Norm and Weak Topologies agree?












5












$begingroup$


so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.






general-topology functional-analysis weak-convergence







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.






    general-topology functional-analysis weak-convergence







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.






      general-topology functional-analysis weak-convergence







      share|cite|improve this question









      $endgroup$




      so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.






      general-topology functional-analysis weak-convergence




      general-topology functional-analysis weak-convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      ScottScott

      34918




      34918






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.



          More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
            $endgroup$
            – Scott
            3 hours ago










          • $begingroup$
            Yes, that's correct.
            $endgroup$
            – Eric Wofsey
            3 hours ago



















          0












          $begingroup$

          Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$



          Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$






          share|cite|improve this answer











          $endgroup$













            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%2f3090244%2fnorm-and-weak-topologies-agree%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.



            More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
              $endgroup$
              – Scott
              3 hours ago










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              3 hours ago
















            5












            $begingroup$

            We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.



            More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
              $endgroup$
              – Scott
              3 hours ago










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              3 hours ago














            5












            5








            5





            $begingroup$

            We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.



            More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance






            share|cite|improve this answer











            $endgroup$



            We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.



            More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 4 hours ago









            Eric WofseyEric Wofsey

            183k13211338




            183k13211338












            • $begingroup$
              So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
              $endgroup$
              – Scott
              3 hours ago










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              3 hours ago


















            • $begingroup$
              So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
              $endgroup$
              – Scott
              3 hours ago










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              3 hours ago
















            $begingroup$
            So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
            $endgroup$
            – Scott
            3 hours ago




            $begingroup$
            So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
            $endgroup$
            – Scott
            3 hours ago












            $begingroup$
            Yes, that's correct.
            $endgroup$
            – Eric Wofsey
            3 hours ago




            $begingroup$
            Yes, that's correct.
            $endgroup$
            – Eric Wofsey
            3 hours ago











            0












            $begingroup$

            Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$



            Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$



              Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$



                Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$






                share|cite|improve this answer











                $endgroup$



                Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$



                Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 11 mins ago

























                answered 20 mins ago









                MatematletaMatematleta

                10.5k2918




                10.5k2918






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090244%2fnorm-and-weak-topologies-agree%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

                    CARDNET

                    Image
                    CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
                    Read more

                    Boot-repair Failure: Unable to locate package grub-common:i386

                    Image
                    0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h...
                    Read more

                    濃尾地震

                    Image
                    濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
                    Read more