Fdtxjr

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Show that $f_nto f$ uniformly on $mathbbR$

Suppose that $f$ is uniformly continuous on $R$ and let $f_n (x)=f(x+frac1n)$. Prove that $f_n$ converges uniformly to $f$.Prove that $int_-infty^infty P_n(x) , dx = pi /n$Is $(epsilon + cos(x))^2k_kinmathbbN$ a family of good kernels?$f_n$ is uniformly integrable if and only if $sup_n int |f_n|,dmu < infty$ and $f_n$ is uniformly absolutely continuous?Suppose $f_n to f $ uniformly on $ R$ what can we say about $ f$?derivative of the Fourier seriesBasic Fourier analysis - Summability kernels over $mathbbR$$sum_n geq 1 |c_n|^2 leq frac 18$Given $P_n(x)=fracn1+n^2x^2,$ Prove $f_n(x)=frac1piint_-infty^infty f(x-t)P_n(t)$ converges uniformlyLimit is $0$ uniformly for fourier coefficients

8

5

$begingroup$

Let $$P_n(x) = fracn1+n^2x^2$$.

First, I had to prove that

$$int_-infty^infty P_n(x) dx = pi$$

And that for any $delta > 0$:

$$lim_ntoinfty int_delta^infty P_n(x) dx = lim_ntoinfty int_-infty^-delta P_n(x) dx = 0$$

I've done that easily.

Now I need to prove that for $f:mathbbRtomathbbC$ which is $2pi$ periodic and continuous and:
$$f_n(x) = frac1pi int_-infty^infty f(x-t)P_n(t) dt$$

$f_nto f$, uniformly on $mathbbR$.

We learned in class about convolution and about Dirichlet/Fejer kernels.
Also, we learned that the trigonometric polynomials, $e^inx_ninmathbbZ$ are a dense set on $C(mathbbT)$ and the density is uniform. Meaning, there's a $P_n(x)=sum c_n e^inx$ converges uniformly to $f$ where $fin C(mathbbT)$.

note: $fin C(mathbbT)$ is a continuous and $2pi$ periodic function (T is for Torus).

calculus integration fourier-analysis fourier-series

share|cite|improve this question

edited Jan 23 '16 at 14:02

Elimination

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

$endgroup$

add a comment | 

8

5

$begingroup$

Let $$P_n(x) = fracn1+n^2x^2$$.

First, I had to prove that

$$int_-infty^infty P_n(x) dx = pi$$

And that for any $delta > 0$:

$$lim_ntoinfty int_delta^infty P_n(x) dx = lim_ntoinfty int_-infty^-delta P_n(x) dx = 0$$

I've done that easily.

Now I need to prove that for $f:mathbbRtomathbbC$ which is $2pi$ periodic and continuous and:
$$f_n(x) = frac1pi int_-infty^infty f(x-t)P_n(t) dt$$

$f_nto f$, uniformly on $mathbbR$.

We learned in class about convolution and about Dirichlet/Fejer kernels.
Also, we learned that the trigonometric polynomials, $e^inx_ninmathbbZ$ are a dense set on $C(mathbbT)$ and the density is uniform. Meaning, there's a $P_n(x)=sum c_n e^inx$ converges uniformly to $f$ where $fin C(mathbbT)$.

note: $fin C(mathbbT)$ is a continuous and $2pi$ periodic function (T is for Torus).

calculus integration fourier-analysis fourier-series

share|cite|improve this question

edited Jan 23 '16 at 14:02

Elimination

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

$endgroup$

add a comment | 

$begingroup$

Let $$P_n(x) = fracn1+n^2x^2$$.

First, I had to prove that

$$int_-infty^infty P_n(x) dx = pi$$

And that for any $delta > 0$:

$$lim_ntoinfty int_delta^infty P_n(x) dx = lim_ntoinfty int_-infty^-delta P_n(x) dx = 0$$

I've done that easily.

Now I need to prove that for $f:mathbbRtomathbbC$ which is $2pi$ periodic and continuous and:
$$f_n(x) = frac1pi int_-infty^infty f(x-t)P_n(t) dt$$

$f_nto f$, uniformly on $mathbbR$.

We learned in class about convolution and about Dirichlet/Fejer kernels.
Also, we learned that the trigonometric polynomials, $e^inx_ninmathbbZ$ are a dense set on $C(mathbbT)$ and the density is uniform. Meaning, there's a $P_n(x)=sum c_n e^inx$ converges uniformly to $f$ where $fin C(mathbbT)$.

note: $fin C(mathbbT)$ is a continuous and $2pi$ periodic function (T is for Torus).

calculus integration fourier-analysis fourier-series

share|cite|improve this question

edited Jan 23 '16 at 14:02

Elimination

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

$endgroup$

share|cite|improve this question

edited Jan 23 '16 at 14:02

Elimination

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

share|cite|improve this question

edited Jan 23 '16 at 14:02

Elimination

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

asked Jan 23 '16 at 13:54

EliminationElimination

1,414923

1 Answer
1

active

oldest

votes

6

$begingroup$

To get you started: $$| f_n(x) - f(x)| =left| (1/pi) int_-infty^infty f(x-t) P_n(t) ; dt - f(x)right| = (1/pi)left| int_-infty^inftyleft[f(x-t)- f(x)right] P_n(t) ; dt right|$$

