A question about the Residue of $h=fg$












3












$begingroup$


Let $f$ and $g$ be two functions (not necessarily analytic) of the complex
variable $z$ such that for some $varepsilon >0$ :



1) $f$ is continuous on $0<leftvert zrightvert <varepsilon $



2) $g$ is continuous on $leftvert zrightvert <varepsilon $ and $%
gleft( 0right) neq 0$



3) $h=fg$ is analytic on $0<leftvert zrightvert <varepsilon .$



Do we have
begin{equation*}
Resleft( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{varepsilon }{2}}fleft( zright) dz
end{equation*}
Or perhaps some other equation ?



Thank you !






integration complex-analysis functional-analysis contour-integration residue-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $0$ is a removable singularity by the assumption that $h$ is continuous and hence bounded.
    $endgroup$
    – Song
    Jan 1 at 14:05












  • $begingroup$
    I will precise the question
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:06










  • $begingroup$
    @Song take a look now please
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:13
















3












$begingroup$


Let $f$ and $g$ be two functions (not necessarily analytic) of the complex
variable $z$ such that for some $varepsilon >0$ :



1) $f$ is continuous on $0<leftvert zrightvert <varepsilon $



2) $g$ is continuous on $leftvert zrightvert <varepsilon $ and $%
gleft( 0right) neq 0$



3) $h=fg$ is analytic on $0<leftvert zrightvert <varepsilon .$



Do we have
begin{equation*}
Resleft( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{varepsilon }{2}}fleft( zright) dz
end{equation*}
Or perhaps some other equation ?



Thank you !






integration complex-analysis functional-analysis contour-integration residue-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $0$ is a removable singularity by the assumption that $h$ is continuous and hence bounded.
    $endgroup$
    – Song
    Jan 1 at 14:05












  • $begingroup$
    I will precise the question
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:06










  • $begingroup$
    @Song take a look now please
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:13














3












3








3


1



$begingroup$


Let $f$ and $g$ be two functions (not necessarily analytic) of the complex
variable $z$ such that for some $varepsilon >0$ :



1) $f$ is continuous on $0<leftvert zrightvert <varepsilon $



2) $g$ is continuous on $leftvert zrightvert <varepsilon $ and $%
gleft( 0right) neq 0$



3) $h=fg$ is analytic on $0<leftvert zrightvert <varepsilon .$



Do we have
begin{equation*}
Resleft( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{varepsilon }{2}}fleft( zright) dz
end{equation*}
Or perhaps some other equation ?



Thank you !






integration complex-analysis functional-analysis contour-integration residue-calculus







share|cite|improve this question











$endgroup$




Let $f$ and $g$ be two functions (not necessarily analytic) of the complex
variable $z$ such that for some $varepsilon >0$ :



1) $f$ is continuous on $0<leftvert zrightvert <varepsilon $



2) $g$ is continuous on $leftvert zrightvert <varepsilon $ and $%
gleft( 0right) neq 0$



3) $h=fg$ is analytic on $0<leftvert zrightvert <varepsilon .$



Do we have
begin{equation*}
Resleft( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{varepsilon }{2}}fleft( zright) dz
end{equation*}
Or perhaps some other equation ?



Thank you !






integration complex-analysis functional-analysis contour-integration residue-calculus




integration complex-analysis functional-analysis contour-integration residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 14:12







Djalal Ounadjela

















asked Jan 1 at 13:40









Djalal OunadjelaDjalal Ounadjela

29818




29818








  • 1




    $begingroup$
    Note that $0$ is a removable singularity by the assumption that $h$ is continuous and hence bounded.
    $endgroup$
    – Song
    Jan 1 at 14:05












  • $begingroup$
    I will precise the question
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:06










  • $begingroup$
    @Song take a look now please
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:13














  • 1




    $begingroup$
    Note that $0$ is a removable singularity by the assumption that $h$ is continuous and hence bounded.
    $endgroup$
    – Song
    Jan 1 at 14:05












  • $begingroup$
    I will precise the question
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:06










  • $begingroup$
    @Song take a look now please
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 14:13








