Total boundedness in a sequence space












0














Let $pin[1,infty)$ Show that a subset $Msubseteq l^{p}(mathbb{K})$ is totally bounded if and only if M:



$underset{xin M}{text{sup}}$ $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$ for $nrightarrowinfty$ and $M$ bounded.



Now I managed to show "$Rightarrow$" (when M is totally bounded) but for the second one I have no idea how to get started. I'm sitting in a introductory class to functional analysis so it would be nice to consider that in your answers.






functional-analysis lp-spaces







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    0














    Let $pin[1,infty)$ Show that a subset $Msubseteq l^{p}(mathbb{K})$ is totally bounded if and only if M:



    $underset{xin M}{text{sup}}$ $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$ for $nrightarrowinfty$ and $M$ bounded.



    Now I managed to show "$Rightarrow$" (when M is totally bounded) but for the second one I have no idea how to get started. I'm sitting in a introductory class to functional analysis so it would be nice to consider that in your answers.






    functional-analysis lp-spaces







    share|cite|improve this question

























      0












      0








      0







      Let $pin[1,infty)$ Show that a subset $Msubseteq l^{p}(mathbb{K})$ is totally bounded if and only if M:



      $underset{xin M}{text{sup}}$ $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$ for $nrightarrowinfty$ and $M$ bounded.



      Now I managed to show "$Rightarrow$" (when M is totally bounded) but for the second one I have no idea how to get started. I'm sitting in a introductory class to functional analysis so it would be nice to consider that in your answers.






      functional-analysis lp-spaces







      share|cite|improve this question













      Let $pin[1,infty)$ Show that a subset $Msubseteq l^{p}(mathbb{K})$ is totally bounded if and only if M:



      $underset{xin M}{text{sup}}$ $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$ for $nrightarrowinfty$ and $M$ bounded.



      Now I managed to show "$Rightarrow$" (when M is totally bounded) but for the second one I have no idea how to get started. I'm sitting in a introductory class to functional analysis so it would be nice to consider that in your answers.






      functional-analysis lp-spaces




      functional-analysis lp-spaces






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      asked Nov 26 at 12:17









      Christian Singer

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          1 Answer
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          Let $epsilon>0$. Since $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$, there exists $N$ such that:



          $sumlimits_{i=N+1}^{infty}|x_{i}|^{p}< dfrac{epsilon}{3}$ for all $xin M$.



          Define the projection $pi:l^p(mathbb{K})rightarrow mathbb{K}^N$, by $pi Big( big( x_ibig)_{iin mathbb{N}} Big)=(x_1, ...,x_N)$.



          Since bounded sets in $mathbb{K}^N$ with the $l^p$ norm are totally bounded, and $Vert pi(x)Vert_p leq Vert xVert_p leq m$ for all $xin M$, there exists a finite set of points ${x^j }_{j=1}^{N_0}$ such that $ { pi(x^j) }_{j=1}^{N_0}$ are an $dfrac{epsilon}{3}$-net for $pi[M]$.



          It remains to show that ${x^j }_{j=1}^{N_0}$ is an $epsilon$-net in $M$. Let $xin M$, we know that:



          $sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} < dfrac{epsilon}{3}$ by choice of ${x^j }_{j=1}^{N_0}$. Then:



          $Vert x^j-xVert_p= sumlimits_{i=1}^{infty}|x^j_{i}-x_i|^{p}= sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p}+ sumlimits_{i=N+1}^{infty}|x^j_{i}-x_i|^{p} leq$



          $leq sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} + sumlimits_{i=N+1}^{infty}|x^j_{i}|^{p}+ sumlimits_{i=N+1}^{infty}|x_i|^{p} overset{x,x^jin M}{leq} dfrac{epsilon}{3}+ dfrac{epsilon}{3}+ dfrac{epsilon}{3} $



          And for a general $epsilon$, you've found a finite cover of $M$ by the balls $B_{l^p}big(x^j, epsilon big)$ (radius $epsilon$).






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
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            active

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            2














            Let $epsilon>0$. Since $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$, there exists $N$ such that:



            $sumlimits_{i=N+1}^{infty}|x_{i}|^{p}< dfrac{epsilon}{3}$ for all $xin M$.



            Define the projection $pi:l^p(mathbb{K})rightarrow mathbb{K}^N$, by $pi Big( big( x_ibig)_{iin mathbb{N}} Big)=(x_1, ...,x_N)$.



            Since bounded sets in $mathbb{K}^N$ with the $l^p$ norm are totally bounded, and $Vert pi(x)Vert_p leq Vert xVert_p leq m$ for all $xin M$, there exists a finite set of points ${x^j }_{j=1}^{N_0}$ such that $ { pi(x^j) }_{j=1}^{N_0}$ are an $dfrac{epsilon}{3}$-net for $pi[M]$.



            It remains to show that ${x^j }_{j=1}^{N_0}$ is an $epsilon$-net in $M$. Let $xin M$, we know that:



            $sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} < dfrac{epsilon}{3}$ by choice of ${x^j }_{j=1}^{N_0}$. Then:



            $Vert x^j-xVert_p= sumlimits_{i=1}^{infty}|x^j_{i}-x_i|^{p}= sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p}+ sumlimits_{i=N+1}^{infty}|x^j_{i}-x_i|^{p} leq$



            $leq sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} + sumlimits_{i=N+1}^{infty}|x^j_{i}|^{p}+ sumlimits_{i=N+1}^{infty}|x_i|^{p} overset{x,x^jin M}{leq} dfrac{epsilon}{3}+ dfrac{epsilon}{3}+ dfrac{epsilon}{3} $



            And for a general $epsilon$, you've found a finite cover of $M$ by the balls $B_{l^p}big(x^j, epsilon big)$ (radius $epsilon$).






            share|cite|improve this answer


























              2














              Let $epsilon>0$. Since $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$, there exists $N$ such that:



              $sumlimits_{i=N+1}^{infty}|x_{i}|^{p}< dfrac{epsilon}{3}$ for all $xin M$.



