Mixing
Convergence of $frac{cos(x)}{nsin(x)}$ an n$rightarrow infty$
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So I have this as my homework in Measure Theory and I don't know how to prove it. The question is the following:
On the domain $Omega = (0, pi)times {rm I!R}$ we have a function defined as:
begin{equation}
phi_{n}(x,y) = frac{cos(x)}{nsin(y)} qquad if,,yin {rm I!R}setminus {kpi | kin {rm I!N} } \ phi_{n}(x,y)=n^2 ,,,,,,qquad if,,yin {kpi | kin {rm I!N} }.
end{equation}
Now I need to prove that the sequence
begin{equation}
{int_{Omega}|phi_{n}(x,y)|dxdy}
end{equation}
is convergent almost everywhere. I know that since ${ kpi | kin {rm I!N}}$ is just a set of disjoint points, its Lebesgue measure is $0$, hence, we don't care about them. I was specifically told that I should use Fatou's lemma which we can apply due to the absolute sign. Thus, since we have sequence of positive functions we have
begin{equation}
int_{Omega}lim inf |phi_{n}| leq lim inf int_{Omega}|phi_{n}|.
end{equation}
Also, we have the inverse inequality for the $lim sup$ as well i.e.
begin{equation}
lim sup int_{Omega}|phi_{n}| leq int_{Omega}lim sup |phi_{n}|.
end{equation}
Then, if the $lim inf$ and $lim sup$ coincide, we have that the sequence is actually converges and we get that
begin{equation}
int_{Omega}lim inf |phi_{n}| = lim int_{Omega} |phi_{n}| = int_{Omega}| lim sup |phi_{n}|.
end{equation}
However, I am stuguling with how to use the properties of $phi_{n}$ to show that $lim inf = lim sup$. Any help would be appriciated.
measure-theory lebesgue-measure
asked 2 days ago
qake4
253
add a comment |
up vote
1
down vote
favorite
So I have this as my homework in Measure Theory and I don't know how to prove it. The question is the following:
On the domain $Omega = (0, pi)times {rm I!R}$ we have a function defined as:
begin{equation}
phi_{n}(x,y) = frac{cos(x)}{nsin(y)} qquad if,,yin {rm I!R}setminus {kpi | kin {rm I!N} } \ phi_{n}(x,y)=n^2 ,,,,,,qquad if,,yin {kpi | kin {rm I!N} }.
end{equation}
Now I need to prove that the sequence
begin{equation}
{int_{Omega}|phi_{n}(x,y)|dxdy}
end{equation}
is convergent almost everywhere. I know that since ${ kpi | kin {rm I!N}}$ is just a set of disjoint points, its Lebesgue measure is $0$, hence, we don't care about them. I was specifically told that I should use Fatou's lemma which we can apply due to the absolute sign. Thus, since we have sequence of positive functions we have
begin{equation}
int_{Omega}lim inf |phi_{n}| leq lim inf int_{Omega}|phi_{n}|.
end{equation}
Also, we have the inverse inequality for the $lim sup$ as well i.e.
begin{equation}
lim sup int_{Omega}|phi_{n}| leq int_{Omega}lim sup |phi_{n}|.
end{equation}
Then, if the $lim inf$ and $lim sup$ coincide, we have that the sequence is actually converges and we get that
begin{equation}
int_{Omega}lim inf |phi_{n}| = lim int_{Omega} |phi_{n}| = int_{Omega}| lim sup |phi_{n}|.
end{equation}
However, I am stuguling with how to use the properties of $phi_{n}$ to show that $lim inf = lim sup$. Any help would be appriciated.
measure-theory lebesgue-measure
asked 2 days ago
qake4
253
Look at the expression for $phi_n(x,y)$. What does it converge to almost everywhere?
– T_M
2 days ago
I know that it converges to 0, but I assumed that I should not know that, and by just using some properties of sin and cos we can prove that limsup and liminf are the same.
– qake4
2 days ago
Why did you assume that? The pointwise a.e. limit doesn't really have anything to do with sin and cos, that's kind of the "trick" with problems like this. At a.e. point $(x,y)$ its just a sequence of the form $const.times 1/n$
– T_M
2 days ago
So then I should just show that liminf = 0 and limsup = 0, if yes then please post it as an answer and when I am done with the proof (probably tomorrow), then I accept as an answer.