Now because $f$ is continuous on $mathbbT$ and $2pi$-periodic, we can essentially deduce its properties by considering its restriction $f_r$ to $[0,4pi]$. As a continuous function on a compact interval, $f_r$ is bounded and uniformly continuous. It's not hard to see that these properties carry over to $f$, meaning that $f$ is bounded and uniformly continous on $mathbbR$. This implies that for every $epsilon > 0$, there is a $delta
> 0$ such that
$$|f(x) - f(x-t)| < epsilon quad forall t in (-delta,delta)forall xin mathbbR$$
Now, given an $epsilon > 0$, we can choose $delta$ accordingly and then split up the integrals giving

$$|f_n(x) - f(x)| leq (1/pi)left[int_-infty^-deltaCcdot P_n(t) ; dt + int_-delta^deltaepsiloncdot P_n(t) ; dt + int_delta^inftyCcdot P_n(t) ; dtright]$$

Because of what you've already shown we know the left and right integral converge to $0$ as $n to infty$. But the middle integral can be estimated by $epsilon$, which concludes the proof.

share|cite|improve this answer

edited yesterday

answered Jan 23 '16 at 14:08

user159517user159517

4,495931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you @user159517. I'm trying to think how to utilize the $f(x-t)-f(x)$ expression but can't think of anything good. Could you help me proceed please? I guess I should rely on what I already found about the integral of $P_n(x)$ but the $f(x-t)-f(x)$ expression is "interfering"
  $endgroup$
  – Elimination
  Jan 23 '16 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer to include a little more information. If you still dont see what I'm getting at, I can make a full answer out of it.
  $endgroup$
  – user159517
  Jan 23 '16 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user159517 I'm sorry but how did you move the $f(x)$ into the integral? On the first line the function is not being integrated, but then on the second line it is?
  $endgroup$
  – TomGrubb
  Jan 23 '16 at 14:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @bburGsamohT note that $f(x)$ does not depend on the integration variable $t$. So because of $int_-infty^infty P_n(t) dt = pi$, we have $(1/pi)int_-infty^inftyf(x) P_n(t) dt = f(x)$
  $endgroup$
  – user159517
  Jan 23 '16 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Elimination no, it's the same $delta$. The left and right integral converge to $0$ no matter what $delta$ we choose, so we can take it to be the $delta$ from the uniform continuity.
  $endgroup$
  – user159517
  Jan 23 '16 at 20:01
  
  

  
  
  
  

 | 
show 3 more comments

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6

$begingroup$

To get you started: $$| f_n(x) - f(x)| =left| (1/pi) int_-infty^infty f(x-t) P_n(t) ; dt - f(x)right| = (1/pi)left| int_-infty^inftyleft[f(x-t)- f(x)right] P_n(t) ; dt right|$$

Now because $f$ is continuous on $mathbbT$ and $2pi$-periodic, we can essentially deduce its properties by considering its restriction $f_r$ to $[0,4pi]$. As a continuous function on a compact interval, $f_r$ is bounded and uniformly continuous. It's not hard to see that these properties carry over to $f$, meaning that $f$ is bounded and uniformly continous on $mathbbR$. This implies that for every $epsilon > 0$, there is a $delta
> 0$ such that
$$|f(x) - f(x-t)| < epsilon quad forall t in (-delta,delta)forall xin mathbbR$$
Now, given an $epsilon > 0$, we can choose $delta$ accordingly and then split up the integrals giving

$$|f_n(x) - f(x)| leq (1/pi)left[int_-infty^-deltaCcdot P_n(t) ; dt + int_-delta^deltaepsiloncdot P_n(t) ; dt + int_delta^inftyCcdot P_n(t) ; dtright]$$

Because of what you've already shown we know the left and right integral converge to $0$ as $n to infty$. But the middle integral can be estimated by $epsilon$, which concludes the proof.

share|cite|improve this answer

edited yesterday

answered Jan 23 '16 at 14:08

user159517user159517

4,495931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you @user159517. I'm trying to think how to utilize the $f(x-t)-f(x)$ expression but can't think of anything good. Could you help me proceed please? I guess I should rely on what I already found about the integral of $P_n(x)$ but the $f(x-t)-f(x)$ expression is "interfering"
  $endgroup$
  – Elimination
  Jan 23 '16 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer to include a little more information. If you still dont see what I'm getting at, I can make a full answer out of it.
  $endgroup$
  – user159517
  Jan 23 '16 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user159517 I'm sorry but how did you move the $f(x)$ into the integral? On the first line the function is not being integrated, but then on the second line it is?
  $endgroup$
  – TomGrubb
  Jan 23 '16 at 14:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @bburGsamohT note that $f(x)$ does not depend on the integration variable $t$. So because of $int_-infty^infty P_n(t) dt = pi$, we have $(1/pi)int_-infty^inftyf(x) P_n(t) dt = f(x)$
  $endgroup$
  – user159517
  Jan 23 '16 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Elimination no, it's the same $delta$. The left and right integral converge to $0$ no matter what $delta$ we choose, so we can take it to be the $delta$ from the uniform continuity.
  $endgroup$
  – user159517
  Jan 23 '16 at 20:01
  