              Define the projection $pi:l^p(mathbb{K})rightarrow mathbb{K}^N$, by $pi Big( big( x_ibig)_{iin mathbb{N}} Big)=(x_1, ...,x_N)$.



              Since bounded sets in $mathbb{K}^N$ with the $l^p$ norm are totally bounded, and $Vert pi(x)Vert_p leq Vert xVert_p leq m$ for all $xin M$, there exists a finite set of points ${x^j }_{j=1}^{N_0}$ such that $ { pi(x^j) }_{j=1}^{N_0}$ are an $dfrac{epsilon}{3}$-net for $pi[M]$.



              It remains to show that ${x^j }_{j=1}^{N_0}$ is an $epsilon$-net in $M$. Let $xin M$, we know that:



              $sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} < dfrac{epsilon}{3}$ by choice of ${x^j }_{j=1}^{N_0}$. Then:



              $Vert x^j-xVert_p= sumlimits_{i=1}^{infty}|x^j_{i}-x_i|^{p}= sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p}+ sumlimits_{i=N+1}^{infty}|x^j_{i}-x_i|^{p} leq$



              $leq sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} + sumlimits_{i=N+1}^{infty}|x^j_{i}|^{p}+ sumlimits_{i=N+1}^{infty}|x_i|^{p} overset{x,x^jin M}{leq} dfrac{epsilon}{3}+ dfrac{epsilon}{3}+ dfrac{epsilon}{3} $



              And for a general $epsilon$, you've found a finite cover of $M$ by the balls $B_{l^p}big(x^j, epsilon big)$ (radius $epsilon$).






              share|cite|improve this answer
























                2












                2








                2






                Let $epsilon>0$. Since $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$, there exists $N$ such that:



                $sumlimits_{i=N+1}^{infty}|x_{i}|^{p}< dfrac{epsilon}{3}$ for all $xin M$.



                Define the projection $pi:l^p(mathbb{K})rightarrow mathbb{K}^N$, by $pi Big( big( x_ibig)_{iin mathbb{N}} Big)=(x_1, ...,x_N)$.



                Since bounded sets in $mathbb{K}^N$ with the $l^p$ norm are totally bounded, and $Vert pi(x)Vert_p leq Vert xVert_p leq m$ for all $xin M$, there exists a finite set of points ${x^j }_{j=1}^{N_0}$ such that $ { pi(x^j) }_{j=1}^{N_0}$ are an $dfrac{epsilon}{3}$-net for $pi[M]$.



                It remains to show that ${x^j }_{j=1}^{N_0}$ is an $epsilon$-net in $M$. Let $xin M$, we know that:



                $sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} < dfrac{epsilon}{3}$ by choice of ${x^j }_{j=1}^{N_0}$. Then:



                $Vert x^j-xVert_p= sumlimits_{i=1}^{infty}|x^j_{i}-x_i|^{p}= sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p}+ sumlimits_{i=N+1}^{infty}|x^j_{i}-x_i|^{p} leq$



                $leq sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} + sumlimits_{i=N+1}^{infty}|x^j_{i}|^{p}+ sumlimits_{i=N+1}^{infty}|x_i|^{p} overset{x,x^jin M}{leq} dfrac{epsilon}{3}+ dfrac{epsilon}{3}+ dfrac{epsilon}{3} $



                And for a general $epsilon$, you've found a finite cover of $M$ by the balls $B_{l^p}big(x^j, epsilon big)$ (radius $epsilon$).






                share|cite|improve this answer












                Let $epsilon>0$. Since $sumlimits_{i=n}^{infty}|x_{i}|^{p}rightarrow 0$, there exists $N$ such that:



                $sumlimits_{i=N+1}^{infty}|x_{i}|^{p}< dfrac{epsilon}{3}$ for all $xin M$.



                Define the projection $pi:l^p(mathbb{K})rightarrow mathbb{K}^N$, by $pi Big( big( x_ibig)_{iin mathbb{N}} Big)=(x_1, ...,x_N)$.



                Since bounded sets in $mathbb{K}^N$ with the $l^p$ norm are totally bounded, and $Vert pi(x)Vert_p leq Vert xVert_p leq m$ for all $xin M$, there exists a finite set of points ${x^j }_{j=1}^{N_0}$ such that $ { pi(x^j) }_{j=1}^{N_0}$ are an $dfrac{epsilon}{3}$-net for $pi[M]$.



                It remains to show that ${x^j }_{j=1}^{N_0}$ is an $epsilon$-net in $M$. Let $xin M$, we know that:



                $sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} < dfrac{epsilon}{3}$ by choice of ${x^j }_{j=1}^{N_0}$. Then:



                $Vert x^j-xVert_p= sumlimits_{i=1}^{infty}|x^j_{i}-x_i|^{p}= sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p}+ sumlimits_{i=N+1}^{infty}|x^j_{i}-x_i|^{p} leq$



                $leq sumlimits_{i=1}^{N}|x^j_{i}-x_i|^{p} + sumlimits_{i=N+1}^{infty}|x^j_{i}|^{p}+ sumlimits_{i=N+1}^{infty}|x_i|^{p} overset{x,x^jin M}{leq} dfrac{epsilon}{3}+ dfrac{epsilon}{3}+ dfrac{epsilon}{3} $



                And for a general $epsilon$, you've found a finite cover of $M$ by the balls $B_{l^p}big(x^j, epsilon big)$ (radius $epsilon$).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 13:41









                Keen-ameteur

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                1,265316






























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