When are homomorphisms between Banach algebras contractions?What are the right categories of finite-dimensional Banach spaces?Which Banach algebras are group algebras?Unbounded representations of Banach algebrasGeneralising Gelfand's spectral theoryAlgebraically simple Banach algebrasReference request for translating from Top to C*-algcharacterization of commutative Banach algebrasBanach-Mazur distance between the cube and the octahedroninduced map on state spaces

When are homomorphisms between Banach algebras contractions?


What are the right categories of finite-dimensional Banach spaces?Which Banach algebras are group algebras?Unbounded representations of Banach algebrasGeneralising Gelfand's spectral theoryAlgebraically simple Banach algebrasReference request for translating from Top to C*-algcharacterization of commutative Banach algebrasBanach-Mazur distance between the cube and the octahedroninduced map on state spaces













4














$begingroup$



When are homomorphisms between Banach algebras contractions?




I recall from my student days that there are results which show that a positive answer to the above question holds under very general conditions (on the algebras). I have tried a literature search but only turned up the fact that it is true for $ast$-homomorphisms between $C^ast$-algebras (Proposition 3.2.2 in Dales et al., “Banach spaces of continuous functions as dual spaces”). Does any forum member have a reference for a definitive account of this theme?



I should add that I know that this not true for ANY pair of algebras, and I know that there are substantial results on classes of algebras for which it IS true. As a non-expert on complete normed algebras, I am looking for some references on the present state of the art.










share|cite|improve this question












$endgroup$










  • 1




    $begingroup$
    I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2(bf C)$ which send $I_2$ to $I_2$ but have norm strictly greater than 1...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:10







  • 1




    $begingroup$
    Characters on commutative Banach algebras are contractive, that much is true...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:11










  • $begingroup$
    Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily.
    $endgroup$
    – Yemon Choi
    Oct 16 at 13:56










  • $begingroup$
    I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject.
    $endgroup$
    – Nik Weaver
    Oct 16 at 16:28















4














$begingroup$



When are homomorphisms between Banach algebras contractions?




I recall from my student days that there are results which show that a positive answer to the above question holds under very general conditions (on the algebras). I have tried a literature search but only turned up the fact that it is true for $ast$-homomorphisms between $C^ast$-algebras (Proposition 3.2.2 in Dales et al., “Banach spaces of continuous functions as dual spaces”). Does any forum member have a reference for a definitive account of this theme?



I should add that I know that this not true for ANY pair of algebras, and I know that there are substantial results on classes of algebras for which it IS true. As a non-expert on complete normed algebras, I am looking for some references on the present state of the art.










share|cite|improve this question












$endgroup$










  • 1




    $begingroup$
    I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2(bf C)$ which send $I_2$ to $I_2$ but have norm strictly greater than 1...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:10







  • 1




    $begingroup$
    Characters on commutative Banach algebras are contractive, that much is true...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:11










  • $begingroup$
    Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily.
    $endgroup$
    – Yemon Choi
    Oct 16 at 13:56










  • $begingroup$
    I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject.
    $endgroup$
    – Nik Weaver
    Oct 16 at 16:28













4












4








4


1



$begingroup$



When are homomorphisms between Banach algebras contractions?




I recall from my student days that there are results which show that a positive answer to the above question holds under very general conditions (on the algebras). I have tried a literature search but only turned up the fact that it is true for $ast$-homomorphisms between $C^ast$-algebras (Proposition 3.2.2 in Dales et al., “Banach spaces of continuous functions as dual spaces”). Does any forum member have a reference for a definitive account of this theme?



I should add that I know that this not true for ANY pair of algebras, and I know that there are substantial results on classes of algebras for which it IS true. As a non-expert on complete normed algebras, I am looking for some references on the present state of the art.










share|cite|improve this question












$endgroup$





When are homomorphisms between Banach algebras contractions?