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

To get you started: $$| f_n(x) - f(x)| =left| (1/pi) int_-infty^infty f(x-t) P_n(t) ; dt - f(x)right| = (1/pi)left| int_-infty^inftyleft[f(x-t)- f(x)right] P_n(t) ; dt right|$$

Now because $f$ is continuous on $mathbbT$ and $2pi$-periodic, we can essentially deduce its properties by considering its restriction $f_r$ to $[0,4pi]$. As a continuous function on a compact interval, $f_r$ is bounded and uniformly continuous. It's not hard to see that these properties carry over to $f$, meaning that $f$ is bounded and uniformly continous on $mathbbR$. This implies that for every $epsilon > 0$, there is a $delta
> 0$ such that
$$|f(x) - f(x-t)| < epsilon quad forall t in (-delta,delta)forall xin mathbbR$$
Now, given an $epsilon > 0$, we can choose $delta$ accordingly and then split up the integrals giving

$$|f_n(x) - f(x)| leq (1/pi)left[int_-infty^-deltaCcdot P_n(t) ; dt + int_-delta^deltaepsiloncdot P_n(t) ; dt + int_delta^inftyCcdot P_n(t) ; dtright]$$

Because of what you've already shown we know the left and right integral converge to $0$ as $n to infty$. But the middle integral can be estimated by $epsilon$, which concludes the proof.

share|cite|improve this answer

edited yesterday

answered Jan 23 '16 at 14:08

user159517user159517

4,495931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you @user159517. I'm trying to think how to utilize the $f(x-t)-f(x)$ expression but can't think of anything good. Could you help me proceed please? I guess I should rely on what I already found about the integral of $P_n(x)$ but the $f(x-t)-f(x)$ expression is "interfering"
  $endgroup$
  – Elimination
  Jan 23 '16 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer to include a little more information. If you still dont see what I'm getting at, I can make a full answer out of it.
  $endgroup$
  – user159517
  Jan 23 '16 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user159517 I'm sorry but how did you move the $f(x)$ into the integral? On the first line the function is not being integrated, but then on the second line it is?
  $endgroup$
  – TomGrubb
  Jan 23 '16 at 14:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @bburGsamohT note that $f(x)$ does not depend on the integration variable $t$. So because of $int_-infty^infty P_n(t) dt = pi$, we have $(1/pi)int_-infty^inftyf(x) P_n(t) dt = f(x)$
  $endgroup$
  – user159517
  Jan 23 '16 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Elimination no, it's the same $delta$. The left and right integral converge to $0$ no matter what $delta$ we choose, so we can take it to be the $delta$ from the uniform continuity.
  $endgroup$
  – user159517
  Jan 23 '16 at 20:01
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

To get you started: $$| f_n(x) - f(x)| =left| (1/pi) int_-infty^infty f(x-t) P_n(t) ; dt - f(x)right| = (1/pi)left| int_-infty^inftyleft[f(x-t)- f(x)right] P_n(t) ; dt right|$$

Now because $f$ is continuous on $mathbbT$ and $2pi$-periodic, we can essentially deduce its properties by considering its restriction $f_r$ to $[0,4pi]$. As a continuous function on a compact interval, $f_r$ is bounded and uniformly continuous. It's not hard to see that these properties carry over to $f$, meaning that $f$ is bounded and uniformly continous on $mathbbR$. This implies that for every $epsilon > 0$, there is a $delta
> 0$ such that
$$|f(x) - f(x-t)| < epsilon quad forall t in (-delta,delta)forall xin mathbbR$$
Now, given an $epsilon > 0$, we can choose $delta$ accordingly and then split up the integrals giving

$$|f_n(x) - f(x)| leq (1/pi)left[int_-infty^-deltaCcdot P_n(t) ; dt + int_-delta^deltaepsiloncdot P_n(t) ; dt + int_delta^inftyCcdot P_n(t) ; dtright]$$

Because of what you've already shown we know the left and right integral converge to $0$ as $n to infty$. But the middle integral can be estimated by $epsilon$, which concludes the proof.

share|cite|improve this answer

edited yesterday

answered Jan 23 '16 at 14:08

user159517user159517

4,495931

$endgroup$

share|cite|improve this answer

edited yesterday

answered Jan 23 '16 at 14:08

user159517user159517

4,495931

answered Jan 23 '16 at 14:08

user159517user159517

4,495931

answered Jan 23 '16 at 14:08

user159517user159517

4,495931