Prove ${fin L^infty : |f|_infty leq 1- epsilon}$ for $epsilonin (0, 1)$ is $w^*$- closed












1














I want to prove the following statement




Prove ${fin L^infty : |f|_infty leq 1- epsilon}$ for $epsilonin (0, 1)$ is $w^*$- closed




I've started:



suppose $f_n xrightarrow{w^*}f$ so tht we have $f_n(g) to f(g)$ for all $fin L_1$ Icould'nt able to continuo from here



Any help is much appreciated! Thank you!






convergence weak-convergence







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  • If $epsilon$ is fixed, this is a norm closed set, and therefore definitely weak star closed.
    – Ashwin Trisal
    Nov 25 at 8:21
















1














I want to prove the following statement




Prove ${fin L^infty : |f|_infty leq 1- epsilon}$ for $epsilonin (0, 1)$ is $w^*$- closed




I've started:



suppose $f_n xrightarrow{w^*}f$ so tht we have $f_n(g) to f(g)$ for all $fin L_1$ Icould'nt able to continuo from here



Any help is much appreciated! Thank you!






convergence weak-convergence







share|cite|improve this question






















  • If $epsilon$ is fixed, this is a norm closed set, and therefore definitely weak star closed.
    – Ashwin Trisal
    Nov 25 at 8:21














1












1








1







I want to prove the following statement




Prove ${fin L^infty : |f|_infty leq 1- epsilon}$ for $epsilonin (0, 1)$ is $w^*$- closed




I've started:



suppose $f_n xrightarrow{w^*}f$ so tht we have $f_n(g) to f(g)$ for all $fin L_1$ Icould'nt able to continuo from here



Any help is much appreciated! Thank you!






convergence weak-convergence







share|cite|improve this question













I want to prove the following statement




Prove ${fin L^infty : |f|_infty leq 1- epsilon}$ for $epsilonin (0, 1)$ is $w^*$- closed




I've started:



suppose $f_n xrightarrow{w^*}f$ so tht we have $f_n(g) to f(g)$ for all $fin L_1$ Icould'nt able to continuo from here



Any help is much appreciated! Thank you!






convergence weak-convergence




convergence weak-convergence






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share|cite|improve this question




share|cite|improve this question










asked Nov 25 at 7:43









user62498

1,894613




1,894613












  • If $epsilon$ is fixed, this is a norm closed set, and therefore definitely weak star closed.
    – Ashwin Trisal
    Nov 25 at 8:21


















  • If $epsilon$ is fixed, this is a norm closed set, and therefore definitely weak star closed.
    – Ashwin Trisal
    Nov 25 at 8:21
















If $epsilon$ is fixed, this is a norm closed set, and therefore definitely weak star closed.
– Ashwin Trisal
Nov 25 at 8:21




If $epsilon$ is fixed, this is a norm closed set, and therefore definitely weak star closed.
– Ashwin Trisal
Nov 25 at 8:21










1 Answer
1






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oldest

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2














Let $r=1-varepsilon$. Let $(f_j)_{jin J}$ be a net (*) in $L^{infty}(Omega)$ with $|f_j|_{infty}leq r$ and $f_jrightharpoonup^* bar{f}$. We then have
$$int_{Omega} f_j gto int_{Omega} bar{f}gqquad forall gin L^1(Omega) qquad (1)$$
We need to prove that $|bar{f}|_{infty}leq r$. Suppose by contradiction that this is false. Then there exists a measurable set $E$ with positive measure ($|E|>0$) such that $|bar{f}|geq r+varepsilon$ on $E$ for some $varepsilon>0$. We may also assume that $E$ has finite measure (as long as $Omega$ is $sigma$-finite, which is a rather standard assumption).



Now apply $(1)$ with $g=chi_E$, which is in $L^1(Omega)$ because $|E|<infty$. Since $|f_j|leq r$ a.e., we have
$$r|E|geqint_E f_jto int_E bar{f}geq (r+varepsilon)|E| $$
which is a contradiction.



Anyway, it is worth to notice that in general




Let $X$ be a normed space. Then the closed ball of radius $r$ centered in the origin
$$ B_r:=left{fin X^*:|f|leq rright}$$
is weakly-star closed.




