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The Fourier transform of the derivative of a function $fin L_1(mathbbR)$

Fourier Transform of DerivativeEquivalence of two ways of expressing Fourier transformderivative of fourier transformQuestion about linear operator and Fourier transformOn the Fourier transform of $f(x)=ln(x^2+a^2)$Fourier transform of the identity function $f(x)=x$Fourier Transform of derivative a vector functionFourier transform of $L^1$ function is continuous functionFourier transform of $frac1sqrt1 + x^2$Find the inverse Fourier transform of $f$How to derive Riesz transform from its Fourier transform?

1

$begingroup$

Let $fin L_1(mathbbR)$ (That is to say $f$ is absolutely integrable over $mathbbR$) with derivative $f'in L_1(mathbbR)$. The Fourier transform of $f$ is given by:
beginalign
hatf(t) = int_-infty^infty f(x) e^itx textdx.
endalign
I want to prove that the Fourier transform of $f'$ is given by:
beginalign
widehatf'(t) = -it hatf(t).
endalign
Here is my own way to give a proof: By definition, we have
beginalign
widehatf'(t) = int_-infty^infty f'(x) e^itx text dx
endalign.
By parts, we get
beginalign
widehatf'(t) = left[f(x)e^itx right]_-infty^infty - it int_-infty^infty f(x) e^itx text dx.
endalign
I wonder why the first term in the preceding equality vanishes? This is my problem. I appreciate any help.

fourier-analysis fourier-transform

share|cite|improve this question

edited Mar 29 at 19:05

Hussein Eid

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe you need $f'$ to be integrable.
  $endgroup$
  – copper.hat
  Mar 29 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f'$ is integrable. I forgot writing this condition.
  $endgroup$
  – Hussein Eid
  Mar 29 at 18:59
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $fin L_1(mathbbR)$ (That is to say $f$ is absolutely integrable over $mathbbR$) with derivative $f'in L_1(mathbbR)$. The Fourier transform of $f$ is given by:
beginalign
hatf(t) = int_-infty^infty f(x) e^itx textdx.
endalign
I want to prove that the Fourier transform of $f'$ is given by:
beginalign
widehatf'(t) = -it hatf(t).
endalign
Here is my own way to give a proof: By definition, we have
beginalign
widehatf'(t) = int_-infty^infty f'(x) e^itx text dx
endalign.
By parts, we get
beginalign
widehatf'(t) = left[f(x)e^itx right]_-infty^infty - it int_-infty^infty f(x) e^itx text dx.
endalign
I wonder why the first term in the preceding equality vanishes? This is my problem. I appreciate any help.

fourier-analysis fourier-transform

share|cite|improve this question

edited Mar 29 at 19:05

Hussein Eid

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe you need $f'$ to be integrable.
  $endgroup$
  – copper.hat
  Mar 29 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f'$ is integrable. I forgot writing this condition.
  $endgroup$
  – Hussein Eid
  Mar 29 at 18:59
  

  
  
  
  

add a comment | 

$begingroup$

Let $fin L_1(mathbbR)$ (That is to say $f$ is absolutely integrable over $mathbbR$) with derivative $f'in L_1(mathbbR)$. The Fourier transform of $f$ is given by:
beginalign
hatf(t) = int_-infty^infty f(x) e^itx textdx.
endalign
I want to prove that the Fourier transform of $f'$ is given by:
beginalign
widehatf'(t) = -it hatf(t).
endalign
Here is my own way to give a proof: By definition, we have
beginalign
widehatf'(t) = int_-infty^infty f'(x) e^itx text dx
endalign.
By parts, we get
beginalign
widehatf'(t) = left[f(x)e^itx right]_-infty^infty - it int_-infty^infty f(x) e^itx text dx.
endalign
I wonder why the first term in the preceding equality vanishes? This is my problem. I appreciate any help.

fourier-analysis fourier-transform

share|cite|improve this question

edited Mar 29 at 19:05

Hussein Eid

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

$endgroup$

share|cite|improve this question

edited Mar 29 at 19:05

Hussein Eid

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

share|cite|improve this question

edited Mar 29 at 19:05

Hussein Eid

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

asked Mar 29 at 16:02

Hussein EidHussein Eid

153

2 Answers
2

active

oldest

votes

2

$begingroup$

We must suppose that $f$ is absolutely continuous for the derivative to really have any meaning. In that case, the fundamental theorem of calculus gives $f(x) = f(0) + int_0^x f'(t),dt$. Since $f'$ is in $L^1$, then as $x to infty$, dominated convergence implies that $int_0^x f'(t),dt to int_0^infty f'(t),dt$; in particular the limit exists. Thus $lim_x to infty f(x)$ exists. Now if this limit equals anything other than zero, $f$ would not be integrable. So we have $lim_x to infty f(x) = 0$, and a similar argument gives $lim_x to -infty f(x) =0$ as well. Since $e^itx$ is bounded, this shows that the term in question does indeed vanish.

