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Interchange between Integral and Sup?
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What is the analysis theorem that gives interchange between Integral and Sup. In particular, let $f_r (t)$ be a function on the circle $U = z $, which condition must check $f_r (t)$ so that we can interchange Integral and sup:
$$sup_0leq r<1int_mathbb Uf_r(t) , dt = int_mathbb U, sup_0leq r<1 f_r(t) , dtquad ?$$
Thank you in advance
real-analysis integration complex-analysis lebesgue-integral
asked Jul 14 at 22:14
Z. Alfata
878413
add a comment |Â
up vote
0
down vote
favorite
What is the analysis theorem that gives interchange between Integral and Sup. In particular, let $f_r (t)$ be a function on the circle $U = z $, which condition must check $f_r (t)$ so that we can interchange Integral and sup:
$$sup_0leq r<1int_mathbb Uf_r(t) , dt = int_mathbb U, sup_0leq r<1 f_r(t) , dtquad ?$$
Thank you in advance
real-analysis integration complex-analysis lebesgue-integral
asked Jul 14 at 22:14
Z. Alfata
878413
"Invert" isn't quite the right word. "Interchange" or "exchange" is what you're looking for.
– Sean Roberson
Jul 14 at 22:22
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the analysis theorem that gives interchange between Integral and Sup. In particular, let $f_r (t)$ be a function on the circle $U = z $, which condition must check $f_r (t)$ so that we can interchange Integral and sup:
$$sup_0leq r<1int_mathbb Uf_r(t) , dt = int_mathbb U, sup_0leq r<1 f_r(t) , dtquad ?$$
Thank you in advance
real-analysis integration complex-analysis lebesgue-integral
asked Jul 14 at 22:14
Z. Alfata
878413
What is the analysis theorem that gives interchange between Integral and Sup. In particular, let $f_r (t)$ be a function on the circle $U = z $, which condition must check $f_r (t)$ so that we can interchange Integral and sup:
$$sup_0leq r<1int_mathbb Uf_r(t) , dt = int_mathbb U, sup_0leq r<1 f_r(t) , dtquad ?$$
Thank you in advance
real-analysis integration complex-analysis lebesgue-integral
asked Jul 14 at 22:14
Z. Alfata
878413
edited Jul 14 at 22:38
asked Jul 14 at 22:14
Z. Alfata
878413
asked Jul 14 at 22:14
Z. Alfata
878413
asked Jul 14 at 22:14
Z. Alfata
878413
878413
"Invert" isn't quite the right word. "Interchange" or "exchange" is what you're looking for.
– Sean Roberson
Jul 14 at 22:22
add a comment |Â
"Invert" isn't quite the right word. "Interchange" or "exchange" is what you're looking for.
– Sean Roberson
Jul 14 at 22:22
"Invert" isn't quite the right word. "Interchange" or "exchange" is what you're looking for.
– Sean Roberson
Jul 14 at 22:22
"Invert" isn't quite the right word. "Interchange" or "exchange" is what you're looking for.
– Sean Roberson
Jul 14 at 22:22
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
It's more common (and equivalent) to work with infima (just put $-$ signs everywhere), so we want to find when
$$ inf_r in A int_X f_r = int_X inf_r in A f_r. $$
Let $inf_r f_r(t) = g(t)$. Then $inf_r g(t) = g(t)$, and the right-hand side of the equality is $int g$, so if we subtract and let $h_r=f_r-g$, equality occurs when
$$ inf_r int_X h_r = 0 $$
But $f_r geq inf_r f_r = g $, so $f_r-g = h_r geq 0$. So the integral is always nonnegative. The infimum is zero if and only if there is a sequence $r_n$ so that $int h_r_n to 0$. In turn, this means that it is possible to extract a subsequence $r_n(k)$ so that $h_r_n(k) to 0$ almost everywhere (see e.g. here) (thanks to zhw. for reminding me that passing to a subsequence is necessary). Indeed, the subsequence can be chosen so that the convergence is uniform almost everywhere.
Therefore equality requires that there is a sequence, which we may call $f_r_n$ so that $f_r_n to g$ uniformly almost everywhere. Is this sufficient? Yes, since in this case $mu(X)<infty$, so Hölder's inequality implies that $int_X (f_r_n-g) < lVert f_r_n-g rVert_infty mu(X) $, so $f_r_n to g$ uniformly almost everywhere if and only if $int_X f_r_n to int_X g $.