I recall from my student days that there are results which show that a positive answer to the above question holds under very general conditions (on the algebras). I have tried a literature search but only turned up the fact that it is true for $ast$-homomorphisms between $C^ast$-algebras (Proposition 3.2.2 in Dales et al., “Banach spaces of continuous functions as dual spaces”). Does any forum member have a reference for a definitive account of this theme?



I should add that I know that this not true for ANY pair of algebras, and I know that there are substantial results on classes of algebras for which it IS true. As a non-expert on complete normed algebras, I am looking for some references on the present state of the art.







reference-request fa.functional-analysis oa.operator-algebras banach-algebras






share|cite|improve this question
















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 16 at 13:52







user131781

















asked Oct 16 at 6:06









user131781user131781

1,1272 silver badges5 bronze badges




1,1272 silver badges5 bronze badges










  • 1




    $begingroup$
    I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2(bf C)$ which send $I_2$ to $I_2$ but have norm strictly greater than 1...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:10







  • 1




    $begingroup$
    Characters on commutative Banach algebras are contractive, that much is true...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:11










  • $begingroup$
    Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily.
    $endgroup$
    – Yemon Choi
    Oct 16 at 13:56










  • $begingroup$
    I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject.
    $endgroup$
    – Nik Weaver
    Oct 16 at 16:28












  • 1




    $begingroup$
    I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2(bf C)$ which send $I_2$ to $I_2$ but have norm strictly greater than 1...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:10







  • 1




    $begingroup$
    Characters on commutative Banach algebras are contractive, that much is true...
    $endgroup$
    – Yemon Choi
    Oct 16 at 7:11










  • $begingroup$
    Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily.
    $endgroup$
    – Yemon Choi
    Oct 16 at 13:56










  • $begingroup$
    I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject.
    $endgroup$
    – Nik Weaver
    Oct 16 at 16:28







1




1




$begingroup$
I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2(bf C)$ which send $I_2$ to $I_2$ but have norm strictly greater than 1...
$endgroup$
– Yemon Choi
Oct 16 at 7:10





$begingroup$
I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2(bf C)$ which send $I_2$ to $I_2$ but have norm strictly greater than 1...
$endgroup$
– Yemon Choi
Oct 16 at 7:10





1




1




$begingroup$
Characters on commutative Banach algebras are contractive, that much is true...
$endgroup$
– Yemon Choi
Oct 16 at 7:11




$begingroup$
Characters on commutative Banach algebras are contractive, that much is true...
$endgroup$
– Yemon Choi
Oct 16 at 7:11












$begingroup$
Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily.
$endgroup$
– Yemon Choi
Oct 16 at 13:56




$begingroup$
Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily.
$endgroup$
– Yemon Choi
Oct 16 at 13:56












$begingroup$
I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject.
$endgroup$
– Nik Weaver
Oct 16 at 16:28




$begingroup$
I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject.
$endgroup$
– Nik Weaver
Oct 16 at 16:28










1 Answer
1






active

oldest

votes


















10
















$begingroup$

About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.



In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms $bf M_2 to bf M_2$, with both sides carrying the natural $rm C^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.



(To get such homomorphisms, let
$$ s_t= left(matrix 1 & t \ 0 & 1 right)$$ and consider the automorphism of $bf M_2$ given by $x mapsto s_t^-1 x s_t$. A simple calculation shows that the norm-1 element
$$p=left(matrix1 & 0 \ 0 & 0 right)$$
satisfies
$$s_t^-1ps_t = left(matrix 1 & t \ 0 & 0 right)$$ and the latter matrix has norm $ > |t|$.)



One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=ell^1(bf Z_+)$ with convolution product (a.k.a. the completion of the polynomial ring $bf C[z]$ in the natural $ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $ain A_+$ definess a continuous unital endomorphism of $A_+$ which sends $delta_1mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see




MR0241980 (39 #3315) Reviewed
D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980




which provides the example $a= (delta_0+delta_1-delta_2)/sqrt5$ among others.



To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B=bf M_2$ or $A=B=ell^1(bf Z_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a $rm C^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.






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    $begingroup$

    About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.



