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If $f:mathbb R^2longrightarrow mathbb R$ continuous, does $lim_hto 0int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx$?
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If $f:mathbb R^2to mathbb R$ continuous, s.t. $int_mathbb Rf(t,x)dx$ exist for all $t$. Does $$lim_hto 0int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
I asked here the question for $$lim_hto 0int_a^b f(t+h,x)dx=int_a^b f(t,x)dx,$$
where $a,binmathbb R$ and as @Surb show, it's indeed true. But what happen when $a=-infty $ and $b=infty $ ? I think it's enough for $a=0$ and $b=infty $, i.e. does
$$lim_hto 0 int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
(we suppose of course that $int_0^infty f(t,x)dx$ exist for all $t$).
My idea was to show that $$F(v,t):=int_0^v f(t,x)dx,$$
is continuous on $[0,infty )times mathbb R$,
and thus
$$beginalignlim_hto 0int_0^infty f(t,x)dx=&lim_hto 0lim_vto infty int_0^v f(t+h,x)dx\
&=lim_vto infty lim_hto 0int_0^v f(t+h,x)dx\
&=lim_vto infty int_0^v f(t,x)dx\
&=int_0^infty f(t,x)dxendalign$$
but unfortunately, I didn't success to show that it's continuous. May be it's not correct. And in this case, do you have a counter example ?
real-analysis integration continuity
edited Jul 17 at 10:21
mechanodroid
22.2k52041
asked Jul 17 at 8:55
Peter
358112
 |Â
show 1 more comment
up vote
3
down vote
favorite
If $f:mathbb R^2to mathbb R$ continuous, s.t. $int_mathbb Rf(t,x)dx$ exist for all $t$. Does $$lim_hto 0int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
I asked here the question for $$lim_hto 0int_a^b f(t+h,x)dx=int_a^b f(t,x)dx,$$
where $a,binmathbb R$ and as @Surb show, it's indeed true. But what happen when $a=-infty $ and $b=infty $ ? I think it's enough for $a=0$ and $b=infty $, i.e. does
$$lim_hto 0 int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
(we suppose of course that $int_0^infty f(t,x)dx$ exist for all $t$).
My idea was to show that $$F(v,t):=int_0^v f(t,x)dx,$$
is continuous on $[0,infty )times mathbb R$,
and thus
$$beginalignlim_hto 0int_0^infty f(t,x)dx=&lim_hto 0lim_vto infty int_0^v f(t+h,x)dx\
&=lim_vto infty lim_hto 0int_0^v f(t+h,x)dx\
&=lim_vto infty int_0^v f(t,x)dx\
&=int_0^infty f(t,x)dxendalign$$
but unfortunately, I didn't success to show that it's continuous. May be it's not correct. And in this case, do you have a counter example ?
real-analysis integration continuity
edited Jul 17 at 10:21
mechanodroid
22.2k52041
asked Jul 17 at 8:55
Peter
358112
You say $f$ is defined on $mathbbR$, but its argument has two variables.
– Lorenzo Quarisa
Jul 17 at 8:57
What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form.
– Kavi Rama Murthy
Jul 17 at 8:59
@LorenzoQuarisa: I corrected it.
– Peter
Jul 17 at 8:59
@KaviRamaMurthy: Yes, I add that $int_mathbb Rf(x,t)dx$ exist for all $t$.
– Peter
Jul 17 at 9:00
Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)longmapsto begincasesfracx^3y^2x^2+y^2&(x,y)neq 0\ 0&(x,y)=0endcases$$ is continuous on $mathbb R^2$ but $ lim_xto infty lim_yto 0g(x,y)=0neq infty =lim_yto 0lim_xto infty g(x,y)$.
– Surb
Jul 17 at 9:04
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $f:mathbb R^2to mathbb R$ continuous, s.t. $int_mathbb Rf(t,x)dx$ exist for all $t$. Does $$lim_hto 0int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
I asked here the question for $$lim_hto 0int_a^b f(t+h,x)dx=int_a^b f(t,x)dx,$$
where $a,binmathbb R$ and as @Surb show, it's indeed true. But what happen when $a=-infty $ and $b=infty $ ? I think it's enough for $a=0$ and $b=infty $, i.e. does
$$lim_hto 0 int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
(we suppose of course that $int_0^infty f(t,x)dx$ exist for all $t$).
My idea was to show that $$F(v,t):=int_0^v f(t,x)dx,$$
is continuous on $[0,infty )times mathbb R$,
and thus
$$beginalignlim_hto 0int_0^infty f(t,x)dx=&lim_hto 0lim_vto infty int_0^v f(t+h,x)dx\
&=lim_vto infty lim_hto 0int_0^v f(t+h,x)dx\
&=lim_vto infty int_0^v f(t,x)dx\
&=int_0^infty f(t,x)dxendalign$$
but unfortunately, I didn't success to show that it's continuous. May be it's not correct. And in this case, do you have a counter example ?
real-analysis integration continuity
edited Jul 17 at 10:21
mechanodroid
22.2k52041
asked Jul 17 at 8:55
Peter
358112
If $f:mathbb R^2to mathbb R$ continuous, s.t. $int_mathbb Rf(t,x)dx$ exist for all $t$. Does $$lim_hto 0int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
I asked here the question for $$lim_hto 0int_a^b f(t+h,x)dx=int_a^b f(t,x)dx,$$
where $a,binmathbb R$ and as @Surb show, it's indeed true. But what happen when $a=-infty $ and $b=infty $ ? I think it's enough for $a=0$ and $b=infty $, i.e. does
$$lim_hto 0 int_0^infty f(t+h,x)dx=int_0^infty f(t,x)dx ?$$
(we suppose of course that $int_0^infty f(t,x)dx$ exist for all $t$).
