The map $L^infty$ to $(L^1)^*$ is not injective in this measure space?
$begingroup$
The question is :
Given $X = {a,b}$, with $M = P(X)$ and $mu({a}) = 1,mu({b}) = infty$, find the dual of $L^1(mu)$. Is it $L^infty(mu)$? (Rudin's Real and Complex Analysis, chapter 6)
For solving this question, I first investigated what $L^1(mu)$ is. For this, we note that every function $f : X to mathbb R$ is measurable, with $int_X f
mathrm d mu = f(a)mu({a}) + f(b) mu({b})$. Under the convention that $0 cdot infty =0$, we see that $L^1(mu)$ consists of all $f$ with $|f(a)| < infty$ and $f(b) =0$.
In other words, let $f_1$ be the indicator of the set ${a}$. Then $L^1(mu) = {cf_1 : |c| < infty}$.
Now, what is $L^infty(mu)$? It consists of functions whose essential supremum is finite. This means that there exists $M$ such that $mu({x : f(x) > M}) = 0$. By definition of $mu$, this forces $f$ to be bounded. Hence, $L^infty(mu)$ is the set of bounded functions on $X$.
We see that for all $g in L^infty(mu)$, the functional $f to int_X fg$ is bounded on $L^1$. So, there is a linear map from $L^infty(mu)$ to $(L^1)^*$.
That it is onto is clear, since every linear functional $T$ is determined by its action on $f_1$, therefore if $Tf_1 = d$, then the function $g = df_1$ is such that $Tf_1 = int f_1 cdot df_1 = d$. So, these linear functionals coincide on the entire of $L^1$.
However, the map is not injective, since if $g(a) = h(a)$ then $g$ and $h$ produce the same functional on $L^1$, since $int gf_1 = int hf_1 = g(a)$. So if $g(b) neq h(b)$ then they produce the same functional despite being different on a set of non-zero measure. The kernel is equal to every function satisfying $g(a) = 0$. Call it $K$.
Therefore, are we right to conclude that $(L^1)^*$ is isomorphic to $L^infty / K$?(first isomorphism theorem)?
My only issue is that I've never seen before, this map being non-injective. For the Lebesgue measure, and others I've seen it has been injective(i.e. if $g$ is such that $int gf = 0$ for all $f in L^1$ then $g=0$ a.e.), and the image has either been the whole or a part of $L^infty$.
measure-theory lp-spaces
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
|
show 1 more comment
$begingroup$
The question is :
Given $X = {a,b}$, with $M = P(X)$ and $mu({a}) = 1,mu({b}) = infty$, find the dual of $L^1(mu)$. Is it $L^infty(mu)$? (Rudin's Real and Complex Analysis, chapter 6)
For solving this question, I first investigated what $L^1(mu)$ is. For this, we note that every function $f : X to mathbb R$ is measurable, with $int_X f
mathrm d mu = f(a)mu({a}) + f(b) mu({b})$. Under the convention that $0 cdot infty =0$, we see that $L^1(mu)$ consists of all $f$ with $|f(a)| < infty$ and $f(b) =0$.
In other words, let $f_1$ be the indicator of the set ${a}$. Then $L^1(mu) = {cf_1 : |c| < infty}$.
Now, what is $L^infty(mu)$? It consists of functions whose essential supremum is finite. This means that there exists $M$ such that $mu({x : f(x) > M}) = 0$. By definition of $mu$, this forces $f$ to be bounded. Hence, $L^infty(mu)$ is the set of bounded functions on $X$.
We see that for all $g in L^infty(mu)$, the functional $f to int_X fg$ is bounded on $L^1$. So, there is a linear map from $L^infty(mu)$ to $(L^1)^*$.
That it is onto is clear, since every linear functional $T$ is determined by its action on $f_1$, therefore if $Tf_1 = d$, then the function $g = df_1$ is such that $Tf_1 = int f_1 cdot df_1 = d$. So, these linear functionals coincide on the entire of $L^1$.
However, the map is not injective, since if $g(a) = h(a)$ then $g$ and $h$ produce the same functional on $L^1$, since $int gf_1 = int hf_1 = g(a)$. So if $g(b) neq h(b)$ then they produce the same functional despite being different on a set of non-zero measure. The kernel is equal to every function satisfying $g(a) = 0$. Call it $K$.
Therefore, are we right to conclude that $(L^1)^*$ is isomorphic to $L^infty / K$?(first isomorphism theorem)?
