Existence of $T$ such that $[(Tcirc T)(f)](n)=f(n+1)$












0














Defined $A$ as the set of all the functions $f:mathbb Zrightarrow mathbb R$ exists a function $T:Arightarrow A$ such that
$$
(T^2f)(n)=f(n+1)text{ for all }ninmathbb Ztext{ and }fin A
$$
or not? And if codomain of $f$ is $mathbb Z$?



Remember that $T^2=Tcirc T$ and $T^2f=Tleft[T(f)right]$.




And if $T$ is also linear on $A$ seen as a vectorial space over $mathbb R$?






functions discrete-mathematics integers natural-numbers







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  • This question is related, but asks for a $T$ that is linear over $mathbb C$ rather than $mathbb R$ and also only considers $ell^1(mathbb Z)$ rather than the full $mathbb C^{mathbb Z}$.
    – Henning Makholm
    Nov 25 at 22:22


















0














Defined $A$ as the set of all the functions $f:mathbb Zrightarrow mathbb R$ exists a function $T:Arightarrow A$ such that
$$
(T^2f)(n)=f(n+1)text{ for all }ninmathbb Ztext{ and }fin A
$$
or not? And if codomain of $f$ is $mathbb Z$?



Remember that $T^2=Tcirc T$ and $T^2f=Tleft[T(f)right]$.




And if $T$ is also linear on $A$ seen as a vectorial space over $mathbb R$?






functions discrete-mathematics integers natural-numbers







share|cite|improve this question
























  • This question is related, but asks for a $T$ that is linear over $mathbb C$ rather than $mathbb R$ and also only considers $ell^1(mathbb Z)$ rather than the full $mathbb C^{mathbb Z}$.
    – Henning Makholm
    Nov 25 at 22:22
















0












0








0







Defined $A$ as the set of all the functions $f:mathbb Zrightarrow mathbb R$ exists a function $T:Arightarrow A$ such that
$$
(T^2f)(n)=f(n+1)text{ for all }ninmathbb Ztext{ and }fin A
$$
or not? And if codomain of $f$ is $mathbb Z$?



Remember that $T^2=Tcirc T$ and $T^2f=Tleft[T(f)right]$.




And if $T$ is also linear on $A$ seen as a vectorial space over $mathbb R$?






functions discrete-mathematics integers natural-numbers







share|cite|improve this question















Defined $A$ as the set of all the functions $f:mathbb Zrightarrow mathbb R$ exists a function $T:Arightarrow A$ such that
$$
(T^2f)(n)=f(n+1)text{ for all }ninmathbb Ztext{ and }fin A
$$
or not? And if codomain of $f$ is $mathbb Z$?



Remember that $T^2=Tcirc T$ and $T^2f=Tleft[T(f)right]$.




And if $T$ is also linear on $A$ seen as a vectorial space over $mathbb R$?






functions discrete-mathematics integers natural-numbers




functions discrete-mathematics integers natural-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 16:42









Henning Makholm

237k16301536




237k16301536










asked Nov 25 at 15:40









P De Donato

3947




3947












  • This question is related, but asks for a $T$ that is linear over $mathbb C$ rather than $mathbb R$ and also only considers $ell^1(mathbb Z)$ rather than the full $mathbb C^{mathbb Z}$.
    – Henning Makholm
    Nov 25 at 22:22




















  • This question is related, but asks for a $T$ that is linear over $mathbb C$ rather than $mathbb R$ and also only considers $ell^1(mathbb Z)$ rather than the full $mathbb C^{mathbb Z}$.
    – Henning Makholm
    Nov 25 at 22:22


















This question is related, but asks for a $T$ that is linear over $mathbb C$ rather than $mathbb R$ and also only considers $ell^1(mathbb Z)$ rather than the full $mathbb C^{mathbb Z}$.
– Henning Makholm
Nov 25 at 22:22






This question is related, but asks for a $T$ that is linear over $mathbb C$ rather than $mathbb R$ and also only considers $ell^1(mathbb Z)$ rather than the full $mathbb C^{mathbb Z}$.
– Henning Makholm
Nov 25 at 22:22












1 Answer
1






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Let $h$ be your favorite bijection $mathbb R^2tomathbb R$, and let $pi_1,pi_2$ be the corresponding projections such that $h(pi_1(x),pi_2(x))=x$. Then take
$$ (Tf)(n) = h(pi_2(f(n)),pi_1(f(n+1))) $$




The same construction works for every infinite codomain for the $f$s, assuming the axiom of choice. Things get trickier for finite codomains. For a codomain whose size is a perfect square, a variant of the above construction will work; on the other hand for codomains of size $4n+2$ or $4n+3$ the task is impossible. I am not yet sure of the remaining cases.




There can't be linear $T$ with this property as long as the field is $mathbb R$ or a subfield of $mathbb R$ such as $mathbb Q$. Consider the functions
$$ f_1(n) = 1 qquad f_{-1}(n) = (-1)^n $$



Since $T^2(Tf_1)= T^3f_1 = T(T^2f_1)=Tf_1$ it must be that $Tf_1$ is something that maps to itself under $T^2$, which means that it is a constant function, that is $Tf_1 = lambda f_1$ for some $lambda$.



