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Showing that $lim_n to infty int_E cos(nx) = lim_n to infty int_E sin(nx) = 0$

1

$begingroup$

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

share|cite|improve this question

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 | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Apr 2 at 2:47

Mark Viola

134k1278177

asked Apr 2 at 2:42

user516079user516079

537311

$endgroup$

share|cite|improve this question

edited Apr 2 at 2:47

Mark Viola

134k1278177

asked Apr 2 at 2:42

user516079user516079

537311

share|cite|improve this question

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

asked Apr 2 at 2:42

user516079user516079

537311

asked Apr 2 at 2:42

user516079user516079

537311

2 Answers
2

active

oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

edited Apr 2 at 16:15

answered Apr 2 at 4:47

kobekobe

35.1k22248

$endgroup$

add a comment | 

1

$begingroup$

$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.

share|cite|improve this answer

answered Apr 2 at 5:13

jawheelejawheele

53139

$endgroup$

add a comment | 

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0

$begingroup$

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.

share|cite|improve this answer

edited Apr 2 at 16:15

answered Apr 2 at 4:47

kobekobe

35.1k22248

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

edited Apr 2 at 16:15

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.

share|cite|improve this answer

edited Apr 2 at 16:15

answered Apr 2 at 4:47

kobekobe

35.1k22248

$endgroup$

share|cite|improve this answer

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

1

$begingroup$

$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.

share|cite|improve this answer

answered Apr 2 at 5:13

jawheelejawheele

53139

$endgroup$

add a comment | 

1

$begingroup$

$mathbfHint$: If you're familiar with this result, use it to reduce to the case that $E$ is an interval, which is nearly immediate.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Apr 2 at 5:13

jawheelejawheele

53139

$endgroup$

share|cite|improve this answer

answered Apr 2 at 5:13

jawheelejawheele

53139

answered Apr 2 at 5:13

jawheelejawheele

53139

answered Apr 2 at 5:13

jawheelejawheele

53139