share|cite|improve this answer

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, I actually proved that the limit $lim_xto +infty f(x)$ exists by the integrability of $f'$. But how can one use the integrability of $f$ to show that the limit must be zero?. This is not clear for me. Anyway, i really appreciate your contribution.
  $endgroup$
  – Hussein Eid
  Mar 29 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HusseinEid: Try drawing a picture of a function with $lim_x to infty f(x) = c ne 0$. It should be immediately clear that $int_-infty^infty |f| = infty$, and it should take only a little bit more work to prove it.
  $endgroup$
  – Nate Eldredge
  Mar 29 at 23:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i will try that.
  $endgroup$
  – Hussein Eid
  Mar 30 at 12:14
  

  
  
  
  

add a comment | 

0

$begingroup$

See:

Fourier Transform of Derivative

share|cite|improve this answer

edited Mar 29 at 16:41

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264

$endgroup$

add a comment | 

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2

$begingroup$

We must suppose that $f$ is absolutely continuous for the derivative to really have any meaning. In that case, the fundamental theorem of calculus gives $f(x) = f(0) + int_0^x f'(t),dt$. Since $f'$ is in $L^1$, then as $x to infty$, dominated convergence implies that $int_0^x f'(t),dt to int_0^infty f'(t),dt$; in particular the limit exists. Thus $lim_x to infty f(x)$ exists. Now if this limit equals anything other than zero, $f$ would not be integrable. So we have $lim_x to infty f(x) = 0$, and a similar argument gives $lim_x to -infty f(x) =0$ as well. Since $e^itx$ is bounded, this shows that the term in question does indeed vanish.

share|cite|improve this answer

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, I actually proved that the limit $lim_xto +infty f(x)$ exists by the integrability of $f'$. But how can one use the integrability of $f$ to show that the limit must be zero?. This is not clear for me. Anyway, i really appreciate your contribution.
  $endgroup$
  – Hussein Eid
  Mar 29 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HusseinEid: Try drawing a picture of a function with $lim_x to infty f(x) = c ne 0$. It should be immediately clear that $int_-infty^infty |f| = infty$, and it should take only a little bit more work to prove it.
  $endgroup$
  – Nate Eldredge
  Mar 29 at 23:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i will try that.
  $endgroup$
  – Hussein Eid
  Mar 30 at 12:14
  

  
  
  
  

add a comment | 

2

$begingroup$

We must suppose that $f$ is absolutely continuous for the derivative to really have any meaning. In that case, the fundamental theorem of calculus gives $f(x) = f(0) + int_0^x f'(t),dt$. Since $f'$ is in $L^1$, then as $x to infty$, dominated convergence implies that $int_0^x f'(t),dt to int_0^infty f'(t),dt$; in particular the limit exists. Thus $lim_x to infty f(x)$ exists. Now if this limit equals anything other than zero, $f$ would not be integrable. So we have $lim_x to infty f(x) = 0$, and a similar argument gives $lim_x to -infty f(x) =0$ as well. Since $e^itx$ is bounded, this shows that the term in question does indeed vanish.

share|cite|improve this answer

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, I actually proved that the limit $lim_xto +infty f(x)$ exists by the integrability of $f'$. But how can one use the integrability of $f$ to show that the limit must be zero?. This is not clear for me. Anyway, i really appreciate your contribution.
  $endgroup$
  – Hussein Eid
  Mar 29 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HusseinEid: Try drawing a picture of a function with $lim_x to infty f(x) = c ne 0$. It should be immediately clear that $int_-infty^infty |f| = infty$, and it should take only a little bit more work to prove it.
  $endgroup$
  – Nate Eldredge
  Mar 29 at 23:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i will try that.
  $endgroup$
  – Hussein Eid
  Mar 30 at 12:14
  

  
  
  
  

add a comment | 

$begingroup$

We must suppose that $f$ is absolutely continuous for the derivative to really have any meaning. In that case, the fundamental theorem of calculus gives $f(x) = f(0) + int_0^x f'(t),dt$. Since $f'$ is in $L^1$, then as $x to infty$, dominated convergence implies that $int_0^x f'(t),dt to int_0^infty f'(t),dt$; in particular the limit exists. Thus $lim_x to infty f(x)$ exists. Now if this limit equals anything other than zero, $f$ would not be integrable. So we have $lim_x to infty f(x) = 0$, and a similar argument gives $lim_x to -infty f(x) =0$ as well. Since $e^itx$ is bounded, this shows that the term in question does indeed vanish.

share|cite|improve this answer

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

$endgroup$

share|cite|improve this answer

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

answered Mar 29 at 23:25

Nate EldredgeNate Eldredge

64.5k682174

0

$begingroup$

See:

Fourier Transform of Derivative

share|cite|improve this answer

edited Mar 29 at 16:41

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264

$endgroup$

add a comment | 

0

$begingroup$

See:

Fourier Transform of Derivative

share|cite|improve this answer

edited Mar 29 at 16:41

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264

$endgroup$

add a comment | 

$begingroup$

See:

Fourier Transform of Derivative

share|cite|improve this answer

edited Mar 29 at 16:41

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264

$endgroup$

share|cite|improve this answer

edited Mar 29 at 16:41

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264

answered Mar 29 at 16:26

George DewhirstGeorge Dewhirst

7264