bbb}, we get
 \begin{eqnarray*}
  I(k, j, s) &\leq & C2^{-j \varepsilon}(1+t)^{n/2}\sum_{\ell>0} (2^{j-\ell(m-1)/m}r)^{(\alpha-n/2)} ( 2^{\ell/m} r )^{-\alpha} 2^{(\alpha-n/2) \ell}\min\{1, (2^{\ell/m} r)^M\}\\
&+&  C2^{-j \varepsilon}(1+t)^{n/2}\sum_{\ell\leq 0} ( 2^{\ell/m} r)^{-\alpha} \min\{1, (2^{\ell/m} r)^M\}\\
&\leq& C2^{-j (\varepsilon-\alpha+n/2)}(1+t)^{n/2}\sum_{\ell>0}  (2^{\ell/m}r )^{-n/2}\min\{1, (2^{\ell/m} r)^M\}\\
&+&  C2^{-j \varepsilon}(1+t)^{n/2}\sum_{\ell\leq 0} ( 2^{\ell/m} r )^{-\alpha} \min\{1, (2^{\ell/m} r)^M\}\\
&\leq& C2^{-j \varepsilon'}(1+t)^{n/2}
\end{eqnarray*}
with $\epsilon'=\epsilon-\alpha+n/2.$

Combining two estimates of $I(k, j, s)$ and $II(k, j, s)$, we obtain \eqref{eddde}.
This, in combination with \eqref{eddd} and \eqref{e1}, shows that
\begin{eqnarray*}
 \sum_{k=0}^{M-1} \| E_k\|_{L^1(X)} \leq  C\sum_{k=0}^{M-1} \sum_{j\geq 0}\sup_{s\in [r,\sqrt[m] 2 r]} \|E(k, j, s)\|_{L^1(X)}
   \leq  C\sum_{j\geq 0}2^{-j \varepsilon'}(1+t)^{n/2} \leq C (1+t)^{n/2}.
\end{eqnarray*}

\medskip

\noindent
{\bf Case 2.} \ $k=M$.

In this case, we write
  \begin{eqnarray}\label{e11000}
\|E_M\|_{L^1(X)}&\leq& C \sum_{j=1}^{(2M-1)M}\left\|(1-e^{-r^mL})^{M}e^{-j r^mL} F(L) (r^{-mM}  b)\right\|_{L^1(X)}.
\end{eqnarray}
Similar to the proof of $E_k$ as in {\bf Case 1 }, we have that
$\| E_M \|_{L^1(X)}
   \leq  C\sum_{j\geq 0}2^{-j \varepsilon'}(1+t)^{n/2} \leq C (1+t)^{n/2}$.  Hence, we have proved
estimate (\ref{e3.3}), and then concluded  the proof of \eqref{e1.6}.

\medskip

For the proof of  \eqref{e1.5} in Theorem~\ref{th1.1},  as mentioned in Introduction,  it
was proved in  \cite[Theorem 1.1]{CDLY} in the   case that when $L$
satisfies  a  Gaussian upper bound.  We now  apply Theorem~5.7 in \cite{DY} to
give   a different proof by using  the complex interpolation method.

\begin{prop}\label{th4.1}
Suppose  that $(X, d, \mu)$ is  a  space of homogeneous type  with a dimension $n$.  Suppose that $L$
satisfies the property \eqref{GE}.
Then  there exists a  constant $C=C(n, m)>0$ independent of $t$ such that
 \begin{eqnarray} \label{e1.66}
   \left\| (I+L)^{-n/2 }e^{itL} f\right\|_{{\rm BMO}_L(X)} \leq C (1+|t|)^{n/2} \|f\|_{L^{\infty}(X)}, \ \ \ t\in{\mathbb R}.
 \end{eqnarray}

As a consequence, estimate  \eqref{e1.5} holds for all $1<p<\infty$.
\end{prop}

 \begin{proof} From Lemma~\ref{th7.5},
estimate \eqref{e1.66}  follows by   the dual argument.
Let us show \eqref{e1.5}. Consider the analytic family of operators
$$
T_z:=e^{z^2}(1+|t|)^{-zn/2}(1+L)^{-zn/2}e^{itL}, \ \ \ \  0\leq {\rm Re}\, z\leq 1.
$$
Note that $T_z$ is a holomorphic function of $z$ in the sense that
$$
z\to \int_X T_z f(x)g(x)d\mu(x)
$$
for $f,g\in L^2(X)$.
If $y\in{\mathbb R}$, then
$$
T_{iy}(L)=e^{-y^2}(1+|t|)^{-iyn/2}(1+L)^{-iny/2}e^{itL}.
 $$
The operators $T_{iy}(L)$ are bounded on $L^2(X)$ and
 $$
\|T_{iy}(f)\|_2=\|e^{-y^2}(1+|t|)^{-iyn/2}(1+L)^{-iny/2}e^{itL}\|_2\leq C\|(1+\lambda)^{-iny/2}e^{it\lambda}\|_{L^\infty}\|f\|_2
\leq C\|f\|_2
$$
with $C$ independent of $t$ and $y$.

