About weak convergence in $L^{infty}$












2














doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces







share|cite|improve this question


















  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28


















2














doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces







share|cite|improve this question


















  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28
















2












2








2







doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces







share|cite|improve this question













doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces




functional-analysis weak-convergence dual-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 at 15:03









James Arten

9910




9910








  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28
















  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28










2




2




You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
– GEdgar
Dec 26 at 15:18




You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
– GEdgar
Dec 26 at 15:18












but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
– James Arten
Dec 26 at 15:21






but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
– James Arten
Dec 26 at 15:21














Correct, it should say $phi$ is in $(L^infty)^*$.
– GEdgar
Dec 26 at 15:25






Correct, it should say $phi$ is in $(L^infty)^*$.
– GEdgar
Dec 26 at 15:25














you mean the definition I've written is wrong? Just to understand
– James Arten
Dec 26 at 15:26




you mean the definition I've written is wrong? Just to understand
– James Arten
Dec 26 at 15:26




1




1




Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
– GEdgar
Dec 26 at 15:28






Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
– GEdgar
Dec 26 at 15:28












1 Answer
1






active

oldest

votes


















2














Ok I think I solved this.



In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
begin{equation}
delta_0(f) = f(0) quad forall f in C[-1,1]
end{equation}



Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



(we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






share|cite|improve this answer























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    1 Answer
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    oldest

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    Ok I think I solved this.



    In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
    begin{equation}
    delta_0(f) = f(0) quad forall f in C[-1,1]
    end{equation}



    Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



    (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






    share|cite|improve this answer




























      2














      Ok I think I solved this.



      In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
      begin{equation}
      delta_0(f) = f(0) quad forall f in C[-1,1]
      end{equation}



      Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



      (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






      share|cite|improve this answer


























        2












        2








        2






        Ok I think I solved this.



        In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
        begin{equation}
        delta_0(f) = f(0) quad forall f in C[-1,1]
        end{equation}



        Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



        (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






        share|cite|improve this answer














        Ok I think I solved this.



        In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
        begin{equation}
        delta_0(f) = f(0) quad forall f in C[-1,1]
        end{equation}



        Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



        (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 27 at 8:53

























        answered Dec 27 at 8:39









        James Arten

        9910




        9910






























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