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







share|cite|improve this question



























    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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 at 12:17









      Christian Singer

      344113




      344113






















          1 Answer
          1






          active

          oldest

          votes


















          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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014238%2ftotal-boundedness-in-a-sequence-space%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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

                1,265316




                1,265316






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014238%2ftotal-boundedness-in-a-sequence-space%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
                    Read more

                    “minimization” problem in Euclidean space related to orthonormal basis

                    Image
                    3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
                    Read more