Isometries on a Banach space converging pointwise











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I'm trying to find a Banach space $V$ with closed unit ball $B$ and a sequence of isometries $(f_n:Vto V)$ such that $(f_n)$ converges pointwise in $B$ but not uniformly in $B$.



My first attempts were to make it kind of simple. If $V=c_0$ (the space with sequences $x_nto 0$ with the sup norm) take for instance the linear functions $T_n(x)=(0,ldots,0,x_{n+1},x_{n+2},ldots)$. It is easy to see that $T_n$ converges pointwise to the zero function, but it doesn't so uniformly in $B$. However, $T_n$ is not an isometry although it satisfies $left|{T(x)}right|leleft|{x}right|$.



Can someone give me a hint?



Thanks.






functional-analysis banach-spaces isometry







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  • Would anything involving the shift operator work? $T(x_1,x_2,ldots) = (0,x_1,x_2,ldots)$ and $T_n:=T^n$. Thinking for a few minutes I can't come up with an appropriate space but perhaps it can work.
    – user25959
    Nov 27 at 5:27










  • @user25959 I thought about that too, but I don't know if it converges pointwise.
    – Tanius
    Nov 27 at 5:29

















up vote
1
down vote

favorite
1












I'm trying to find a Banach space $V$ with closed unit ball $B$ and a sequence of isometries $(f_n:Vto V)$ such that $(f_n)$ converges pointwise in $B$ but not uniformly in $B$.



My first attempts were to make it kind of simple. If $V=c_0$ (the space with sequences $x_nto 0$ with the sup norm) take for instance the linear functions $T_n(x)=(0,ldots,0,x_{n+1},x_{n+2},ldots)$. It is easy to see that $T_n$ converges pointwise to the zero function, but it doesn't so uniformly in $B$. However, $T_n$ is not an isometry although it satisfies $left|{T(x)}right|leleft|{x}right|$.



Can someone give me a hint?



Thanks.






functional-analysis banach-spaces isometry







share|cite|improve this question






















  • Would anything involving the shift operator work? $T(x_1,x_2,ldots) = (0,x_1,x_2,ldots)$ and $T_n:=T^n$. Thinking for a few minutes I can't come up with an appropriate space but perhaps it can work.
    – user25959
    Nov 27 at 5:27










  • @user25959 I thought about that too, but I don't know if it converges pointwise.
    – Tanius
    Nov 27 at 5:29















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I'm trying to find a Banach space $V$ with closed unit ball $B$ and a sequence of isometries $(f_n:Vto V)$ such that $(f_n)$ converges pointwise in $B$ but not uniformly in $B$.



My first attempts were to make it kind of simple. If $V=c_0$ (the space with sequences $x_nto 0$ with the sup norm) take for instance the linear functions $T_n(x)=(0,ldots,0,x_{n+1},x_{n+2},ldots)$. It is easy to see that $T_n$ converges pointwise to the zero function, but it doesn't so uniformly in $B$. However, $T_n$ is not an isometry although it satisfies $left|{T(x)}right|leleft|{x}right|$.



Can someone give me a hint?



Thanks.






functional-analysis banach-spaces isometry







share|cite|improve this question













I'm trying to find a Banach space $V$ with closed unit ball $B$ and a sequence of isometries $(f_n:Vto V)$ such that $(f_n)$ converges pointwise in $B$ but not uniformly in $B$.



My first attempts were to make it kind of simple. If $V=c_0$ (the space with sequences $x_nto 0$ with the sup norm) take for instance the linear functions $T_n(x)=(0,ldots,0,x_{n+1},x_{n+2},ldots)$. It is easy to see that $T_n$ converges pointwise to the zero function, but it doesn't so uniformly in $B$. However, $T_n$ is not an isometry although it satisfies $left|{T(x)}right|leleft|{x}right|$.



Can someone give me a hint?



Thanks.






functional-analysis banach-spaces isometry




functional-analysis banach-spaces isometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 at 5:05









Tanius

48827




48827












  • Would anything involving the shift operator work? $T(x_1,x_2,ldots) = (0,x_1,x_2,ldots)$ and $T_n:=T^n$. Thinking for a few minutes I can't come up with an appropriate space but perhaps it can work.
    – user25959
    Nov 27 at 5:27










  • @user25959 I thought about that too, but I don't know if it converges pointwise.
    – Tanius
    Nov 27 at 5:29




















  • Would anything involving the shift operator work? $T(x_1,x_2,ldots) = (0,x_1,x_2,ldots)$ and $T_n:=T^n$. Thinking for a few minutes I can't come up with an appropriate space but perhaps it can work.
    – user25959
    Nov 27 at 5:27










  • @user25959 I thought about that too, but I don't know if it converges pointwise.
    – Tanius
    Nov 27 at 5:29


















Would anything involving the shift operator work? $T(x_1,x_2,ldots) = (0,x_1,x_2,ldots)$ and $T_n:=T^n$. Thinking for a few minutes I can't come up with an appropriate space but perhaps it can work.
– user25959
Nov 27 at 5:27




