Mixing
Different ( Is it equivalent?) hypothesis for stronger version of Goursat's theorem
$begingroup$
Ahlfors, states that
Theorem: Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ (interior and on the boundary of the rectangle) by omitting a finite number of interior points $a_i$. If it is true that
$lim_{zto a_i}(z-a_i)f(z)=0$ then the integration of $f(z)$ along $R$ is zero.
Some other books use the hypothesis, Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite number of interior points $a_i$ and f(z) is continuous on $R$ (inside and on the boundary of the rectangle), then integration of $f(z)$ over the rectangle R is zero.
My question is : "Is it equivalent that "$f(z)$ is continuous at $a_i$ and analytic everywhere expect a_i" and "$lim_{zto a_i}(z-a_i)f(z)=0$ and analytic everywhere expect a_i" ?
complex-analysis complex-integration analytic-functions
asked Jan 14 at 3:49
BabaiBabai
2,62621540
$endgroup$
add a comment |
$begingroup$
Ahlfors, states that
Theorem: Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ (interior and on the boundary of the rectangle) by omitting a finite number of interior points $a_i$. If it is true that
$lim_{zto a_i}(z-a_i)f(z)=0$ then the integration of $f(z)$ along $R$ is zero.
Some other books use the hypothesis, Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite number of interior points $a_i$ and f(z) is continuous on $R$ (inside and on the boundary of the rectangle), then integration of $f(z)$ over the rectangle R is zero.
My question is : "Is it equivalent that "$f(z)$ is continuous at $a_i$ and analytic everywhere expect a_i" and "$lim_{zto a_i}(z-a_i)f(z)=0$ and analytic everywhere expect a_i" ?
complex-analysis complex-integration analytic-functions
asked Jan 14 at 3:49
BabaiBabai
2,62621540
$endgroup$
add a comment |
$begingroup$
Ahlfors, states that
Theorem: Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ (interior and on the boundary of the rectangle) by omitting a finite number of interior points $a_i$. If it is true that
$lim_{zto a_i}(z-a_i)f(z)=0$ then the integration of $f(z)$ along $R$ is zero.
Some other books use the hypothesis, Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite number of interior points $a_i$ and f(z) is continuous on $R$ (inside and on the boundary of the rectangle), then integration of $f(z)$ over the rectangle R is zero.
My question is : "Is it equivalent that "$f(z)$ is continuous at $a_i$ and analytic everywhere expect a_i" and "$lim_{zto a_i}(z-a_i)f(z)=0$ and analytic everywhere expect a_i" ?
complex-analysis complex-integration analytic-functions
asked Jan 14 at 3:49
BabaiBabai
2,62621540
$endgroup$
Ahlfors, states that
Theorem: Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ (interior and on the boundary of the rectangle) by omitting a finite number of interior points $a_i$. If it is true that
$lim_{zto a_i}(z-a_i)f(z)=0$ then the integration of $f(z)$ along $R$ is zero.
Some other books use the hypothesis, Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite number of interior points $a_i$ and f(z) is continuous on $R$ (inside and on the boundary of the rectangle), then integration of $f(z)$ over the rectangle R is zero.
My question is : "Is it equivalent that "$f(z)$ is continuous at $a_i$ and analytic everywhere expect a_i" and "$lim_{zto a_i}(z-a_i)f(z)=0$ and analytic everywhere expect a_i" ?
complex-analysis complex-integration analytic-functions
complex-analysis complex-integration analytic-functions
asked Jan 14 at 3:49
BabaiBabai
2,62621540
asked Jan 14 at 3:49
BabaiBabai
2,62621540
asked Jan 14 at 3:49
BabaiBabai
2,62621540
asked Jan 14 at 3:49
BabaiBabai
2,62621540
asked Jan 14 at 3:49
BabaiBabai
2,62621540
2,62621540
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes. Since all $a_i$ are isolated in $R$, those statements equivalent, due to Riemann's theorem on removable singularities.
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
$endgroup$
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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votes
$begingroup$
Yes. Since all $a_i$ are isolated in $R$, those statements equivalent, due to Riemann's theorem on removable singularities.
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
$endgroup$
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
add a comment |
$begingroup$
Yes. Since all $a_i$ are isolated in $R$, those statements equivalent, due to Riemann's theorem on removable singularities.
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
$endgroup$
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
add a comment |
$begingroup$
Yes. Since all $a_i$ are isolated in $R$, those statements equivalent, due to Riemann's theorem on removable singularities.
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
$endgroup$
Yes. Since all $a_i$ are isolated in $R$, those statements equivalent, due to Riemann's theorem on removable singularities.
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
answered Jan 14 at 6:06
Martin RMartin R
28.9k33356
28.9k33356
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
add a comment |
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
I think Riemann's Removable Singularity theorem says $lim_{zto a_i}(z-a_i)f(z)=0 implies lim_{zto a}f(z)$ exists, it need not be equal to $f(a)$, that is need not be continuous. @Martin R
$endgroup$
– Babai
Jan 14 at 15:53
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
@Babai: If $f$ is analytic on $R'$ then it is not defined in the $a_j$. If $lim_{zto a_i}(z-a_i)f(z)=0 $ then $lim_{zto a_j}f(z)$ exists, to that $f(a_j)$ can be defined as that limit, making $f$ continuous. – I'll edit the answer later, when I have more time, to make it more precise.
$endgroup$
– Martin R
Jan 14 at 16:00
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
$begingroup$
Make sense now. Thank you.
$endgroup$
– Babai
Jan 14 at 17:39
add a comment |
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