Mixing
Show that $cos(2pi nx) nrightarrow 0$ almost everywhere on $[0, 1]$
$begingroup$
I am stuck on trying to show that the sequence ${cos(2pi n x)}$ does not converge to $0$ almost everywhere on $[0, 1]$.
I have already shown that ${cos(2pi n x)}$ is an orthonormal sequence in $L^2([0,1])$ equipped with Lebesgue measure and hence converges to $0$ weakly in $L^2$.
I also have shown that $int_0^1 cos^2(2pi nx) = frac{1}{2}$ for each $n in mathbb{N}$.
Trying to utilize the above fact, I came up with something which is probably wrong:
By way of contradiction, suppose it is true. Then, $ space f_n(x) rightarrow 0$ for all $x in E subset [0,1] $ such that $E^c$ has measure $0$. Then there is an $N in mathbb{N}$ such that $|cos(2pi nx)| < frac{1}{2}$ whenever $n geq N$ and $x in E$. This implies that $cos^2(2pi nx) < frac{1}{4}$ for $n geq N$. But then, by properties of the Lebesgue integral, we must have:
$$int_E cos^2(2pi nx) = int_0^1 cos^2(2pi nx) < frac{1}{4}m(E) = frac{1}{4} $$
which is a contradiction.
If anyone could tell me where my proof goes wrong and point me in the direction of a correct solution, I would be highly appreciative.
real-analysis sequences-and-series
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
$endgroup$
add a comment |
$begingroup$
I am stuck on trying to show that the sequence ${cos(2pi n x)}$ does not converge to $0$ almost everywhere on $[0, 1]$.
I have already shown that ${cos(2pi n x)}$ is an orthonormal sequence in $L^2([0,1])$ equipped with Lebesgue measure and hence converges to $0$ weakly in $L^2$.
I also have shown that $int_0^1 cos^2(2pi nx) = frac{1}{2}$ for each $n in mathbb{N}$.
Trying to utilize the above fact, I came up with something which is probably wrong:
By way of contradiction, suppose it is true. Then, $ space f_n(x) rightarrow 0$ for all $x in E subset [0,1] $ such that $E^c$ has measure $0$. Then there is an $N in mathbb{N}$ such that $|cos(2pi nx)| < frac{1}{2}$ whenever $n geq N$ and $x in E$. This implies that $cos^2(2pi nx) < frac{1}{4}$ for $n geq N$. But then, by properties of the Lebesgue integral, we must have:
$$int_E cos^2(2pi nx) = int_0^1 cos^2(2pi nx) < frac{1}{4}m(E) = frac{1}{4} $$
which is a contradiction.
If anyone could tell me where my proof goes wrong and point me in the direction of a correct solution, I would be highly appreciative.
real-analysis sequences-and-series
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
$endgroup$
$begingroup$
You are assuming that this $N$ is uniform: this is not true, this $N$ is in fact depending on $x$. More precisely, you say that for all $xin E$ it is $f_n(x)to 0$. That means that for EACH $xin E$ there exists a $N_xinmathbb{N}$ (i added $_x$ to point out dependence) such that for all $ngeq N_x$ it is $|f_n(x)|<1/2$.
$endgroup$
– JustDroppedIn
Jan 21 at 18:44
$begingroup$
@JustDroppedIn Right, I see. Thank you.
$endgroup$
– Nicholas Roberts
Jan 21 at 18:47
add a comment |
$begingroup$
I am stuck on trying to show that the sequence ${cos(2pi n x)}$ does not converge to $0$ almost everywhere on $[0, 1]$.
I have already shown that ${cos(2pi n x)}$ is an orthonormal sequence in $L^2([0,1])$ equipped with Lebesgue measure and hence converges to $0$ weakly in $L^2$.
I also have shown that $int_0^1 cos^2(2pi nx) = frac{1}{2}$ for each $n in mathbb{N}$.
Trying to utilize the above fact, I came up with something which is probably wrong:
By way of contradiction, suppose it is true. Then, $ space f_n(x) rightarrow 0$ for all $x in E subset [0,1] $ such that $E^c$ has measure $0$. Then there is an $N in mathbb{N}$ such that $|cos(2pi nx)| < frac{1}{2}$ whenever $n geq N$ and $x in E$. This implies that $cos^2(2pi nx) < frac{1}{4}$ for $n geq N$. But then, by properties of the Lebesgue integral, we must have:
$$int_E cos^2(2pi nx) = int_0^1 cos^2(2pi nx) < frac{1}{4}m(E) = frac{1}{4} $$
which is a contradiction.