My only issue is that I've never seen before, this map being non-injective. For the Lebesgue measure, and others I've seen it has been injective(i.e. if $g$ is such that $int gf = 0$ for all $f in L^1$ then $g=0$ a.e.), and the image has either been the whole or a part of $L^infty$.
measure-theory lp-spaces
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
1
$begingroup$
just a side note: strictly speaking $0cdotinfty=0$ is a convention just in notation but it represent a logical truth in measure theory: any sequence of integrals of simple functions on a set of measure zero is zero, hence it point-wise limit is zero regardless of the values it take
$endgroup$
– Masacroso
Nov 30 '18 at 11:11
$begingroup$
@Masacroso Thank you for your input.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:12
2
$begingroup$
$L^{1}$ is one dimensional and so is $(L^{1})^{*}$ but $L^{infty}$ is two dimensional.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:30
$begingroup$
@KaviRamaMurthy Thank you. So that confirms my suspicion. I just want you to tell me if the above characterization is correct. This is , therefore an example of the case where the map $L^infty$ to $(L^1)^*$ is not injective, by the rank nullity theorem. Is that right?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:32
1
$begingroup$
Yes, you have identified the spaces correctly and the dual of $L^{1}$ is not $L^{infty}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:43
|
show 1 more comment
$begingroup$
The question is :
Given $X = {a,b}$, with $M = P(X)$ and $mu({a}) = 1,mu({b}) = infty$, find the dual of $L^1(mu)$. Is it $L^infty(mu)$? (Rudin's Real and Complex Analysis, chapter 6)
For solving this question, I first investigated what $L^1(mu)$ is. For this, we note that every function $f : X to mathbb R$ is measurable, with $int_X f
mathrm d mu = f(a)mu({a}) + f(b) mu({b})$. Under the convention that $0 cdot infty =0$, we see that $L^1(mu)$ consists of all $f$ with $|f(a)| < infty$ and $f(b) =0$.
In other words, let $f_1$ be the indicator of the set ${a}$. Then $L^1(mu) = {cf_1 : |c| < infty}$.
Now, what is $L^infty(mu)$? It consists of functions whose essential supremum is finite. This means that there exists $M$ such that $mu({x : f(x) > M}) = 0$. By definition of $mu$, this forces $f$ to be bounded. Hence, $L^infty(mu)$ is the set of bounded functions on $X$.
We see that for all $g in L^infty(mu)$, the functional $f to int_X fg$ is bounded on $L^1$. So, there is a linear map from $L^infty(mu)$ to $(L^1)^*$.
That it is onto is clear, since every linear functional $T$ is determined by its action on $f_1$, therefore if $Tf_1 = d$, then the function $g = df_1$ is such that $Tf_1 = int f_1 cdot df_1 = d$. So, these linear functionals coincide on the entire of $L^1$.
However, the map is not injective, since if $g(a) = h(a)$ then $g$ and $h$ produce the same functional on $L^1$, since $int gf_1 = int hf_1 = g(a)$. So if $g(b) neq h(b)$ then they produce the same functional despite being different on a set of non-zero measure. The kernel is equal to every function satisfying $g(a) = 0$. Call it $K$.
Therefore, are we right to conclude that $(L^1)^*$ is isomorphic to $L^infty / K$?(first isomorphism theorem)?
My only issue is that I've never seen before, this map being non-injective. For the Lebesgue measure, and others I've seen it has been injective(i.e. if $g$ is such that $int gf = 0$ for all $f in L^1$ then $g=0$ a.e.), and the image has either been the whole or a part of $L^infty$.
measure-theory lp-spaces
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
$endgroup$
The question is :
Given $X = {a,b}$, with $M = P(X)$ and $mu({a}) = 1,mu({b}) = infty$, find the dual of $L^1(mu)$. Is it $L^infty(mu)$? (Rudin's Real and Complex Analysis, chapter 6)
For solving this question, I first investigated what $L^1(mu)$ is. For this, we note that every function $f : X to mathbb R$ is measurable, with $int_X f
mathrm d mu = f(a)mu({a}) + f(b) mu({b})$. Under the convention that $0 cdot infty =0$, we see that $L^1(mu)$ consists of all $f$ with $|f(a)| < infty$ and $f(b) =0$.
In other words, let $f_1$ be the indicator of the set ${a}$. Then $L^1(mu) = {cf_1 : |c| < infty}$.
Now, what is $L^infty(mu)$? It consists of functions whose essential supremum is finite. This means that there exists $M$ such that $mu({x : f(x) > M}) = 0$. By definition of $mu$, this forces $f$ to be bounded. Hence, $L^infty(mu)$ is the set of bounded functions on $X$.
We see that for all $g in L^infty(mu)$, the functional $f to int_X fg$ is bounded on $L^1$. So, there is a linear map from $L^infty(mu)$ to $(L^1)^*$.
That it is onto is clear, since every linear functional $T$ is determined by its action on $f_1$, therefore if $Tf_1 = d$, then the function $g = df_1$ is such that $Tf_1 = int f_1 cdot df_1 = d$. So, these linear functionals coincide on the entire of $L^1$.
However, the map is not injective, since if $g(a) = h(a)$ then $g$ and $h$ produce the same functional on $L^1$, since $int gf_1 = int hf_1 = g(a)$. So if $g(b) neq h(b)$ then they produce the same functional despite being different on a set of non-zero measure. The kernel is equal to every function satisfying $g(a) = 0$. Call it $K$.
Therefore, are we right to conclude that $(L^1)^*$ is isomorphic to $L^infty / K$?(first isomorphism theorem)?
