3. Transfer of the properties of the Murshidabad Estate to the State Government.
4. General effect of transfer.
5. Appointment of Estate Manager.
6. Management and administration of Estate properties.
8. Issue of notice to show cause against order of eviction of unauthorised occupants of any premises included in Estate properties.
9. Eviction of unauthorised occupants.
10. Alternative accommodation for the surviving sons and daughter of the late Nawab Bahadur.
11. Power of the Estate Manager to dispose of immovable property.
12. Application of income from Estate properties.
13. Management of Hazar Duari Palace and Imambarah.
14A. Bar of jurisdiction of Civil Courts.
Assent of the President was first published in the Calcutta Gazette, Extraordinary, dated the 11th March, 1981.
An Act to provide for the management of properties of the Murshidabad Estate and for matters connected therewith or incidental thereto.
1. Short title and commencement. - (1) This Act may be called the Murshidabad Estate (Management of Properties) and Miscellaneous Provisions Act, 1980.
(2) It shall come into force on such date as the State Government may by notification appoint.
[(i) "unauthorised occupation"], in relation to any premises or other immovable properties, means the occupation by any person of the premises or other immovable properties without authority for such occupation, and includes the continuance in occupation by any person of the premises or other immovable properties after the authority (whether by way of grant or any other mode of transfer) under which he was allowed to occupy the premises or other immovable properties has expired or has been determined for any reason whatsoever.
3. Transfer of the properties of the Murshidabad Estate to the State Government. - (1) On and from the appointed day, the properties of the Murshidabad Estate (hereinafter referred to as the Estate properties) shall, by virtue of this Act, stand transferred to, and vest in, the State Government.
(2) Upon the transfer of the Estate properties to, and vesting thereof in, the State Government under sub-section (1), the State Government may take, or cause to be taken, such steps as it considers necessary for securing the possession thereof.
4. General effect of transfer. - (1) For the purpose of managing and administering the Estate properties (including any income therefrom), the State Government may take such measures as it considers necessary or expedient and do all acts and incur all expenditure necessary or incidental thereto.
(2) Any payment made to the State Government in the discharge of any obligation in relation to any of the Estate properties shall be a full and valid discharge to the person making the payment from all liabilities in respect thereof.
(3) Any payment made otherwise than in accordance with sub-section (2) shall not be deemed to discharge the person paying it from his obligation to make the payment and shall not affect the right of the State Government to enforce such obligation against that person.
(4) Any suit or proceeding pending before the appointed day in relation to any of the Estate properties may be continued by or against the State Government, as the case may be, and" not otherwise.
(5) All valid contracts subsisting immediately before the appointed day in relation to any of the Estate properties shall be of as full force and effect against or in favour of the State Government and shall be enforceable as fully and effectively against or by the State Government as if the State Government had always been named therein and had been a party thereto.
5. Appointment of Estate Manager. - (1) As soon as may be after the appointed day, the State Government shall by notification appoint such person as it thinks fit to be Estate Manager on such terms and conditions as the State Government may determine.
(2) The Estate Manager shall exercise the powers conferred, and perform the duties imposed, on him by or under this Act.
6. Management and administration of Estate properties. - The Estate properties which stand transferred to and vest in the State Government under sub-section (1) of section 3 shall, subject to the provisions of this Act and the rules made thereunder, be managed and administered on behalf of the State Government by the Estate Manager.
Provided that the Estate Manager may, with the previous approval of the State Government, re-employ any such person in connection with the management and administration of the Estate properties on such terms and conditions as the State Government may determine.
8. Issue of notice to show cause against order of eviction of unauthorised occupants of any premises included in Estate properties. - (1) If the Estate Manager is of opinion that any persons are in unauthorised occupation of [any premises or other immovable properties included in the Estate properties (hereinafter referred to as the premises or other immovable properties)] and that they should be evicted, the Estate Manager shall issue in the manner hereinafter provided a notice in writing calling upon all persons concerned to show cause why an order of eviction should not be made.
(b) require all persons concerned, that is to say, all persons who are, or may be, in occupation of, or claim interest in, 5[the premises or other immovable properties,] to show cause, if any, against the proposed order on or before such date as is specified in the notice, being a date not earlier than ten days from the date of issue thereof.
(3) The Estate Manager shall cause the notice to be served by having it affixed on the outer door or some other conspicuous part of [the premises or other immovable properties,] and in such other manner as may be prescribed, whereupon the notice shall be deemed to have been duly given to all persons concerned.
(4) Where the Estate Manager knows or has reasons to believe that any persons are [in unauthorised occupation of the premises or other immovable properties,] then, without prejudice to the provisions of sub-section (3), he shall cause a copy of the notice to be served on every such person by post or by delivering or tendering it to that person or in such other manner as may be prescribed.
9. Eviction of unauthorised occupants. - (1) If, after considering the cause, if any, shown by any person in pursuance of a notice under section 8 and any evidence he may produce in support of the same and after giving him a reasonable opportunity of being heard, the Estate Manager is satisfied that the premises are in unauthorised occupation, the Estate Manager may. on a date to be fixed for the purpose, make an order of eviction, for reasons to be recorded there in, directing that the premises shall be vacated by all persons who may be in occupation thereof or any part thereof, and cause a copy of the order to be affixed on the outer door or some other conspicuous part of the premises.
(2) If any person refuses or fails to comply with the order of eviction within thirty days of the date of its publication under sub-section (1), the Estate Manager or any other officer duly authorised by the Estate Manager in this behalf may evict that person from, and take possession of, the premises and may, for that purpose, use such force as may be necessary.
(2) If any person refuses or fails to comply with the order of eviction made under sub-section (1) within the period specified in the order or within fifteen days of its publication under sub-section (1), whichever is later, the Estate Manager or any other officer duly authorised by the Estate Manager in this behalf may evict that person from, and take possession of, the premises or other immovable properties and may, for that purpose, use such force as may be necessary.
(3) The cost of taking possession of the premises of other immovable properties under sub-section (2) shall be payable by the person from whom possession of the premises or other immovable properties has been taken within such period as the Estate Manager may by order direct.
[9A. Disposal of property.]- (1) Where any persons have been evicted from any premises or other immovable properties under section 9, the Estate Manager may, after giving fourteen days� notice to the persons from whom possession of the premises or other immovable properties has been taken, remove or cause to be removed or dispose of by public auction any property remaining on such premises.
(2) Where any property is sold under sub-section (1), the sale proceeds shall, after deducting the expenses of the sale and the amount, if any, due to the Estate Manager on account of arrears of rent or damages or costs, be paid to such person or persons as may appear to the Estate Manager to be entitled to the same.
[9B. Appeal.-] (1) An appeal shall lie from every order of the Estate Manager made under section 9 to an appellate officer who shall be the district judge of the district in which the premises or other immovable properties are situate or such other judicial officer in that district of not less than ten years� standing as the district judge may designate in this behalf.
Provided that the appellate officer may entertain the appeal after the expiry of the said period of fifteen days, if he is satisfied that the appellant was prevented by sufficient cause from filing the appeal in time.
(3) Where an appeal is preferred from an order of the Estate Manager, the appellate officer may stay the enforcement of that order for such period and on such conditions as he deems fit.
(4) Every appeal under this section shall be disposed of by the appellate officer as expeditiously as possible.
(5) The costs of any appeal under this section shall be in the discretion of the appellate officer.
(6) For the purposes of this section, the City of Calcutta as defined in clause (3) of section 2 of the City Civil Court Act, 1953, shall be deemed to be a district and the Chief Judge or the principal judge of the City Civil Court therein shall be deemed to be the district judge of the district.
Provided that such surviving son or daughter does not own any pacca house in West Bengal at the date of commencement of this Act.
11. Power of the Estate Manager to dispose of immovable property. - The Estate Manager may, with the previous approval of the State Government, dispose of by sale or public auction any immovable property forming part of the Estate properties for the purpose of repayment of debts and discharge of liabilities, if any, being the first charge on the Estate properties under any enactment in force for the time being, and invest the balance of the proceeds thereof, if any, after such repayment and discharge, in such manner as the State Government may direct.
Provided that no such allowances shall be paid where the annual income of such surviving son or daughter exceeds ten thousand rupees.
13. Management of Hazar Duari Palace and Imambarah. - (1) Notwithstanding anything contained elsewhere in this Act, the Director of Archaeology, Government of West Bengal, shall be responsible for the management of, and for the preservation of art objects, books, manuscripts, records, oil paintings and any other like objects kept in, the Hazar Duari Palace and Imambarah (which have been devlared by the Central Government under the Ancient Monuments and Archaeological Sites and Remains Act, 1958 to be of national importance) till the said Hazar Duari Palace and Imambarah are taken over by the Central Government under that Act.
(2) The expenses for the preservation of art objects, book, manuscripts, records, oil paintings and any other like objects kept in the Hazar Duari Palace and Imambarah shall be borne by the State Government till the Hazar Duari Palace and Imambarah are taken over by the Cental Government.
14. Protection of action taken in good faith. - No suit, prosecution or other legal proceeding shall lie against the State Government or the Director of Archaeology, Government of West Bengal or the Estate Manager or any other officer for anything which is in good faith done or intended to be done under this Act or the rules made thereunder.
[14A. Bar of jurisdiction of Civil Courts.] - No Civil Court shall have jurisdiction to deal with or to decide any question which is required to be dealt with or to be decided under the provisions of this Act.
15. Power to make rules. - (1) The State Government may by notification make rules for carrying out the purposes of this Act.
(2) In particular, and without prejudice to the generality of the foregoing power, such rules may provide for all or any of the matters which may be, or are required to be, prescribed.
16. Repeal. - (1) The Murshidabad Estate (Trust) Act, 1963 is hereby repealed.
(4) the Murshidabad Estate Administration (Amendment) Act, 1959, repealed under the Murshidabad Estate (Trust) Act, 1963.


Did you make any health-related resolutions at the New Year? New Seasons Market on Mercer Island is all set to help you establish healthy eating habits and pump up your nutrition knowledge!
Every Saturday this month, you can jumpstart your January by attending a free talk given by NSM nutritionist Tran Hoang. Registration is required, but other than that be ready to support your healthy resolutions with helpful information. Below is a schedule of the January Jumpstart topics, with links to the event registration pages.
You can also stop by New Seasons Market for Talk N' Taste Thursdays. Swing by the Mercer Island store's Solutions counter between 1-5pm on Thursdays to discuss with Ms. Hoang your nutrition questions, and sample some healthy foods that are tasty as well!
In addition to these two AWESOME opportunities to make headway on eating healthier, you can schedule a Talk N ' Shop appointment with Ms. Hoang throughout the month . The calendar is accessible here on the News Seasons Market website. Toggle to Mercer Island on the location drop-down menu, and pick an available date and time from the calendar. Then fill out the form to submit your booking request. This service is also free (how cool is that?!).
Here's the January calendar of events for New Seasons. They have a bunch of other fabulous events sprinkled throughout the month. World Tour Wednesday sounds a lot of fun, and it would be easy to stop by on your way home from work! Click on the photo to access the pdf of the calendar, or view it online here.


