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Using dominated convergence theorem to asset Leibniz rule when $partialf/partialx(x,y)$ is bounded and $f$ is integrable
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I was thinking about the following problem
Let $f(x,y)$, $0 leq x, y leq 1$, be a function such that for each $x$, $f(x,y)$ is integrable, and $partialf/partialx(x,y)$ is bounded function of $(x,y)$. Show that $partialf/partialx(x,y)$ is a measurable function of $y$ for each $x$ and
$$ fracpartialpartialxint^1_0 f(x,y)dy = int^1_0 fracpartialfpartialx(x,y)dy
$$
Assume that $partialf/partialx(x,y)$ exists everywhere for now (I'm not sure if we can assume this):
I think the problem is asking us to justify using Leibniz rule by applying dominated convergence theorem to the sequence of functions
$$phi_h(x,y) := fracf(x+h,y)-f(x,y)h.$$
To do this, we need to show that the sequence is dominated by an integrable function, by the mean value theorem, there is an $x_0 in (x, x+h)$ where
$phi_h(x,y)= fracpartialfpartialx(x_0,y)$
But I am a bit stuck relating $fracpartialfpartialx(x_0,y)$ to the integrable function $f$. I think I am missing some estimate.
real-analysis measure-theory
asked Aug 6 at 21:53
user135520
909718
add a comment |Â
up vote
0
down vote
favorite
I was thinking about the following problem
Let $f(x,y)$, $0 leq x, y leq 1$, be a function such that for each $x$, $f(x,y)$ is integrable, and $partialf/partialx(x,y)$ is bounded function of $(x,y)$. Show that $partialf/partialx(x,y)$ is a measurable function of $y$ for each $x$ and
$$ fracpartialpartialxint^1_0 f(x,y)dy = int^1_0 fracpartialfpartialx(x,y)dy
$$
Assume that $partialf/partialx(x,y)$ exists everywhere for now (I'm not sure if we can assume this):
I think the problem is asking us to justify using Leibniz rule by applying dominated convergence theorem to the sequence of functions
$$phi_h(x,y) := fracf(x+h,y)-f(x,y)h.$$
To do this, we need to show that the sequence is dominated by an integrable function, by the mean value theorem, there is an $x_0 in (x, x+h)$ where
$phi_h(x,y)= fracpartialfpartialx(x_0,y)$
But I am a bit stuck relating $fracpartialfpartialx(x_0,y)$ to the integrable function $f$. I think I am missing some estimate.
real-analysis measure-theory
asked Aug 6 at 21:53
user135520
909718
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was thinking about the following problem
Let $f(x,y)$, $0 leq x, y leq 1$, be a function such that for each $x$, $f(x,y)$ is integrable, and $partialf/partialx(x,y)$ is bounded function of $(x,y)$. Show that $partialf/partialx(x,y)$ is a measurable function of $y$ for each $x$ and
$$ fracpartialpartialxint^1_0 f(x,y)dy = int^1_0 fracpartialfpartialx(x,y)dy
$$
Assume that $partialf/partialx(x,y)$ exists everywhere for now (I'm not sure if we can assume this):
I think the problem is asking us to justify using Leibniz rule by applying dominated convergence theorem to the sequence of functions
$$phi_h(x,y) := fracf(x+h,y)-f(x,y)h.$$
To do this, we need to show that the sequence is dominated by an integrable function, by the mean value theorem, there is an $x_0 in (x, x+h)$ where
$phi_h(x,y)= fracpartialfpartialx(x_0,y)$
But I am a bit stuck relating $fracpartialfpartialx(x_0,y)$ to the integrable function $f$. I think I am missing some estimate.
real-analysis measure-theory
asked Aug 6 at 21:53
user135520
909718
I was thinking about the following problem
Let $f(x,y)$, $0 leq x, y leq 1$, be a function such that for each $x$, $f(x,y)$ is integrable, and $partialf/partialx(x,y)$ is bounded function of $(x,y)$. Show that $partialf/partialx(x,y)$ is a measurable function of $y$ for each $x$ and
$$ fracpartialpartialxint^1_0 f(x,y)dy = int^1_0 fracpartialfpartialx(x,y)dy
$$
Assume that $partialf/partialx(x,y)$ exists everywhere for now (I'm not sure if we can assume this):
I think the problem is asking us to justify using Leibniz rule by applying dominated convergence theorem to the sequence of functions
$$phi_h(x,y) := fracf(x+h,y)-f(x,y)h.$$
To do this, we need to show that the sequence is dominated by an integrable function, by the mean value theorem, there is an $x_0 in (x, x+h)$ where
$phi_h(x,y)= fracpartialfpartialx(x_0,y)$
But I am a bit stuck relating $fracpartialfpartialx(x_0,y)$ to the integrable function $f$. I think I am missing some estimate.
