Rudin's functional analysis theorem 3.12












0












$begingroup$



Suppose $E$ is a convex subset of a locally convex space $X$. Then the weak closure $overline{E}_w$ of $E$ is equal to its original closure $overline{E}$.




The proof starts as follows




$overline{E}_w$ is weakly closed, hence originally closed, so that $overline{E} subset overline{E}_w$.




I don't get that inclusion, could anyone expound it please?



The proof also continues for the opposite inclusion, but I don't get that either




To obtain the poosite inclusion, choose $x_0 in X, x_0 notin overline{E}$. Part (b) of the separation theorem 3.4 shows that there exists $Lambda in X^*$ and $gamma in mathbb{R}$ such that, for every $x in overline{E}$
$$
Re ; Lambda x_0 < gamma < Re ; Lambda x
$$
The set $left{ x : Re ; Lambda x < gamma right}$ is therefore a weak neighborhood of $x_0$ that does not intersect $E$, thus $x_0 notin overline{E}_w$. This proves $overline{E}_w subset overline{E}$







functional-analysis proof-explanation topological-vector-spaces weak-topology







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Suppose $E$ is a convex subset of a locally convex space $X$. Then the weak closure $overline{E}_w$ of $E$ is equal to its original closure $overline{E}$.




    The proof starts as follows




    $overline{E}_w$ is weakly closed, hence originally closed, so that $overline{E} subset overline{E}_w$.




    I don't get that inclusion, could anyone expound it please?



    The proof also continues for the opposite inclusion, but I don't get that either




    To obtain the poosite inclusion, choose $x_0 in X, x_0 notin overline{E}$. Part (b) of the separation theorem 3.4 shows that there exists $Lambda in X^*$ and $gamma in mathbb{R}$ such that, for every $x in overline{E}$
    $$
    Re ; Lambda x_0 < gamma < Re ; Lambda x
    $$
    The set $left{ x : Re ; Lambda x < gamma right}$ is therefore a weak neighborhood of $x_0$ that does not intersect $E$, thus $x_0 notin overline{E}_w$. This proves $overline{E}_w subset overline{E}$







    functional-analysis proof-explanation topological-vector-spaces weak-topology







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Suppose $E$ is a convex subset of a locally convex space $X$. Then the weak closure $overline{E}_w$ of $E$ is equal to its original closure $overline{E}$.




      The proof starts as follows




      $overline{E}_w$ is weakly closed, hence originally closed, so that $overline{E} subset overline{E}_w$.




      I don't get that inclusion, could anyone expound it please?



      The proof also continues for the opposite inclusion, but I don't get that either




      To obtain the poosite inclusion, choose $x_0 in X, x_0 notin overline{E}$. Part (b) of the separation theorem 3.4 shows that there exists $Lambda in X^*$ and $gamma in mathbb{R}$ such that, for every $x in overline{E}$
      $$
      Re ; Lambda x_0 < gamma < Re ; Lambda x
      $$
      The set $left{ x : Re ; Lambda x < gamma right}$ is therefore a weak neighborhood of $x_0$ that does not intersect $E$, thus $x_0 notin overline{E}_w$. This proves $overline{E}_w subset overline{E}$







      functional-analysis proof-explanation topological-vector-spaces weak-topology







      share|cite|improve this question











      $endgroup$





      Suppose $E$ is a convex subset of a locally convex space $X$. Then the weak closure $overline{E}_w$ of $E$ is equal to its original closure $overline{E}$.




      The proof starts as follows




      $overline{E}_w$ is weakly closed, hence originally closed, so that $overline{E} subset overline{E}_w$.




      I don't get that inclusion, could anyone expound it please?



      The proof also continues for the opposite inclusion, but I don't get that either




      To obtain the poosite inclusion, choose $x_0 in X, x_0 notin overline{E}$. Part (b) of the separation theorem 3.4 shows that there exists $Lambda in X^*$ and $gamma in mathbb{R}$ such that, for every $x in overline{E}$
      $$
      Re ; Lambda x_0 < gamma < Re ; Lambda x
      $$
      The set $left{ x : Re ; Lambda x < gamma right}$ is therefore a weak neighborhood of $x_0$ that does not intersect $E$, thus $x_0 notin overline{E}_w$. This proves $overline{E}_w subset overline{E}$







      functional-analysis proof-explanation topological-vector-spaces weak-topology




      functional-analysis proof-explanation topological-vector-spaces weak-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 12:02







      user8469759

















      asked Jan 10 at 11:51









      user8469759user8469759

      1,5681618




      1,5681618






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          First, the weak topology is weaker than the original topology
          (it has fewer open and closed sets).
          This means that every weakly closed set is also a closed set.