As you may have heard, Netflix started to block the usage of VPN and Proxy servers to be able to watch the USA Netflix (which has more content) from another country.
Unfortunately there are still a lot of licensing issues for Netflix to allow them to have the same content at a global scale. They are aware of the problem and they are working on a solution, but this will take years.
I have been testing different VPN & DNS Services that would allow you to not only protect your privacy by using a VPN to connect to the Internet, but would also work for watching USA Netflix wherever you are.
In some simple words a VPN is a Virtual Private Network that will allow you to connect to the Internet through an third party server to browse the internet. The general benefit of using a VPN is that whatever you do online will be through a another provider that usually offers complete anonymity and also encryption from your house to their servers.
VPN´s are also useful when you need to access content that is only available in a specific country or region. Thats why these services allow you to for example watch USA Netflix from wherever you are. Some time ago, Netflix started to filter some providers preventing the end users behind these services to watch their content through a VPN provider, thats why on this review, we will focus on the services that do work with Netflix.
The downside of using a VPN Server is that usually the connections are slower. For example in my house in Switzerland my Internet provider gives me 500 Mbps Downstream and 50 Mbps upstream, but if I would use a VPN to browse the Internet, then in most cases I wont be able to browse at higher speed than 30 or 40 Mbps downstream.
What is a DNS or Smart DNS Provider?
A DNS Server translates your request of a website into the real address of the server which is an IP Address. Another alternative to VPN servers is to use a SmartDNS provider, this is more simple and in general does not affect the performance of your Internet connection.
The downside is that this is useful only if you need change the country from where you are browsing to the destination website, for example Netflix. These type of services do not offer any encryption and should not be used for security purposes but only to unlock content available in a particular region.
There are a lot of providers offering different VPN services. I tested several of them and the following are the ones that do work with Netflix, or at least at the time of testing.
PureVPN offers a very simple interface through Windows and Mac applications among others.
The service works pretty well and offers a non technical way to achieve activities, this is great for end users who are not tech savvy.
Speed wise connecting to the US server was OK but not super fast, it was enough to test USA Netflix from my computer though.
When I tried to do some tests with the mobile application (which looks very nice by the way) I was able to connect to the US VPN server only after several retries. Once connected unfortunately Netflix through the mobile App did not work. I did this test a couple of times and also failed.
PureVPN also offers a SmartDNS add-on service (you need to pay more), but to be honest I there is very little documentation on this service and doesn't seem to be working fine for the tests I`ve done as I was unable to register my IP address to use their service although I set the DNS manually on my MacBook Pro laptop. I contacted support over chat and they send me a link to documentation that actually was broken, chatted for more than 40 minutes and the support agent was unable to properly guide me how to register an IP address on their SmartDNS service.
I pinged their support over Chat a couple of times to ask them some questions but to be honest it was slow and at some point the support agent dropped my chat session due to inactivity in one opportunity. Not cool.
I like the interface and flexibility that PureVPN offers, but they need to work harder on the mobile part. If you dont need mobile, it can be a good alternative, but they need to still polish some aspects of their service.
LifeTime Subscription (pay only once) for USD 99: Referral Link (Ends April 30th 2016 approx.).
PIA is on the VPN business since several years, its hard not to find references to them from all over the world. I remember buying a 2 years contract with them back in December 2015. But then Netflix was not working properly.
I did tests today and Netflix was working back again, so I guess that when they find out that Netflix is being blocked, they try to fix it by changing their blocked IP for another one.
I pinged them to understand what is their policy when their IP gets blocked by Netflix and this is their answer over Chat support: ¨To be clear about the Netflix issue. We do not block Netflix and the service itself should have no issues on the VPN. However that being said Netflix does block our IPs and if this happens there's nothing we can do to assist you.¨ I like that they are being straight forward.
I confirmed with their chat support that you can use their service from up to 5 devices at a time no matter from which IP address they are connected. So lets imagine you have someone at your house using the service, you are travelling and 2 of your kids are also on their universities all can theoretically use the service with one single account.
I must say the experience with support which I contacted several times to clarify questions for this review post was very good. I felt that the person on the other side actually understood first time what I was asking and gave me detailed links and explanations.
They are in the market since quite a while, so they have a lot of experience and seem to offer a pretty solid service. They dont offer a SmartDNS service, but if you do require VPN it might be a good option to go for PIA.
GetFlix is the sister company from Smart DNS Proxy (see below), they also primarily offer a SmartDNS Service, but also will give a VPN Service to their subscribers. I paid 1 month to be able to test and fairly compare against the other contenders too.
The configuration was pretty straight forward for the DNS part. And for the VPN when I tried to configure it on my Mac, I was unable to authenticate so I couldn't make any further tests until I reach support.
They are the only ones that do not offer an online Chat support service, so I just had to wait until my ticket gets responded. Due to the time that responding to my emails took, it looks like they do not offer 24/7 support.
They offer a similar service as Smart DNS Proxy does, if its not exactly the same service, because when I configured the VPN connection I was using the same server names as Smart DNS Proxy.
They openly acknowledge the issues that you may face due to newest Netflix´s restrictions so they seem to be constantly working on providing workaround for the same, something good if you are focusing your usage of the SmartDNS or VPN service to watch US Netflix.
GetFlix Lifetime Subscription Promo: Referral Link (ends April 28th 2016 approx).
Smart DNS Proxy is the parent company of GetFlix according to what it has been confirmed to me by one of their support agents. Although they are primary focused in offering SmartDNS services as their own name indicates, if you buy a subscription you can also use their VPN services. Unfortunately the VPN part cannot be enabled during the 14 days trial period.
So to be fair among all the contenders I have acquired a one month account to make the same tests that you could see on the published table on this same post.
I tried to configure TunnelBlick Open VPN client on my MacBook Pro but it was not possible to import the configuration files. Then I noticed that there are other methods to create a VPN directly in OSX network settings and that they were supporting so I did their tests in this way.
They also support automatic updating of your own IP address through a DynDNS or NO-IP account that you can configure on your profile on their website and they would update the IPAddress every 5 minutes so you can use their SmartDNS servers without you having to worry about updating your IP manually, in case it changes frequently of course.
They also offer something very nice which is connection of the VPN Service through OpenVPN for Asus and DDRWT routers which is a nice catch if you want to make your whole house traffic to go through a VPN without having to configure it on each device. Remember it has it bad side too, all your traffic speed will be restricted to your VPN connection capacity, which as you can see from the tests table I made, none of them offer really a bandwidth close to your ISP capacity.
I like the idea of simplicity offered through the DNS Servers and it does work well with Netflix and TVPlayer too without affecting your ISP provided bandwidth.
The price is pretty affordable too as you can see on the table. Unfortunately I couldn't find any special offers for this service.
Each service has its benefits and its disadvantages. For example PrivateInternetAccess (PIA) seems to be a great stable service for VPN, but they dont offer a DNS Service like GetFlix or SmartDNSProxy does so all your traffic goes through them, maybe making your ISP provided Internet link crawl a bit.
So if your main objective to use a VPN or SmartDNS is to bypass the country or regional restriction from Netflix or others, you might be better with only a Smart DNS Solution like the last 2 options.
If you need to use it on mobile, most of the tests I made actually failed. It seems the fact that there is a VPN client configured on your mobile makes the Netflix App to know you are doing this. I made a quick test by loading Netflix on a browser session and seemed to work though, so take this in consideration if you need to use on mobile from outside US.
Also, there are other options like Unlocator.com, Ublock.us among others but I didn't have the opportunity to test them.
I hope you found this reviews interesting, if you do, please share them on your favourite social networks. Please leave your comment below if you want to add something or comment about another service you have tested, also if you find another offer for any of these services, I can update the post.
I must say that you should add "Hide-My-IP" and HMA VPN in your list. Both of these are really effective and useful. Cost of HMA is relatively high but Hide-My-ip is offering good service in a cheap rate. Hide My IP VPN is a better deal than all the providers mentioned in the article, $2.95/mo includes VPN + Smart DNS to unblock any streaming site, up to 5 devices allowed per account.
Hi BryanT88 thank for sharing this other option!
You can also test ActiVPN. I use it to watch Netflix US outside US. It works perfectly !


This is an authentic, pre-owned HERMES Herline Sac Cabas Tote Bag. It features a brown canvas with silver-tone hardware, canvas handles, an open pocket, and a zip closure. It's a chic Hermes tote for every day.


text:
Q: How to detect individual phonetic alphabet like ʌ,ɪ,ʊ,ɒ,ə on Android speech recognition? Recently, I started to learn voice recognition on Unity. Had tried voice command on both Windows 10 and Android, it works perfectly. But now I am trying to do a project where it uses voice command like "Ah", "Ee", "Oo" instead of "Move forward", " Turn Left"...
For Windows 10, it have a roughly 60% success rate where it recognize what I had said and do the correct pattern.
While for Android, it doesn't seem so bright as is failed all the time when I tried to give voice command(Had tested out it can only detect words not phonetic alphabet)
For Android currently, I built my project based on a github project
https://github.com/gsssrao/UnityAndroidSpeechRecognition
While for Windows 10, I am using Keyword Recognizer
For both the logic is:

*

*Player call out a certain words

*Then it will pass to a game object which hold script that check whether user input is same as the keyword that I had set in my project

*If it got a same match, then it will perform a corresponding action else do nothing.

For Windows 10, I am fine with its results as it works most of the time.
but for Android it doesn't recognize any "sound" (Ah, Ee, Ou, Oh, Er... etc) is there a better way to do it or I had misuse a certain function?

redpajama_set_name: RedPajamaStackExchange


Leading the way into the new era of Formula 1, Liberty Media boss Chase Carey has said that the sport needs a fresh start, without former chief Bernie Ecclestone.
In a deal worth around $8 billion, Carey acquired motorsport's most popular brand which resulted in Ecclestone being relieved of his duties as Formula 1's supremo.
Although wishing Ecclestone well for his future and admitting he made the sport what it is today, the American businessman feels that it needs a new direction, saying Formula 1 was run like a dictatorship.
"He has run this sport for his entire adult life and I respect completely that this is a difficult change," Carey said.
"We have tried to deal with him with the respect he's due, which is why we offered him the chairman emeritus title.
"I have been sincere in saying I value his help and advice as we go forward.
"But he has a lot to continue to offer and he will always be part of the F1 family."


text:
Allison, Mose (John, Jr. )
Patti Jones
(John, Jr. )
(b Tippo, MI, Nov 11, 1927). American jazz and blues pianist, singer and songwriter. His style was influenced by the blues music he heard on the juke box at his father's general store. Primarily self taught on piano and trumpet, Allison began playing professionally in Delta roadhouses and attended the University of Mississippi, Oxford. However, he left to enlist in the US Army in 1946, and during his service he played trumpet and piano and wrote arrangements for an army band. After completing a degree in English at Louisiana State University, he moved to New York in 1956 and attracted attention nationally playing piano with such leaders as Chet Baker, Al Cohn, Zoot Sims, Gerry Mulligan, and Stan Getz.
Allison created a hybrid style that integrated country blues with urbane jazz; it can be heard on his first album, Back Country Suite (1959, Prst.), which includes what became his signature tune, "Young Man's Blues." In the 1960s Allison's music influenced British rock musicians, and this tune was covered as a generational anthem by The Who. During the same period Allison recorded for Atlantic and wrote pithy lyrics about public service and social commentary ("Everybody Cryin' Mercy") and personal crisis ("Hello There, Universe"), some with a playful sense of humor ("Your Mind's on Vacation"). Later songs such as "Ever Since the World Ended" and "Certified Senior Citizen" focused on contemporary culture and aging. Allison has also interpreted blues and jazz standards such as Willie Dixon's "Seventh Son," Muddy Waters' "Rollin' Stone," and Duke Ellington's "I ain't got nothin' but the blues." His elaborate piano instrumentals and improvisations draw upon the music of Charles Ives and Alexander Scriabin and reflect his experimentation with conventional ideas of time....

redpajama_set_name: RedPajamaCommonCrawl


text:
(-) NSW
(-) SA
(-) Tunnel
Selected Filters: 2013 , Tunnel , Publications , NSW, SA
NSW Pacific Highway Corridor Upgrade
This assessment brief outlines the New South Wales Government's proposal to complete the remaining 40 per cent of works to upgrade the Pacific Highway to a minimum four-lane divided highway between Sydney and Brisbane.
SA Northern Connector
This assessment brief outlines the South Australian Government's plan to construct a 15.6 kilometre road (the Northern Connector) between Port Adelaide and the Northern Expressway.
NSW F3 Widening Tuggerah to Doyalson
This assessment brief outlines the New South Wales Government's proposal to upgrade a 10.5 kilometre section of the F3 freeway from Tuggerah to Doyalson, including pavement strengthening and widening the freeway from four to six lanes. The upgrade will increase the productivity of the freeway and enhance connectivity for passenger and freight vehicles between the central coast and Sydney by improving the existing road network and reducing congestion. The New South Wales Government is seeking $100 million in Commonwealth funding for the project.


text:
\section{Introduction}
\vspace{0.5cm}
Let $E\subset {\mathbb C}^{\mathbb Z}$ be a Banach space of sequences. Denote by $S: {\mathbb C}^{\mathbb Z} \longrightarrow {\mathbb C}^{\mathbb Z}$, the shift operator defined by $Sx=(x({n-1}))_{n \in {\mathbb Z}}$, for $x=(x(n))_{n \in {\mathbb Z}} \in {\mathbb C}^{\mathbb Z}$, so that $S^{-1}x=(x({n+1}))_{n \in {\mathbb Z}}$. Let $F({\mathbb Z})$ be the set of sequences on ${\mathbb Z}$, which have a finite number of non-zero coefficients and assume that $F({\mathbb Z})$ is dense in $ E$. The elements of $F({\mathbb Z})$ will be said finite sequences. We will call multiplier on $E$ every bounded operator $M$ on $E$ such that $MSa=SMa$, for every $a \in F({\mathbb Z})$. Denote by ${\mu}(E)$ the space of multipliers on $E$. For $z\in {\mathbb T}=\{z \in {\mathbb C}\:|\:|z|=1\}$, set $\psi_z(x)=(x(n)z^n)_{n \in {\mathbb Z}}$, for $x=(x(n))_{n \in {\mathbb Z}}$. Notice that if for $z \in {\mathbb T}$, $\psi_z(E)\subset E$ and if for all $n \in {\mathbb Z}$, the map $p_n: x\longrightarrow x(n)$ is continuous from $E$ into ${\mathbb C}$, then from the closed graph theorem it follows that $\psi_z$ is bounded on $E$. In this paper, we deal with Banach spaces of sequences on ${\mathbb Z}$ satisfying only the following very natural hypothesis:\\