($*$) You should use nets because the weak-star topology is not metrizable in general, unless by weakly*-closed you mean sequentially weakly-star closed. In this specific case, though, this is not necessary, as it can be shown that the restriction to a closed ball of the weak-star topology in the dual of a separable space (and $L^{infty}$ is the dual of $L^1$ which is separable) is metrizable, but this is a more advanced result.






share|cite|improve this answer























  • @Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
    – user62498
    Nov 25 at 9:36










  • @user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
    – Lorenzo Quarisa
    Nov 25 at 10:05











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Let $r=1-varepsilon$. Let $(f_j)_{jin J}$ be a net (*) in $L^{infty}(Omega)$ with $|f_j|_{infty}leq r$ and $f_jrightharpoonup^* bar{f}$. We then have
$$int_{Omega} f_j gto int_{Omega} bar{f}gqquad forall gin L^1(Omega) qquad (1)$$
We need to prove that $|bar{f}|_{infty}leq r$. Suppose by contradiction that this is false. Then there exists a measurable set $E$ with positive measure ($|E|>0$) such that $|bar{f}|geq r+varepsilon$ on $E$ for some $varepsilon>0$. We may also assume that $E$ has finite measure (as long as $Omega$ is $sigma$-finite, which is a rather standard assumption).



Now apply $(1)$ with $g=chi_E$, which is in $L^1(Omega)$ because $|E|<infty$. Since $|f_j|leq r$ a.e., we have
$$r|E|geqint_E f_jto int_E bar{f}geq (r+varepsilon)|E| $$
which is a contradiction.



Anyway, it is worth to notice that in general




Let $X$ be a normed space. Then the closed ball of radius $r$ centered in the origin
$$ B_r:=left{fin X^*:|f|leq rright}$$
is weakly-star closed.




($*$) You should use nets because the weak-star topology is not metrizable in general, unless by weakly*-closed you mean sequentially weakly-star closed. In this specific case, though, this is not necessary, as it can be shown that the restriction to a closed ball of the weak-star topology in the dual of a separable space (and $L^{infty}$ is the dual of $L^1$ which is separable) is metrizable, but this is a more advanced result.






share|cite|improve this answer























  • @Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
    – user62498
    Nov 25 at 9:36










  • @user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
    – Lorenzo Quarisa
    Nov 25 at 10:05
















2














Let $r=1-varepsilon$. Let $(f_j)_{jin J}$ be a net (*) in $L^{infty}(Omega)$ with $|f_j|_{infty}leq r$ and $f_jrightharpoonup^* bar{f}$. We then have
$$int_{Omega} f_j gto int_{Omega} bar{f}gqquad forall gin L^1(Omega) qquad (1)$$
We need to prove that $|bar{f}|_{infty}leq r$. Suppose by contradiction that this is false. Then there exists a measurable set $E$ with positive measure ($|E|>0$) such that $|bar{f}|geq r+varepsilon$ on $E$ for some $varepsilon>0$. We may also assume that $E$ has finite measure (as long as $Omega$ is $sigma$-finite, which is a rather standard assumption).



Now apply $(1)$ with $g=chi_E$, which is in $L^1(Omega)$ because $|E|<infty$. Since $|f_j|leq r$ a.e., we have
$$r|E|geqint_E f_jto int_E bar{f}geq (r+varepsilon)|E| $$
which is a contradiction.



Anyway, it is worth to notice that in general




Let $X$ be a normed space. Then the closed ball of radius $r$ centered in the origin
$$ B_r:=left{fin X^*:|f|leq rright}$$
is weakly-star closed.




