Dominated convergence theorem - what sequence?
$begingroup$
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 :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
integration limits
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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 :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
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 :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
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 :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
integration limits
integration limits
asked 5 hours ago
Ivan V.Ivan V.
931216
931216
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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(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
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
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 $pmb{a_nto 0^+}$ 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
add a comment |
$begingroup$
Let's look at it in a silly 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 sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
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
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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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(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
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
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 $pmb{a_nto 0^+}$ 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(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
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
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 $pmb{a_nto 0^+}$ 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(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
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
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 $pmb{a_nto 0^+}$ 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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(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
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
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 $pmb{a_nto 0^+}$ 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 4 hours ago
answered 4 hours ago
Alex OrtizAlex Ortiz
11.2k21441
11.2k21441
$begingroup$
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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
$begingroup$
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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
$begingroup$
Alright, thank you, much appreciated!
$endgroup$
– Ivan V.
13 mins ago
add a comment |
$begingroup$
Let's look at it in a silly 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 sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
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 silly 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 sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
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 silly 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 sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
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 silly 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 sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
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 $e^{-x}$.
answered 4 hours ago
Saucy O'PathSaucy O'Path
6,2141627
6,2141627
add a comment |
add a comment |
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