(H1) The set $F({\mathbb Z})$ is dense in $E$.\\

(H2) For every $n \in {\mathbb Z}$, $p_n$ is continuous from $E$ into ${\mathbb C}$. \\

(H3) We have $\psi_z(E)\subset E, \:\forall z \in {\mathbb T}$ and $\sup_{z \in {\mathbb T}}\|\psi_z\|<+\infty$.\\

It is easy to see that if $S(E)\subset E$, then by the closed graph theorem the restriction $S\vert_E$ of $S$ to $E$ is bounded from $E$ into $E$. From now we will say that $S$ (resp. $S^{-1}$) is bounded when $S(E)\subset E$ (resp. $S^{-1}(E)\subset E$).
If $S(E)\subset E$, we denote by $spec(S)$ the spectrum of the operator $S$ with domain $E$. If $S$ is not bounded, denote by $spec(S)$ the spectrum of $\overline{S}$. Here $\overline{S}$ is the smallest extension of $S\vert_{F({\mathbb Z})}$ as a closed operator. Recall that the domain of $\overline{S}$ is
$$D(\overline{S})=\{x\in E,\:\exists (x_n)_{n\in {\mathbb Z}}\subset F({\mathbb Z})\:s.t. \:x_n\longrightarrow x\: and\: Sx_n\longrightarrow y\in E\}$$
and $\overline{S}x=y$.
Our aim is to prove that every multiplier on $E$ is associated to a $L^\infty$-function on $spec(S)$, which is holomorphic on $\overset{\circ}{spec(S)}$, if $\overset{\circ}{spec(S)} \neq \emptyset$.
The multipliers on $l^2_{\omega}({\mathbb Z})$, when the weight ${\omega}$ is such that $S$ and $S^{-1}$ are bounded and $\overset{\circ}{spec(S)}\neq \emptyset$ were investigated by Shields in \cite{S} (see \cite{L} for the case of the spaces $l^p({\mathbb Z})$, $p\geq 1$). Gellar considered in \cite{G} a larger class of Banach spaces with Schauder basis. On the other hand, the case of spaces $l^2_{\omega}({\mathbb Z})$ when $\overset{\circ}{spec(S)}=\emptyset$ was treated only recently by Esterle in \cite{E}. The case of weighted spaces $L_\delta^2({\mathbb R})$ has been examined in \cite{P1}. Nevertheless, to our best knowledge it seems that this problem for general Banach space of sequences satisfying only hypothesis (H1), (H2) and (H3) has been not yet considered in the literature.\\

Let $e_k$ be the sequence such that ${e_k}(n)=0$ if $n \neq k$ and ${e_k}(k)=1$. Define for $r>0$, $$C_r:=\{z \in {\mathbb C}\:|\:|z|=r\}.$$
From now, $E$ is a Banach space of sequences on ${\mathbb Z}$ satisfying (H1)-(H3).
For $M \in{\mu}(E)$, set $\widehat{M}=M(e_0)$ and for $z \in {\mathbb C}$, denote by ${\widetilde{M}}(z)$ the formal Laurent series
$${\widetilde{M}}(z)=\sum_{n \in {\mathbb Z}}\widehat{M}(n)z^n.$$ For $M \in {\mu}(E)$, we call ${\widetilde{M}}$ the symbol of $M$.
Given $a \in E$, set ${\widetilde{a}}(z)=\sum_{n \in {\mathbb Z}}a(n)z^n$, for $z \in {\mathbb C}$. It is easy to see that $Ma=\widehat{M}*a$, for $a \in F({\mathbb Z})$ and that we have on the space of formal Laurent series
$$\widetilde{Ma}(z)={\widetilde{M}}(z){\widetilde{a}}(z),\:\:\forall z \in {\mathbb C},\: \forall a \in F({\mathbb Z}),$$ but it is more difficult to determine when ${\widetilde{M}}(z)$ converges. It is natural to conjecture that ${\widetilde{M}}(z)$ is a convergent series for $z\in spec(S)$ and that the function ${\widetilde{M}}$ is holomorphic on $\overset{\circ}{spec(S)}$, if $\overset{\circ}{spec(S)} \neq \emptyset.$ Notice that the main difficulty is to show that ${\widetilde{M}}$ is essentially bounded on $spec(S)$ when ${spec(S)}$ is a circle.
For a closed operator $A$ with dense domain, denote by $\rho(A)$ the spectral radius of $A$ defined by $\rho(A)= \sup\{|\lambda|\:,\:\lambda \in spec(A)\}.$ We suppose that at least one of the operators $S$ and $S^{-1}$ is bounded.
Our main result is the following.
\begin{thm}
$1)$ If $S$ is not bounded, but $S^{-1}$ is bounded, $\rho(S)=+\infty$ and if $S$ is bounded, but $S^{-1}$ is not bounded,
$\rho(S^{-1})=+\infty.$\\
$2)$ We have $spec(S)=\Bigl\{ \frac{1}{\rho(S^{-1})}\leq |z|\leq \rho(S)\Bigr\}$.\\
$3)$ Let $M\in {\mu}(E)$. For $r> 0$ such that $C_r\subset spec(S)$, we have ${\widetilde{M}} \in L^\infty(C_r)$ and
$$|{\widetilde{M}}(z)|\leq \|M\|,$$ a.e. on $C_r$.\\
$4)$ If $\rho(S)>\frac{1}{\rho(S^{-1})}$, ${\widetilde{M}}$ is holomorphic on $\overset{\circ}{spec(S)}.$
\end{thm}
If $\rho(S^{-1})=+\infty$, here $\frac{1}{\rho(S^{-1})}$ denotes $0$.

The class of Banach spaces of sequences on ${\mathbb Z}$ that we consider in this paper is very general. For example (H3) is satisfied for every Banach space $E$ with a norm such that $\|(x(n))_{n \in {\mathbb Z}}\|=\|(|x(n)|)_{n \in {\mathbb Z}}\|$. We will see later that our hypothesis imply that for $x \in E$, we have
$$\lim_{k \to +\infty}\Bigl\|x-\sum_{p=0}^k\frac{1}{k+1}\Bigl ( \sum_{n=-p}^p {x(n)}\:e_n \Bigr)\Bigr\|=0,$$ but not necessary $\lim_{p \to +\infty}\|x-\sum_{n=-p}^p {x(n)}\:e_n \|$ as it has been assumed in \cite{G}. The Example 5 below shows that this situation appears and this makes obvious the generality of our considerations.
We will give some classical examples of Banach spaces satisfying the conditions (H1), (H2) and (H3). \\

{\bf Example 1. }\\
Let ${\omega}$ be a weight on ${\mathbb Z}$, i.e. ${\omega}$ is a positive sequence on ${\mathbb Z}$. Set
$$l_{\omega}^p({\mathbb Z})=\Bigl\{(x(n))_{n \in {\mathbb Z}}\in {\mathbb C}^{\mathbb Z}\:|\:\sum_{n \in {\mathbb Z}}|x(n)|^p{\omega}(n)^p<+\infty\Bigr\}, \:1\leq p<+ \infty $$
and $$\|x\|_{{\omega},p}=\Bigl(\sum_{n \in {\mathbb Z}}|x(n)|^p{\omega}(n)^p\Bigr)^{\frac{1}{p}}.$$
It is easy to see that the Banach space $l_{\omega}^p({\mathbb Z})$ satisfies our hypothesis. \\

{\bf Example 2.} \\
For every two weights ${\omega}_1$ and ${\omega}_2$ and $ 1\leq p <+\infty$, $1\leq q <+\infty$, the space $l_{{\omega}_1}^p({\mathbb Z})\cap l_{{\omega}_2}^q({\mathbb Z})$ with the norm $\|x\|=\max\{\|x\|_{{\omega}_1,p},\:\|x\|_{{\omega}_2,q}\}$ satisfies also our conditions.\\

{\bf Example 3.}\\
Let ${\mathcal K}$ be a convex, non-decreasing, continuous function on ${\mathbb R}^+$ such that ${\mathcal K}(0)=0$ and ${\mathcal K}(x)>0$, for $x >0.$ For example, $\mathcal {K}$ may be $x^p$, for
$\:1\leq p< +\infty$ or $x ^{p+sin(\log(-\log( x))},\: p>1+\sqrt{2}$. Let ${\omega}$ be a weight on ${\mathbb Z}$. Set
$$l_{{\mathcal K},{\omega}}({\mathbb Z})=\Bigl\{(x(n))_{n \in {\mathbb Z}}\in {\mathbb C}^{\mathbb Z} \:|\:\sum_{n \in {\mathbb Z}}{\mathcal K}\Bigl(\frac{|x(n)|}{t}\Bigr){\omega}(n)<+\infty, \:\rm {for\: some\: t>0}\Bigr\}$$
and
$$\|x\|=\inf\Bigl\{ t>0 \:|\:\sum_{n \in {\mathbb Z}}{\mathcal K}\Bigl(\frac{|x(n)|}{t}\Bigr){\omega}(n)\leq 1\Bigr\}.$$
The space $l_{{\mathcal K},{\omega}}({\mathbb Z})$, called a weighted Orlicz space (see \cite{R}, \cite{I}), is a Banach space satisfying our hypothesis.
We can apply Theorem 1 to the multipliers on $l_{{\mathcal K},{\omega}}({\mathbb Z})$ as well as to the spectrum of the shift on $l_{{\mathcal K},{\omega}}({\mathbb Z})$. It seems that in the literature there are no complete results concerning the spectrum of the shift on $l_{{\mathcal K},{\omega}}({\mathbb Z})$.
\\

{\bf Example 4.}\\
Let $(q(n))_{n \in {\mathbb Z}}$ be a real sequence such that $q(n) \geq 1$, for all $n \in {\mathbb Z}$. For $a =(a(n))_{n \in {\mathbb Z}}\in {\mathbb C}^{\mathbb Z}$, set
$$\|a\|_{\{q\}}=\inf\Bigl\{ t>0\:|\:\sum_{n\in {\mathbb Z}}\Bigl|\frac{a(n)}{t}\Bigr|^{q(n)}\leq 1\Bigr\}.$$
Consider the space $l^{\{q\}}=\{a\in {\mathbb C}^{\mathbb Z}\:|\:\|a\|_{\{q\}}<+\infty\},$ which is a Banach space (see \cite{Ed}) satisfying our hypothesis. Notice that if $\lim_{n \to +\infty}|q({n+1})-q(n)|\neq 0$ and if $\sup_{n \in {\mathbb Z}} q(n)<+\infty$, then either $S$ or $S^{-1}$ is not bounded (see \cite{N}). \\

{\bf Example 5.}\\
Denote by $C_{[0,2\pi]}$ the space of continuous, $2\pi$-periodic, complex-valued functions on ${\mathbb R}$. For $f\in C_{[0,2\pi]}$, we denote by $\hat{f}$ the sequence of Fourier coefficients of $f$. Set ${\mathcal C}=\{\hat{f}\:|\:f \in C_{[0,2\pi]}\}$ and $\|\hat{f}\|=\|f\|_\infty,$ for $f \in C_{[0,2\pi]}$. It is easy to check that the hypothesis (H1) and (H2) are satisfied by ${\mathcal C}$. For $ \alpha \in {\mathbb R}$ and $f\in C_{[0,2\pi]}$, $\psi_{{e^{i\alpha}}}(\hat{f})$ is the sequence of Fourier coefficients of the function $t \longrightarrow f(t+\alpha)$. So it is clear that (H3) is satisfied by $\mathcal{C}$. Notice that in $\mathcal{C}$, $\hat{f}$ is not the limit of $\sum_{|n|\leq k}\hat{f}(n)e_n$ as $k \to +\infty$ and the space $\mathcal{C}$ is not included in the class of Banach spaces treated in \cite{G}. \\

{\bf Remark 1.}
If both $S$ and $S^{-1}$ are unbounded then Theorem 1 is not valid in general. For example, if $E=l_{\omega}^2({\mathbb Z})$, where ${\omega}(2n)=1$ and ${\omega}(2n+1)=|n|+1$, for $n \in {\mathbb Z}$, $S$ and $S^{-1}$ are not bounded. It is easy to see that $spec(S)={\mathbb C}$ and $S^2\in {\mu}(E)$, but $\widetilde{S^2}(z)=z^2$ is obviously not bounded on ${\mathbb C}$. \\