1




1




$begingroup$
Note that $0$ is a removable singularity by the assumption that $h$ is continuous and hence bounded.
$endgroup$
– Song
Jan 1 at 14:05






$begingroup$
Note that $0$ is a removable singularity by the assumption that $h$ is continuous and hence bounded.
$endgroup$
– Song
Jan 1 at 14:05














$begingroup$
I will precise the question
$endgroup$
– Djalal Ounadjela
Jan 1 at 14:06




$begingroup$
I will precise the question
$endgroup$
– Djalal Ounadjela
Jan 1 at 14:06












$begingroup$
@Song take a look now please
$endgroup$
– Djalal Ounadjela
Jan 1 at 14:13




$begingroup$
@Song take a look now please
$endgroup$
– Djalal Ounadjela
Jan 1 at 14:13










1 Answer
1






active

oldest

votes


















3












$begingroup$

The answer to the question as you stated is no. This is why I doubt it is true. If it were true, then it must be that for all $eta<epsilon$ $$text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{eta}{2}}fleft( zright) dz.$$ The right-hand side may depend on $eta$ if $f$ is not analytic but the left-hand side remains constant. So I guess more meaningful (and weaker) version of your question is to ask whether
$$
text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz
$$ holds. Since $text{Res}left( h,0right) =frac{1}{2pi i}oint_{leftvert
zrightvert =eta}fleft(zright)g(z) dz$ for all $eta<epsilon$, it is asking whether we can treat $g(z)$ as if it is constant $g(0)$ as $eta to 0$. But it turns out that it is also false even if we assume $f$ and $g$ are analytic. A counterexample is $g(z) = frac{1}{1+z}$ and $h(z) = frac{1}{z}+frac{1}{z^2}$. Then $text{Res}left( h,0right) = 1$ while
$$
frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz = frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}(1+z)(frac{1}{z}+frac{1}{z^2}) dz =2.
$$



Note: However, if we assume that $h$ has a simple pole at $z=0$ and $frac{1}{g(z)} = frac{1}{g(0)}+mathcal{O}(|z|^{epsilon'})$ for some $epsilon'>0$, then $text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz$ must be true.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 15:15











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1 Answer
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1 Answer
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3












$begingroup$

The answer to the question as you stated is no. This is why I doubt it is true. If it were true, then it must be that for all $eta<epsilon$ $$text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{eta}{2}}fleft( zright) dz.$$ The right-hand side may depend on $eta$ if $f$ is not analytic but the left-hand side remains constant. So I guess more meaningful (and weaker) version of your question is to ask whether
$$
text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz
$$ holds. Since $text{Res}left( h,0right) =frac{1}{2pi i}oint_{leftvert
zrightvert =eta}fleft(zright)g(z) dz$ for all $eta<epsilon$, it is asking whether we can treat $g(z)$ as if it is constant $g(0)$ as $eta to 0$. But it turns out that it is also false even if we assume $f$ and $g$ are analytic. A counterexample is $g(z) = frac{1}{1+z}$ and $h(z) = frac{1}{z}+frac{1}{z^2}$. Then $text{Res}left( h,0right) = 1$ while
$$
frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz = frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}(1+z)(frac{1}{z}+frac{1}{z^2}) dz =2.
$$



Note: However, if we assume that $h$ has a simple pole at $z=0$ and $frac{1}{g(z)} = frac{1}{g(0)}+mathcal{O}(|z|^{epsilon'})$ for some $epsilon'>0$, then $text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz$ must be true.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 15:15
















