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Is the application of Lebesgue DCT valid here?
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Problem:
Let $f_n: [0,1] to mathbbR$ be a sequence of measurable functions.
Suppose that $int_0^1|f_n(x)|^2 ~ dx le 1$ for $n in mathbbN$ and $f_n$ converges to $0$ a.e.
Show that $lim_n to infty int_0^1 f_n(x) ~ dx = 0$.
Question: Is the following solution correct? If not, how can it be fixed?
Proposed Solution:
We have $|f_n|_L^2 le 1$ $forall n in mathbbN$ and the measure space is sigma-finite. So $|f_n|_L^1 le |f_n|_L^2 le 1$ and $f_n$ is bounded above by the integrable function $g(x) = 1$ for every $n$. By the pointwise-a.e. convergence, $f equiv 0$, and $|f| le g$. So by DCT, the result follows.
measure-theory proof-verification
asked Jul 27 at 2:44
Mathemagica
85922763
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up vote
2
down vote
favorite
Problem:
Let $f_n: [0,1] to mathbbR$ be a sequence of measurable functions.
Suppose that $int_0^1|f_n(x)|^2 ~ dx le 1$ for $n in mathbbN$ and $f_n$ converges to $0$ a.e.
Show that $lim_n to infty int_0^1 f_n(x) ~ dx = 0$.
Question: Is the following solution correct? If not, how can it be fixed?
Proposed Solution:
We have $|f_n|_L^2 le 1$ $forall n in mathbbN$ and the measure space is sigma-finite. So $|f_n|_L^1 le |f_n|_L^2 le 1$ and $f_n$ is bounded above by the integrable function $g(x) = 1$ for every $n$. By the pointwise-a.e. convergence, $f equiv 0$, and $|f| le g$. So by DCT, the result follows.
measure-theory proof-verification
asked Jul 27 at 2:44
Mathemagica
85922763
1
This application of DCT is invalid. You need a function $g$ such that $lvert f_n rvert le g$ and you do not have this; instead you have $|f_n|_L^2 le 1$, but $f_n$ could still get large in places.
– User8128
Jul 27 at 2:52
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem:
Let $f_n: [0,1] to mathbbR$ be a sequence of measurable functions.
Suppose that $int_0^1|f_n(x)|^2 ~ dx le 1$ for $n in mathbbN$ and $f_n$ converges to $0$ a.e.
Show that $lim_n to infty int_0^1 f_n(x) ~ dx = 0$.
Question: Is the following solution correct? If not, how can it be fixed?
Proposed Solution:
We have $|f_n|_L^2 le 1$ $forall n in mathbbN$ and the measure space is sigma-finite. So $|f_n|_L^1 le |f_n|_L^2 le 1$ and $f_n$ is bounded above by the integrable function $g(x) = 1$ for every $n$. By the pointwise-a.e. convergence, $f equiv 0$, and $|f| le g$. So by DCT, the result follows.
measure-theory proof-verification
asked Jul 27 at 2:44
Mathemagica
85922763
Problem:
Let $f_n: [0,1] to mathbbR$ be a sequence of measurable functions.
Suppose that $int_0^1|f_n(x)|^2 ~ dx le 1$ for $n in mathbbN$ and $f_n$ converges to $0$ a.e.
Show that $lim_n to infty int_0^1 f_n(x) ~ dx = 0$.
Question: Is the following solution correct? If not, how can it be fixed?
Proposed Solution:
We have $|f_n|_L^2 le 1$ $forall n in mathbbN$ and the measure space is sigma-finite. So $|f_n|_L^1 le |f_n|_L^2 le 1$ and $f_n$ is bounded above by the integrable function $g(x) = 1$ for every $n$. By the pointwise-a.e. convergence, $f equiv 0$, and $|f| le g$. So by DCT, the result follows.
measure-theory proof-verification
asked Jul 27 at 2:44
Mathemagica
85922763
asked Jul 27 at 2:44
Mathemagica
85922763
asked Jul 27 at 2:44
Mathemagica
85922763
asked Jul 27 at 2:44
Mathemagica
85922763
85922763
1
This application of DCT is invalid. You need a function $g$ such that $lvert f_n rvert le g$ and you do not have this; instead you have $|f_n|_L^2 le 1$, but $f_n$ could still get large in places.
– User8128
Jul 27 at 2:52
add a comment |Â
1
This application of DCT is invalid. You need a function $g$ such that $lvert f_n rvert le g$ and you do not have this; instead you have $|f_n|_L^2 le 1$, but $f_n$ could still get large in places.
– User8128
Jul 27 at 2:52
1
1
This application of DCT is invalid. You need a function $g$ such that $lvert f_n rvert le g$ and you do not have this; instead you have $|f_n|_L^2 le 1$, but $f_n$ could still get large in places.
– User8128
Jul 27 at 2:52
This application of DCT is invalid. You need a function $g$ such that $lvert f_n rvert le g$ and you do not have this; instead you have $|f_n|_L^2 le 1$, but $f_n$ could still get large in places.