In Section 3, we investigate Toeplitz operators on a general Banach space of sequences on ${\mathbb Z}^+ = {\mathbb N}$, which will be defined precisely in Definition 2 below. There are many similarities between multipliers and Toeplitz operators. We are motivated by the recent results in \cite{E} about Toeplitz operators on $l_{\omega}^2({\mathbb Z}^+)$, where ${\omega}$ is a weight on ${\mathbb Z}^+$ and the results of the author (see \cite{P2}) concerning Wiener-Hopf operators on weighted spaces $L_\delta^2({\mathbb R}^+)$. Let ${\bf E}\subset {\mathbb C}^{{\mathbb Z}^+}$ be a Banach space.
Let $F({\mathbb Z}^+)$ (resp. $F({\mathbb Z}^-)$) be the space of the sequences on ${\mathbb Z}^+$ (resp. ${\mathbb Z}^-$) which have a finite number of non-zero coefficients. By convention, we will say that $x\in F({\mathbb Z})$ is a sequence of $F({\mathbb Z}^+)$ (resp. $F({\mathbb Z}^-)$) if $x(n)=0$, for $n<0$ (resp. $n>0$).
We will assume that ${\bf E}$ is satisfying the following hypothesis:\\

(${\mathcal H}$1) The set ${F({\mathbb Z}^+)}$ is dense in $ {\bf E}$.\\

(${\mathcal H}$2) For every $n \in {\mathbb Z}^+$, the application $p_n:x\longrightarrow x(n)$ is continuous from ${\bf E}$ into ${\mathbb C}$. \\

(${\mathcal H}$3) For $x=(x(n))_{n \in {\mathbb Z}+}\in {\bf E}$, we have $\gamma_z(x)=(z^n x(n))_{n \in {\mathbb Z}^+}\in {\bf E}$, for every $z \in {\mathbb T}$ and $\sup_{z\in{\mathbb T}}\|\gamma_z\|<+\infty.$\\

Notice again that if $\gamma_z(x)\in {\bf E}$, for every $x \in {\bf E}$, then $\gamma_z:{\bf E}\longrightarrow {\bf E}$ is bounded.

\begin{deff}
We define on ${\mathbb C}^{{\mathbb Z}^+}$ the operators ${\bf S}_1$ and ${\bf S}_{-1}$ as follows.
$$ For\: u\in {\mathbb C}^{{\mathbb Z}^+},\:({\bf S}_1(u))(n)=0,\:if \: n = 0\:and \:({\bf S}_1(u))(n)=u(n-1),\:if \:n\geq 1$$
$$({\bf S}_{-1}(u))(n)=u(n+1),\: for\: n \geq 0.$$

\end{deff}
For simplicity, we note ${\bf S}$ instead of ${\bf S}_1$. Remark that we have ${\bf S}_{-1}{\bf S}=I$, however we do not have ${\bf S}\Ss_{-1}=I$ and this is the main technical difficulty in the analysis of the case of Toeplitz operators. It is easy to see that if ${\bf S}({\bf E})\subset {\bf E}$, then by the closed graph theorem the restriction ${\bf S}\vert_{{\bf E}}$ of ${\bf S}$ to ${\bf E}$ is bounded from ${\bf E}$ into ${\bf E}$. We will say that ${\bf S}$ (resp. ${\bf S}_{-1}$) is bounded when ${\bf S}({\bf E})\subset{\bf E}$ (resp ${\bf S}_{-1}({\bf E})\subset {\bf E}$).
 Next, if ${\bf S}\vert_{{\bf E}}$ (resp. ${\bf S}_{-1}\vert_{{\bf E}})$ is bounded, $spec({\bf S})$ (resp. $spec({\bf S}_{-1})$) denotes the spectrum of ${\bf S}\vert_{{\bf E}}$ (resp. ${\bf S}_{-1}\vert_{{\bf E}})$.
 If ${\bf S}$ (resp. ${\bf S}_{-1}$) is not bounded, $spec({\bf S})$ (resp. $spec({\bf S}_{-1})$) denotes the spectrum of the smallest closed extension of  ${\bf S}\vert_{F({\mathbb Z}^+)}$ (resp. ${\bf S}_{-1}\vert_{F({\mathbb Z}^+)}$).

\begin{deff}
A bounded operator on ${\bf E}$ is called a Toeplitz operator, if we have:
$$({\bf S}_{-1}T{\bf S})u=Tu, \: \forall u \in F({\mathbb Z}^+).$$
\end{deff}
For $u \in l^2({\mathbb Z}^-)\oplus {\bf E} $ introduce
$$(P^+(u))(n)=u(n),\:\forall n\geq 0\:{\rm and}\: (P^+(u))(n)=0,\:\forall n<0.$$
Given a Toeplitz operator T, set $\widehat{T}(n)=<Te_0,e_{-n}>$ and $\widehat{T}({-n})=<Te_n,e_0>$, for $n \geq 0$. Define $\widehat{T}=(\widehat{T}(n))_{n \in {\mathbb Z}}.$ It is easy to check that
$$ Tu=P^+(\widehat{T}*u),\:\forall u \in F({\mathbb Z}^+).$$
Set
$${\widetilde{T}}(z)=\sum_{n \in {\mathbb Z}}\widehat{T}(n)z^n,$$
for $z \in {\mathbb C}$. Notice that the series $\tilde{T}(z)$ could diverge.\\

Taking into account the similarities between multipliers and Toeplitz operators, it is natural to obtain analogous results for Toeplitz operators and to conjecture that ${\widetilde{T}}(z)$ converges for $z \in spec({\bf S})\cap (spec({\bf S}_{-1}))^{-1}$. It is clear that if $M$ is a multiplier on ${{\bf E}}^-\oplus {\bf E}$, where ${\bf E}^-$ and ${\bf E}$ are Banach spaces of sequences respectively on ${\mathbb Z}^-$ and ${\mathbb Z}^+$, then $P^+M$ is a Toeplitz operator on ${\bf E}$. However, despite the extensive literature related to Toeplitz operators, it seems that it is not known if every Toeplitz operator is induced by a multiplier on some suitable Banach space of sequences on ${\mathbb Z}$. Thus we cannot use our results for the multipliers on spaces of sequences on ${\mathbb Z}$ to prove similar ones for Toeplitz operators. In this way, we apply the methods of Section 2 and we obtain the following theorem, when at least one of the operators ${\bf S}$ and ${\bf S}_{-1}$ is bounded.
\begin{thm}
Let $T$ be a Toeplitz operator on ${\bf E}$.\\
$1)$ For $r \in \Bigl[\frac{1}{\rho({{\bf S}}_{-1})}, \rho({\bf S})\Bigr]$, if $\rho({\bf S})<+\infty$ or for $r \in \Bigl[\frac{1}{\rho({{\bf S}}_{-1})}, +\infty\Bigr[$, if $\rho({\bf S})=+\infty$ we have $\widetilde{T}\in L^\infty(C_r)$ and $|\widetilde{T}(z)|\leq \|T\|$, a.e. on $C_r$.\\
$2)$ If ${\bf S}$ and ${\bf S}_{-1}$ are bounded and if $\frac{1}{\rho({ {\bf S}}_{-1})}<\rho({\bf S})$, then we have ${\widetilde{T}} \in\mathcal{H}^\infty(\overset{\circ}{\bf {\Omega}})$, where ${\bf \Omega}:=\Bigl\{ z \in {\mathbb C}\:|\:\frac{1}{\rho({{\bf S}}_{-1})}\leq |z|\leq \rho({\bf S})\Bigr\}$.\\
$3)$ If ${\bf S}$ is not bounded, but ${{\bf S}}_{-1}$ is bounded, ${\widetilde{T}} \in \mathcal{H}^\infty(\overset{\circ}{U})$, where $U:=\Bigl\{ z \in {\mathbb C}\:|\:\frac{1}{\rho({{\bf S}}_{-1})}\leq |z|\Bigr\}.$
\\
$4)$ If ${\bf S}$ is bounded, but ${{\bf S}}_{-1}$ is not bounded, ${\widetilde{T}} \in \mathcal{H}^\infty(\overset{\circ}{V}),$ where $V:=\Bigl\{z\in {\mathbb C}\:|\: |z|\leq \rho({\bf S})\Bigr\}.$\\

\end{thm}

\vspace{0.5cm}
\section{Multipliers}
\vspace{0.5cm}
In this section, we prove Theorem 1. We denote by $E^*$ the dual space of $E$, by $\|\:.\: \| $ the norm of $E$ and by $\|\:.\:\|_* $ the norm of $E^*$. For $y \in E^*$ and $x \in E$, define $<x,y>:=y(x)$. For $k \in {\mathbb Z}$, setting $<x, e_k>=x(-k)$, we will consider $e_k$ as an element of $ E^*$. Notice that if we set $|||x|||=\sup_{z\in {\mathbb T}}\|\psi_z(x)\|$, the norm $|||.|||$ is equivalent to the norm $\|.\|$. We have $\sup\{|||\psi_z(x)|||, \:x\in E,\: |||x|||=1\}=1$, so without losing generality we can assume that $\psi_z$ is an isometry from E into E, for every $z\in {\mathbb T}$.
We start with the following lemma.
\begin{lem} For $x \in E$, we have
$$\lim_{k \to +\infty}\Bigl\|\sum_{p=0}^k\frac{1}{k+1}\Bigl ( \sum_{n=-p}^p {x(n)}\:e_n \Bigr)-x\Bigr\|=0.$$
\end{lem}
{\bf Proof.}
Fix $x\in E$. The function $\Psi: z\longrightarrow \psi_z(x)$ is continuous from ${\mathbb T}$ into $E$. If $x \in F({\mathbb Z}) $, this is obvious and if $x\in E$, the continuity follows immediately from (H1) and (H3). Consider the Fejer kernels $(g_k)_{k \in {\mathbb N}}\subset L^1({\mathbb T})$ defined by the formula
$$g_k(e^{it}):=\sum_{p=0}^k\frac{1}{k+1}\sum_{|m|\leq p}e^{imt}$$
$$=\frac{1}{k+1}\Bigr(\frac{\sin(\frac{(k+1)t}{2})}{\sin\frac {t}{2}}\Bigr)^2,\:{\rm for}\: t \in {\mathbb R}.$$ We have $\|g_k\|_{L^1({\mathbb T})}=1$, for $k \in {\mathbb N}$ and $\lim_{k \to +\infty}\int_{\delta \leq |t|\leq \pi}g_k(e^{it}) dt=0$, for $\delta >0$. Moreover, for $|n|\leq k$,
$$\hat{g_k}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}g_k(e^{it})e^{-int}\:dt=1-\frac{|n|}{k+1}$$
and for $|n|>k,$
$$\hat{g_k}(n)=0.$$
We have $$\lim_{k \to +\infty}\|(g_k*\Psi)(1)-\Psi(1)\|=0.$$
Below we write $dz$ instead of $dm(z)$, where $m$ is the Haar measure on ${\mathbb T}$ such that $m({\mathbb T})=1$. For $n \in {\mathbb Z}$, we have
$$ \Bigl((g_k*\Psi)(1)\Bigr)(n)=\Bigl(\int_{\mathbb T} g_k(z)\psi_{z^{-1}}(x)dz\Bigr)(n)=\int_{\mathbb T} g_k(z)z^{-n}x(n) dz=\hat{g_k}(n)x(n).$$
So we obtain $$(g_k*\Psi)(1)=\sum_{n=-k}^k\Bigl( 1-\frac{|n|}{k+1}\Bigr)x(n)e_n=\sum_{p=0}^k\frac{1}{k+1}\Bigl ( \sum_{n=-p}^p {x(n)}\:e_n \Bigr)$$ and since $\Psi(1)=x$, the proof is complete.
$\Box$
\begin{lem}
For $x\in E$ and $M\in{\mu}(E)$, the function ${\mathcal M}_{x}:{\mathbb T} \longrightarrow E$, defined by $${\mathcal M}_{x}(z)=(\psi_{z}\circ M \circ \psi_{z^{-1}})(x)$$ is continuous.
\end{lem}
{\bf Proof.}
Fix $x$ in $F({\mathbb Z})$ and $M\in {\mu}(E)$.
It is easy to see that
\begin{equation}\label{eq:f}
{\mathcal M}_{x}(z)=(\psi_z\circ M \circ \psi_{z^{-1}})(x)=\psi_{z}({\widehat{M}})*x, \:\forall z \in {\mathbb T}.\end{equation}
Indeed, for some $k\in {\mathbb N}$, we have
$$\Bigl((\psi_z\circ M \circ \psi_{z^{-1}})(x)\Bigr)(n)=z^n\sum_{|p|\leq k}\widehat{M}(n-p)z^{-p}x(p),\:\forall n \in {\mathbb Z}. $$
Thus, for every $x \in F({\mathbb Z})$, the function $z \longrightarrow (\psi_z\circ M \circ \psi_{z^{-1}})(x)$ is continuous from ${\mathbb T}$ into $E$. Since ${F({\mathbb Z})}$ is dense in $E$ and $\|\psi_z\circ M \circ \psi_{z^{-1}}\|\leq \|M\|$, for $z \in {\mathbb T}$, we deduce that ${\mathcal M}_{x}$ is continuous from ${\mathbb T}$ into $E$, for every $x\in E$. $\Box$\\

Denote by $M_\phi$ the operator of convolution with $\phi \in F({\mathbb Z})$, when $\phi*E\subset E$. Then it is clear that $\widehat{M_\phi}=\phi$. We need the following.

\begin{lem} Let $ M\in{\mu}(E)$, $x\in E$.\\
$1)$ We have $$\lim_{k \to +\infty}\|{\bf M_k} x-Mx\|=0,$$ where for $k\in {\mathbb N}$,
$${\bf M_k}=\sum_{p=0}^k\frac{1}{k+1}\Bigl ( \sum_{n=-p}^p \widehat{M}(n)\:S^n \Bigr)=\sum_{n=-k}^{k}\Bigl(1-\frac{|n|}{k+1}\Bigr) \widehat{M}(n)S^n.$$
$2)$ $\|{\bf M_k}\|\leq \|M\|,$ $\forall k \in {\mathbb N}$.\\
$3)$ If $S^{-1}$ is not bounded, but $S$ is bounded, $\widehat{M}(n)=0,$ for $n<0$, while if $S^{-1}$ is bounded, but $S$ is not bounded, $\widehat{M}(n)=0,$ for $n>0$.