On the other hand,  it follows from Proposition~\ref{prop2.7} that
$$
\|(1+L)^{-iny/2}\|_{H_L^1(X)\to H_L^1(X)}\leq C(1+|y|)^{n/2+1}.
$$
This, together with    Theorem 1.1, shows that $(1+L)^{-n/2}e^{itL}$ is   bounded from
$H^1_L(X)$ to $L^1(X)$ and
$$
\|T_{1+iy}(f)\|_{L^1(X)}\leq Ce^{1-y^2}(1+|t|)^{-n/2}\|(1+L)^{-n/2}e^{itL}\|_{H_L^1(X)\to L^1(X)} \|(1+L)^{-iny/2}f\|_{H_L^1(X)}
\leq C\|f\|_{H_L^1(X)}
$$
with $C$ independent of $t$ and $y$, and so
$$
\|T_{1+iy}(f)\|_{{\rm BMO}_L(X)}\leq C\|f\|_{L^\infty(X)}
$$
by duality.

Then by the complex interpolation theorem \cite[Theorem 5.7]{DY}, we have for $\theta=1-2/p$
and $s=n(1/2-1/p)$
$$
\|(1+L)^{-s}e^{itL}f\|_{p}=\|e^{-\theta^2}(1+|t|)^{\theta n/2}T_\theta f\|_{p}\leq C(1+|t|)^{s}\|f\|_p
$$
as desired for $2<p<\infty$. By duality, estimate~\eqref{e1.5} holds for $1<p<2$. This  completes the proof of Proposition~\ref{th4.1}.
\end{proof}

Finally, we achieve the proof of Theorem~\ref{th1.1}.

\vskip 0.08in
\noindent{\bf Proof of Theorem~\ref{th1.1}.} From  the proof of \eqref{e1.6}
and Proposition~\ref{th4.1}, Theorem~\ref{th1.1} follows readily.
\hfill{}$\Box$

   \medskip

  \begin{rem} When $L$ is the Laplace operator $\Delta$ on the Euclidean spaces ${\mathbb R}^n$,
    it is well known (\cite{FS}) that the Hardy space $H^1({\mathbb R}^n)$ has
	a characterization  in terms of Riesz transforms, that is,
  $f\in H^1({\mathbb R}^n)$ if and only if $R_j f\in L^1({\mathbb R}^n)$ for all $j=1, 2, \cdots, n$, where
 $R_j$ is the $j$-th Riesz transform given by
  $$
  {\widehat R_j f}(\xi)= -i{\xi_j\over |\xi|} {\widehat f}(\xi), \ \ \ \ \ \xi\in \mathbb{R}^n,\,\, j=1, 2, \cdots, n.
  $$
 Applying our Theorem~\ref{th1.1}, we obtain for all $t\in \mathbb R$
  \begin{eqnarray*}
   \left\|  (1+\Delta)^{-n/2} e^{it\Delta}  f\right\|_{H^1(\mathbb R^n)}
   &\leq &  C\sum_{j=1}^n
 \left\|  R_j (1+\Delta)^{-n/2} e^{it\Delta}  f\right\|_{L^1(\mathbb R^n)}\\
  &=&  C\sum_{j=1}^n
 \left\|   (1+\Delta)^{-n/2} e^{it\Delta}  R_jf\right\|_{L^1(\mathbb R^n)}\\
  &\leq &   C(1+|t|)^{n/2}\sum_{j=1}^n\|R_jf\|_{H^1 (\mathbb R^n)}\\
  &\leq &   C(1+|t|)^{n/2} \| f\|_{H^1 (\mathbb R^n)},
\end{eqnarray*}
where in the last inequality we used the fact that   Riesz transforms $R_j$ is bounded on the space $H^1({\mathbb R}^n).$
Hence, our Theorem~\ref{th1.1} gives a new proof of the result of   Miyachi (\cite{Mi1, Mi})
for the Schr\"odinger group  $e^{it\Delta}$ on the Euclidean spaces ${\mathbb R}^n$.