          Second, the closure of $E$ is the smallest closed set that contains $E$.
          Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
            $endgroup$
            – user8469759
            Jan 10 at 12:06










          • $begingroup$
            the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
            $endgroup$
            – supinf
            Jan 10 at 12:07






          • 1




            $begingroup$
            How about the other inclusion?
            $endgroup$
            – user8469759
            Jan 10 at 12:09



















          2












          $begingroup$

          Second part: ${x: Re Lambda x <gamma}$ is the inverse image under $Lambda$ of ${zin mathbb C: Re z <gamma}$ which is open in $mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $Re Lambda x >gamma$ for all $x in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_onotin overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $overline {E}$.






          share|cite|improve this answer









          $endgroup$














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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            First, the weak topology is weaker than the original topology
            (it has fewer open and closed sets).
            This means that every weakly closed set is also a closed set.



            Second, the closure of $E$ is the smallest closed set that contains $E$.
            Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
              $endgroup$
              – user8469759
              Jan 10 at 12:06










            • $begingroup$
              the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
              $endgroup$
              – supinf
              Jan 10 at 12:07






            • 1




              $begingroup$
              How about the other inclusion?
              $endgroup$
              – user8469759
              Jan 10 at 12:09
















            3












            $begingroup$

            First, the weak topology is weaker than the original topology
            (it has fewer open and closed sets).
            This means that every weakly closed set is also a closed set.



            Second, the closure of $E$ is the smallest closed set that contains $E$.
            Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
              $endgroup$
              – user8469759
              Jan 10 at 12:06










            • $begingroup$
              the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
              $endgroup$
              – supinf
              Jan 10 at 12:07






            • 1




              $begingroup$
              How about the other inclusion?
              $endgroup$
              – user8469759
              Jan 10 at 12:09














            3












            3








            3





            $begingroup$

            First, the weak topology is weaker than the original topology
            (it has fewer open and closed sets).
            This means that every weakly closed set is also a closed set.



            Second, the closure of $E$ is the smallest closed set that contains $E$.
            Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.






            share|cite|improve this answer









            $endgroup$



            First, the weak topology is weaker than the original topology
            (it has fewer open and closed sets).
            This means that every weakly closed set is also a closed set.



            Second, the closure of $E$ is the smallest closed set that contains $E$.
            Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 10 at 12:02









            supinfsupinf

            6,7561129




            6,7561129












            • $begingroup$
              I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
              $endgroup$
              – user8469759
              Jan 10 at 12:06










            • $begingroup$
              the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
              $endgroup$
              – supinf
              Jan 10 at 12:07






            • 1




              $begingroup$
              How about the other inclusion?
              $endgroup$
              – user8469759
              Jan 10 at 12:09


















            • $begingroup$
              I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
              $endgroup$
              – user8469759
              Jan 10 at 12:06










            • $begingroup$
              the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
              $endgroup$
              – supinf
              Jan 10 at 12:07






            • 1




              $begingroup$
              How about the other inclusion?
              $endgroup$
              – user8469759
              Jan 10 at 12:09
















            $begingroup$
            I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
            $endgroup$
            – user8469759
            Jan 10 at 12:06




            $begingroup$
            I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy.
            $endgroup$
            – user8469759
            Jan 10 at 12:06












            $begingroup$
            the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
            $endgroup$
            – supinf
            Jan 10 at 12:07




            $begingroup$
            the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$.
            $endgroup$
            – supinf
            Jan 10 at 12:07




            1




            1




            $begingroup$
            How about the other inclusion?
            $endgroup$
            – user8469759
            Jan 10 at 12:09




            $begingroup$
            How about the other inclusion?
            $endgroup$
            – user8469759
            Jan 10 at 12:09











            2












            $begingroup$

            Second part: ${x: Re Lambda x <gamma}$ is the inverse image under $Lambda$ of ${zin mathbb C: Re z <gamma}$ which is open in $mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $Re Lambda x >gamma$ for all $x in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_onotin overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $overline {E}$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Second part: ${x: Re Lambda x <gamma}$ is the inverse image under $Lambda$ of ${zin mathbb C: Re z <gamma}$ which is open in $mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $Re Lambda x >gamma$ for all $x in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_onotin overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $overline {E}$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Second part: ${x: Re Lambda x <gamma}$ is the inverse image under $Lambda$ of ${zin mathbb C: Re z <gamma}$ which is open in $mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $Re Lambda x >gamma$ for all $x in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_onotin overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $overline {E}$.






                share|cite|improve this answer









                $endgroup$



                Second part: ${x: Re Lambda x <gamma}$ is the inverse image under $Lambda$ of ${zin mathbb C: Re z <gamma}$ which is open in $mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $Re Lambda x >gamma$ for all $x in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_onotin overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $overline {E}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 12:31









                Kavi Rama MurthyKavi Rama Murthy

                72.9k53170




                72.9k53170






























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