\end{lem}
{\bf Proof.} It is immediate to see that $\Bigl(1-\frac{|n|}{k+1}\Bigr) \widehat{M}(n)$ converges to $\widehat{M}(n)$, for $n \in{\mathbb Z}$ and assertion 1) follows from the density of $F({\mathbb Z})$ in $E$. However the control of the norm of ${\bf M_k}$ is less obvious. The proof follows with some modifications the arguments of \cite{S} in our more general case. Consider the Fejer kernels $(g_k)_{k \in {\mathbb N}}\subset L^1({\mathbb T})$ defined in the proof of Lemma 1.
Fix $ M\in{\mu}(E)$.
Since ${\mathcal M}_{x}$ is continuous from ${\mathbb T}$ into $E$, for every $x \in E$ and ${\mathcal M}_{x}(1)=Mx$, we have
$$\lim_{k \to +\infty}\|(g_k*{\mathcal M}_{x})(1)-Mx\|=0,\:\forall x \in E.$$
Fix $x \in F({\mathbb Z}).$
For $k \in {\mathbb N}$, we obtain
$$(g_k*{\mathcal M}_{x})(1)=\int_{{\mathbb T}}g_k(z){\mathcal M}_{x}(z^{-1})dz$$
$$=\int_{{\mathbb T}}g_k(z)\psi_{z^{-1}}(M\psi_z (x))dz
=\int_{\mathbb T} g_k(z)(\psi_{z^{-1}}({\widehat{M}})*x)dz,$$
taking into account (\ref{eq:f}).
Then we have
$$(g_k*{\mathcal M}_{x})(1)=\Bigl(\int_{{\mathbb T}}g_k(z)\psi_{z^{-1}}({\widehat{M}})dz\Bigr)* x.$$
We observe that, for $|n|\leq k$, we have
$$\Big( \int_{\mathbb T} g_k(z) \psi_{z^{-1}}({\widehat{M}}) dz \Big) (n)=\int_{\mathbb T} g_k(z)z^{-n}\widehat{M}(n)dz=
\widehat{g_k}(n){\widehat M}(n)=\Bigl(1-\frac{|n|}{k+1}\Bigr) \widehat{M}(n),$$
while for $|n|>k$, we get
$$\Big( \int_{\mathbb T} g_k(z) \psi_{z^{-1}}({\widehat{M}}) dz \Big) (n)=0.$$
Since
$$\widehat{{\bf M_k}}=\sum_{n=-k}^{k}\Bigl(1-\frac{|n|}{k+1}\Bigr) \widehat{M}(n)e_n,$$
it follows that
$$\widehat{{\bf M_k}}=\Big( \int_{\mathbb T} g_k(z) \psi_{z^{-1}}({\widehat{M}}) dz \Big).$$
Now it is clear that

$$\|{\bf M_k} a\| =\|\widehat{{\bf M_k}}*a\|=\Bigl\|\int_{\mathbb T} g_k(z)(\psi_{z^{-1}}({\widehat{M}})*a)dz\Bigr\|=\Bigl\|\int_{{\mathbb T}}g_k(z)(\psi_{z^{-1}}\circ M\circ\psi_z)(a)dz\Bigr\|$$
$$\leq \int_{\mathbb T} |g_k(z)| \:\|\psi_{z^{-1}}\|\|M\|\|\psi_{z}\|\: \|a\|dz \leq \|M\|\|a\|,\:\forall a\in F({\mathbb Z}) $$
and, since ${F({\mathbb Z})}$ is dense in $E$, we obtain $\|{\bf M_k}\| \leq \|M\|, \: \forall k \in {\mathbb N}.$\\

Suppose that $S^{-1}$ is not bounded, but $S$ is bounded. Fix $k \in {\mathbb N}.$ Since
$${\bf M_k}=\sum_{n=-k}^{k}\Bigl(1-\frac{|n|}{k+1}\Bigr)\widehat{M}(n)S^n$$ is bounded, the operator $S^{k-1}{\bf M_k}$ is bounded. We have the equality
$$ S^{k-1}{\bf M_k}= \Bigl(1-\frac{k}{k+1}\Bigr)\widehat{M}({-k})S^{-1}+\sum_{n=-k+1}^{k}\Bigl(1-\frac{|n|}{k+1}\Bigr)\widehat{M}(n)S^{n+k-1}$$
and taking into account that the operator $\sum_{n=-k+1}^{k}\Bigl(1-\frac{|n|}{k+1}\Bigr)\widehat{M}(n)S^{n+k-1}$ is bounded and that $S^{-1}$ is not bounded, it is clear that $\widehat{M}({-k})=0.$ In the same way, composing ${\bf M_k}$ and $S^p$, for $p=k-2,\:k-3,\:...., 1$, we obtain that $\widehat{M}({-n})=0$, for $n>0$.
We can use the same argument if $S^{-1}$ is bounded but $S$ is not bounded. Thus the proof is complete.
$\Box$\\

\begin{lem} Let $\phi\in F({\mathbb Z})$ be such that $\phi*E\subset E.$\\
$1)$ If $S$ and $S^{-1}$ are bounded, then
$$|{\widetilde{\phi}}(z)|\leq \|M_\phi\|,\:\forall z \in \Omega:=\Bigl\{ \frac{1}{\rho(S^{-1})}\leq |z|\leq \rho(S)\Bigr\}.$$
$2)$ If $S$ is not bounded, but $S^{-1}$ is bounded and $\phi \in F({\mathbb Z}^{-})$, then
$$|{\widetilde{\phi}}(z)|\leq \|M_\phi\|,\:\forall z\in \mathcal{O}:=\Bigl\{z \in {\mathbb C}\:|\:|z|\geq \frac{1}{\rho(S^{-1})}\Bigr\} .$$
$3)$ If $S$ is bounded, but $S^{-1}$ is not bounded and $\phi \in F({\mathbb Z}^+)$, we have
$$|{\widetilde{\phi}}(z)|\leq \|M_\phi\|,\:\forall z\in W:=\Bigl\{z \in {\mathbb C}\:|\:|z|\leq \rho(S)\Bigr\} .$$
\end{lem}
{\bf Proof.} Suppose that $S$ and $S^{-1}$ are bounded.
For $z\in spec (S)$, we have three cases:\\
Case 1.
The operator $S-zI$ is not injective. Then there exists $x \in E\backslash \{0\}$ such that $Sx=zx$.\\
Case 2.
The operator $S^*-zI$ is injective. Then the range of $S-zI$ is dense in $E$ and it is not closed. Consequently, there exists a sequence $(f_p)_{p \in {\mathbb N}} \subset E$ such that
$$\lim_{p \to +\infty}\Bigl\|(S-zI) \frac {f_p}{\|f_p\|}\Bigr\|=0.$$
Case 3. The operator $S^*-zI$ is not injective. Then there exists $y \in E^*\backslash \{0\}$ such that $S^*y=zy.$ \\

Fix $z \in spec(S)$. First, assume that there exists $(h_p)_{p \in {\mathbb N}}\subset E$ such that
$$\lim_{p \to +\infty}\|S h_p-z h_p\|=0\:\:{\rm and}\:\: \|h_p\|=1, \: \forall p \in {\mathbb N}.$$ It follows immediately that
$$\lim_{p \to +\infty}\|S^kh_p-z^kh_p\|=0,\:\forall k \in {\mathbb Z}.$$
Then for $\phi\in F({\mathbb Z})$, we have for some $N>0$,
$$\|\phi*h_p-{\widetilde{\phi}}(z)h_p\| \leq \sum_{k=-N}^{N}(\sup_{|k|\leq N}|\phi(k)|) \|S^k h_p-z^kh_p\|$$ and we obtain
$$\lim_{p \to +\infty}\|\phi*h_p-{\widetilde{\phi}}(z)h_p\|=0.$$
Since $$|{\widetilde{\phi}}(z)|=\|{\widetilde{\phi}}(z)h_p\|=\|{\widetilde{\phi}}(z)h_p-\phi*h_p\|+\|M_\phi h_p\|,$$ it follows that $|{\widetilde{\phi}}(z)|\leq \|M_\phi\|.$\\

Now assume that there exists $y \in E^*\backslash \{0\} $ such that $S^*y=zy$. We obtain in the same way
$$|{\widetilde{\phi}}(z)|\leq \|M^*_\phi\|=\|M_\phi\|$$
and we conclude that for $\phi\in F({\mathbb Z})$, we have
$$|{\widetilde{\phi}}(z)|\leq \|M_\phi\|,\:\forall z \in spec(S).$$
If $S$ is bounded and $S^{-1}$ is not bounded, the proof is similar. If $S$ is not bounded and $S^{-1}$ is bounded, we use the spectrum of $S^{-1}$ and the same arguments. Thus in the case when $\frac{1}{\rho(S^{-1})}=\rho(S)$ the proof is complete.\\

Suppose again that $S$ and $S^{-1}$ are bounded and that
$\frac{1}{\rho(S^{-1})}<\rho(S).$ Fix $\phi \in F({\mathbb Z})$.
Let $R_1>0,\:R_2>0$ be such that $R_1<R_2$ and such that the circles $C_{R_1}$ and $C_{R_2}$ with radius respectively $R_1$ and $R_2$ are included in $spec(S)$. Since ${\widetilde{\phi}}$ is holomorphic on ${\mathbb C}\backslash\{0\}$ and $|{\widetilde{\phi}}(z)|\leq \|M_{\phi}\|$, for $z \in C_{R_1}\cup C_{R_2}$, by the maximum modulus theorem we obtain
$$|{\widetilde{\phi}}(z)|\leq \|M_{\phi}\|, \:\forall z \in \Omega_{R_1,R_2}:=\Bigl\{z \in {\mathbb C}\:|\:R_1 \leq |z|\leq R_2\Bigr\}.$$ The inclusions $ C_{\rho(S)}\subset spec(S)$ and $C_{\frac{1}{\rho(S^{-1})}}\subset spec(S)$ imply
$$|{\widetilde{\phi}}(z)| \leq \|M_\phi\|,\:{\rm for}\:z \in \Omega.$$
We complete the proof of 2) and 3) with similar arguments taking into account that if $\phi \in F({\mathbb Z}^-)$, the function $z \longrightarrow {\widetilde{\phi}}(z^{-1})$ is holomorphic on ${\mathbb C}$ and if $\phi \in F({\mathbb Z}^+)$, ${\widetilde{\phi}}$ is holomorphic on ${\mathbb C}$.
$\:\:\:\Box$\\