  It would be interesting to know that under the assumption \eqref{GE} of  the operator $L$ in Theorem~\ref{th1.1},
    whether  the operator $(I+L)^{-n/2 }e^{itL}$
  is bounded on the Hardy space $H^1_L(X),$
  that is,
  \begin{eqnarray*}
 \left\| (I+L)^{-n/2 }e^{itL} f\right\|_{H^1_L(X)} \leq C (1+|t|)^{n/2} \|f\|_{H^1_L(X)}, \ \ \ t\in{\mathbb R}
\end{eqnarray*}
for some constant $C=C(n, m)>0$ independent of $t.$
  \end{rem}

\bigskip

\section{An application to Riesz means of the solutions of the Schr\"odinger equations}
\setcounter{equation}{0}

 The aim of this section is to prove Theorem~\ref{th1.2}.
  To show  it,
	we need following result.

	\begin{thm}\label{th5.1}
Suppose  that $(X, d, \mu)$ is  a  space of homogeneous type  with a dimension $n$.  Suppose that $L$
satisfies  \eqref{GE}.
Then  there exists a  constant $C=C(n)>0$  such that for all $t\in {\mathbb R}\backslash \{0\},$
\begin{eqnarray} \label{e5.5}
 \left\| (I+|t|L)^{-n/2 }e^{itL} f\right\|_{L^1(X)} \leq C  \|f\|_{H^1_L(X)}.
\end{eqnarray}
\end{thm}

\begin{proof}
We prove this theorem by following the approach as in the proof of  Theorem~\ref{th1.1}.
 For the details, we leave to the reader.
\end{proof}

 \medskip

\begin{proof}[Proof of Theorem~\ref{th1.2} ]  The proof   is inspired by the idea of \cite{Sj}.
Take  a function $\Phi\in C^\infty(\R)$ such that $\Phi(t)=0$ if $t<1/2$ and $\Phi(t)=1$ if $t>1$.
Fix $s=n/2$. Define function $F$ by
$$
F(u)=I_s(1)(u)-C_s\Phi(u)u^{-s}e^{-i u},
$$
where $C_s$ is defined by
$$
s\int_{-\infty}^1 (1-\lambda)^{s-1}e^{i\lambda u}d\lambda=C_s u^{-s}e^{iu}, \quad u>0.
$$
It is seen that for $0<u\leq 1$ and $k\in \mathbb{N}$,
$$
\frac{d^k}{d u^k}F(u)\leq C,
$$
and   for $u>1$ and $k\in \mathbb{N},$
$$
\frac{d^k}{d u^k}F(u)\leq Cu^{-k},
$$
see   \cite[Lemma 2.1]{Sj}.
Hence,  for every $\beta>(n+1)/2$  we have that $
\sup_{R>0}\|\eta\delta_R F\|_{C^\beta}\leq C,
 $
 and so
 $
\sup_{R>0}\|\eta\delta_R F(t\cdot)\|_{C^\beta}\leq C
 $
with a constant $C>0$ independent of $t>0$. Then we apply Proposition~\ref{prop2.7}
to know that $F(tL)$ is bounded on  $H_L^1(X)$. Notice that for every $t>0,$
\begin{eqnarray} \label{vvc}
F(tL)=I_s(t)(L)-C_s\Phi(tL)(tL)^{-s}e^{-i tL}.
\end{eqnarray}
This yields  that for every $t>0,$
\begin{eqnarray} \label{vvcc}
\|I_s(t)(L)\|_{H^1_L\to L^1}&\leq& \|F(tL)\|_{H^1_L\to L^1}+C\|\Phi(tL)(tL)^{-s}e^{-itL}\|_{H^1_L\to L^1}\nonumber\\
&\leq& C+C\|\Phi(tL)(tL)^{-s}(1+tL)^s\|_{H^1_L\to H^1_L}\|(1+tL)^{-s} e^{-itL}\|_{H^1_L\to L^1}.
\end{eqnarray}
Applying Proposition~\ref{prop2.7} again, we have that $ \|\Phi(tL)(tL)^{-s}(1+tL)^s\|_{H^1_L\to H^1_L}  \leq C. $
This, in combination with  \eqref{e5.5} in Theorem~\ref{th5.1},  implies  $
\|I_s(t)(L)\|_{H^1_L\to L^1}\leq C
$
for $t>0$.
Since $I_{s}(t)(L)={\overline I}_{s}(-t)(L)$  for $t<0$, we have  that $\|I_s(t)(L)\|_{H^1_L\to L^1}\leq C
$
for $t<0$. This proves \eqref{nnnw}.