3












$begingroup$

The answer to the question as you stated is no. This is why I doubt it is true. If it were true, then it must be that for all $eta<epsilon$ $$text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{eta}{2}}fleft( zright) dz.$$ The right-hand side may depend on $eta$ if $f$ is not analytic but the left-hand side remains constant. So I guess more meaningful (and weaker) version of your question is to ask whether
$$
text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz
$$ holds. Since $text{Res}left( h,0right) =frac{1}{2pi i}oint_{leftvert
zrightvert =eta}fleft(zright)g(z) dz$ for all $eta<epsilon$, it is asking whether we can treat $g(z)$ as if it is constant $g(0)$ as $eta to 0$. But it turns out that it is also false even if we assume $f$ and $g$ are analytic. A counterexample is $g(z) = frac{1}{1+z}$ and $h(z) = frac{1}{z}+frac{1}{z^2}$. Then $text{Res}left( h,0right) = 1$ while
$$
frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz = frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}(1+z)(frac{1}{z}+frac{1}{z^2}) dz =2.
$$



Note: However, if we assume that $h$ has a simple pole at $z=0$ and $frac{1}{g(z)} = frac{1}{g(0)}+mathcal{O}(|z|^{epsilon'})$ for some $epsilon'>0$, then $text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz$ must be true.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 15:15














3












3








3





$begingroup$

The answer to the question as you stated is no. This is why I doubt it is true. If it were true, then it must be that for all $eta<epsilon$ $$text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{eta}{2}}fleft( zright) dz.$$ The right-hand side may depend on $eta$ if $f$ is not analytic but the left-hand side remains constant. So I guess more meaningful (and weaker) version of your question is to ask whether
$$
text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz
$$ holds. Since $text{Res}left( h,0right) =frac{1}{2pi i}oint_{leftvert
zrightvert =eta}fleft(zright)g(z) dz$ for all $eta<epsilon$, it is asking whether we can treat $g(z)$ as if it is constant $g(0)$ as $eta to 0$. But it turns out that it is also false even if we assume $f$ and $g$ are analytic. A counterexample is $g(z) = frac{1}{1+z}$ and $h(z) = frac{1}{z}+frac{1}{z^2}$. Then $text{Res}left( h,0right) = 1$ while
$$
frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz = frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}(1+z)(frac{1}{z}+frac{1}{z^2}) dz =2.
$$



Note: However, if we assume that $h$ has a simple pole at $z=0$ and $frac{1}{g(z)} = frac{1}{g(0)}+mathcal{O}(|z|^{epsilon'})$ for some $epsilon'>0$, then $text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz$ must be true.






share|cite|improve this answer











$endgroup$



The answer to the question as you stated is no. This is why I doubt it is true. If it were true, then it must be that for all $eta<epsilon$ $$text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}oint_{leftvert
zrightvert =frac{eta}{2}}fleft( zright) dz.$$ The right-hand side may depend on $eta$ if $f$ is not analytic but the left-hand side remains constant. So I guess more meaningful (and weaker) version of your question is to ask whether
$$
text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz
$$ holds. Since $text{Res}left( h,0right) =frac{1}{2pi i}oint_{leftvert
zrightvert =eta}fleft(zright)g(z) dz$ for all $eta<epsilon$, it is asking whether we can treat $g(z)$ as if it is constant $g(0)$ as $eta to 0$. But it turns out that it is also false even if we assume $f$ and $g$ are analytic. A counterexample is $g(z) = frac{1}{1+z}$ and $h(z) = frac{1}{z}+frac{1}{z^2}$. Then $text{Res}left( h,0right) = 1$ while
$$
frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz = frac{g(0)}{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}(1+z)(frac{1}{z}+frac{1}{z^2}) dz =2.
$$



Note: However, if we assume that $h$ has a simple pole at $z=0$ and $frac{1}{g(z)} = frac{1}{g(0)}+mathcal{O}(|z|^{epsilon'})$ for some $epsilon'>0$, then $text{Res}left( h,0right) =frac{gleft( 0right) }{2pi i}lim_{eta to 0}oint_{leftvert
zrightvert =eta}fleft( zright) dz$ must be true.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 1 at 15:09

























answered Jan 1 at 15:03









SongSong

18.5k21651




18.5k21651












  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 15:15


















  • $begingroup$
    Thank you so much
    $endgroup$
    – Djalal Ounadjela
    Jan 1 at 15:15
















$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Jan 1 at 15:15




$begingroup$
Thank you so much
$endgroup$
– Djalal Ounadjela
Jan 1 at 15:15


















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