{\bf Proof of Theorem 1.}
Suppose that $S$ and $S^{-1}$ are bounded and let $M \in {\mu}(E)$. Let $({\bf M_k})_{k \in {\mathbb N}}$ be the sequence constructed in Lemma 2 so that
\begin{equation}\label{eq:h}\lim_{k \to +\infty}\|{\bf M_k} x-Mx\|=0 ,\: \forall x \in E \end{equation}
and $\|{\bf M_k}\|\leq \|M\|, \:\forall k \in {\mathbb N}.$ Set $\phi_k=\widehat{{\bf M_k}},$ for $k \in {\mathbb N}$, so that ${\bf M_k}=M_{\phi_k}.$
For $r>0$ and $a=(a(n))_{n \in {\mathbb Z}}\in E $, denote
$(a)_r(n)=a(n)r^n$. Fix $r \in [\frac{1}{\rho(S^{-1})},\rho(S)]$.
We have
$$|{{\widetilde{(\phi_k)_r}}}(z)|\leq \|M_{\phi_k}\|\leq \|M\|, \:\forall z \in {\mathbb T},\:\forall k \in {\mathbb N}.$$
We can extract from $\Bigl({{\widetilde{(\phi_k)_r}}}\Bigr)_{k \in {\mathbb N}}$ a subsequence which converges with respect to the weak topology $\sigma (L^{\infty}({\mathbb T}), L^1({\mathbb T})) $ to a function $\nu_r \in L^\infty({\mathbb T}).$ For simplicity, this subsequence will be denoted also by $\Big({{\widetilde{(\phi_k)_r}}}\Big)_{k \in {\mathbb N}}$. We obtain
$$\lim_{k \to +\infty}\int_{\mathbb T} \Big({{\widetilde{(\phi_k)_r}}}(z)g(z)-\nu_r(z)g(z)\Bigl)dz=0, \:\forall g \in L^1({\mathbb T})$$
and $\|\nu_r\|_{\infty}\leq \|M\|.$
It is clear that
$$\lim_{k \to +\infty}\int_{\mathbb T} \Big({{\widetilde{(\phi_k)_r}}}(z){\widetilde{(a)_r}}(z)g(z)-\nu_r(z){\widetilde{(a)_r}}(z)g(z)\Bigl)dz=0, \:\forall g \in L^2({\mathbb T}),\:\: \forall a \in F({\mathbb Z}).$$
We conclude that, for $a \in F({\mathbb Z})$, $\Bigl( {{\widetilde{(\phi_k)_r}}}{\widetilde{(a)_r}}\Bigr)_{k \in {\mathbb N}}$ converges with respect to the weak topology of $L^2({\mathbb T})$ to $\nu_r {\widetilde{(a)_r}}.$
Set $\widehat{\nu_r}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\nu_r(e^{it})e^{-itn}dt$, for $n \in {\mathbb Z}$ and let $\widehat{\nu_r}=(\widehat{\nu_r}(n))_{n \in {\mathbb Z}}$ be the sequence of the Fourier coefficients of $\nu_r$.
The Fourier transform from $l^2({\mathbb Z})$ to $L^2({\mathbb T})$ defined by
$$\mathcal{F}:l^2({\mathbb Z})\ni (a(n))_{n \in {\mathbb Z}} \longrightarrow \tilde{a}\vert_{\mathbb T} \in L^2({\mathbb T})$$
is unitary, so the sequence $\Bigl( (M_{\phi_k}a)_r \Bigr)_{k \in {\mathbb N}} =\Bigl((\phi_k)_r*(a)_r\Bigr)_{k \in {\mathbb N}}$ converges to $\widehat{\nu_r}*(a)_r$ with respect to the weak topology of $l^2({\mathbb Z})$. Taking into account (\ref{eq:h}), we obtain
$$\lim_{k \to +\infty}|<(M_{\phi_k}a)_r-(Ma)_r,\:b>|\leq\lim_{k \to +\infty}\|M_{\phi_k}a-Ma\|\:\|(b)_{r^{-1}}\|_*=0,\:\forall b \in F({\mathbb Z}).$$
We deduce that
$$(Ma)_r(n) =(\widehat{\nu_r}*(a)_r)(n),\: \forall n \in {\mathbb Z},\: \forall a \in F({\mathbb Z}).$$
This implies
$$({\widehat{M}})_r*(a)_r=\widehat{\nu_r}*(a)_r, \forall a \in F({\mathbb Z})$$
and
$$({\widehat{M}})_r=\widehat{\nu_r}.$$
Thus, we have
$${\widetilde{M}}(rz)=\sum_{n \in {\mathbb Z}}\widehat{M}(n)r^nz^n=\sum_{n \in {\mathbb Z}}\widehat{\nu_r}(n)z^n=\nu_r(z), \:\forall z \in {\mathbb T}.$$
Since $\|\nu_r\|_\infty\leq \|M\|$, it follows that the function ${\widetilde{M}}$ is essentially bounded by $\|M\| $ on every circle included in $\Omega$. If $\rho(S)=\frac{1}{\rho(S^{-1})}$, it is clear that $spec(S)=C_{\rho(S)}=\Omega$.\\

We assume below that $\rho(S)>\frac{1}{\rho(S^{-1})}$. Since $( {\widetilde{\phi_k}})_{ k \in {\mathbb N}}$ is an uniformly bounded sequence of holomorphic functions on $\overset{\circ}{\Omega} $, we can replace $({\widetilde{\phi_k}})_{ k \in {\mathbb N}}$ by a subsequence which converges to a function $\nu \in {\mathcal H}^{\infty}(\overset{\circ}{\Omega} )$ uniformly on every compact subset of $\overset{\circ}{\Omega}$. Thus, for $r\in ]\frac{1}{\rho(S^{-1})},\rho(S)[$,
the sequence $( {{\widetilde{(\phi_k)_r}}})_{ k \in {\mathbb N}}$ converges uniformly on ${\mathbb T}$ to the function $z \longrightarrow \nu(rz)$ and we obtain $\nu(rz)=\nu_r(z).$ We conclude that $\nu(rz)={\widetilde{M}}(rz)$, for $z \in {\mathbb T}$ and we get
$$\nu(z)={\widetilde{M}}(z)= \sum_{n \in {\mathbb Z}}\widehat{M}(n)z^n, \:{\rm for}\: z \in\overset{\circ}{\Omega}.$$
Consequently, ${\widetilde{M}}$ is holomorphic on ${\overset{\circ}{\Omega}}$.\\

Now we will prove that $spec(S)=\Omega$.
Let $\alpha \not\in spec(S)$. Then $(S-\alpha I)^{-1}\in {\mu}(E)$ and for $r>0$, if $C_r \subset \Omega$, there exists $\nu_r \in L^{\infty}({\mathbb T})$ such that
$$\mathcal{F}\Bigl({((S-\alpha I)^{-1}a)_r}\Bigr)(z)=\nu_r(z){\widetilde{(a)_r}}(z),\:\forall z \in {\mathbb T},\:\forall a \in F({\mathbb Z}).$$
Replacing $a$ by $(S-\alpha I)a$, it follows that
$${\widetilde{(a)_r}}(z)=\nu_r(z)\mathcal{F}\Bigl({((S-\alpha I)a)_r}\Bigr)(z)=\nu_r(z)(rz-\alpha){\widetilde{(a)_r}}(z),\:\forall z \in {\mathbb T},\:\forall a \in F({\mathbb Z}),$$
and we get $(rz-\alpha)\nu_r(z)=1$. Suppose that $\alpha \in C_r$ i.e. $\alpha= rz_0,\:z_0\in {\mathbb T}$. For $\epsilon >0$, there exists $z_\epsilon \in {\mathbb T}$ such that $|rz_\epsilon-rz_0|\leq \epsilon$ and $|\nu_r(z_\epsilon)|\leq \|\nu_r\|_\infty$. This implies $1 \leq \epsilon \|\nu_r\|_\infty$ and we obtain a contradiction. We deduce that $C_r \subset spec(S),\:\Omega \subset spec(S)$ and $spec(S)=\Omega$.
If we suppose that $S$ or $S^{-1}$ is not bounded, we obtain the same results by the same argument replacing $\Omega$ by $\mathcal{O}$ and $W$, where $\mathcal{O}$ and $W$ are introduced in Lemma 4. Notice that when $spec(S)=\mathcal{O}$, we deduce that $\rho(S)=+\infty$ and when $spec(S)=W$, we conclude that $\rho(S^{-1})=+\infty$.
$\Box$\\

\vspace{1cm}
\section{Toeplitz operators}

\vspace{1cm}
In this section, we prove Theorem 2. In the same way, as in the proof of Lemma 1, for $x \in {\bf E}$, we obtain
$$\lim_{k \to +\infty}\Bigl\|\sum_{n=0}^k\frac{1}{k+1}\sum_{p=0}^nx(p)e_p-x\Bigr\|=0.$$

If $\phi\in F({\mathbb Z})$ is such that $P^+(\phi*{\bf E})\subset {\bf E}$, we denote by $T_\phi$ the operator on ${\bf E}$ defined by $T_\phi x=P^+(\phi*x),$ for $x\in {\bf E}.$
By the same method, as in Section 2, we obtain the following lemma.

\begin{lem}
$1)$ Given a Toeplitz operator $T$ on ${\bf E}$, the sequence $(\phi_n)_{n \in {\mathbb N}}$, where $$\phi_n =\sum_{p=0}^n\frac{1}{n+1}\Bigl ( \sum_{k=-p}^p \widehat{T}(k)e_k \Bigr)$$
has the properties
$$\lim_{n \to +\infty}\|T_{\phi_n}x-Tx\|, \:\forall x \in {\bf E},\:and \: \|T_{\phi_n}\|\leq \|T\|,\:\forall n \in {\mathbb N}.$$\\
$2)$ If ${\bf S}$ is bounded, but ${{\bf S}}_{-1}$ is not bounded, $\widehat{T}(k)=0$, for $k<0$.\\
$3)$ If ${\bf S}$ is not bounded, but ${{\bf S}}_{-1}$ is bounded, $\widehat{T}(k)=0$, for $k>0$.\\

\end{lem}

\begin{lem}
$1)$ If ${\bf S}$ and ${{\bf S}}_{-1}$ are bounded, for $\phi\in F({\mathbb Z})$, we have
$$|\widetilde{\phi}(z)|\leq \|T_\phi\|, \:\forall z \in {\bf \Omega}:=\Bigl\{ z \in {\mathbb C}\:|\:\frac{1}{\rho({{\bf S}}_{-1})}\leq |z|\leq \rho({\bf S})\Bigr\}.$$
$2)$ If ${\bf S}$ is not bounded, but ${{\bf S}}_{-1}$ is bounded, for $\phi\in F({\mathbb Z}^-)$, we have
$$|\widetilde{\phi}(z)|\leq \|T_\phi\|, \:\forall z \in V:=\Bigl\{ z \in {\mathbb C}\:|\:\frac{1}{\rho({{\bf S}}_{-1})}\leq |z|\Bigr\}.$$
$3)$ If ${\bf S}$ is bounded, but ${{\bf S}}_{-1}$ is not bounded, for $\phi\in F({\mathbb Z}^+)$, we have
$$|\widetilde{\phi}(z)|\leq \|T_\phi\|, \:\forall z \in U:=\Bigl\{z\in {\mathbb C}\:|\: |z|\leq \rho({\bf S})\Bigr\}.$$

\end{lem}

{\bf Proof of Lemma 6.} We will present only the proof of 1). The proofs of 2) and 3) are very similar. Suppose that ${\bf S}$ and ${{\bf S}}_{-1}$ are bounded. Let $\lambda \in spec({\bf S})\cap(spec({{\bf S}}_{-1}))^{-1}.$
Since $\lambda \in spec({\bf S})$,
there exists a sequence $(f_n)_{n \in {\mathbb N}}$, $f_n\in {\bf E}$ such that
\begin{equation}\label{eq:1} \lim_{n \to +\infty}\|{\bf S} f_n-\lambda f_n\|=0 \: {\rm and}\: \|f_n\|=1,\: \forall n \in {\mathbb N}
\end{equation}
or there exists \begin{equation}\label{eq:2}a \in {\bf E}^*\backslash\{0\},\:{\bf S}^*a=\lambda a.\end{equation}
If (\ref{eq:1}) holds, we obtain
\begin{equation}\label{eq:3}
\lim_{n \to +\infty}\|{\bf S}^kf_n-\lambda^k f_n\|=0 \:{\rm and}\:\lim_{n \to +\infty}\|{\bf S}_{-1}^kf_n-\lambda^{-k} f_n\|=0,\:\forall k \in {\mathbb N}.
\end{equation}
Since $\lambda^{-1} \in spec({{\bf S}}^*_{-1})$, there exists a sequence $(g_n)_{n \in {\mathbb N}}$, $g_n\in {\bf E}^*$ such that