The proof of \eqref{nnnww} is similar to that  of Proposition~\ref{th4.1}
   by using  the complex interpolation theorem, and we leave the detail to the reader. Hence, the proof of Theorem~\ref{th1.2} is complete.
\end{proof}

\medskip
 \section{Extensions}
\setcounter{equation}{0}

 In Section 4, we proved estimate \eqref{e1.6} of Theorem~\ref{th1.1} under assumption that $L$ satisfies the Gaussian estimate
 \eqref{GE} with $m\geq 2$.
 However,  the assumption
 \eqref{GE} of the operator $L$ does not play a fundamental role in the estimates.

Consider  a non-negative self-adjoint operator   $L$ on $L^2({X})$ and  number $m\geq 2$.
We say that  $L$ satisfies   {\it $m$-th order Davies-Gaffney  estimates},
that is,  there exist constants $C, c>0$ such that
for all $t>0$,  and all $x,y\in X,$
 \begin{equation*}
 \label{DG}
 \tag{${\rm DG}_m$}
\big\|P_{B(x, t^{1/m})} e^{-tL} P_{B(y, t^{1/m})}\big\|_{2\to {2}}\leq
C   \exp\left(-c\left({d(x,y) \over    t^{1/m}}\right)^{m\over m-1}\right).
\end{equation*}
There are numbers of operators which satisfy Davies-Gaffney  estimates \eqref{DG} and, among them,
 there exist many for which classical Gaussian estimates \eqref{GE}  fail.
  This happens, e.g., for Schr\"odinger operators with rough
 potentials \cite{ScV}, second order elliptic operators with rough  lower order terms \cite{LSV}, or
 higher order elliptic operators with bounded measurable coefficients
 \cite{D2}.

\iffalse
It is seen that the Gaussian estimate
 \eqref{GE} implies   estimate \eqref{e7.1}. Note that if
 $L$ satisfies the   condition \eqref{e7.1}, then the Davies-Gaffney estimate
   \eqref{DG}  holds.
For the second order like operators (i.e., $m=2$),   it is shown in   \cite[Section 3]{CouS})  that  both  estimates
 ${\rm (DG)_2}$ and \eqref{e7.1}  are equivalent.  However,
  it    is  not clear for us whether or not the Davies-Gaffney estimate
   \eqref{DG}    is equivalent to  \eqref{e7.1}  for general $m$-order operators.
\fi

The Hardy spaces $H_{L}^p(X), 1\leq p<\infty$ and ${\rm BMO}_L(X)$ for operators satisfying the Davies-Gaffney estimate \eqref{DG}
were introduced and studied in \cite{DL, DY3, HLMMY, KU} and the references therein,
 and they are  known to possess nice properties, for example, they form a complex interpolation scale, and
 the Hardy space  $H^1_L(X)$ has a molecular characterization.

To  extend   Theorem~\ref{th1.1}, we need to introduce a slightly different  form of the Davies-Gaffney estimate:
 there exist constants $C$, $c>0$ such that for any open subsets
$U_1,\,U_2\subset X$,
\begin{eqnarray}\label{e7.1}
|\langle e^{-tL}f_1, f_2\rangle|
\leq C\exp\left(-c\left({{\rm dist}(U_1,U_2) \over c\,t^{1/m}}\right)^{m\over m-1}\right)
\|f_1\|_{L^2(X)}\|f_2\|_{L^2(X)},\quad\forall\,t>0,
\end{eqnarray}
 for every $f_i\in L^2(X)$ with
$\mbox{supp}\,f_i\subset U_i$, $i=1,2$,
where ${\rm dist}(U_1,U_2):=\inf_{x\in U_1, y\in U_2} d(x,y).$

 It is seen that the Gaussian estimate
 \eqref{GE} implies   estimate \eqref{e7.1}. Note that if
 $L$ satisfies the   condition \eqref{e7.1}, then the Davies-Gaffney estimate
   \eqref{DG}  holds.
For the second order like operators (i.e., $m=2$),   it is shown in   \cite[Section 3]{CouS})  that  both  estimates
 ${\rm (DG)_2}$ and \eqref{e7.1}  are equivalent.

