Prove ${fin L^infty : |f|_infty leq 1- epsilon}$ for $epsilonin (0, 1)$ is $w^*$- closed
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
asked Nov 25 at 7:43
user62498
1,894613
add a comment |
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
asked Nov 25 at 7:43
user62498
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
add a comment |
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
asked Nov 25 at 7:43
user62498
1,894613
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
asked Nov 25 at 7:43
user62498
1,894613
asked Nov 25 at 7:43
user62498
1,894613
asked Nov 25 at 7:43
user62498
1,894613
asked Nov 25 at 7:43
user62498
1,894613
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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.
answered Nov 25 at 9:19
Lorenzo Quarisa
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
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
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.
answered Nov 25 at 9:19
Lorenzo Quarisa
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
add a comment |
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.
answered Nov 25 at 9:19
Lorenzo Quarisa
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
add a comment |
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.
answered Nov 25 at 9:19
Lorenzo Quarisa
3,156316
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.
answered Nov 25 at 9:19
Lorenzo Quarisa
3,156316
edited Nov 25 at 9:25
answered Nov 25 at 9:19
Lorenzo Quarisa
3,156316
answered Nov 25 at 9:19
Lorenzo Quarisa
3,156316
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
add a comment |
@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
add a comment |
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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