Showing that $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Approximating a Lebesgue measurable set by a finite union of intervalsSequences for which $lim_n sin(a_nx)$ exists on a set of positive measure$lim_x to infty xsin(frac1x) = 1$ (epsilon-delta like condition)Compute $lim_ntoinfty int_E sin^n(x)dx$$f_n geq 0$ and $int f_n = 1$ implies $limsup_n left( f_n(x) right)^frac1n leq 1$ for a.e. $x$Find $lim_n to infty n int_0^1 (cos x - sin x)^n dx$Another way of showing that $int_0^inftyfracsin xxdx = fracpi2$Prove that $lim_nto infty [int_E f^n (x) dx)]^1/n = M$ where $M = sup_xin E f (x)$For E of finite Lebesgue measure, is $f(t) = int_E cos(tx) dx$Fatou's Lemma proof from Royden 4thProve that if $f$ is nonnegative, measurable and $E_k nearrow E$, then $lim_k int_E_k f = int_E f$.
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Showing that $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Approximating a Lebesgue measurable set by a finite union of intervalsSequences for which $lim_n sin(a_nx)$ exists on a set of positive measure$lim_x to infty xsin(frac1x) = 1$ (epsilon-delta like condition)Compute $lim_ntoinfty int_E sin^n(x)dx$$f_n geq 0$ and $int f_n = 1$ implies $limsup_n left( f_n(x) right)^frac1n leq 1$ for a.e. $x$Find $lim_n to infty n int_0^1 (cos x - sin x)^n dx$Another way of showing that $int_0^inftyfracsin xxdx = fracpi2$Prove that $lim_nto infty [int_E f^n (x) dx)]^1/n = M$ where $M = sup_xin E f (x)$For E of finite Lebesgue measure, is $f(t) = int_E cos(tx) dx$Fatou's Lemma proof from Royden 4thProve that if $f$ is nonnegative, measurable and $E_k nearrow E$, then $lim_k int_E_k f = int_E f$.
$begingroup$
So, I have to show that if $E subset [0, 2pi]$ is a set of finite measure, then $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$
My original plan of action was to observe the following:
$$left|int_E e^inx ,dxright| = left|frace^inxinright| = frac1$$
The integral would then go to $0$ as $|n| to infty$, and since $cos(nx), sin(nx)$ are the real and imaginary parts of $e^inx$ we would be done. However, I realized that I didn't really use the hypothesis regarding $E$ being a set of finite measure, so I don't think my proof is correct. However, I would like some advice as to how to proceed in actually proving the above statements.
real-analysis integration lebesgue-integral
edited Apr 2 at 2:47
Mark Viola
134k1278177
asked Apr 2 at 2:42
user516079user516079
537311
$endgroup$
add a comment |
$begingroup$
So, I have to show that if $E subset [0, 2pi]$ is a set of finite measure, then $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$
My original plan of action was to observe the following:
$$left|int_E e^inx ,dxright| = left|frace^inxinright| = frac1$$
The integral would then go to $0$ as $|n| to infty$, and since $cos(nx), sin(nx)$ are the real and imaginary parts of $e^inx$ we would be done. However, I realized that I didn't really use the hypothesis regarding $E$ being a set of finite measure, so I don't think my proof is correct. However, I would like some advice as to how to proceed in actually proving the above statements.
real-analysis integration lebesgue-integral
edited Apr 2 at 2:47
Mark Viola
134k1278177
asked Apr 2 at 2:42
user516079user516079
537311
$endgroup$
1
$begingroup$
$$left|int_E sin(nx),dxright|le int_0^pisin(nx),dx-int_pi^2pisin(nx),dx$$
$endgroup$
– Mark Viola
Apr 2 at 2:46
1
$begingroup$
Have you heard of Riemann Lebesgue Lemma? Also note that any measurable subset of $[0,2pi]$ has finite measure.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 6:00
add a comment |
$begingroup$
So, I have to show that if $E subset [0, 2pi]$ is a set of finite measure, then $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$
My original plan of action was to observe the following:
$$left|int_E e^inx ,dxright| = left|frace^inxinright| = frac1$$
The integral would then go to $0$ as $|n| to infty$, and since $cos(nx), sin(nx)$ are the real and imaginary parts of $e^inx$ we would be done. However, I realized that I didn't really use the hypothesis regarding $E$ being a set of finite measure, so I don't think my proof is correct. However, I would like some advice as to how to proceed in actually proving the above statements.