 Our result can be stated as follows.

\begin{thm}\label{th6.1}
Suppose  that $(X, d, \mu)$ is  a  space of homogeneous type  with a dimension $n$.  Suppose that $L$
satisfies the   estimate \eqref{e7.1}.
Then  there exists a  constant $C=C(m, n)>0$ independent of $t$ such that
\begin{eqnarray} \label{e6.2}
  \left\| (I+L)^{-n/2 }e^{itL} f\right\|_{L^1(X)} \leq C (1+|t|)^{n/2} \|f\|_{H_L^1(X)}, \ \ \ t\in{\mathbb R}
\end{eqnarray}
and  by duality,
\begin{eqnarray}\label{e6.3}
\|(1+L)^{-n/2}e^{itL}f\|_{{\rm BMO}_L(X)}\leq C(1+|t|)^{n/2}\|f\|_{L^\infty(X)}, \ \ \ t\in{\mathbb R}.
\end{eqnarray}
\end{thm}

\begin{proof}
Since $L$ has the property \eqref{e7.1}, Lemma~\ref{le2.2} and Proposition~\ref{le2.3} hold. Then we follow
an argument
 in the proof of Theorem~\ref{th1.1} to
show that for every
$(1,2,M)$-molecular $a$ associated to  an operator $L$,
\begin{eqnarray}\label{e6.4}
\|(1+L)^{-n/2}e^{itL}a\|_{L^1(X)}\leq C(1+|t|)^{n/2},
\end{eqnarray}
to obtain the proof of \eqref{e6.2}.
 For the detail, we leave   to the reader.  This proves   Theorem~\ref{th6.1}.
\end{proof}

  \begin{rem}

   For the second order like operators (e.g. $m=2$),
  assume  that  $L$   satisfies the generalized Gaussian estimate (${\rm GGE_{p_0, p'_0, 2} }$)
  for some $ 1\leq p_0\leq 2 $, that is,
    there exist constants $C, c>0$ such that
 the semigroup $e^{-tL}$  generated by $L$,  satisfies
 \begin{equation*}
  \label{GGE}
 \tag{${\rm GGE_{p_0,p'_0, 2} }$}
 \big\|P_{B(x, t^{1/2})} e^{-tL} P_{B(y, t^{1/2})}\big\|_{p_0\to {p'_0}}\leq
 C V(x,t^{1/2})^{-({\frac{1}{ p_0}}-{1\over p'_0})} \exp\left(-c {d(x,y)^2 \over    t } \right)
 \end{equation*}
 for  every $t>0$ and $x, y\in X$.
 Note that condition \eqref{GGE} for the special case $p_0=1$ is equivalent to
the Gaussian upper bounds  ${\rm (GE)_2}$ (see   for example, \cite{BK2}).

    It is known (see for example, \cite{DY}) that $H^p_L(X)=L^p(X)$ for $p_0<p\leq 2$.
  From Theorem~\ref{th6.1} (i.e. $m=2$),  it would be  interesting whether or not we   establish
  a  complex interpolation theorem to  obtain
 \begin{eqnarray*}
 \left\| (I+L)^{-s }e^{itL} f\right\|_{L^p(X)} \leq C (1+|t|)^{s} \|f\|_{H^p(X)}, \ \ \ t\in{\mathbb R}, \ \
  \ s\geq n\big|{1\over  2}-{1\over  p}\big|
\end{eqnarray*}
for every  $1<p\leq 2$, which gives  a new proof of the result \eqref{e1.5} for   $p\in (p_0, p'_0)$ obtained   in  \cite[Theorem 1.1]{CDLY}
in the case that when $L$ satisfies the property (${\rm GGE_{p_0,p'_0, 2}}$) for some $1\leq p_0<2$.
  \end{rem}

 \medskip
\noindent
{\bf Acknowledgements}:
P. Chen was supported by NNSF of China 11501583, Guangdong Natural Science Foundation
2016A030313351.
Duong and Li were supported by the Australian Research Council (ARC) through the
research grant DP160100153 and by Macquarie University Research Seeding Grant.
 Yan was supported by the NNSF
of China, Grant No. ~11521101 and  ~11871480, and Guangdong Special Support Program.
 L. Yan would like to thank A. Sikora for helpful discussioons.