My only issue is that I've never seen before, this map being non-injective. For the Lebesgue measure, and others I've seen it has been injective(i.e. if $g$ is such that $int gf = 0$ for all $f in L^1$ then $g=0$ a.e.), and the image has either been the whole or a part of $L^infty$.
measure-theory lp-spaces
measure-theory lp-spaces
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
asked Nov 30 '18 at 11:04
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.4k33376
37.4k33376
1
$begingroup$
just a side note: strictly speaking $0cdotinfty=0$ is a convention just in notation but it represent a logical truth in measure theory: any sequence of integrals of simple functions on a set of measure zero is zero, hence it point-wise limit is zero regardless of the values it take
$endgroup$
– Masacroso
Nov 30 '18 at 11:11
$begingroup$
@Masacroso Thank you for your input.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:12
2
$begingroup$
$L^{1}$ is one dimensional and so is $(L^{1})^{*}$ but $L^{infty}$ is two dimensional.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:30
$begingroup$
@KaviRamaMurthy Thank you. So that confirms my suspicion. I just want you to tell me if the above characterization is correct. This is , therefore an example of the case where the map $L^infty$ to $(L^1)^*$ is not injective, by the rank nullity theorem. Is that right?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:32
1
$begingroup$
Yes, you have identified the spaces correctly and the dual of $L^{1}$ is not $L^{infty}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:43
|
show 1 more comment
1
$begingroup$
just a side note: strictly speaking $0cdotinfty=0$ is a convention just in notation but it represent a logical truth in measure theory: any sequence of integrals of simple functions on a set of measure zero is zero, hence it point-wise limit is zero regardless of the values it take
$endgroup$
– Masacroso
Nov 30 '18 at 11:11
$begingroup$
@Masacroso Thank you for your input.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:12
2
$begingroup$
$L^{1}$ is one dimensional and so is $(L^{1})^{*}$ but $L^{infty}$ is two dimensional.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:30
$begingroup$
@KaviRamaMurthy Thank you. So that confirms my suspicion. I just want you to tell me if the above characterization is correct. This is , therefore an example of the case where the map $L^infty$ to $(L^1)^*$ is not injective, by the rank nullity theorem. Is that right?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:32
1
$begingroup$
Yes, you have identified the spaces correctly and the dual of $L^{1}$ is not $L^{infty}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:43
1
1
$begingroup$
just a side note: strictly speaking $0cdotinfty=0$ is a convention just in notation but it represent a logical truth in measure theory: any sequence of integrals of simple functions on a set of measure zero is zero, hence it point-wise limit is zero regardless of the values it take
$endgroup$
– Masacroso
Nov 30 '18 at 11:11
$begingroup$
just a side note: strictly speaking $0cdotinfty=0$ is a convention just in notation but it represent a logical truth in measure theory: any sequence of integrals of simple functions on a set of measure zero is zero, hence it point-wise limit is zero regardless of the values it take
$endgroup$
– Masacroso
Nov 30 '18 at 11:11
$begingroup$
@Masacroso Thank you for your input.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:12
$begingroup$
@Masacroso Thank you for your input.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:12
2
2
$begingroup$
$L^{1}$ is one dimensional and so is $(L^{1})^{*}$ but $L^{infty}$ is two dimensional.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:30
$begingroup$
$L^{1}$ is one dimensional and so is $(L^{1})^{*}$ but $L^{infty}$ is two dimensional.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:30
$begingroup$
@KaviRamaMurthy Thank you. So that confirms my suspicion. I just want you to tell me if the above characterization is correct. This is , therefore an example of the case where the map $L^infty$ to $(L^1)^*$ is not injective, by the rank nullity theorem. Is that right?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:32
$begingroup$
@KaviRamaMurthy Thank you. So that confirms my suspicion. I just want you to tell me if the above characterization is correct. This is , therefore an example of the case where the map $L^infty$ to $(L^1)^*$ is not injective, by the rank nullity theorem. Is that right?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:32
1
1
$begingroup$
Yes, you have identified the spaces correctly and the dual of $L^{1}$ is not $L^{infty}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:43
$begingroup$
Yes, you have identified the spaces correctly and the dual of $L^{1}$ is not $L^{infty}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:43
|
show 1 more comment
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1
$begingroup$
just a side note: strictly speaking $0cdotinfty=0$ is a convention just in notation but it represent a logical truth in measure theory: any sequence of integrals of simple functions on a set of measure zero is zero, hence it point-wise limit is zero regardless of the values it take
$endgroup$
– Masacroso
Nov 30 '18 at 11:11
$begingroup$
@Masacroso Thank you for your input.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:12
2
$begingroup$
$L^{1}$ is one dimensional and so is $(L^{1})^{*}$ but $L^{infty}$ is two dimensional.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:30
$begingroup$
@KaviRamaMurthy Thank you. So that confirms my suspicion. I just want you to tell me if the above characterization is correct. This is , therefore an example of the case where the map $L^infty$ to $(L^1)^*$ is not injective, by the rank nullity theorem. Is that right?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 11:32
1
$begingroup$
Yes, you have identified the spaces correctly and the dual of $L^{1}$ is not $L^{infty}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 11:43