If anyone could tell me where my proof goes wrong and point me in the direction of a correct solution, I would be highly appreciative.
real-analysis sequences-and-series
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
$endgroup$
I am stuck on trying to show that the sequence ${cos(2pi n x)}$ does not converge to $0$ almost everywhere on $[0, 1]$.
I have already shown that ${cos(2pi n x)}$ is an orthonormal sequence in $L^2([0,1])$ equipped with Lebesgue measure and hence converges to $0$ weakly in $L^2$.
I also have shown that $int_0^1 cos^2(2pi nx) = frac{1}{2}$ for each $n in mathbb{N}$.
Trying to utilize the above fact, I came up with something which is probably wrong:
By way of contradiction, suppose it is true. Then, $ space f_n(x) rightarrow 0$ for all $x in E subset [0,1] $ such that $E^c$ has measure $0$. Then there is an $N in mathbb{N}$ such that $|cos(2pi nx)| < frac{1}{2}$ whenever $n geq N$ and $x in E$. This implies that $cos^2(2pi nx) < frac{1}{4}$ for $n geq N$. But then, by properties of the Lebesgue integral, we must have:
$$int_E cos^2(2pi nx) = int_0^1 cos^2(2pi nx) < frac{1}{4}m(E) = frac{1}{4} $$
which is a contradiction.
If anyone could tell me where my proof goes wrong and point me in the direction of a correct solution, I would be highly appreciative.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
asked Jan 21 at 18:16
Nicholas RobertsNicholas Roberts
117112
117112
$begingroup$
You are assuming that this $N$ is uniform: this is not true, this $N$ is in fact depending on $x$. More precisely, you say that for all $xin E$ it is $f_n(x)to 0$. That means that for EACH $xin E$ there exists a $N_xinmathbb{N}$ (i added $_x$ to point out dependence) such that for all $ngeq N_x$ it is $|f_n(x)|<1/2$.
$endgroup$
– JustDroppedIn
Jan 21 at 18:44
$begingroup$
@JustDroppedIn Right, I see. Thank you.
$endgroup$
– Nicholas Roberts
Jan 21 at 18:47
add a comment |
$begingroup$
You are assuming that this $N$ is uniform: this is not true, this $N$ is in fact depending on $x$. More precisely, you say that for all $xin E$ it is $f_n(x)to 0$. That means that for EACH $xin E$ there exists a $N_xinmathbb{N}$ (i added $_x$ to point out dependence) such that for all $ngeq N_x$ it is $|f_n(x)|<1/2$.
$endgroup$
– JustDroppedIn
Jan 21 at 18:44
$begingroup$
@JustDroppedIn Right, I see. Thank you.
$endgroup$
– Nicholas Roberts
Jan 21 at 18:47
$begingroup$
You are assuming that this $N$ is uniform: this is not true, this $N$ is in fact depending on $x$. More precisely, you say that for all $xin E$ it is $f_n(x)to 0$. That means that for EACH $xin E$ there exists a $N_xinmathbb{N}$ (i added $_x$ to point out dependence) such that for all $ngeq N_x$ it is $|f_n(x)|<1/2$.
$endgroup$
– JustDroppedIn
Jan 21 at 18:44
$begingroup$
You are assuming that this $N$ is uniform: this is not true, this $N$ is in fact depending on $x$. More precisely, you say that for all $xin E$ it is $f_n(x)to 0$. That means that for EACH $xin E$ there exists a $N_xinmathbb{N}$ (i added $_x$ to point out dependence) such that for all $ngeq N_x$ it is $|f_n(x)|<1/2$.
$endgroup$
– JustDroppedIn
Jan 21 at 18:44
$begingroup$
@JustDroppedIn Right, I see. Thank you.
$endgroup$
– Nicholas Roberts
Jan 21 at 18:47
$begingroup$
@JustDroppedIn Right, I see. Thank you.
$endgroup$
– Nicholas Roberts
Jan 21 at 18:47
add a comment |
1 Answer
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$begingroup$
Your argument is not quite right, since the number $N$ that you choose may depend on $xin E$.
One thinkable solution is to use that $f_n(x):=|cos(2pi n x)|le 1$ for all $xin[0,1]$ and hence, if $f_n$ did converge to $0$ almost everywhere, dominated convergence would yield the contradiction $frac12=int_0^1f_n^2(x)dx to 0$.
I am not quite sure that this is the question, though. Maybe you actually want to show the stronger statement that for almost every $xin[0,1]$ the term $f_n(x)$ does not converge to $0$ (which is also true).