 \vskip 1cm

redpajama_set_name: RedPajamaArXiv


A Lil' Duck, handsewn duck, 3 Ultra Mini Uglies (pattern coming later this weekend), and a Lil Ugly about to fly away. They'll all be released at the Sheep to Shawl Festival at Philipsburg Manor in Sleepy Hollow, NY as soon as they put on their Little Yellow Duck tags. How could they not be released at an event that features sheep shearing, yarn spinning, AND duck herding!


text:
There Are No Small Parts: Minor Characters in David Copperfield
Three 50-minute sessions
Scott Filkins
Capitalizing on the enormous cast of characters in Dickens' novel-and students' interest in the phenomenon of social networking, this lesson asks students to "fill in the gaps" of the life story of a minor character from David Copperfield. Small groups choose a minor character from the novel and determine everything they already know about him or her. They then use Web research and imaginative thinking to create a "back story" for the character. Groups share their new perspective on the character with a social networking page for him or her, with options for extending the activity in several different directions.
This lesson was adapted from material originally developed for Teaching Dickens: A Masterpiece Guide by Katherine Schulten and Georgia Scurletis, published by WBGH Boston. This companion guide accompanies the 2009 Masterpiece presentation of "The Tales of Charles Dickens."
Profile Publisher: Students can use this online tool to draft online social networking profiles, yearbook profiles, and newspaper or magazine profiles for themselves, other real or fictional characters.
You Gotta BE the Book: Teaching Engaged and Reflective Reading With Adolescents (Wilhelm)
In "You Gotta BE the Book": Teaching Engaged and Reflective Reading with Adolescents, Jeffrey Wilhelm outlines ten dimensions of response that he sees students using as they work with a literary text. Wilhelm calls one of those dimensions "elaborating on the story world" and explains that students engaging in that connective dimension "build up clues from throughout the story to make meaning; play detective, flesh out clues, and fill in story gaps, often creating meaning that goes well beyond that suggested by the text" (68). This lesson invites student to engage in that dimension of creative and inferential response, giving them a structure for such imaginative connection and allowing them to match imaginations with Charles Dickens.
Grade Select9101112
Copies of David Copperfield or another novel by Charles Dickens
Large piece of chart paper and markers
Major and Minor Characters chart
Filling in the Gaps in a Minor Character Planning Sheet
Analyzing Character and Elaborating on the Story World
Fictional Character Profile sample
Minor Character Online Profile Rubric
Web Resources for Exploring the Times of Charles Dickens
Students should have read David Copperfield, another novel by Charles Dickens, or another novel with a similarly rich cast of minor characters.
Although this lesson is a culminating activity and is intended to come after reading the novel, you may wish to prepare students by having them track a character as they read the book.
While this lesson includes some instruction and materials related to characterization and the Victorian Era, it assumes that these topics will have been part of the instruction during the reading of the novel. ReadWriteThink lessons that support instruction related to characterization include Become a Character: Adjectives, Character Traits, and Perspective and Beyond the Story: A Dickens of a Party. Use the links from the Dickens/Victorian Era Web resources to support learning about the historical era associated with Dickens's work.
Arrange for Internet access in a computer lab during Session Three.
Test the Profile Publisher tool and familiarize yourself with its contents and layout. Ensure that you have the proper Flash plug-ins installed. You can download the plug-in from the technical support page.
select a minor character and investigate what is already known about him or her through close review of the text.
determine what is unknown about the character that might be useful in developing a "back story" for the character.
use imaginative thinking to develop a plausible but creative "back story" for the character.
Post the large piece of chart paper prominently in the room.
Prompt a review of David Copperfield by asking individual students to write the name of any character from the book on the paper. The first student volunteer should write a name on the paper and pass the marker on to another student.
As students add characters' names to the paper, ask them to draw lines between characters indicating connections that they have. They should put a brief descriptor on the line indicating the nature of the relationship between the characters. For example, the line between David Copperfield and Clara might say "mother, died when DC was young" while the line between David and Mr. Micawber might say "friendly landlord." There would be no line connecting Clara Copperfield and Mr. Micawber.
The paper will fill up quickly, and it will become more and more difficult for students to make complete and accurate connections (or find space for doing so).
After every student has had a turn, ask students if there are characters that have been missed. Add these names to the chart paper. Do not worry about completing connections at this point. Simply providing a visual representation of the richness of the novel's cast of characters is the goal of this part of the activity.
When the class is reasonably satisfied that the list is complete (a complete list is available), arrange students into small groups and ask each group to use the Major and Minor Characters chart to classify the characters on the chart paper into two categories: major and minor.
As students are doing so, prompt them to think about how they are making their choices. Is a character minor because he or she appears only briefly? Because his or her influence is not substantial? Or is there some other way to determine who readers consider major or minor?
Facilitate a brief discussion of the students' lists and, on the board or overhead, create a list of characters that the class agrees are "minor."