    In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms $bf M_2 to bf M_2$, with both sides carrying the natural $rm C^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.



    (To get such homomorphisms, let
    $$ s_t= left(matrix 1 & t \ 0 & 1 right)$$ and consider the automorphism of $bf M_2$ given by $x mapsto s_t^-1 x s_t$. A simple calculation shows that the norm-1 element
    $$p=left(matrix1 & 0 \ 0 & 0 right)$$
    satisfies
    $$s_t^-1ps_t = left(matrix 1 & t \ 0 & 0 right)$$ and the latter matrix has norm $ > |t|$.)



    One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=ell^1(bf Z_+)$ with convolution product (a.k.a. the completion of the polynomial ring $bf C[z]$ in the natural $ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $ain A_+$ definess a continuous unital endomorphism of $A_+$ which sends $delta_1mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see




    MR0241980 (39 #3315) Reviewed
    D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980




    which provides the example $a= (delta_0+delta_1-delta_2)/sqrt5$ among others.



    To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B=bf M_2$ or $A=B=ell^1(bf Z_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a $rm C^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.






    share|cite|improve this answer












    $endgroup$



















      10
















      $begingroup$

      About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.



      In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms $bf M_2 to bf M_2$, with both sides carrying the natural $rm C^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.



      (To get such homomorphisms, let
      $$ s_t= left(matrix 1 & t \ 0 & 1 right)$$ and consider the automorphism of $bf M_2$ given by $x mapsto s_t^-1 x s_t$. A simple calculation shows that the norm-1 element
      $$p=left(matrix1 & 0 \ 0 & 0 right)$$
      satisfies
      $$s_t^-1ps_t = left(matrix 1 & t \ 0 & 0 right)$$ and the latter matrix has norm $ > |t|$.)



      One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=ell^1(bf Z_+)$ with convolution product (a.k.a. the completion of the polynomial ring $bf C[z]$ in the natural $ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $ain A_+$ definess a continuous unital endomorphism of $A_+$ which sends $delta_1mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see




      MR0241980 (39 #3315) Reviewed
      D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980




      which provides the example $a= (delta_0+delta_1-delta_2)/sqrt5$ among others.



      To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B=bf M_2$ or $A=B=ell^1(bf Z_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a $rm C^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.






      share|cite|improve this answer












      $endgroup$

















        10














        10










        10







        $begingroup$

        About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.



        In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms $bf M_2 to bf M_2$, with both sides carrying the natural $rm C^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.



        (To get such homomorphisms, let
        $$ s_t= left(matrix 1 & t \ 0 & 1 right)$$ and consider the automorphism of $bf M_2$ given by $x mapsto s_t^-1 x s_t$. A simple calculation shows that the norm-1 element
        $$p=left(matrix1 & 0 \ 0 & 0 right)$$
        satisfies
        $$s_t^-1ps_t = left(matrix 1 & t \ 0 & 0 right)$$ and the latter matrix has norm $ > |t|$.)



        One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=ell^1(bf Z_+)$ with convolution product (a.k.a. the completion of the polynomial ring $bf C[z]$ in the natural $ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $ain A_+$ definess a continuous unital endomorphism of $A_+$ which sends $delta_1mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see




        MR0241980 (39 #3315) Reviewed
        D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980




        which provides the example $a= (delta_0+delta_1-delta_2)/sqrt5$ among others.



        To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B=bf M_2$ or $A=B=ell^1(bf Z_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a $rm C^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.






        share|cite|improve this answer












        $endgroup$



        About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.



        In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms $bf M_2 to bf M_2$, with both sides carrying the natural $rm C^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.



        (To get such homomorphisms, let
        $$ s_t= left(matrix 1 & t \ 0 & 1 right)$$ and consider the automorphism of $bf M_2$ given by $x mapsto s_t^-1 x s_t$. A simple calculation shows that the norm-1 element
        $$p=left(matrix1 & 0 \ 0 & 0 right)$$
        satisfies
        $$s_t^-1ps_t = left(matrix 1 & t \ 0 & 0 right)$$ and the latter matrix has norm $ > |t|$.)