– User8128
Jul 27 at 2:52
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Here is a way to solve the problem. Since $f_nto 0$ a.e, and the measure space is finite, you have $f_nto 0$ in probability. Given $epsilon>0$, let $A_n=$. Then
$$
int_0^1 |f_n|,dx=int_A_n |f_n|,dx + int_[0,1]setminus A_n|f_n|,dxle int_0^1 f_n(x)bf 1_A_n(x),dx + epsilon
$$
Now apply Cauchy-Schwarz to that last integral.
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
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up vote
0
down vote
The proof is false. DCT requires that the function be dominated in the sense that $|f_n| leq g$.
Also, the inequalities that you use don't apply in a sigma finite measure space. They apply in a finite measure space! They are false over the entire real line!
answered Jul 27 at 2:51
Wraith1995
470314
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
1
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Here is a way to solve the problem. Since $f_nto 0$ a.e, and the measure space is finite, you have $f_nto 0$ in probability. Given $epsilon>0$, let $A_n=$. Then
$$
int_0^1 |f_n|,dx=int_A_n |f_n|,dx + int_[0,1]setminus A_n|f_n|,dxle int_0^1 f_n(x)bf 1_A_n(x),dx + epsilon
$$
Now apply Cauchy-Schwarz to that last integral.
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
add a comment |Â
up vote
3
down vote
accepted
Here is a way to solve the problem. Since $f_nto 0$ a.e, and the measure space is finite, you have $f_nto 0$ in probability. Given $epsilon>0$, let $A_n=$. Then
$$
int_0^1 |f_n|,dx=int_A_n |f_n|,dx + int_[0,1]setminus A_n|f_n|,dxle int_0^1 f_n(x)bf 1_A_n(x),dx + epsilon
$$
Now apply Cauchy-Schwarz to that last integral.
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Here is a way to solve the problem. Since $f_nto 0$ a.e, and the measure space is finite, you have $f_nto 0$ in probability. Given $epsilon>0$, let $A_n=$. Then
$$
int_0^1 |f_n|,dx=int_A_n |f_n|,dx + int_[0,1]setminus A_n|f_n|,dxle int_0^1 f_n(x)bf 1_A_n(x),dx + epsilon
$$
Now apply Cauchy-Schwarz to that last integral.
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
Here is a way to solve the problem. Since $f_nto 0$ a.e, and the measure space is finite, you have $f_nto 0$ in probability. Given $epsilon>0$, let $A_n=$. Then
$$
int_0^1 |f_n|,dx=int_A_n |f_n|,dx + int_[0,1]setminus A_n|f_n|,dxle int_0^1 f_n(x)bf 1_A_n(x),dx + epsilon
$$
Now apply Cauchy-Schwarz to that last integral.
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
answered Jul 27 at 3:10
Mike Earnest
14.9k11644
14.9k11644
add a comment |Â
add a comment |Â
up vote
0
down vote
The proof is false. DCT requires that the function be dominated in the sense that $|f_n| leq g$.
Also, the inequalities that you use don't apply in a sigma finite measure space. They apply in a finite measure space! They are false over the entire real line!
answered Jul 27 at 2:51
Wraith1995
470314
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
1
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
add a comment |Â
up vote
0
down vote
The proof is false. DCT requires that the function be dominated in the sense that $|f_n| leq g$.
Also, the inequalities that you use don't apply in a sigma finite measure space. They apply in a finite measure space! They are false over the entire real line!
answered Jul 27 at 2:51
Wraith1995
470314
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
1
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The proof is false. DCT requires that the function be dominated in the sense that $|f_n| leq g$.
Also, the inequalities that you use don't apply in a sigma finite measure space. They apply in a finite measure space! They are false over the entire real line!
answered Jul 27 at 2:51
Wraith1995
470314
The proof is false. DCT requires that the function be dominated in the sense that $|f_n| leq g$.
Also, the inequalities that you use don't apply in a sigma finite measure space. They apply in a finite measure space! They are false over the entire real line!
answered Jul 27 at 2:51
Wraith1995
470314
answered Jul 27 at 2:51
Wraith1995
470314
answered Jul 27 at 2:51
Wraith1995
470314
answered Jul 27 at 2:51
Wraith1995
470314
470314
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
1
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
add a comment |Â
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
1
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
Yes, but we're in a finite measure space here, so the $L^1$-$L^2$ inequality is fine.
– User8128
Jul 27 at 3:01
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
I agree. But in the proof, you (EDIT: the op) justify this via sigma finite-ness. The issue is that you used the wrong justification.
– Wraith1995
Jul 27 at 3:02
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sure, not me, but I agree the OP used the wrong statement. Still I'm pretty sure this property holds (i.e., it is true that $lim_nto infty int^1_0 f_n(x) dx =0$), just the proof is incorrect.
– User8128
Jul 27 at 3:04
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
Sorry. I assumed that you were OP. Sorry I didn't read before posting. I'll edit to clarify.
– Wraith1995
Jul 27 at 3:05
1
1
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
As for the proof: I think that I have the idea. I'll post it after thinking for ab it.
– Wraith1995
Jul 27 at 3:07
add a comment |Â
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1
This application of DCT is invalid. You need a function $g$ such that $lvert f_n rvert le g$ and you do not have this; instead you have $|f_n|_L^2 le 1$, but $f_n$ could still get large in places.
– User8128
Jul 27 at 2:52