Then share the following quote from an interview with Simon Curtis, director of the Masterpiece film adaptation of David Copperfield:
Every single character in Dickens has something about him or her that's very actable. Even if it's just a one-line part, every character has a back story and a life. That's what draws actors to these parts.
Ask students to consider Curtis's comment about the "back story" and "life" of the minor characters on the list. Give groups time to look over the list and choose one minor character they would like to investigate further and "fill in the gaps" through creative thinking. Have them record and explain their choice on the Major and Minor Characters chart.
Begin the session by asking students to recall the character their group chose. Distribute the Filling in the Gaps in a Minor Character Planning Sheet and ask students to complete independently the first column for their character with everything they already know or remember about him or her.
Then ask students to get back into their groups and share their brainstormed individual lists. At this point, they should return to the text to get a complete and accurate picture of the character as he or she is presented in the novel.
When students have completed a close look at the novel, ask them to start thinking about the second column on the planning sheet: What information does Dickens not give about their character? What do they still want to know? What part of that character's story would they still like to hear?
Share some basic information on characterization by distributing the Analyzing Character and Elaborating on the Story World handout and prompting students to think about areas they would like to explore further.
By the end of the session, the groups should have consensus on the elements they plan to develop for their character.
Make students aware of the Dickens/Victorian Era Web resources if they need to do research to remain historically accurate or to spark some ideas.
As homework for the next session, each member of the group should complete the last column of the chart with creative ideas that "fill in the gaps" about their character's life and personality.
Begin the session by sharing a completed example of the project for this lesson: an online social networking profile for a minor character. Though this kind of networking technology certainly did not exist in the Victorian era, ask students to consider the attraction it might have held for Dickens, especially its ability to make connections among disparate people (just as the chart from Session One demonstrates.)
Distribute copies or share an overhead of a completed profile for the character of Herbert Pocket from Dickens' novel Great Expectations. Share the Minor Character Online Profile Rubric to clarify expectations for the assignment, pointing out that they will be completing a profile for the character at one specific point of the book.
Have students return to their groups and give them time to share the creative ideas they generated from the third column of their Filling in the Gaps in a Minor Character Planning Sheet.
Direct students to the Profile Publisher and ask them to complete the sections of the tool for their character. They should combine information from the text with the imaginative work they did.
Have students present their profiles and facilitate a conversation about their choices. Given the characters students chose for their profiles, have them consider who would likely be online "friends" with one another.
At the close of that conversation, ask students to complete the Reflection Questions.
Though this lesson invites students to focus on minor characters for further exploration, the richness of Dickens's major characters would allow for similar imaginative exploration as well.
Have students write a letter to an actor who has been cast to play the part of their character. Watch the Masterpiece Theater version of David Copperfield and have students reflect on how well their imagined back story seems to compare with the choices made by the actor in the part.
Let students further develop their characters' back stories using the online Bio-cube tool.
Ask students to write the character's back story in a chapter written in Dickens' style. Use resources from the ReadWriteThink lesson Style: Defining and Exploring an Author's Stylistic Choices to support students in this activity.
Though this lesson takes a creative approach to thinking about a minor character in the novel, the initial thinking can give rise to a more traditional essay in which students explore the significance of a character they initially considered minor or insignificant.
Find other teaching ideas for David Copperfield and other works by Charles Dickens at Teaching Dickens: A Masterpiece Guide.
To provide context for David Copperfield, have students read about Charles Dickens' life and work at A Charles Dickens Journal or Dickens on PBS.
Explore how Dickens's skill in creating memorable secondary characters help make his works not only memorable, but has kept them popular for more than 160 years in this video excerpt from the MASTERPIECE adaptation of Charles Dickens' David Copperfield.
Explore Dickens's skill in creating memorable secondary characters by watching this video excerpt from the PBS MASTERPIECE adaptation of David Copperfield.
Students should respond to the Reflection Questions, discussing their creative choices and their work within the group.
Take student rationales in to account as you respond to their projects using the Minor Character Online Profile Rubric.
Profile Publisher
Students use the Profile Publisher to draft online social networking profiles, yearbook profiles, and newspaper or magazine profiles for themselves, other real or fictional characters.
From Text to Film: Exploring Classic Literature Adaptations
Students create storyboards to compare and contrast a book and its film adaptation.
A Christmas Carol was the first book transmitted over radio!
Students write a script for A Christmas Carol on a level that primary students can read and perform. Older students record the performance and create a website to showcase photographs.