real-analysis measure-theory
asked Aug 6 at 21:53
user135520
909718
asked Aug 6 at 21:53
user135520
909718
asked Aug 6 at 21:53
user135520
909718
asked Aug 6 at 21:53
user135520
909718
909718
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
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You're overthinking this. Since $partialf/partialx(x,y)$ is bounded on all of $[0,1]times[0,1]$, just let $M$ be a constant bound for it and apply the dominated convergence theorem with $M$ as your bound.
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
add a comment |Â
up vote
2
down vote
$leftlvertfracpartial fpartial x(xi_h,y)rightrvertle suplimits_textdomainleftlvertfracpartial fpartial xrightrvert=M$. Since constant functions are integrable on the finite interval $(0,1)$, by DCT you can assert that for all $x$ $$lim_hto 0int_0^1phi_h(x,y),dy=int_0^1lim_hto 0phi_h(x,y),dy$$
answered Aug 6 at 21:59
Saucy O'Path
2,821220
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're overthinking this. Since $partialf/partialx(x,y)$ is bounded on all of $[0,1]times[0,1]$, just let $M$ be a constant bound for it and apply the dominated convergence theorem with $M$ as your bound.
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
add a comment |Â
up vote
2
down vote
accepted
You're overthinking this. Since $partialf/partialx(x,y)$ is bounded on all of $[0,1]times[0,1]$, just let $M$ be a constant bound for it and apply the dominated convergence theorem with $M$ as your bound.
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're overthinking this. Since $partialf/partialx(x,y)$ is bounded on all of $[0,1]times[0,1]$, just let $M$ be a constant bound for it and apply the dominated convergence theorem with $M$ as your bound.
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
You're overthinking this. Since $partialf/partialx(x,y)$ is bounded on all of $[0,1]times[0,1]$, just let $M$ be a constant bound for it and apply the dominated convergence theorem with $M$ as your bound.
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
answered Aug 6 at 22:00
Eric Wofsey
163k12190301
163k12190301
add a comment |Â
add a comment |Â
up vote
2
down vote
$leftlvertfracpartial fpartial x(xi_h,y)rightrvertle suplimits_textdomainleftlvertfracpartial fpartial xrightrvert=M$. Since constant functions are integrable on the finite interval $(0,1)$, by DCT you can assert that for all $x$ $$lim_hto 0int_0^1phi_h(x,y),dy=int_0^1lim_hto 0phi_h(x,y),dy$$
answered Aug 6 at 21:59
Saucy O'Path
2,821220
add a comment |Â
up vote
2
down vote
$leftlvertfracpartial fpartial x(xi_h,y)rightrvertle suplimits_textdomainleftlvertfracpartial fpartial xrightrvert=M$. Since constant functions are integrable on the finite interval $(0,1)$, by DCT you can assert that for all $x$ $$lim_hto 0int_0^1phi_h(x,y),dy=int_0^1lim_hto 0phi_h(x,y),dy$$
answered Aug 6 at 21:59
Saucy O'Path
2,821220
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$leftlvertfracpartial fpartial x(xi_h,y)rightrvertle suplimits_textdomainleftlvertfracpartial fpartial xrightrvert=M$. Since constant functions are integrable on the finite interval $(0,1)$, by DCT you can assert that for all $x$ $$lim_hto 0int_0^1phi_h(x,y),dy=int_0^1lim_hto 0phi_h(x,y),dy$$
answered Aug 6 at 21:59
Saucy O'Path
2,821220
$leftlvertfracpartial fpartial x(xi_h,y)rightrvertle suplimits_textdomainleftlvertfracpartial fpartial xrightrvert=M$. Since constant functions are integrable on the finite interval $(0,1)$, by DCT you can assert that for all $x$ $$lim_hto 0int_0^1phi_h(x,y),dy=int_0^1lim_hto 0phi_h(x,y),dy$$
answered Aug 6 at 21:59
Saucy O'Path
2,821220
answered Aug 6 at 21:59
Saucy O'Path
2,821220
answered Aug 6 at 21:59
Saucy O'Path
2,821220
answered Aug 6 at 21:59
Saucy O'Path
2,821220
2,821220
add a comment |Â
add a comment |Â
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