\begin{equation}\label{eq:4} \lim_{n \to +\infty}\|{{\bf S}}^*_{-1}g_n-\lambda^{-1} g_n\|_*=0 \: {\rm and}\: \|g_n\|_*=1,\: \forall n \in {\mathbb N}
\end{equation}
or there exists \begin{equation}\label{eq:5}b \in {\bf E}\backslash\{0\},\:({{\bf S}}^*_{-1})^*b={\bf S}_{-1}b=\lambda^{-1} b.\end{equation}
Next if (\ref{eq:4}) holds, we get
\begin{equation}\label{eq:6}
\lim_{n \to +\infty}\|({{\bf S}}^*)^{k}g_n-\lambda^k g_n\|_*=0 \:{\rm and}\:\lim_{n \to +\infty}\|({{\bf S}}_{-1}^*)^{k}g_n-\lambda^{-k} g_n\|_*=0 , \:\forall k \in {\mathbb N}.
\end{equation}
Suppose that we have (\ref{eq:2}) and (\ref{eq:5}). Let $a\in {\bf E}^*\backslash\{0\}$ be such that ${{\bf S}}^*(a)=\lambda a$. Set
$$x(-n)=<e_n,x>=<{\bf S}^ne_0,x>=<e_0,{\bf S}^{*n}x>,\:\forall x\in E^*.$$
Since $F({\mathbb Z}^+)$ is dense in ${\bf E}$, the map
$$E^*\ni x\longrightarrow (x(-n))_{n\geq 0}$$
is injective. We have $$a(-n)=\lambda^n a(0),\:\:n\geq 0.$$
Let $b\in {\bf E}\backslash\{0\}$ be such that ${\bf S}_{-1}b=\lambda^{-1} b$. We obtain
$$b(n)\lambda^n=b(0), \:n\geq 0.$$
Since $a\neq 0$ and $b \neq 0$, we have $a(0)\neq 0$, $b(0)\neq 0$.
For $k \in {\mathbb N}$, define $u_k \in F({\mathbb Z}^+)$ by
$$u_k=\sum_{n=0}^k\frac{1}{k+1}\sum_{p=0}^nb(p)e_p=\sum_{n=0}^k\Bigl(1-\frac {n}{k+1}\Bigr)b(n)e_n.$$
We have $\lim_{k \to +\infty}\|u_k-b\|=0$ and so $\lim_{k \to +\infty}<u_k,a>=<b,a>$. On the other hand,
$$\lim_{k \to +\infty}<u_k,a>=\lim_{k \to +\infty}\sum_{n=0}^k\Bigl(1-\frac {n}{k+1}\Bigr)\lambda^{-n}b(0)\lambda^na(0)=\lim_{k \to +\infty}\Bigl(\frac{k}{2}+1\Bigr)a(0)b(0)=+\infty.$$ We obtain an obvious contradiction and
we conclude that we cannot have in the same time (\ref{eq:2}) and (\ref{eq:5}), hence we have (\ref{eq:3}) or (\ref{eq:6}). Using the same arguments as in the proof of Lemma 4 and (\ref{eq:3}) or (\ref{eq:6}), we deduce
$$|\widetilde{\phi}(\lambda)|\leq \|T_\phi\|,\:\forall \phi \in F({\mathbb Z}), \:\forall \lambda \in spec({\bf S})\cap(spec({{\bf S}}_{-1}))^{-1}.$$
By the maximum modulus theorem we obtain

\begin{equation}\label{eq:7}
|\widetilde{\phi}(\lambda)|\leq \|T_\phi\|,\:\forall \phi \in F({\mathbb Z}), \:\forall \lambda \in {\bf \Omega}.
\end{equation}
If ${\bf S}$ is bounded and ${{\bf S}}_{-1}$ is not bounded, then for $\lambda \in spec({\bf S})$ there exists a sequence $(h_n)_{n \in {\mathbb N}}$, $h_n\in {\bf E}$ such that $\lim_{n \to +\infty}\|{\bf S} h_n-\lambda h_n\|=0 $ and $\|h_n\|=1$ or there exists $c \in {\bf E}^*\backslash \{0\}$ such that ${\bf S}^*c=\lambda c$. Using the same arguments as in the proof of Lemma 4, we obtain
$$|\widetilde{\phi}(\lambda)|\leq \|T_\phi\|,\:\forall \phi \in F({\mathbb Z}^+), \:\forall \lambda \in spec({\bf S}).$$
If ${{\bf S}}_{-1}$ is bounded, we use the spectrum of ${{\bf S}}_{-1}$. In the both situations, we obtain the result by using the maximum modulus theorem. $\Box$\\

Now we will prove the main result in this section.\\

{\bf Proof of Theorem 2.} The proof of Theorem 2 goes by using the same arguments as the proof of Theorem 1 with minor modifications. For the convenience of the reader we will give the main steps.
First, assume that ${\bf S}$ and ${{\bf S}}_{-1}$ are bounded.
Let $T$ be a Toeplitz operator on ${\bf E}$ and let $(\phi_k)_{k \in {\mathbb N}}\subset F({\mathbb Z})$ be such that
$$\lim_{k \to +\infty} \|T_{\phi_k}a-Ta\|=0, \:\forall a \in {\bf E}$$
and
$$\|T_{\phi_k}\|\leq \|T\|, \:\forall k \in {\mathbb N}.$$
For $r>0$ and $a \in {\bf E}$, denote
$(a)_r(n)=a(n)r^n.$ Fix $r \in [\frac{1}{\rho({{\bf S}}_{-1})},\rho({\bf S})]$.
We have
$$|\widetilde{(\phi_k)_r}(z)|\leq \|T_{\phi_k}\|\leq \|T\|, \:\forall z \in {\mathbb T},\:\forall k \in {\mathbb N}.$$
We can extract from $\Bigl(\widetilde{(\phi_k)_r}\Bigr)_{k \in {\mathbb N}}$ a subsequence which converges with respect to the weak topology $\sigma(L^{\infty}({\mathbb T}), L^1({\mathbb T})) $ to a function $\nu_r \in L^\infty({\mathbb T}).$ For simplicity, this subsequence will be denoted also by $\Big(\widetilde{(\phi_k)_r}\Big)_{k \in {\mathbb N}}$.

We conclude that, for $a \in F({\mathbb Z})$, $\Bigl( \widetilde{(\phi_k)_r}\widetilde{(a)_r}\Bigr)_{k \in {\mathbb N}}$ converges with respect to the weak topology of $L^2({\mathbb T})$ to $\nu_r \widetilde{(a)_r}.$
Denote by $\widehat{\nu_r}=(\widehat{\nu_r}(n))_{n \in {\mathbb Z}}$ the sequence of the Fourier coefficients of $\nu_r$.
Since the Fourier transform from $l^2({\mathbb Z})$ to $L^2({\mathbb T})$ is an isometry, the sequence $(\phi_k)_r*(a)_r$ converges to $\widehat{\nu_r}*(a)_r$ with respect to the weak topology of $l^2({\mathbb Z})$. On the other hand, $\Big( T_{\phi_k}a \Big)_{k \in {\mathbb N}} $ converges to $Ta$ with respect to the topology of ${\bf E}$. Consequently, we have
$$\lim_{k \to +\infty}|<(T_{\phi_k}a)_r-(Ta)_r,\:e_{-n}>|$$
$$\leq\lim_{k \to +\infty}\|T_{\phi_k}a-Ta\|\:\|(e_{-n})_{r^{-1}}\|_*=0,\:\forall n \in{\mathbb N},\:\forall a \in F({\mathbb Z}^+).$$
We conclude that
$$(Ta)_r =P^+(\widehat{\nu_r}*(a)_r),\: \forall a \in F({\mathbb Z}^+).$$
Since
$$(Ta)_r =P^+(({\widehat{T}}*a)_r),\: \forall a \in F({\mathbb Z}^+),$$ it follows that
$\widehat{T}(n)r^n=\widehat{\nu_r}(n),\:\forall n \in {\mathbb Z}.$
From the estimation $\|\nu_r\|_\infty\leq \|T\|$, we deduce that the function ${\widetilde{T}}$ is essentially bounded by $\|T\| $ on every circle included in ${\bf \Omega}$. \\

If we assume that $\rho({\bf S})>\frac{1}{\rho({{\bf S}}_{-1})}$, as in the proof of Theorem 1, we conclude that ${\widetilde{T}}$ is holomorphic on ${\overset{\circ}{{\bf \Omega}}}$.\\

Replacing ${\bf \Omega}$ by $U$ and $V$ and using the same arguments, we obtain the results when one of the operators ${\bf S}$ and ${{\bf S}}_{-1}$ is not bounded.
$\Box$\\
\vspace{0.3cm}

{\bf Acknowledgments.} The author thanks Jean Esterle for his useful advices and encouragements.\\
\vspace{0.3cm}