My idea was to show that $$F(v,t):=int_0^v f(t,x)dx,$$
is continuous on $[0,infty )times mathbb R$,
and thus
$$beginalignlim_hto 0int_0^infty f(t,x)dx=&lim_hto 0lim_vto infty int_0^v f(t+h,x)dx\
&=lim_vto infty lim_hto 0int_0^v f(t+h,x)dx\
&=lim_vto infty int_0^v f(t,x)dx\
&=int_0^infty f(t,x)dxendalign$$
but unfortunately, I didn't success to show that it's continuous. May be it's not correct. And in this case, do you have a counter example ?
real-analysis integration continuity
edited Jul 17 at 10:21
mechanodroid
22.2k52041
asked Jul 17 at 8:55
Peter
358112
edited Jul 17 at 10:21
mechanodroid
22.2k52041
edited Jul 17 at 10:21
mechanodroid
22.2k52041
edited Jul 17 at 10:21
mechanodroid
22.2k52041
22.2k52041
asked Jul 17 at 8:55
Peter
358112
asked Jul 17 at 8:55
Peter
358112
asked Jul 17 at 8:55
Peter
358112
358112
You say $f$ is defined on $mathbbR$, but its argument has two variables.
– Lorenzo Quarisa
Jul 17 at 8:57
What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form.
– Kavi Rama Murthy
Jul 17 at 8:59
@LorenzoQuarisa: I corrected it.
– Peter
Jul 17 at 8:59
@KaviRamaMurthy: Yes, I add that $int_mathbb Rf(x,t)dx$ exist for all $t$.
– Peter
Jul 17 at 9:00
Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)longmapsto begincasesfracx^3y^2x^2+y^2&(x,y)neq 0\ 0&(x,y)=0endcases$$ is continuous on $mathbb R^2$ but $ lim_xto infty lim_yto 0g(x,y)=0neq infty =lim_yto 0lim_xto infty g(x,y)$.
– Surb
Jul 17 at 9:04
 |Â
show 1 more comment
You say $f$ is defined on $mathbbR$, but its argument has two variables.
– Lorenzo Quarisa
Jul 17 at 8:57
What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form.
– Kavi Rama Murthy
Jul 17 at 8:59
@LorenzoQuarisa: I corrected it.
– Peter
Jul 17 at 8:59
@KaviRamaMurthy: Yes, I add that $int_mathbb Rf(x,t)dx$ exist for all $t$.
– Peter
Jul 17 at 9:00
Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)longmapsto begincasesfracx^3y^2x^2+y^2&(x,y)neq 0\ 0&(x,y)=0endcases$$ is continuous on $mathbb R^2$ but $ lim_xto infty lim_yto 0g(x,y)=0neq infty =lim_yto 0lim_xto infty g(x,y)$.
– Surb
Jul 17 at 9:04
You say $f$ is defined on $mathbbR$, but its argument has two variables.
– Lorenzo Quarisa
Jul 17 at 8:57
You say $f$ is defined on $mathbbR$, but its argument has two variables.
– Lorenzo Quarisa
Jul 17 at 8:57
What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form.
– Kavi Rama Murthy
Jul 17 at 8:59
What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form.
– Kavi Rama Murthy
Jul 17 at 8:59
@LorenzoQuarisa: I corrected it.
– Peter
Jul 17 at 8:59
@LorenzoQuarisa: I corrected it.
– Peter
Jul 17 at 8:59
@KaviRamaMurthy: Yes, I add that $int_mathbb Rf(x,t)dx$ exist for all $t$.
– Peter
Jul 17 at 9:00
@KaviRamaMurthy: Yes, I add that $int_mathbb Rf(x,t)dx$ exist for all $t$.
– Peter
Jul 17 at 9:00
Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)longmapsto begincasesfracx^3y^2x^2+y^2&(x,y)neq 0\ 0&(x,y)=0endcases$$ is continuous on $mathbb R^2$ but $ lim_xto infty lim_yto 0g(x,y)=0neq infty =lim_yto 0lim_xto infty g(x,y)$.
– Surb
Jul 17 at 9:04
Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)longmapsto begincasesfracx^3y^2x^2+y^2&(x,y)neq 0\ 0&(x,y)=0endcases$$ is continuous on $mathbb R^2$ but $ lim_xto infty lim_yto 0g(x,y)=0neq infty =lim_yto 0lim_xto infty g(x,y)$.