So a necessary and sufficient condition in your case is that there is a sequence $r_n$ so that $f_r_n to sup_r f_r$ uniformly.
answered Jul 14 at 23:41
Chappers
55k74191
1
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
add a comment |Â
up vote
1
down vote
We have from the property of $sup$:
$$forall rin[0,1):~~~f_r(t)=sup_0leq r<1f_r(t)$$
in your question you suppose that the supremum of integrable functions is integrable, then with integration over $U$
$$forall rin[0,1):~~~int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
this gives
$$sup_0leq r<1int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
because $int_Usup_0leq r<1f_r(t)dt$ is an upper bound for $int_Uf_r(t)dt$ where $0leq r<1$.
answered Jul 15 at 4:34
Nosrati
20k41644
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
It's more common (and equivalent) to work with infima (just put $-$ signs everywhere), so we want to find when
$$ inf_r in A int_X f_r = int_X inf_r in A f_r. $$
Let $inf_r f_r(t) = g(t)$. Then $inf_r g(t) = g(t)$, and the right-hand side of the equality is $int g$, so if we subtract and let $h_r=f_r-g$, equality occurs when
$$ inf_r int_X h_r = 0 $$
But $f_r geq inf_r f_r = g $, so $f_r-g = h_r geq 0$. So the integral is always nonnegative. The infimum is zero if and only if there is a sequence $r_n$ so that $int h_r_n to 0$. In turn, this means that it is possible to extract a subsequence $r_n(k)$ so that $h_r_n(k) to 0$ almost everywhere (see e.g. here) (thanks to zhw. for reminding me that passing to a subsequence is necessary). Indeed, the subsequence can be chosen so that the convergence is uniform almost everywhere.
Therefore equality requires that there is a sequence, which we may call $f_r_n$ so that $f_r_n to g$ uniformly almost everywhere. Is this sufficient? Yes, since in this case $mu(X)<infty$, so Hölder's inequality implies that $int_X (f_r_n-g) < lVert f_r_n-g rVert_infty mu(X) $, so $f_r_n to g$ uniformly almost everywhere if and only if $int_X f_r_n to int_X g $.
So a necessary and sufficient condition in your case is that there is a sequence $r_n$ so that $f_r_n to sup_r f_r$ uniformly.
answered Jul 14 at 23:41
Chappers
55k74191
1
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
add a comment |Â
up vote
3
down vote
accepted
It's more common (and equivalent) to work with infima (just put $-$ signs everywhere), so we want to find when
$$ inf_r in A int_X f_r = int_X inf_r in A f_r. $$
Let $inf_r f_r(t) = g(t)$. Then $inf_r g(t) = g(t)$, and the right-hand side of the equality is $int g$, so if we subtract and let $h_r=f_r-g$, equality occurs when
$$ inf_r int_X h_r = 0 $$
But $f_r geq inf_r f_r = g $, so $f_r-g = h_r geq 0$. So the integral is always nonnegative. The infimum is zero if and only if there is a sequence $r_n$ so that $int h_r_n to 0$. In turn, this means that it is possible to extract a subsequence $r_n(k)$ so that $h_r_n(k) to 0$ almost everywhere (see e.g. here) (thanks to zhw. for reminding me that passing to a subsequence is necessary). Indeed, the subsequence can be chosen so that the convergence is uniform almost everywhere.
Therefore equality requires that there is a sequence, which we may call $f_r_n$ so that $f_r_n to g$ uniformly almost everywhere. Is this sufficient? Yes, since in this case $mu(X)<infty$, so Hölder's inequality implies that $int_X (f_r_n-g) < lVert f_r_n-g rVert_infty mu(X) $, so $f_r_n to g$ uniformly almost everywhere if and only if $int_X f_r_n to int_X g $.
So a necessary and sufficient condition in your case is that there is a sequence $r_n$ so that $f_r_n to sup_r f_r$ uniformly.
answered Jul 14 at 23:41
Chappers
55k74191
1
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It's more common (and equivalent) to work with infima (just put $-$ signs everywhere), so we want to find when
$$ inf_r in A int_X f_r = int_X inf_r in A f_r. $$
Let $inf_r f_r(t) = g(t)$. Then $inf_r g(t) = g(t)$, and the right-hand side of the equality is $int g$, so if we subtract and let $h_r=f_r-g$, equality occurs when
$$ inf_r int_X h_r = 0 $$
But $f_r geq inf_r f_r = g $, so $f_r-g = h_r geq 0$. So the integral is always nonnegative. The infimum is zero if and only if there is a sequence $r_n$ so that $int h_r_n to 0$. In turn, this means that it is possible to extract a subsequence $r_n(k)$ so that $h_r_n(k) to 0$ almost everywhere (see e.g. here) (thanks to zhw. for reminding me that passing to a subsequence is necessary). Indeed, the subsequence can be chosen so that the convergence is uniform almost everywhere.
Therefore equality requires that there is a sequence, which we may call $f_r_n$ so that $f_r_n to g$ uniformly almost everywhere. Is this sufficient? Yes, since in this case $mu(X)<infty$, so Hölder's inequality implies that $int_X (f_r_n-g) < lVert f_r_n-g rVert_infty mu(X) $, so $f_r_n to g$ uniformly almost everywhere if and only if $int_X f_r_n to int_X g $.