real-analysis integration lebesgue-integral
edited Apr 2 at 2:47
Mark Viola
134k1278177
asked Apr 2 at 2:42
user516079user516079
537311
$endgroup$
So, I have to show that if $E subset [0, 2pi]$ is a set of finite measure, then $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$
My original plan of action was to observe the following:
$$left|int_E e^inx ,dxright| = left|frace^inxinright| = frac1$$
The integral would then go to $0$ as $|n| to infty$, and since $cos(nx), sin(nx)$ are the real and imaginary parts of $e^inx$ we would be done. However, I realized that I didn't really use the hypothesis regarding $E$ being a set of finite measure, so I don't think my proof is correct. However, I would like some advice as to how to proceed in actually proving the above statements.
real-analysis integration lebesgue-integral
real-analysis integration lebesgue-integral
edited Apr 2 at 2:47
Mark Viola
134k1278177
asked Apr 2 at 2:42
user516079user516079
537311
edited Apr 2 at 2:47
Mark Viola
134k1278177
asked Apr 2 at 2:42
user516079user516079
537311
edited Apr 2 at 2:47
Mark Viola
134k1278177
edited Apr 2 at 2:47
Mark Viola
134k1278177
edited Apr 2 at 2:47
Mark Viola
134k1278177
134k1278177
asked Apr 2 at 2:42
user516079user516079
537311
asked Apr 2 at 2:42
user516079user516079
537311
asked Apr 2 at 2:42
user516079user516079
537311
537311
1
$begingroup$
$$left|int_E sin(nx),dxright|le int_0^pisin(nx),dx-int_pi^2pisin(nx),dx$$
$endgroup$
– Mark Viola
Apr 2 at 2:46
1
$begingroup$
Have you heard of Riemann Lebesgue Lemma? Also note that any measurable subset of $[0,2pi]$ has finite measure.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 6:00
add a comment |
1
$begingroup$
$$left|int_E sin(nx),dxright|le int_0^pisin(nx),dx-int_pi^2pisin(nx),dx$$
$endgroup$
– Mark Viola
Apr 2 at 2:46
1
$begingroup$
Have you heard of Riemann Lebesgue Lemma? Also note that any measurable subset of $[0,2pi]$ has finite measure.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 6:00
1
1
$begingroup$
$$left|int_E sin(nx),dxright|le int_0^pisin(nx),dx-int_pi^2pisin(nx),dx$$
$endgroup$
– Mark Viola
Apr 2 at 2:46
$begingroup$
$$left|int_E sin(nx),dxright|le int_0^pisin(nx),dx-int_pi^2pisin(nx),dx$$
$endgroup$
– Mark Viola
Apr 2 at 2:46
1
1
$begingroup$
Have you heard of Riemann Lebesgue Lemma? Also note that any measurable subset of $[0,2pi]$ has finite measure.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 6:00
$begingroup$
Have you heard of Riemann Lebesgue Lemma? Also note that any measurable subset of $[0,2pi]$ has finite measure.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 6:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's redundant to say $E$ has finite measure since its measure is no greater than the measure of $[0,2pi]$, which is $2pi$. In any case, since the indicator function $1_E$ is Lebesgue integrable on $[0,2pi]$ you can fix a positive number $epsilon$ and choose a smooth function $g$ on $[0,2pi]$ for which $sup_xin [0,2pi] lvert 1_E(x) - g(x)rvert < fracepsilon2pi$. Let $h(x) = 1_E(x) - g(x)$ and show that $int_0^2pi h(x) e^inx, dx$ is bounded by $epsilon$. By integration by parts, show that $int_0^2pi g(x)e^inx, dx = O(frac1n)$ as $nto infty$. Using those two estimates show that $int_E e^inx, dx = int_0^2pi 1_E(x)e^inx, dx to 0$ as $nto infty$. The result follows from taking real and imaginary parts.
answered Apr 2 at 4:47
kobekobe
35.1k22248
$endgroup$
add a comment |
$begingroup$
$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.
answered Apr 2 at 5:13
jawheelejawheele
53139
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It's redundant to say $E$ has finite measure since its measure is no greater than the measure of $[0,2pi]$, which is $2pi$. In any case, since the indicator function $1_E$ is Lebesgue integrable on $[0,2pi]$ you can fix a positive number $epsilon$ and choose a smooth function $g$ on $[0,2pi]$ for which $sup_xin [0,2pi] lvert 1_E(x) - g(x)rvert < fracepsilon2pi$. Let $h(x) = 1_E(x) - g(x)$ and show that $int_0^2pi h(x) e^inx, dx$ is bounded by $epsilon$. By integration by parts, show that $int_0^2pi g(x)e^inx, dx = O(frac1n)$ as $nto infty$. Using those two estimates show that $int_E e^inx, dx = int_0^2pi 1_E(x)e^inx, dx to 0$ as $nto infty$. The result follows from taking real and imaginary parts.
