Dominated convergence theorem - what sequence? The 2019 Stack Overflow Developer Survey Results Are InLebesgue Convergence TheoremWhat are some good integration problems where you can use some of the function convergence theorem of Lesbegue integrals?Find Limit Using Lebesgue Dominated ConvergenceSolving these types of integrals, using Monotone convergence theorem and Dominated convergence theorem.Dominated Convergence Theorem - variable in set over which we integrateConvergence of an uncountable sequence of functionsGeneralization of the dominated convergence theoremuniform or dominated convergence of sequence of functions which are boundedCalculate the limit using dominated or monotone convergence theoremArzelà's Dominated Convergence Theorem for improper integrals?
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Dominated convergence theorem - what sequence?
The 2019 Stack Overflow Developer Survey Results Are InLebesgue Convergence TheoremWhat are some good integration problems where you can use some of the function convergence theorem of Lesbegue integrals?Find Limit Using Lebesgue Dominated ConvergenceSolving these types of integrals, using Monotone convergence theorem and Dominated convergence theorem.Dominated Convergence Theorem - variable in set over which we integrateConvergence of an uncountable sequence of functionsGeneralization of the dominated convergence theoremuniform or dominated convergence of sequence of functions which are boundedCalculate the limit using dominated or monotone convergence theoremArzelà's Dominated Convergence Theorem for improper integrals?
$begingroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
integration limits
$endgroup$
add a comment |
$begingroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
integration limits
$endgroup$
add a comment |
$begingroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
integration limits
$endgroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
integration limits
integration limits
edited Mar 31 at 14:44
Ivan V.
asked Mar 30 at 21:07
Ivan V.Ivan V.
981216
981216
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
$$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.
$endgroup$
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
add a comment |
$begingroup$
Let's look at it in a sample case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$
This is the case if and only if for all sequences $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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oldest
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$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
$$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.
$endgroup$
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
add a comment |
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
$$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.
$endgroup$
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
add a comment |
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
$$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.
$endgroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
$$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.
edited Mar 30 at 21:35
answered Mar 30 at 21:29
Alex OrtizAlex Ortiz
11.4k21442
11.4k21442
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
add a comment |
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
Mar 30 at 23:45
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
Mar 31 at 0:23
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
Mar 31 at 2:00
add a comment |
$begingroup$
Let's look at it in a sample case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$
This is the case if and only if for all sequences $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.
$endgroup$
add a comment |
$begingroup$
Let's look at it in a sample case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$
This is the case if and only if for all sequences $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.
$endgroup$
add a comment |
$begingroup$
Let's look at it in a sample case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$
This is the case if and only if for all sequences $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.
$endgroup$
Let's look at it in a sample case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$
This is the case if and only if for all sequences $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.
edited Mar 31 at 6:49
answered Mar 30 at 21:18
Saucy O'PathSaucy O'Path
6,5251627
6,5251627
add a comment |
add a comment |
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