So a necessary and sufficient condition in your case is that there is a sequence $r_n$ so that $f_r_n to sup_r f_r$ uniformly.
answered Jul 14 at 23:41
Chappers
55k74191
It's more common (and equivalent) to work with infima (just put $-$ signs everywhere), so we want to find when
$$ inf_r in A int_X f_r = int_X inf_r in A f_r. $$
Let $inf_r f_r(t) = g(t)$. Then $inf_r g(t) = g(t)$, and the right-hand side of the equality is $int g$, so if we subtract and let $h_r=f_r-g$, equality occurs when
$$ inf_r int_X h_r = 0 $$
But $f_r geq inf_r f_r = g $, so $f_r-g = h_r geq 0$. So the integral is always nonnegative. The infimum is zero if and only if there is a sequence $r_n$ so that $int h_r_n to 0$. In turn, this means that it is possible to extract a subsequence $r_n(k)$ so that $h_r_n(k) to 0$ almost everywhere (see e.g. here) (thanks to zhw. for reminding me that passing to a subsequence is necessary). Indeed, the subsequence can be chosen so that the convergence is uniform almost everywhere.
Therefore equality requires that there is a sequence, which we may call $f_r_n$ so that $f_r_n to g$ uniformly almost everywhere. Is this sufficient? Yes, since in this case $mu(X)<infty$, so Hölder's inequality implies that $int_X (f_r_n-g) < lVert f_r_n-g rVert_infty mu(X) $, so $f_r_n to g$ uniformly almost everywhere if and only if $int_X f_r_n to int_X g $.
So a necessary and sufficient condition in your case is that there is a sequence $r_n$ so that $f_r_n to sup_r f_r$ uniformly.
answered Jul 14 at 23:41
Chappers
55k74191
edited Jul 16 at 12:13
answered Jul 14 at 23:41
Chappers
55k74191
answered Jul 14 at 23:41
Chappers
55k74191
answered Jul 14 at 23:41
Chappers
55k74191
55k74191
1
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
add a comment |Â
1
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
1
1
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
Only partly right in one direction, and not right in the other direction. If each $h_n$ is nonnegative and $int h_n to 0,$ then $h_n$ need not $to 0$ pointwise everywhere, but there will be some subsequence $h_n_k$ that does (this is not obvious). And if $h_nto 0$ pointwise a.e., then $int h_n$ need not $to 0,$ nor need any subsequence. In fact $int h_n to infty$ is possible.
– zhw.
Jul 15 at 6:18
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
@zhw. You're quite right, I must have been tireder than I thought when I wrote that! I think it's correct now.
– Chappers
Jul 15 at 20:21
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
Not sure what you mean by "the convergence can be forced to be uniform".
– zhw.
Jul 16 at 2:18
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
The convergence cannot be forced to be uniform in general...
– Sangchul Lee
Jul 16 at 3:28
add a comment |Â
up vote
1
down vote
We have from the property of $sup$:
$$forall rin[0,1):~~~f_r(t)=sup_0leq r<1f_r(t)$$
in your question you suppose that the supremum of integrable functions is integrable, then with integration over $U$
$$forall rin[0,1):~~~int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
this gives
$$sup_0leq r<1int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
because $int_Usup_0leq r<1f_r(t)dt$ is an upper bound for $int_Uf_r(t)dt$ where $0leq r<1$.
answered Jul 15 at 4:34
Nosrati
20k41644
add a comment |Â
up vote
1
down vote
We have from the property of $sup$:
$$forall rin[0,1):~~~f_r(t)=sup_0leq r<1f_r(t)$$
in your question you suppose that the supremum of integrable functions is integrable, then with integration over $U$
$$forall rin[0,1):~~~int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
this gives
$$sup_0leq r<1int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
because $int_Usup_0leq r<1f_r(t)dt$ is an upper bound for $int_Uf_r(t)dt$ where $0leq r<1$.
answered Jul 15 at 4:34
Nosrati
20k41644
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have from the property of $sup$:
$$forall rin[0,1):~~~f_r(t)=sup_0leq r<1f_r(t)$$
in your question you suppose that the supremum of integrable functions is integrable, then with integration over $U$
$$forall rin[0,1):~~~int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
this gives
$$sup_0leq r<1int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
because $int_Usup_0leq r<1f_r(t)dt$ is an upper bound for $int_Uf_r(t)dt$ where $0leq r<1$.
answered Jul 15 at 4:34
Nosrati
20k41644
We have from the property of $sup$:
$$forall rin[0,1):~~~f_r(t)=sup_0leq r<1f_r(t)$$
in your question you suppose that the supremum of integrable functions is integrable, then with integration over $U$
$$forall rin[0,1):~~~int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
this gives
$$sup_0leq r<1int_Uf_r(t)dt=int_Usup_0leq r<1f_r(t)dt$$
because $int_Usup_0leq r<1f_r(t)dt$ is an upper bound for $int_Uf_r(t)dt$ where $0leq r<1$.
answered Jul 15 at 4:34
Nosrati
20k41644
answered Jul 15 at 4:34
Nosrati
20k41644
answered Jul 15 at 4:34
Nosrati
20k41644
answered Jul 15 at 4:34
Nosrati
20k41644
20k41644
add a comment |Â
add a comment |Â
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"Invert" isn't quite the right word. "Interchange" or "exchange" is what you're looking for.
– Sean Roberson
Jul 14 at 22:22