answered Apr 2 at 4:47
kobekobe
35.1k22248
$endgroup$
add a comment |
$begingroup$
It's redundant to say $E$ has finite measure since its measure is no greater than the measure of $[0,2pi]$, which is $2pi$. In any case, since the indicator function $1_E$ is Lebesgue integrable on $[0,2pi]$ you can fix a positive number $epsilon$ and choose a smooth function $g$ on $[0,2pi]$ for which $sup_xin [0,2pi] lvert 1_E(x) - g(x)rvert < fracepsilon2pi$. Let $h(x) = 1_E(x) - g(x)$ and show that $int_0^2pi h(x) e^inx, dx$ is bounded by $epsilon$. By integration by parts, show that $int_0^2pi g(x)e^inx, dx = O(frac1n)$ as $nto infty$. Using those two estimates show that $int_E e^inx, dx = int_0^2pi 1_E(x)e^inx, dx to 0$ as $nto infty$. The result follows from taking real and imaginary parts.
answered Apr 2 at 4:47
kobekobe
35.1k22248
$endgroup$
add a comment |
$begingroup$
It's redundant to say $E$ has finite measure since its measure is no greater than the measure of $[0,2pi]$, which is $2pi$. In any case, since the indicator function $1_E$ is Lebesgue integrable on $[0,2pi]$ you can fix a positive number $epsilon$ and choose a smooth function $g$ on $[0,2pi]$ for which $sup_xin [0,2pi] lvert 1_E(x) - g(x)rvert < fracepsilon2pi$. Let $h(x) = 1_E(x) - g(x)$ and show that $int_0^2pi h(x) e^inx, dx$ is bounded by $epsilon$. By integration by parts, show that $int_0^2pi g(x)e^inx, dx = O(frac1n)$ as $nto infty$. Using those two estimates show that $int_E e^inx, dx = int_0^2pi 1_E(x)e^inx, dx to 0$ as $nto infty$. The result follows from taking real and imaginary parts.
answered Apr 2 at 4:47
kobekobe
35.1k22248
$endgroup$
It's redundant to say $E$ has finite measure since its measure is no greater than the measure of $[0,2pi]$, which is $2pi$. In any case, since the indicator function $1_E$ is Lebesgue integrable on $[0,2pi]$ you can fix a positive number $epsilon$ and choose a smooth function $g$ on $[0,2pi]$ for which $sup_xin [0,2pi] lvert 1_E(x) - g(x)rvert < fracepsilon2pi$. Let $h(x) = 1_E(x) - g(x)$ and show that $int_0^2pi h(x) e^inx, dx$ is bounded by $epsilon$. By integration by parts, show that $int_0^2pi g(x)e^inx, dx = O(frac1n)$ as $nto infty$. Using those two estimates show that $int_E e^inx, dx = int_0^2pi 1_E(x)e^inx, dx to 0$ as $nto infty$. The result follows from taking real and imaginary parts.
answered Apr 2 at 4:47
kobekobe
35.1k22248
edited Apr 2 at 16:15
answered Apr 2 at 4:47
kobekobe
35.1k22248
answered Apr 2 at 4:47
kobekobe
35.1k22248
answered Apr 2 at 4:47
kobekobe
35.1k22248
35.1k22248
add a comment |
add a comment |
$begingroup$
$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.
answered Apr 2 at 5:13
jawheelejawheele
53139
$endgroup$
add a comment |
$begingroup$
$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.
answered Apr 2 at 5:13
jawheelejawheele
53139
$endgroup$
add a comment |
$begingroup$
$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.
answered Apr 2 at 5:13
jawheelejawheele
53139
$endgroup$
$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.
answered Apr 2 at 5:13
jawheelejawheele
53139
answered Apr 2 at 5:13
jawheelejawheele
53139
answered Apr 2 at 5:13
jawheelejawheele
53139
answered Apr 2 at 5:13
jawheelejawheele
53139
53139
add a comment |
add a comment |
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1
$begingroup$
$$left|int_E sin(nx),dxright|le int_0^pisin(nx),dx-int_pi^2pisin(nx),dx$$
$endgroup$
– Mark Viola
Apr 2 at 2:46
1
$begingroup$
Have you heard of Riemann Lebesgue Lemma? Also note that any measurable subset of $[0,2pi]$ has finite measure.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 6:00