– qake4
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So I have this as my homework in Measure Theory and I don't know how to prove it. The question is the following:
On the domain $Omega = (0, pi)times {rm I!R}$ we have a function defined as:
begin{equation}
phi_{n}(x,y) = frac{cos(x)}{nsin(y)} qquad if,,yin {rm I!R}setminus {kpi | kin {rm I!N} } \ phi_{n}(x,y)=n^2 ,,,,,,qquad if,,yin {kpi | kin {rm I!N} }.
end{equation}
Now I need to prove that the sequence
begin{equation}
{int_{Omega}|phi_{n}(x,y)|dxdy}
end{equation}
is convergent almost everywhere. I know that since ${ kpi | kin {rm I!N}}$ is just a set of disjoint points, its Lebesgue measure is $0$, hence, we don't care about them. I was specifically told that I should use Fatou's lemma which we can apply due to the absolute sign. Thus, since we have sequence of positive functions we have
begin{equation}
int_{Omega}lim inf |phi_{n}| leq lim inf int_{Omega}|phi_{n}|.
end{equation}
Also, we have the inverse inequality for the $lim sup$ as well i.e.
begin{equation}
lim sup int_{Omega}|phi_{n}| leq int_{Omega}lim sup |phi_{n}|.
end{equation}
Then, if the $lim inf$ and $lim sup$ coincide, we have that the sequence is actually converges and we get that
begin{equation}
int_{Omega}lim inf |phi_{n}| = lim int_{Omega} |phi_{n}| = int_{Omega}| lim sup |phi_{n}|.
end{equation}
However, I am stuguling with how to use the properties of $phi_{n}$ to show that $lim inf = lim sup$. Any help would be appriciated.
measure-theory lebesgue-measure
asked 2 days ago
qake4
253
So I have this as my homework in Measure Theory and I don't know how to prove it. The question is the following:
On the domain $Omega = (0, pi)times {rm I!R}$ we have a function defined as:
begin{equation}
phi_{n}(x,y) = frac{cos(x)}{nsin(y)} qquad if,,yin {rm I!R}setminus {kpi | kin {rm I!N} } \ phi_{n}(x,y)=n^2 ,,,,,,qquad if,,yin {kpi | kin {rm I!N} }.
end{equation}
Now I need to prove that the sequence
begin{equation}
{int_{Omega}|phi_{n}(x,y)|dxdy}
end{equation}
is convergent almost everywhere. I know that since ${ kpi | kin {rm I!N}}$ is just a set of disjoint points, its Lebesgue measure is $0$, hence, we don't care about them. I was specifically told that I should use Fatou's lemma which we can apply due to the absolute sign. Thus, since we have sequence of positive functions we have
begin{equation}
int_{Omega}lim inf |phi_{n}| leq lim inf int_{Omega}|phi_{n}|.
end{equation}
Also, we have the inverse inequality for the $lim sup$ as well i.e.
begin{equation}
lim sup int_{Omega}|phi_{n}| leq int_{Omega}lim sup |phi_{n}|.
end{equation}
Then, if the $lim inf$ and $lim sup$ coincide, we have that the sequence is actually converges and we get that
begin{equation}
int_{Omega}lim inf |phi_{n}| = lim int_{Omega} |phi_{n}| = int_{Omega}| lim sup |phi_{n}|.
end{equation}
However, I am stuguling with how to use the properties of $phi_{n}$ to show that $lim inf = lim sup$. Any help would be appriciated.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked 2 days ago
qake4
253
asked 2 days ago
qake4
253
asked 2 days ago
qake4
253
asked 2 days ago
qake4
253
asked 2 days ago
qake4
253
253
Look at the expression for $phi_n(x,y)$. What does it converge to almost everywhere?
– T_M
2 days ago
I know that it converges to 0, but I assumed that I should not know that, and by just using some properties of sin and cos we can prove that limsup and liminf are the same.
– qake4
2 days ago
Why did you assume that? The pointwise a.e. limit doesn't really have anything to do with sin and cos, that's kind of the "trick" with problems like this. At a.e. point $(x,y)$ its just a sequence of the form $const.times 1/n$
– T_M
2 days ago
So then I should just show that liminf = 0 and limsup = 0, if yes then please post it as an answer and when I am done with the proof (probably tomorrow), then I accept as an answer.
– qake4
2 days ago
add a comment |
Look at the expression for $phi_n(x,y)$. What does it converge to almost everywhere?
– T_M
2 days ago
I know that it converges to 0, but I assumed that I should not know that, and by just using some properties of sin and cos we can prove that limsup and liminf are the same.
– qake4
2 days ago
Why did you assume that? The pointwise a.e. limit doesn't really have anything to do with sin and cos, that's kind of the "trick" with problems like this. At a.e. point $(x,y)$ its just a sequence of the form $const.times 1/n$
– T_M
2 days ago
So then I should just show that liminf = 0 and limsup = 0, if yes then please post it as an answer and when I am done with the proof (probably tomorrow), then I accept as an answer.
– qake4
2 days ago
Look at the expression for $phi_n(x,y)$. What does it converge to almost everywhere?
– T_M
2 days ago
Look at the expression for $phi_n(x,y)$. What does it converge to almost everywhere?
– T_M
2 days ago
I know that it converges to 0, but I assumed that I should not know that, and by just using some properties of sin and cos we can prove that limsup and liminf are the same.
– qake4
2 days ago
I know that it converges to 0, but I assumed that I should not know that, and by just using some properties of sin and cos we can prove that limsup and liminf are the same.