Would anything involving the shift operator work? $T(x_1,x_2,ldots) = (0,x_1,x_2,ldots)$ and $T_n:=T^n$. Thinking for a few minutes I can't come up with an appropriate space but perhaps it can work.
– user25959
Nov 27 at 5:27












@user25959 I thought about that too, but I don't know if it converges pointwise.
– Tanius
Nov 27 at 5:29






@user25959 I thought about that too, but I don't know if it converges pointwise.
– Tanius
Nov 27 at 5:29












1 Answer
1






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up vote
4
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accepted










Let $V=l^{2}, f_n(x_1,x_2,...))=(x_1,x_2,...,x_n,0,x_{n+1},...)$ (where a $0$ is inserted at the n-th position). Then $f_n(x) to x$ for every $x in B$ but $|f_n(x)-x|^{2}geq |x_{n+1}|^{2}$ so the convergence is not uniform on $B$.






share|cite|improve this answer























  • $f_n$ are supposed to be isometries
    – user25959
    Nov 27 at 5:37










  • @user25959 I have revised my answer.
    – Kavi Rama Murthy
    Nov 27 at 7:21










  • Thank you so much
    – Tanius
    Nov 27 at 13:23






  • 1




    It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
    – gerw
    Nov 27 at 15:01











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up vote
4
down vote



accepted










Let $V=l^{2}, f_n(x_1,x_2,...))=(x_1,x_2,...,x_n,0,x_{n+1},...)$ (where a $0$ is inserted at the n-th position). Then $f_n(x) to x$ for every $x in B$ but $|f_n(x)-x|^{2}geq |x_{n+1}|^{2}$ so the convergence is not uniform on $B$.






share|cite|improve this answer























  • $f_n$ are supposed to be isometries
    – user25959
    Nov 27 at 5:37










  • @user25959 I have revised my answer.
    – Kavi Rama Murthy
    Nov 27 at 7:21










  • Thank you so much
    – Tanius
    Nov 27 at 13:23






  • 1




    It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
    – gerw
    Nov 27 at 15:01















up vote
4
down vote



accepted










Let $V=l^{2}, f_n(x_1,x_2,...))=(x_1,x_2,...,x_n,0,x_{n+1},...)$ (where a $0$ is inserted at the n-th position). Then $f_n(x) to x$ for every $x in B$ but $|f_n(x)-x|^{2}geq |x_{n+1}|^{2}$ so the convergence is not uniform on $B$.






share|cite|improve this answer























  • $f_n$ are supposed to be isometries
    – user25959
    Nov 27 at 5:37










  • @user25959 I have revised my answer.
    – Kavi Rama Murthy
    Nov 27 at 7:21










  • Thank you so much
    – Tanius
    Nov 27 at 13:23






  • 1




    It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
    – gerw
    Nov 27 at 15:01













up vote
4
down vote



accepted







up vote
4
down vote



accepted






Let $V=l^{2}, f_n(x_1,x_2,...))=(x_1,x_2,...,x_n,0,x_{n+1},...)$ (where a $0$ is inserted at the n-th position). Then $f_n(x) to x$ for every $x in B$ but $|f_n(x)-x|^{2}geq |x_{n+1}|^{2}$ so the convergence is not uniform on $B$.






share|cite|improve this answer














Let $V=l^{2}, f_n(x_1,x_2,...))=(x_1,x_2,...,x_n,0,x_{n+1},...)$ (where a $0$ is inserted at the n-th position). Then $f_n(x) to x$ for every $x in B$ but $|f_n(x)-x|^{2}geq |x_{n+1}|^{2}$ so the convergence is not uniform on $B$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 5:46

























answered Nov 27 at 5:31









Kavi Rama Murthy

46.3k31854




46.3k31854












  • $f_n$ are supposed to be isometries
    – user25959
    Nov 27 at 5:37










  • @user25959 I have revised my answer.
    – Kavi Rama Murthy
    Nov 27 at 7:21










  • Thank you so much
    – Tanius
    Nov 27 at 13:23






  • 1




    It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
    – gerw
    Nov 27 at 15:01


















  • $f_n$ are supposed to be isometries
    – user25959
    Nov 27 at 5:37










  • @user25959 I have revised my answer.
    – Kavi Rama Murthy
    Nov 27 at 7:21










  • Thank you so much
    – Tanius
    Nov 27 at 13:23






  • 1




    It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
    – gerw
    Nov 27 at 15:01
















$f_n$ are supposed to be isometries
– user25959
Nov 27 at 5:37




$f_n$ are supposed to be isometries
– user25959
Nov 27 at 5:37












@user25959 I have revised my answer.
– Kavi Rama Murthy
Nov 27 at 7:21




@user25959 I have revised my answer.
– Kavi Rama Murthy
Nov 27 at 7:21












Thank you so much
– Tanius
Nov 27 at 13:23




Thank you so much
– Tanius
Nov 27 at 13:23




1




1




It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
– gerw
Nov 27 at 15:01




It is even possible to construct isometric bijections -- just swap $x_n$ and $x_{2n}$ in the sequence $x_n$ to define $f_n$.
– gerw
Nov 27 at 15:01


















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