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
$endgroup$
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
|
show 1 more comment
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$begingroup$
Your argument is not quite right, since the number $N$ that you choose may depend on $xin E$.
One thinkable solution is to use that $f_n(x):=|cos(2pi n x)|le 1$ for all $xin[0,1]$ and hence, if $f_n$ did converge to $0$ almost everywhere, dominated convergence would yield the contradiction $frac12=int_0^1f_n^2(x)dx to 0$.
I am not quite sure that this is the question, though. Maybe you actually want to show the stronger statement that for almost every $xin[0,1]$ the term $f_n(x)$ does not converge to $0$ (which is also true).
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
$endgroup$
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
|
show 1 more comment
$begingroup$
Your argument is not quite right, since the number $N$ that you choose may depend on $xin E$.
One thinkable solution is to use that $f_n(x):=|cos(2pi n x)|le 1$ for all $xin[0,1]$ and hence, if $f_n$ did converge to $0$ almost everywhere, dominated convergence would yield the contradiction $frac12=int_0^1f_n^2(x)dx to 0$.
I am not quite sure that this is the question, though. Maybe you actually want to show the stronger statement that for almost every $xin[0,1]$ the term $f_n(x)$ does not converge to $0$ (which is also true).
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
$endgroup$
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
|
show 1 more comment
$begingroup$
Your argument is not quite right, since the number $N$ that you choose may depend on $xin E$.
One thinkable solution is to use that $f_n(x):=|cos(2pi n x)|le 1$ for all $xin[0,1]$ and hence, if $f_n$ did converge to $0$ almost everywhere, dominated convergence would yield the contradiction $frac12=int_0^1f_n^2(x)dx to 0$.
I am not quite sure that this is the question, though. Maybe you actually want to show the stronger statement that for almost every $xin[0,1]$ the term $f_n(x)$ does not converge to $0$ (which is also true).
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
$endgroup$
Your argument is not quite right, since the number $N$ that you choose may depend on $xin E$.
One thinkable solution is to use that $f_n(x):=|cos(2pi n x)|le 1$ for all $xin[0,1]$ and hence, if $f_n$ did converge to $0$ almost everywhere, dominated convergence would yield the contradiction $frac12=int_0^1f_n^2(x)dx to 0$.
I am not quite sure that this is the question, though. Maybe you actually want to show the stronger statement that for almost every $xin[0,1]$ the term $f_n(x)$ does not converge to $0$ (which is also true).
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
edited Jan 21 at 18:54
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
answered Jan 21 at 18:49
Mars PlasticMars Plastic
1,09720
1,09720
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
|
show 1 more comment
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
So we would apply DCT to the sequence $f_n^2$ actually?
$endgroup$
– Nicholas Roberts
Jan 21 at 18:53
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
Yes, thanks for pointing out that typo. :-) It has been corrected now.
$endgroup$
– Mars Plastic
Jan 21 at 18:55
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
You say that youre not sure that this is the question, but why not? Does proof by contradiction not work here?
$endgroup$
– Nicholas Roberts
Jan 21 at 19:02
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
My proof shows that the statement "$f_n$ converges to 0 a.e." is wrong. In other words: for any measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ there are $xin E$ such that $f_n(x)$ does not converge to 0. A similar, but stronger (and more interesting) statement is the following: There is a measurable set $Esubset[0,1]$ with $lambda(E^c)=0$ such that for all $xin E$ the term $f_n(x)$ does not converge to 0. Note the different position of "there is" and "for all". Maybe the actual question on your mind was the second one. The opening sentence of your post is slightly ambiguous.
$endgroup$
– Mars Plastic
Jan 21 at 19:14
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
$begingroup$
Thank you @Mars Plastic. Very helpful responses.
$endgroup$
– Nicholas Roberts
Jan 21 at 19:22
|
show 1 more comment
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$begingroup$
You are assuming that this $N$ is uniform: this is not true, this $N$ is in fact depending on $x$. More precisely, you say that for all $xin E$ it is $f_n(x)to 0$. That means that for EACH $xin E$ there exists a $N_xinmathbb{N}$ (i added $_x$ to point out dependence) such that for all $ngeq N_x$ it is $|f_n(x)|<1/2$.
$endgroup$
– JustDroppedIn
Jan 21 at 18:44
$begingroup$
@JustDroppedIn Right, I see. Thank you.
$endgroup$
– Nicholas Roberts
Jan 21 at 18:47