Then how about $Tf_{-1}$? It will need to be something with period $2$ under $T^2$, which means that it is a linear combination $af_1 + bf_{-1}$. But the equation for $T^2 f_{-1}$ becomes $T^2 f_{-1} = -f_{-1}$, and the preceding boils down to the matrix equation
$$ begin{pmatrix} lambda & a \ 0 & b end{pmatrix}^2 = begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} $$
But this is impossible because it would require $b^2=-1$.






share|cite|improve this answer























  • But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
    – P De Donato
    Nov 25 at 16:18












  • @PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
    – Henning Makholm
    Nov 25 at 16:22










  • Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
    – P De Donato
    Nov 25 at 16:24












  • @PDeDonato: Answer extended to show that a linear $T$ is impossible.
    – Henning Makholm
    Nov 25 at 16:57










  • It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
    – Henning Makholm
    Nov 25 at 22:28











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Let $h$ be your favorite bijection $mathbb R^2tomathbb R$, and let $pi_1,pi_2$ be the corresponding projections such that $h(pi_1(x),pi_2(x))=x$. Then take
$$ (Tf)(n) = h(pi_2(f(n)),pi_1(f(n+1))) $$




The same construction works for every infinite codomain for the $f$s, assuming the axiom of choice. Things get trickier for finite codomains. For a codomain whose size is a perfect square, a variant of the above construction will work; on the other hand for codomains of size $4n+2$ or $4n+3$ the task is impossible. I am not yet sure of the remaining cases.




There can't be linear $T$ with this property as long as the field is $mathbb R$ or a subfield of $mathbb R$ such as $mathbb Q$. Consider the functions
$$ f_1(n) = 1 qquad f_{-1}(n) = (-1)^n $$



Since $T^2(Tf_1)= T^3f_1 = T(T^2f_1)=Tf_1$ it must be that $Tf_1$ is something that maps to itself under $T^2$, which means that it is a constant function, that is $Tf_1 = lambda f_1$ for some $lambda$.



Then how about $Tf_{-1}$? It will need to be something with period $2$ under $T^2$, which means that it is a linear combination $af_1 + bf_{-1}$. But the equation for $T^2 f_{-1}$ becomes $T^2 f_{-1} = -f_{-1}$, and the preceding boils down to the matrix equation
$$ begin{pmatrix} lambda & a \ 0 & b end{pmatrix}^2 = begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} $$
But this is impossible because it would require $b^2=-1$.






share|cite|improve this answer























  • But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
    – P De Donato
    Nov 25 at 16:18












  • @PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
    – Henning Makholm
    Nov 25 at 16:22










  • Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
    – P De Donato
    Nov 25 at 16:24












  • @PDeDonato: Answer extended to show that a linear $T$ is impossible.
    – Henning Makholm
    Nov 25 at 16:57










  • It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
    – Henning Makholm
    Nov 25 at 22:28
















2














Let $h$ be your favorite bijection $mathbb R^2tomathbb R$, and let $pi_1,pi_2$ be the corresponding projections such that $h(pi_1(x),pi_2(x))=x$. Then take
$$ (Tf)(n) = h(pi_2(f(n)),pi_1(f(n+1))) $$




The same construction works for every infinite codomain for the $f$s, assuming the axiom of choice. Things get trickier for finite codomains. For a codomain whose size is a perfect square, a variant of the above construction will work; on the other hand for codomains of size $4n+2$ or $4n+3$ the task is impossible. I am not yet sure of the remaining cases.




There can't be linear $T$ with this property as long as the field is $mathbb R$ or a subfield of $mathbb R$ such as $mathbb Q$. Consider the functions
$$ f_1(n) = 1 qquad f_{-1}(n) = (-1)^n $$



Since $T^2(Tf_1)= T^3f_1 = T(T^2f_1)=Tf_1$ it must be that $Tf_1$ is something that maps to itself under $T^2$, which means that it is a constant function, that is $Tf_1 = lambda f_1$ for some $lambda$.



Then how about $Tf_{-1}$? It will need to be something with period $2$ under $T^2$, which means that it is a linear combination $af_1 + bf_{-1}$. But the equation for $T^2 f_{-1}$ becomes $T^2 f_{-1} = -f_{-1}$, and the preceding boils down to the matrix equation
$$ begin{pmatrix} lambda & a \ 0 & b end{pmatrix}^2 = begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} $$
But this is impossible because it would require $b^2=-1$.






share|cite|improve this answer























  • But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
    – P De Donato
    Nov 25 at 16:18












  • @PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
    – Henning Makholm
    Nov 25 at 16:22










  • Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
    – P De Donato
    Nov 25 at 16:24












  • @PDeDonato: Answer extended to show that a linear $T$ is impossible.
    – Henning Makholm
    Nov 25 at 16:57










  • It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
    – Henning Makholm
    Nov 25 at 22:28














2












2








2






Let $h$ be your favorite bijection $mathbb R^2tomathbb R$, and let $pi_1,pi_2$ be the corresponding projections such that $h(pi_1(x),pi_2(x))=x$. Then take
$$ (Tf)(n) = h(pi_2(f(n)),pi_1(f(n+1))) $$




The same construction works for every infinite codomain for the $f$s, assuming the axiom of choice. Things get trickier for finite codomains. For a codomain whose size is a perfect square, a variant of the above construction will work; on the other hand for codomains of size $4n+2$ or $4n+3$ the task is impossible. I am not yet sure of the remaining cases.




