Norm and Weak Topologies agree?












6












$begingroup$


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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$endgroup$

















    6












    $begingroup$


    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









    $endgroup$















      6












      6








      6


      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 Jan 27 at 23:06









      ScottScott

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          2 Answers
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          7












          $begingroup$

          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
            Jan 27 at 23:16










          • $begingroup$
            Yes, that's correct.
            $endgroup$
            – Eric Wofsey
            Jan 27 at 23:19



















          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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            active

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            7












            $begingroup$

            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
              Jan 27 at 23:16










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              Jan 27 at 23:19
















            7












            $begingroup$

            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
              Jan 27 at 23:16










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              Jan 27 at 23:19














            7












            7








            7





            $begingroup$

            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 Jan 27 at 23:23

























            answered Jan 27 at 23:11









            Eric WofseyEric Wofsey

            185k13213339




            185k13213339












            • $begingroup$
              So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
              $endgroup$
              – Scott
              Jan 27 at 23:16










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              Jan 27 at 23:19


















            • $begingroup$
              So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
              $endgroup$
              – Scott
              Jan 27 at 23:16










            • $begingroup$
              Yes, that's correct.
              $endgroup$
              – Eric Wofsey
              Jan 27 at 23:19
















            $begingroup$
            So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
            $endgroup$
            – Scott
            Jan 27 at 23:16




            $begingroup$
            So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
            $endgroup$
            – Scott
            Jan 27 at 23:16












            $begingroup$
            Yes, that's correct.
            $endgroup$
            – Eric Wofsey
            Jan 27 at 23:19




            $begingroup$
            Yes, that's correct.
            $endgroup$
            – Eric Wofsey
            Jan 27 at 23:19











            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 Jan 28 at 3:02

























                answered Jan 28 at 2:53









                MatematletaMatematleta

                10.7k2918




                10.7k2918






























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