– Surb
Jul 17 at 9:04
 |Â
show 1 more comment
2 Answers
2
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Consider $f : mathbbR^2 to mathbbR$ given by $f(t,x) = t^2e^-t^2x$.
We have $$int_0^infty f(0,x),dx = int_0^infty 0,dx = 0$$
but
$$lim_hto0 int_0^infty f(h,x),dx = lim_hto0 int_0^infty h^2e^-h^2x,dx = lim_hto 0 left[-e^-h^2xright]_0^infty = 1$$
answered Jul 17 at 10:18
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.
You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have
$$
int_0^infty f(t, x),mathrm dx = 1,
$$
so that
$$
lim_h to 0int_0^infty f(h, x),mathrm dx = 1 neq 0 = int_0^infty f(0, x) ,mathrm dx.
$$
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
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
Consider $f : mathbbR^2 to mathbbR$ given by $f(t,x) = t^2e^-t^2x$.
We have $$int_0^infty f(0,x),dx = int_0^infty 0,dx = 0$$
but
$$lim_hto0 int_0^infty f(h,x),dx = lim_hto0 int_0^infty h^2e^-h^2x,dx = lim_hto 0 left[-e^-h^2xright]_0^infty = 1$$
answered Jul 17 at 10:18
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
accepted
Consider $f : mathbbR^2 to mathbbR$ given by $f(t,x) = t^2e^-t^2x$.
We have $$int_0^infty f(0,x),dx = int_0^infty 0,dx = 0$$
but
$$lim_hto0 int_0^infty f(h,x),dx = lim_hto0 int_0^infty h^2e^-h^2x,dx = lim_hto 0 left[-e^-h^2xright]_0^infty = 1$$
answered Jul 17 at 10:18
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Consider $f : mathbbR^2 to mathbbR$ given by $f(t,x) = t^2e^-t^2x$.
We have $$int_0^infty f(0,x),dx = int_0^infty 0,dx = 0$$
but
$$lim_hto0 int_0^infty f(h,x),dx = lim_hto0 int_0^infty h^2e^-h^2x,dx = lim_hto 0 left[-e^-h^2xright]_0^infty = 1$$
answered Jul 17 at 10:18
mechanodroid
22.2k52041
Consider $f : mathbbR^2 to mathbbR$ given by $f(t,x) = t^2e^-t^2x$.
We have $$int_0^infty f(0,x),dx = int_0^infty 0,dx = 0$$
but
$$lim_hto0 int_0^infty f(h,x),dx = lim_hto0 int_0^infty h^2e^-h^2x,dx = lim_hto 0 left[-e^-h^2xright]_0^infty = 1$$
answered Jul 17 at 10:18
mechanodroid
22.2k52041
answered Jul 17 at 10:18
mechanodroid
22.2k52041
answered Jul 17 at 10:18
mechanodroid
22.2k52041
answered Jul 17 at 10:18
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
up vote
3
down vote
This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.
You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have
$$
int_0^infty f(t, x),mathrm dx = 1,
$$
so that
$$
lim_h to 0int_0^infty f(h, x),mathrm dx = 1 neq 0 = int_0^infty f(0, x) ,mathrm dx.
$$
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
add a comment |Â
up vote
3
down vote
This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.
You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have
$$
int_0^infty f(t, x),mathrm dx = 1,
$$
so that
$$
lim_h to 0int_0^infty f(h, x),mathrm dx = 1 neq 0 = int_0^infty f(0, x) ,mathrm dx.
$$
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.
You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have
$$
int_0^infty f(t, x),mathrm dx = 1,
$$
so that
$$
lim_h to 0int_0^infty f(h, x),mathrm dx = 1 neq 0 = int_0^infty f(0, x) ,mathrm dx.
$$
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.
You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have
$$
int_0^infty f(t, x),mathrm dx = 1,
$$
so that
$$
lim_h to 0int_0^infty f(h, x),mathrm dx = 1 neq 0 = int_0^infty f(0, x) ,mathrm dx.
$$
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
answered Jul 17 at 9:28
Mees de Vries
13.7k12345
13.7k12345
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
add a comment |Â
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible).
– Peter
Jul 17 at 9:36
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately.
– Mees de Vries
Jul 17 at 9:39
add a comment |Â
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You say $f$ is defined on $mathbbR$, but its argument has two variables.
– Lorenzo Quarisa
Jul 17 at 8:57
What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form.
– Kavi Rama Murthy
Jul 17 at 8:59
@LorenzoQuarisa: I corrected it.
– Peter
Jul 17 at 8:59
@KaviRamaMurthy: Yes, I add that $int_mathbb Rf(x,t)dx$ exist for all $t$.
– Peter
Jul 17 at 9:00
Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)longmapsto begincasesfracx^3y^2x^2+y^2&(x,y)neq 0\ 0&(x,y)=0endcases$$ is continuous on $mathbb R^2$ but $ lim_xto infty lim_yto 0g(x,y)=0neq infty =lim_yto 0lim_xto infty g(x,y)$.
– Surb
Jul 17 at 9:04