– qake4
2 days ago
Why did you assume that? The pointwise a.e. limit doesn't really have anything to do with sin and cos, that's kind of the "trick" with problems like this. At a.e. point $(x,y)$ its just a sequence of the form $const.times 1/n$
– T_M
2 days ago
Why did you assume that? The pointwise a.e. limit doesn't really have anything to do with sin and cos, that's kind of the "trick" with problems like this. At a.e. point $(x,y)$ its just a sequence of the form $const.times 1/n$
– T_M
2 days ago
So then I should just show that liminf = 0 and limsup = 0, if yes then please post it as an answer and when I am done with the proof (probably tomorrow), then I accept as an answer.
– qake4
2 days ago
So then I should just show that liminf = 0 and limsup = 0, if yes then please post it as an answer and when I am done with the proof (probably tomorrow), then I accept as an answer.
– qake4
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Let $Y={kpi : kinmathbb N}$ then $|Y|=0$ so by Fubini theorem
$$
int_{Omega} |phi_n(x, y)|dxdy = frac 1nint_{Omegasetminus Y}leftlvertfrac{cos x}{sin y}rightrvert dxdy=frac 1nint^pi_0lvertcos xrvert dxint_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=
$$
Now
$$
int_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{(i+1)pi}_{ipi}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{pi}_{0}leftlvertfrac{1}{sin y}rightrvert dy=+infty
$$
and doesn't converge
answered 2 days ago
P De Donato
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $Y={kpi : kinmathbb N}$ then $|Y|=0$ so by Fubini theorem
$$
int_{Omega} |phi_n(x, y)|dxdy = frac 1nint_{Omegasetminus Y}leftlvertfrac{cos x}{sin y}rightrvert dxdy=frac 1nint^pi_0lvertcos xrvert dxint_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=
$$
Now
$$
int_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{(i+1)pi}_{ipi}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{pi}_{0}leftlvertfrac{1}{sin y}rightrvert dy=+infty
$$
and doesn't converge
answered 2 days ago
P De Donato
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
add a comment |
up vote
1
down vote
Let $Y={kpi : kinmathbb N}$ then $|Y|=0$ so by Fubini theorem
$$
int_{Omega} |phi_n(x, y)|dxdy = frac 1nint_{Omegasetminus Y}leftlvertfrac{cos x}{sin y}rightrvert dxdy=frac 1nint^pi_0lvertcos xrvert dxint_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=
$$
Now
$$
int_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{(i+1)pi}_{ipi}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{pi}_{0}leftlvertfrac{1}{sin y}rightrvert dy=+infty
$$
and doesn't converge
answered 2 days ago
P De Donato
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $Y={kpi : kinmathbb N}$ then $|Y|=0$ so by Fubini theorem
$$
int_{Omega} |phi_n(x, y)|dxdy = frac 1nint_{Omegasetminus Y}leftlvertfrac{cos x}{sin y}rightrvert dxdy=frac 1nint^pi_0lvertcos xrvert dxint_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=
$$
Now
$$
int_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{(i+1)pi}_{ipi}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{pi}_{0}leftlvertfrac{1}{sin y}rightrvert dy=+infty
$$
and doesn't converge
answered 2 days ago
P De Donato
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $Y={kpi : kinmathbb N}$ then $|Y|=0$ so by Fubini theorem
$$
int_{Omega} |phi_n(x, y)|dxdy = frac 1nint_{Omegasetminus Y}leftlvertfrac{cos x}{sin y}rightrvert dxdy=frac 1nint^pi_0lvertcos xrvert dxint_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=
$$
Now
$$
int_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{(i+1)pi}_{ipi}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{pi}_{0}leftlvertfrac{1}{sin y}rightrvert dy=+infty
$$
and doesn't converge
answered 2 days ago
P De Donato
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
P De Donato
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
P De Donato
2136
answered 2 days ago
P De Donato
2136
2136
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
add a comment |
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
That is true, $int |1/sin(y)|$ $rightarrow infty$ but as was pointed out by @T_M, we can consider $int |1/sin(y)|$ as a constant for all $y in Rsetminus Y$ which is less than $infty$ and as n $rightarrow infty$ we indeed get that $int |phi_{n}(x,y)|$ converges to 0, don't we?
– qake4
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
The integral doesn't converge for all $n$, if $frac{1}{lvertsin yrvert}$ is integrated on a set smaller than $mathbb R$ (like compact set) the integral may be considered as a finite constant
– P De Donato
2 days ago
add a comment |
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Look at the expression for $phi_n(x,y)$. What does it converge to almost everywhere?
– T_M
2 days ago
I know that it converges to 0, but I assumed that I should not know that, and by just using some properties of sin and cos we can prove that limsup and liminf are the same.
– qake4
2 days ago
Why did you assume that? The pointwise a.e. limit doesn't really have anything to do with sin and cos, that's kind of the "trick" with problems like this. At a.e. point $(x,y)$ its just a sequence of the form $const.times 1/n$
– T_M
2 days ago
So then I should just show that liminf = 0 and limsup = 0, if yes then please post it as an answer and when I am done with the proof (probably tomorrow), then I accept as an answer.
– qake4
2 days ago