{\footnotesize

redpajama_set_name: RedPajamaArXiv


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text:
Home Essays Analysis of Better by Atul...
Analysis of Better by Atul Gawande
Topics: Physician, Medicine, Insurance Pages: 3 (876 words) Published: October 8, 2008
ow much do you really want to make? Many would love to make millions of dollars and do nothing to earn it. However, there are many who could care less about how much they make, and view it as a necessity for living and not as a luxury. Doctors come to a standstill when they are asked how much they make or would like to make. In a section of Atul Gawande's novel, "Piecework," he gives the reader a life lesson on how doing right in the medical field can be extensive and expensive. One must consider all the factors that go into a medical decision before assuming the worse.
Atul Gawande sees the medical profession more as a business rather than actual healing. Today doctors get so caught up in mess of how much a particular surgery should cost that many forget about the patient's care. At the beginning of "Piecework," Gawande recently finished his residency and is looking to become an independent doctor. However, he was conflicted about how much they should pay him. He never thought or acted upon the annual salary of a new doctor before, because most doctors never boast about their yearly income. However, when he did ask certain doctors the conversation "turned out to be awkward…and they'd [mumble] as if their mouths were full of crackers" (Gawande 113). Gawande states that doctors should not have to respond, because their main goal is to take care and save the patients. The author explains certain things have a definite cost and one must follow those costs no matter how extreme they may be.
Gawande describes how the government and many insurers have made some conclusions to the average cost of every surgery. This extensive book is called the "master fee schedule" (114). It lists costs of every surgery down to the exact penny. Gawande explains that the surgeries listed ranges from trimming one's nails which can cost $10.15 to reconstructing a new born baby who was born without a diaphragm, which can cost at least $5,366.98 (114). This basic idea for a list of...
Cited: Page
1.	Gawande, Atul. Better, "Piecework." New York, New York, Henry Holt and Company:
2007. Pages 112-129.
Atul Gawande: Letting Go Essay
..."Letting Go" that was being published in The New Yorker, Atul Gawande addresses the issues regarding to the current medical care system that fails to meet the needs of the patients with terminal illness. Gawande points out that the patients want to spend more quality time with their family members and having some special last moments rather than struggling to stay alive when they know that the chances are thin. Knowing the time to let go was one of the crucial part of the art of dying which people nowadays has forgotten. Gawande argues that choosing the hospice care would sometimes be a better choice for the terminally ill patients. He uses statistics such that the patients that choose hospice lives longer or than other patients and they tend to suffer lesser to support his argument. Hospice care tends to go with less pain treatment and focuses on the needs of the patients. It increases the quality of life of the patients during the last moments of their live. The patients' family members are less likely to suffer from depression when they have chosen to go with the hospice care mainly because they have prepared themselves to face the death of the patient when time has come. Gawande's argument is that the medical care system nowadays fails to meet the needs of the patients. His argument is convincing because he appeals to the emotions of his reader through both his own and others experience and...
Complications, By Atul Gawande Essay
...surgeon's notes on imperfect science is a scientific book that successfully published at the beginning of the 21st century, specifically in 2002. Atul Gawande discussed a plenty of cases that he dealt with in his first years of training. It seems to be a surgeon's diary because he included details about his training and the ones who affected him. Complications is mainly divided into three parts and all of them described the message of the book. The first part in Complications, "Fallibility" gives the reader an impression of its content before he begins to read. This part discusses a plenty of mistakes that Atul did when he starts his training. Some of these medical errors were under control, but a lot of them was out of control. For the ones that Atul cannot handle, a complication starts to develop in the patient. When a complication occurs, Dr. S the chief resident takes the shift to end the procedure safely. Atul mentioned important characteristic that a medical student, a medical resident and a doctor need which are: tenacity, self-confidence, and patience. He also mentioned that doctors attend conventions to get more knowledge and to learn from others mistakes to avoid having the same ones. Atul talked to one of the good doctors who operated badly after years of success. Dr. Goodman was this doctor who shared his story with Atul to let medical students and doctors learn...
Essay about Atul Gawande
...AStandardization is good, it saves lives Agpar score Standardization resulted in good things such as neonatal intensive units, prenatal ultrasound, obstetrics package. "And that package has produced dramatic results". Gawande highlights some flaws of Obstetrics such as their lack of double blind, controlled randomized trials (obstetrics uses the least double blind controlled trials in all of medicine), use of fetal heart rate monitoring device, forceps eliminated even though better for mothers Gawande though absolves them of blame because they are so successful at saving lives. Gawande even complements obstetrics for not relying on clinical trials, but trying things on the fly. Even admits that it is unclear whether many aspects of the obstetrics package are beneficial. Gawande seems to be satisfied that obstetrics is saving lives, not concerned that there are some alarming practices that could be improved. Because Obstetrics still is concerned primarily with safety, "reliability [is chosen] over the possibility of occasional perfection". Gawande postulates whether medicine is a craft or industry, with craft being about perfection, and industry being about safety and reliability. Gawande's view is black and white, should be a craft where if you can do forceps you do that and you perfect c-section if you can do that. Gawande's view leads to a reliance in c-sections. A system...
Being Mortal: Medicine And What Matters In The End, By Atul Gawande: Analysis Essay
...The Author Atul Gawande is a surgeon, staff writer for The New Yorker and a professor at the Harvard Medical School. Being Mortal: Medicine and What Matters in the End was an inspiring book that unwrap people's mind for discussion and question our current practice of medicine and care. It is easy for audiences of all ages to relate to this book even if the young do not think about the process of death. It has a comprehensive coverage of medical sociology, where it deliberates on the evolution, controversial conversation of medicine and issues after medicine becomes impotent to people's health. Gawande uses recounts of people (patients) and his own reflections on the stories to illustrate the dilemmas of the two facet of medicine: to attempt to prolong individuals' life or promote their quality of life at the end of life care through palliative care and assisted living, which attempts to discontinue all treatment and alleviate suffering, and the contentious problems that entails. He expresses how his patient and his own intimate encounter with his father shape his perspective of dying. They had enlightened him about the end of life care, moving from the lens of a doctor to the lens of a family member to his frail father. It also echoes what transpires and what could be done when we have exhausted the use of medicine to support our health particularly in the terminal stage of life. In that stage, the dispute of cure and to prolong...
Atul Essay
...Case 1 A cafeteria at GDGWI has one special dish it serves like clockwork every Thursday at noon. This supposedly tasty dish is a casserole that contains sautéed onions, boiled sliced potatoes, green beans, and cream of mushroom soup. Unfortunately, students fail to see the special quality of this dish, and they loathingly refer to it as the Killer Casserole. The students reluctantly eat the casserole, however, because the cafeteria provides only a limited selection of dishes for Thursday's lunch (namely, the casserole). Parag, the cafeteria manager, is looking to cut costs for the coming year, and he believes that one sure way to cut costs is to buy less expensive and perhaps lower-quality ingredients. Because the casserole is a weekly staple of the cafeteria menu, he concludes that if he can cut costs on the ingredients purchased for the casserole, he can significantly reduce overall cafeteria operating costs. He therefore decides to invest time in determining how to minimize the costs of the casserole while maintaining nutritional and taste requirements. Maria focuses on reducing the costs of the two main ingredients in the casserole, the potatoes and green beans. These two ingredients are responsible for the greatest costs, nutritional content, and taste of the dish. Parag buys the potatoes and green beans from a wholesaler each week. Potatoes cost Rs. 40 per kg, and green beans cost Rs. 100 per kg. GDGWI has established nutritional requirements that each main...
Better Essay
... The Chinese Exclusion Act The United States of America is a typical country that stresses fairer for people. Americans advocate that everyone should be free and equal. They emphasize the basic rights of humans. In the Declaration of Independence, it says people have three unalienable rights, life, liberty, and the pursuit of happiness. Nobody can despoil these rights without suitable reason. But, after a hundred years that the Declaration of Independence was passed, the American government passed a very unfair federal law. It unreasonably deprived these unalienable rights of the group of people who live in America. They lost liberty and were treated unfairly. The act was named The Chinese Exclusion Act. The Chinese Exclusion Act was signed by Chester A. Arthur on May 6, 1882, following revisions made in 1880 to the Burlingame Treaty of 1868. These laws allowed the U.S. to suspend Chinese immigration, a ban that was intended to last 10 years. The background of the act During 1848 to 1865, the first group of Chinese immigrated to America with the California Gold Rush. They continued to immigrate with the large labor projects, such as the First Transcontinental Railroad. At that time, a lot of Chinese immigrated to America. During the early stages of the gold rush, surface gold was abundant, and Americans tolerated the Chinese took the gold. But later, the gold was less and less, the white men began to hate Chinese labors. They said the gold was belonged...
Nike Innovate For A Better World Analysis Essay
...Nike one of the biggest sports brand ever, have a slogan "Innovate for a better world". Mark Parker, CEO and President of the global corporation for Nike said "'Innovate for a better world' is more than just a catchphrase for being a responsible corporation. It's about reaching our potential" one of the most important thing for Nike, is to think more innovative, Mark Parker believes that that they now see challenges as opportunities that will make the company growing any further, they believe they can make better products for the athletes and the planet, and have the best conditions for their employees. Which did not go particularly well for Nike. in the 90's in Asia were Nike have several factories , Nike was accused for using cheap labor in Asia, where they can get away from the tighter administration and rules in USA and Europe. The have also been exposed for using child labour, as well. Which forced them to lead the company in a different way than they did before the incident, As Hannah Jones, VP of sustainable business and innovation in Nike, says: "it was the most important thing that has happened to us". In the mid 90's there was a shift on the expectation on the company, and on that time, they started to develop a corporate social responsibility policy, just because they had to get a better repetition, buy looking into their supply they understand how important their environment and social impact is,...
Chapter summaries for Better by Atule Gawande Essay
... Biology Enriched Extra Credit Better: A Surgeon's Notes on Performance Chapter 1: On Washing Hands Mr. Gawande starts his literature on washing hands. He introduces two friends a microbiologist and an infectious disease specialist. Both work hard and diligently against the spread of diseases just like Semmelweis who is mentioned in the chapter. Something I learned, that not many realize, is that each year two million people acquire an infection while they are in the hospital. Mainly because the clinicians only wash their hands one-third to one-half as many times as they should. Semmelweis, mentioned earlier, concluded in 1847 that doctors themselves were to blame for childbed fever, which was the leading cause of maternal death in childbirth. The best solutions are apparently the sanitizing gels that have only recently caught on in the U.S. Then there was an initiative to make the sanitizing easier for all. The engineer Perreiah came up with solutions that gave the staff more time which was revolutionary in itself but the format worked only under his supervision. After he left it all went down the drain, so, Lloyd a surgeon who had helped Perreiah decided to do more research and was excited when he encountered the positive deviance idea, the idea of building on people's capabilities instead of trying to change them. The idea worked and even got funding for ten more hospitals across the country. At the end of the...
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text:
Wal-Mart Unwraps Holiday Toy Wishes
Leading Toy Retailer Predicts the Season's Top 12 Toys
and Introduces Magical Characters at the 2006 Toy Wishes Holiday Preview
BENTONVILLE, Ark., October 4, 2006–Boasting life-like dolls, adventurous ride-on vehicles and advanced digital games, the 2006 holiday season is chock-full of thrilling toys. And Wal-Mart, the world's leading toy retailer, is the place for the hottest toys at the coolest prices, allowing shoppers to get and give more. This week, Wal-Mart releases its predictions for the top 12 toys of the season as it provides a sneak peak into its in-store "Toyland" at the fifth annual Toy Wishes Holiday Preview in New York. Additionally, Wal-Mart introduces delightful namesake characters Wally & Marty, "renegade" elves, which will soon be synonymous with holiday spirit.
Top 12 Toys of Christmas and Holidays '06
Wal-Mart's list of top 12 toys reflects the most prevalent trends in the industry, including high-tech digital toys, imaginative toys, experiential toys, characters that come to life and the modernization of classic favorites. The top 12 toys for Christmas and Holidays 2006 are:
Monopoly® Here & Now™ Edition from Hasbro® ($28.96) – Ages 8 and up
Zoombox™ ($248) – For kids of all ages
V.Flash™ Edutainment System from VTech® ($99.77) – Ages 6 -10
BABY ALIVE® doll from Hasbro® ($49.97) – Ages 3 and up
Power Wheels® Ford Mustang from Fisher-Price® ($249.94), red Mustang available exclusively at Wal-Mart and Walmart.com – Ages 3 and up
TMX™ Elmo from Fisher-Price® ($39.97) – Ages 18 months and up
Kid-Tough™ Digital Camera from Fisher-Price® ($59.88) – Ages 3 and up
ESPN® Fast Action Football™ from Fisher-Price® ($198.88) – Ages 6 and up
"BARBIE™ IN THE 12 DANCING PRINCESSES" PRINCESS GENEVIEVE™ DOLL™ from Mattel® ($16.88) – Ages 3 and up
Bratz™ Forever Diamondz™: Vinessa ($19.97), available exclusively at Wal-Mart and Walmart.com – Ages 6 and up
Littlest Pet Shop Virtual Pet (Dog or Cat) from Hasbro® ($12.84) – Ages 6 and up
Hot Wheels® Radar Gun from Mattel® ($29.74) – Ages 7 and up
"Wal-Mart is committed to offering the most imaginative and innovative toys during the holiday season and year-round," says Scott McCall, Vice President and Divisional Merchandise Manager of toys at Wal-Mart. "We stay on top of trends year-round and offer a one-stop destination to find the hottest toys at a great value."
Wally & Marty
This Holiday for the first time ever, Wal-Mart introduces Wally & Marty, a pair of 3-D animated elves. These "renegade elves" have started their own business, Toyland, a magical place where Wal-Mart's most innovative toys are tested, built and delivered. Wally & Marty are Wal-Mart's "guardians of amazing toys" and will come to life for children this season through animated cinema and television advertisements, a comic book, various print executions and the highly interactive www.walmart.com/toyland microsite, where children can build a "Toyland Wishlist" of their favorite toys.
About Wal-Mart
Wal-Mart Stores, Inc. operates Wal-Mart Stores, Supercenters, Neighborhood Markets and SAM'S CLUB locations in the United States. Internationally, the company operates in Argentina, Brazil, Canada, China, Costa Rica, El Salvador, Germany, Guatemala, Honduras, Japan, Mexico, Nicaragua, Puerto Rico, and the United Kingdom. The company's securities are listed on the New York and Pacific stock exchanges under the symbol WMT. More information about Wal-Mart can be found by visiting www.walmartfacts.com. Online merchandise sales are available at www.walmart.com.


text:
Home » Homepage Featured Story Slider »
Nigeria's president urged to sign African trade deal
Celestine Hendricks Celestine Hendricks
28 June 2019, 11:05 AM | AFP	| @SABCNewsOnline
Tags: Nigeria, Muhammadu Buhari, African trade deal
Image: AFP
A special government panel was formed by President Muhammadu Buhari in March last year to assess the impact of the African Continental Free Trade Agreement (AfCTA), after Nigeria abruptly pulled out of signing the deal adopted by a majority of African nations.
A special government panel has recommended that President Muhammadu Buhari sign Nigeria up for a landmark African free trade agreement, ahead of a key meeting of African leaders.
The panel was formed by Buhari in March last year to assess the impact of the African Continental Free Trade Agreement (AfCFTA) after Nigeria abruptly pulled out of signing the deal adopted by a majority of African nations.
"Our reports shows that, on balance, Nigeria should consider joining the AfCFTA", the panel's chair, Desmond Guobadia, said in a statement to the president on Thursday, after submitting the report.
The trade deal "provides immense opportunities for Nigeria's manufacturing and service companies to expand to Africa", Guobadia said.
The panel, however, recommended Buhari delay "liberalising" and ratifying the deal which would open Nigeria's markets to the continent, in order to manage the agreement's "negative impacts."
"AfCFTA is not without major risks", it said, warning that the deal could lead to increased smuggling and abuse of 'rule of origin' labels on products.
"The risk is that it will provide incentive for traders to disguise goods imported from outside the continent as made-in-Africa goods", the report warned.
The continental pact which formally came into force last month represents an ambitious African Union(AU) plan that seeks to integrate and boost intra-African trade.
Nigeria had been a key backer of the plan for several years yet suddenly pulled out last March, days before other African countries endorsed the document at a summit in Rwanda.
Calling for more time to address concerns from business and labour unions, Buhari instead formed a panel to engage industries across the country on the impact of the deal on Nigeria.
"Our position is very simple, we support free trade", Buhari said, "as long as it is fair and conducted on an equitable basis".
Ahead of an AU summit of African leaders in Niger Republic next week, the AfCFTA will again be on the agenda with Nigeria widely expected by industries and labour unions in Nigeria to at least partially adopt the agreement.
The report "will form part of the consideration in our decision on the next steps on the AfCFTA", said Buhari.
Nigeria Muhammadu Buhari African trade deal
This entry was posted on 28 Jun 2019,11:05AM at 11:05 AM and is filed under Africa, Homepage Featured Story Slider. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.
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Home / News / MFractor 3.3 out now!
Today Tom and I are excited to announce the release of MFractor v3.3 for Visual Studio for Mac!
This release is jammed packed with features; you can now customise the code MFractor creates using code snippets, generate interface implementations, create effects and easily localise C# and XAML.
Let's go through code snippets first.
Code snippets allow you to define templated code and then tell MFractor to use it for features like view model property or command generation.
Using code snippets, it's now possible to make MFractor generate code that is specific to your project.
So, how do you use your own code snippets in MFractor? Let's learn by customising the code generated when we generate a new view model property!
Firstly, add a file named app.mfc.xml to the top level of your project; this is a project level MFractor configuration file.
Next, create a directory in your project named Snippets, this is where we will place our customised snippets so they can be reused between developers.
We will use this snippet to change the behaviour of the Snippet property of the View Model Property code generator.
MFractor can also use code snippets defined within Visual Studio Macs code snippets editor. To learn more, check out the code snippets tutorial on our documentation.
So what else have we added?
In addition to all these shiny new features, you can expect a tonne of bug fixes, performance enhancements and better product stability. You can check out the full release notes here.
Interested in upgrading to MFractor Professional? Feel free to email hello@mfractor.com at any time to have a chat about how MFractor can make your Xamarin development easier.