        One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=ell^1(bf Z_+)$ with convolution product (a.k.a. the completion of the polynomial ring $bf C[z]$ in the natural $ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $ain A_+$ definess a continuous unital endomorphism of $A_+$ which sends $delta_1mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see




        MR0241980 (39 #3315) Reviewed
        D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980




        which provides the example $a= (delta_0+delta_1-delta_2)/sqrt5$ among others.



        To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B=bf M_2$ or $A=B=ell^1(bf Z_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a $rm C^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.







        share|cite|improve this answer















        share|cite|improve this answer




        share|cite|improve this answer








        edited Oct 16 at 14:16

























        answered Oct 16 at 10:33









        Yemon ChoiYemon Choi

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            Tom Holland Mục lục Đầu đời và giáo dục | Sự nghiệp | Cuộc sống cá nhân | Phim tham gia | Giải thưởng và đề cử | Chú thích | Liên kết ngoài | Trình đơn chuyển hướngProfile“Person Details for Thomas Stanley Holland, "England and Wales Birth Registration Index, 1837-2008" — FamilySearch.org”"Meet Tom Holland... the 16-year-old star of The Impossible""Schoolboy actor Tom Holland finds himself in Oscar contention for role in tsunami drama"“Naomi Watts on the Prince William and Harry's reaction to her film about the late Princess Diana”lưu trữ"Holland and Pflueger Are West End's Two New 'Billy Elliots'""I'm so envious of my son, the movie star! British writer Dominic Holland's spent 20 years trying to crack Hollywood - but he's been beaten to it by a very unlikely rival"“Richard and Margaret Povey of Jersey, Channel Islands, UK: Information about Thomas Stanley Holland”"Tom Holland to play Billy Elliot""New Billy Elliot leaving the garage"Billy Elliot the Musical - Tom Holland - Billy"A Tale of four Billys: Tom Holland""The Feel Good Factor""Thames Christian College schoolboys join Myleene Klass for The Feelgood Factor""Government launches £600,000 arts bursaries pilot""BILLY's Chapman, Holland, Gardner & Jackson-Keen Visit Prime Minister""Elton John 'blown away' by Billy Elliot fifth birthday" (video with John's interview and fragments of Holland's performance)"First News interviews Arrietty's Tom Holland"“33rd Critics' Circle Film Awards winners”“National Board of Review Current Awards”Bản gốc"Ron Howard Whaling Tale 'In The Heart Of The Sea' Casts Tom Holland"“'Spider-Man' Finds Tom Holland to Star as New Web-Slinger”lưu trữ“Captain America: Civil War (2016)”“Film Review: ‘Captain America: Civil War’”lưu trữ“‘Captain America: Civil War’ review: Choose your own avenger”lưu trữ“The Lost City of Z reviews”“Sony Pictures and Marvel Studios Find Their 'Spider-Man' Star and Director”“‘Mary Magdalene’, ‘Current War’ & ‘Wind River’ Get 2017 Release Dates From Weinstein”“Lionsgate Unleashing Daisy Ridley & Tom Holland Starrer ‘Chaos Walking’ In Cannes”“PTA's 'Master' Leads Chicago Film Critics Nominations, UPDATED: Houston and Indiana Critics Nominations”“Nominaciones Goya 2013 Telecinco Cinema – ENG”“Jameson Empire Film Awards: Martin Freeman wins best actor for performance in The Hobbit”“34th Annual Young Artist Awards”Bản gốc“Teen Choice Awards 2016—Captain America: Civil War Leads Second Wave of Nominations”“BAFTA Film Award Nominations: ‘La La Land’ Leads Race”“Saturn Awards Nominations 2017: 'Rogue One,' 'Walking Dead' Lead”Tom HollandTom HollandTom HollandTom Hollandmedia.gettyimages.comWorldCat Identities300279794no20130442900000 0004 0355 42791085670554170004732cb16706349t(data)XX5557367