Conditions for the differentiation of Fourier Transform of a function.












2












$begingroup$


Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12
















2












$begingroup$


Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12














2












2








2





$begingroup$


Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform







share|cite|improve this question











$endgroup$




Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform




real-analysis fourier-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 0:01







ersh

















asked Dec 18 '18 at 23:48









ershersh

389113




389113








  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12














  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12








1




1




$begingroup$
Integrability of $xf(x)$ is a sufficient condition.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:55




$begingroup$
Integrability of $xf(x)$ is a sufficient condition.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:55












$begingroup$
@KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
$endgroup$
– ersh
Dec 19 '18 at 0:04




$begingroup$
@KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
$endgroup$
– ersh
Dec 19 '18 at 0:04




2




2




$begingroup$
You are right and you get these by an application of DCT.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 0:07




$begingroup$
You are right and you get these by an application of DCT.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 0:07




1




1




$begingroup$
The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
$endgroup$
– reuns
Dec 19 '18 at 8:12




$begingroup$
The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
$endgroup$
– reuns
Dec 19 '18 at 8:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28













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1 Answer
1






active

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active

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oldest

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1












$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28


















1












$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28
















1












1








1





$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$



This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 0:11









MalikMalik

1018




1018












  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28




















  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28


















$begingroup$
Can you please provide me reference to this result?
$endgroup$
– ersh
Dec 19 '18 at 1:41




$begingroup$
Can you please provide me reference to this result?
$endgroup$
– ersh
Dec 19 '18 at 1:41




1




1




$begingroup$
Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
$endgroup$
– Malik
Dec 19 '18 at 2:28






$begingroup$
Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
$endgroup$
– Malik
Dec 19 '18 at 2:28




















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