text:
'Made to Move': Time to Act!
Love You Bike is calling on the Greater Manchester Combined Authority to support Chris Boardman's (Greater Manchester cycling and walking commissioner) 'Made to Move: 15 steps to transform Greater Manchester' plan at the GMCA meeting on Friday 15th December. [1]
Made to Move aims to put "people first, creating world class streets for walking, building one of the world's best cycle networks within a generation" and outlines 15 steps to achieve this goal including a commitment to establish a "ring-fenced, 10 year, £1.5 billion infrastructure fund and immediate short term Mayor's Active Streets Fund to maintain delivery for walking and cycling."
Love Your Bike strongly supports the Made to Move plan and is encouraging people across Greater Manchester to contact their Council leaders and urge them to support the plan. [2]
The recently published Bike Life Greater Manchester 2017 report highlighted that:
* 77% of residents support building more protected roadside cycle lanes, even when this could mean less space for other road traffic.
* £70 million annual benefit to Greater Manchester from all trips currently made by bicycle.
* 86p per mile net benefit for each mile cycled instead of driven. [3]
More than one million cycle journeys have been made on the Oxford Road and Wilmslow Road Cycleway in Manchester since recording began in September 2016. [4]
Pete Abel, from Love Your Bike said: "The Oxford Road route has demonstrated that when we provide safe and convenient cycle routes more people will cycle and more than 75% people in Greater Manchester support building more protected cycle lanes."
"The benefits from increased cycling and walking are clear, it is time for the Greater Manchester Combined Authority to show leadership and support the Made to Move plan and provide the funds required to deliver high quality cycle schemes such as the Manchester – Chorlton Cycle Way."
Love Your Bike welcomes Andy Burnham's comment that the £243 million for transport improvements in Greater Manchester announced in the Budget could be used to improve cycling infrastructure in Greater Manchester [5]
Enabling more people to choose to walk or cycle can help Greater Manchester tackle congestion, physical inactivity, climate change and air pollution. [6]
In 2012, Love Your Bike published the 'Getting moving: a cycling manifesto for Greater Manchester' [7] supported by British Cycling, Cycling UK, Sustrans and 30 other businesses/organisations in Greater Manchester. It contains 5 headline actions:
* Political leadership & governance that facilitate the transition to a cycling city
* High standards of design to enable all people to cycle confidently and safely
* Sustained promotion of cycling as a mode of transport
* Safe cycling and safe driving through facilitation and enforcement
* Integration of cycling with other modes of transport
and called for funding of a minimum of £20 per person in Greater Manchester.
Contact for comments:
Pete Abel, Manchester Friends of the Earth. Mobile: 07951 642858
Graeme Sherriff, Manchester Friends of the Earth, Sustainable Transport campaign co-ordinator 07948 405071
[1] Link to GMCA agenda on website.
[2] See https://www.manchesterfoe.org.uk/made-to-move-1-5-billion-reasons-to-act/
[3] Bike Life Greater Manchester 2017
[4] https://tfgm.com/news/cycleway-surpasses-one-million-milestone
[5] See 1:48 of the video clip. Mayor Andy Burnham (@MayorofGM) tweeted at 4:30 pm on Wed, Nov 22, 2017: Watch what Andy thinks about #Budget2017 #BudgetDay https://t.co/uyZudkUbE5
[6] See 'The role of walking and cycling in solving the UK's air quality crisis. Sustrans / Eunomia.
[7 ] Love Your Bike https://www.manchesterfoe.org.uk/loveyourbike/cyclingmanifesto/
[8] The Love Your Bike campaign is based in Greater Manchester. We aim to promote cycling and to help make it an attractive, accessible and fun way to get around. Cycling is far more than a leisure activity: it is the low-carbon transport of the future. Co-ordinated by Manchester Friends of the Earth, we aim to work closely with other like-minded organisations. Launched in March 2006, we celebrated our 10th anniversary in 2016.
[9] Manchester Friends of the Earth is an award-winning environmental campaign group, raising awareness and lobbying for policy changes at a local, regional, national and international level. The group consists entirely of volunteers, and its campaigns are funded by membership fees and individual donations. Up-to-date information is available on the group's website: www.manchesterfoe.org.uk. Manchester Friends of the Earth is a Licensed Local Group of Friends of the Earth, England, Wales and Northern Ireland. www.foe.co.uk