($*$) You should use nets because the weak-star topology is not metrizable in general, unless by weakly*-closed you mean sequentially weakly-star closed. In this specific case, though, this is not necessary, as it can be shown that the restriction to a closed ball of the weak-star topology in the dual of a separable space (and $L^{infty}$ is the dual of $L^1$ which is separable) is metrizable, but this is a more advanced result.






share|cite|improve this answer























  • @Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
    – user62498
    Nov 25 at 9:36










  • @user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
    – Lorenzo Quarisa
    Nov 25 at 10:05














2












2








2






Let $r=1-varepsilon$. Let $(f_j)_{jin J}$ be a net (*) in $L^{infty}(Omega)$ with $|f_j|_{infty}leq r$ and $f_jrightharpoonup^* bar{f}$. We then have
$$int_{Omega} f_j gto int_{Omega} bar{f}gqquad forall gin L^1(Omega) qquad (1)$$
We need to prove that $|bar{f}|_{infty}leq r$. Suppose by contradiction that this is false. Then there exists a measurable set $E$ with positive measure ($|E|>0$) such that $|bar{f}|geq r+varepsilon$ on $E$ for some $varepsilon>0$. We may also assume that $E$ has finite measure (as long as $Omega$ is $sigma$-finite, which is a rather standard assumption).



Now apply $(1)$ with $g=chi_E$, which is in $L^1(Omega)$ because $|E|<infty$. Since $|f_j|leq r$ a.e., we have
$$r|E|geqint_E f_jto int_E bar{f}geq (r+varepsilon)|E| $$
which is a contradiction.



Anyway, it is worth to notice that in general




Let $X$ be a normed space. Then the closed ball of radius $r$ centered in the origin
$$ B_r:=left{fin X^*:|f|leq rright}$$
is weakly-star closed.




($*$) You should use nets because the weak-star topology is not metrizable in general, unless by weakly*-closed you mean sequentially weakly-star closed. In this specific case, though, this is not necessary, as it can be shown that the restriction to a closed ball of the weak-star topology in the dual of a separable space (and $L^{infty}$ is the dual of $L^1$ which is separable) is metrizable, but this is a more advanced result.






share|cite|improve this answer














Let $r=1-varepsilon$. Let $(f_j)_{jin J}$ be a net (*) in $L^{infty}(Omega)$ with $|f_j|_{infty}leq r$ and $f_jrightharpoonup^* bar{f}$. We then have
$$int_{Omega} f_j gto int_{Omega} bar{f}gqquad forall gin L^1(Omega) qquad (1)$$
We need to prove that $|bar{f}|_{infty}leq r$. Suppose by contradiction that this is false. Then there exists a measurable set $E$ with positive measure ($|E|>0$) such that $|bar{f}|geq r+varepsilon$ on $E$ for some $varepsilon>0$. We may also assume that $E$ has finite measure (as long as $Omega$ is $sigma$-finite, which is a rather standard assumption).



Now apply $(1)$ with $g=chi_E$, which is in $L^1(Omega)$ because $|E|<infty$. Since $|f_j|leq r$ a.e., we have
$$r|E|geqint_E f_jto int_E bar{f}geq (r+varepsilon)|E| $$
which is a contradiction.



Anyway, it is worth to notice that in general




Let $X$ be a normed space. Then the closed ball of radius $r$ centered in the origin
$$ B_r:=left{fin X^*:|f|leq rright}$$
is weakly-star closed.




($*$) You should use nets because the weak-star topology is not metrizable in general, unless by weakly*-closed you mean sequentially weakly-star closed. In this specific case, though, this is not necessary, as it can be shown that the restriction to a closed ball of the weak-star topology in the dual of a separable space (and $L^{infty}$ is the dual of $L^1$ which is separable) is metrizable, but this is a more advanced result.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 9:25

























answered Nov 25 at 9:19









Lorenzo Quarisa

3,156316




3,156316












  • @Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
    – user62498
    Nov 25 at 9:36










  • @user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
    – Lorenzo Quarisa
    Nov 25 at 10:05


















  • @Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
    – user62498
    Nov 25 at 9:36










  • @user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
    – Lorenzo Quarisa
    Nov 25 at 10:05
















@Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
– user62498
Nov 25 at 9:36




@Dear Lorenzo Quarisa, Thank you!! It makes perfect sense now
– user62498
Nov 25 at 9:36












@user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
– Lorenzo Quarisa
Nov 25 at 10:05




@user62498 You're welcome! Please consider accepting my answer so that your question doesn't show up in the unanswered questions queue.
– Lorenzo Quarisa
Nov 25 at 10:05


















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