There can't be linear $T$ with this property as long as the field is $mathbb R$ or a subfield of $mathbb R$ such as $mathbb Q$. Consider the functions
$$ f_1(n) = 1 qquad f_{-1}(n) = (-1)^n $$



Since $T^2(Tf_1)= T^3f_1 = T(T^2f_1)=Tf_1$ it must be that $Tf_1$ is something that maps to itself under $T^2$, which means that it is a constant function, that is $Tf_1 = lambda f_1$ for some $lambda$.



Then how about $Tf_{-1}$? It will need to be something with period $2$ under $T^2$, which means that it is a linear combination $af_1 + bf_{-1}$. But the equation for $T^2 f_{-1}$ becomes $T^2 f_{-1} = -f_{-1}$, and the preceding boils down to the matrix equation
$$ begin{pmatrix} lambda & a \ 0 & b end{pmatrix}^2 = begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} $$
But this is impossible because it would require $b^2=-1$.






share|cite|improve this answer














Let $h$ be your favorite bijection $mathbb R^2tomathbb R$, and let $pi_1,pi_2$ be the corresponding projections such that $h(pi_1(x),pi_2(x))=x$. Then take
$$ (Tf)(n) = h(pi_2(f(n)),pi_1(f(n+1))) $$




The same construction works for every infinite codomain for the $f$s, assuming the axiom of choice. Things get trickier for finite codomains. For a codomain whose size is a perfect square, a variant of the above construction will work; on the other hand for codomains of size $4n+2$ or $4n+3$ the task is impossible. I am not yet sure of the remaining cases.




There can't be linear $T$ with this property as long as the field is $mathbb R$ or a subfield of $mathbb R$ such as $mathbb Q$. Consider the functions
$$ f_1(n) = 1 qquad f_{-1}(n) = (-1)^n $$



Since $T^2(Tf_1)= T^3f_1 = T(T^2f_1)=Tf_1$ it must be that $Tf_1$ is something that maps to itself under $T^2$, which means that it is a constant function, that is $Tf_1 = lambda f_1$ for some $lambda$.



Then how about $Tf_{-1}$? It will need to be something with period $2$ under $T^2$, which means that it is a linear combination $af_1 + bf_{-1}$. But the equation for $T^2 f_{-1}$ becomes $T^2 f_{-1} = -f_{-1}$, and the preceding boils down to the matrix equation
$$ begin{pmatrix} lambda & a \ 0 & b end{pmatrix}^2 = begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} $$
But this is impossible because it would require $b^2=-1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 3:27

























answered Nov 25 at 15:59









Henning Makholm

237k16301536




237k16301536












  • But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
    – P De Donato
    Nov 25 at 16:18












  • @PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
    – Henning Makholm
    Nov 25 at 16:22










  • Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
    – P De Donato
    Nov 25 at 16:24












  • @PDeDonato: Answer extended to show that a linear $T$ is impossible.
    – Henning Makholm
    Nov 25 at 16:57










  • It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
    – Henning Makholm
    Nov 25 at 22:28


















  • But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
    – P De Donato
    Nov 25 at 16:18












  • @PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
    – Henning Makholm
    Nov 25 at 16:22










  • Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
    – P De Donato
    Nov 25 at 16:24












  • @PDeDonato: Answer extended to show that a linear $T$ is impossible.
    – Henning Makholm
    Nov 25 at 16:57










  • It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
    – Henning Makholm
    Nov 25 at 22:28
















But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
– P De Donato
Nov 25 at 16:18






But in this case we obtain $$ T^2f(n)=hleft{pi_2left[h(pi_2f(n), pi_1f(n+1))right], pi_1left[h(pi_2f(n+1), pi_1f(n+2))right]right} $$ and seems it doesn't work
– P De Donato
Nov 25 at 16:18














@PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
– Henning Makholm
Nov 25 at 16:22




@PDeDonato: Yes; now simplify using $pi_1(h(x,y))=x$ and $pi_2(h(x,y)=y$.
– Henning Makholm
Nov 25 at 16:22












Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
– P De Donato
Nov 25 at 16:24






Ah now I understand, $pi_1$ and $pi_2$ form the inverse of $h$
– P De Donato
Nov 25 at 16:24














@PDeDonato: Answer extended to show that a linear $T$ is impossible.
– Henning Makholm
Nov 25 at 16:57




@PDeDonato: Answer extended to show that a linear $T$ is impossible.
– Henning Makholm
Nov 25 at 16:57












It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
– Henning Makholm
Nov 25 at 22:28




It looks a lot like a linear $T$ over $mathbb C$ ought to be possible, but the details are subtle and elusive.
– Henning Makholm
Nov 25 at 22:28


















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