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







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$endgroup$

















    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






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      asked 4 hours ago









      ScottScott

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          2 Answers
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          $begingroup$

          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











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            2 Answers
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            2 Answers
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            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 3 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 5 mins ago

























                answered 13 mins ago









                MatematletaMatematleta

                10.5k2918




                10.5k2918






























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