Being holomorphic at a single point does not imply to be $C^{infty}$ at the point
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I read a claim such that "being holomorphic at a single point does not imply to be $C^{infty}$ at the point" which is the second answer in the following post Holomorphic functions and real functions: continuity of partial derivatives
The definition is from Wiki: "holomorphic at a point $z_0$" means not just differentiable at $z_0$, but differentiable everywhere within some neighbourhood of $z_0$ in the complex plane.
Can you give me an example for this claim?
complex-analysis
asked Jan 22 at 23:14
user315531user315531
14014
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add a comment |
$begingroup$
I read a claim such that "being holomorphic at a single point does not imply to be $C^{infty}$ at the point" which is the second answer in the following post Holomorphic functions and real functions: continuity of partial derivatives
The definition is from Wiki: "holomorphic at a point $z_0$" means not just differentiable at $z_0$, but differentiable everywhere within some neighbourhood of $z_0$ in the complex plane.
Can you give me an example for this claim?
complex-analysis
asked Jan 22 at 23:14
user315531user315531
14014
$endgroup$
1
$begingroup$
There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{infty}$ in that neighborhood.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 23:24
add a comment |
$begingroup$
I read a claim such that "being holomorphic at a single point does not imply to be $C^{infty}$ at the point" which is the second answer in the following post Holomorphic functions and real functions: continuity of partial derivatives
The definition is from Wiki: "holomorphic at a point $z_0$" means not just differentiable at $z_0$, but differentiable everywhere within some neighbourhood of $z_0$ in the complex plane.
Can you give me an example for this claim?
complex-analysis
asked Jan 22 at 23:14
user315531user315531
14014
$endgroup$
I read a claim such that "being holomorphic at a single point does not imply to be $C^{infty}$ at the point" which is the second answer in the following post Holomorphic functions and real functions: continuity of partial derivatives
The definition is from Wiki: "holomorphic at a point $z_0$" means not just differentiable at $z_0$, but differentiable everywhere within some neighbourhood of $z_0$ in the complex plane.
Can you give me an example for this claim?
complex-analysis
complex-analysis
asked Jan 22 at 23:14
user315531user315531
14014
asked Jan 22 at 23:14
user315531user315531
14014
asked Jan 22 at 23:14
user315531user315531
14014
asked Jan 22 at 23:14
user315531user315531
14014
asked Jan 22 at 23:14
user315531user315531
14014
14014
1
$begingroup$
There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{infty}$ in that neighborhood.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 23:24
add a comment |
1
$begingroup$
There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{infty}$ in that neighborhood.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 23:24
1
1
$begingroup$
There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{infty}$ in that neighborhood.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 23:24
$begingroup$
There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{infty}$ in that neighborhood.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 23:24
add a comment |
1 Answer
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oldest
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This is just a matter of conflicting definitions. If a holomorphic function at a point is defined to be complex-differentiable in a neighborhood of the point (which is the standard definition), then a holomorphic function is always $C^infty$. On the other hand, if a holomorphic function at a point is just required to be complex-differentiable at the point itself, then it need not be $C^infty$. The claim which you linked is using the second definition, not the first.
For a simple example, let $g:mathbb{R}tomathbb{R}$ be any function which is differentiable at $0$ with $g'(0)=0$, but not $C^infty$ in any neighborhood of $0$ (for instance, $g(x)=x|x|$). Then $f:mathbb{C}tomathbb{C}$ defined by $f(z)=g(operatorname{Re}(z))$ is complex-differentiable at any $z$ with $operatorname{Re}(z)=0$ (and $f'(z)=0$ at those points), but is not $C^infty$ in any neighborhood of such a point.
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
add a comment |
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This is just a matter of conflicting definitions. If a holomorphic function at a point is defined to be complex-differentiable in a neighborhood of the point (which is the standard definition), then a holomorphic function is always $C^infty$. On the other hand, if a holomorphic function at a point is just required to be complex-differentiable at the point itself, then it need not be $C^infty$. The claim which you linked is using the second definition, not the first.
For a simple example, let $g:mathbb{R}tomathbb{R}$ be any function which is differentiable at $0$ with $g'(0)=0$, but not $C^infty$ in any neighborhood of $0$ (for instance, $g(x)=x|x|$). Then $f:mathbb{C}tomathbb{C}$ defined by $f(z)=g(operatorname{Re}(z))$ is complex-differentiable at any $z$ with $operatorname{Re}(z)=0$ (and $f'(z)=0$ at those points), but is not $C^infty$ in any neighborhood of such a point.
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
add a comment |
$begingroup$
This is just a matter of conflicting definitions. If a holomorphic function at a point is defined to be complex-differentiable in a neighborhood of the point (which is the standard definition), then a holomorphic function is always $C^infty$. On the other hand, if a holomorphic function at a point is just required to be complex-differentiable at the point itself, then it need not be $C^infty$. The claim which you linked is using the second definition, not the first.
For a simple example, let $g:mathbb{R}tomathbb{R}$ be any function which is differentiable at $0$ with $g'(0)=0$, but not $C^infty$ in any neighborhood of $0$ (for instance, $g(x)=x|x|$). Then $f:mathbb{C}tomathbb{C}$ defined by $f(z)=g(operatorname{Re}(z))$ is complex-differentiable at any $z$ with $operatorname{Re}(z)=0$ (and $f'(z)=0$ at those points), but is not $C^infty$ in any neighborhood of such a point.
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
add a comment |
$begingroup$
This is just a matter of conflicting definitions. If a holomorphic function at a point is defined to be complex-differentiable in a neighborhood of the point (which is the standard definition), then a holomorphic function is always $C^infty$. On the other hand, if a holomorphic function at a point is just required to be complex-differentiable at the point itself, then it need not be $C^infty$. The claim which you linked is using the second definition, not the first.
For a simple example, let $g:mathbb{R}tomathbb{R}$ be any function which is differentiable at $0$ with $g'(0)=0$, but not $C^infty$ in any neighborhood of $0$ (for instance, $g(x)=x|x|$). Then $f:mathbb{C}tomathbb{C}$ defined by $f(z)=g(operatorname{Re}(z))$ is complex-differentiable at any $z$ with $operatorname{Re}(z)=0$ (and $f'(z)=0$ at those points), but is not $C^infty$ in any neighborhood of such a point.
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
This is just a matter of conflicting definitions. If a holomorphic function at a point is defined to be complex-differentiable in a neighborhood of the point (which is the standard definition), then a holomorphic function is always $C^infty$. On the other hand, if a holomorphic function at a point is just required to be complex-differentiable at the point itself, then it need not be $C^infty$. The claim which you linked is using the second definition, not the first.
For a simple example, let $g:mathbb{R}tomathbb{R}$ be any function which is differentiable at $0$ with $g'(0)=0$, but not $C^infty$ in any neighborhood of $0$ (for instance, $g(x)=x|x|$). Then $f:mathbb{C}tomathbb{C}$ defined by $f(z)=g(operatorname{Re}(z))$ is complex-differentiable at any $z$ with $operatorname{Re}(z)=0$ (and $f'(z)=0$ at those points), but is not $C^infty$ in any neighborhood of such a point.
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
edited Jan 23 at 7:24
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
answered Jan 22 at 23:25
Eric WofseyEric Wofsey
187k14216344
187k14216344
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
add a comment |
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
$begingroup$
Can you give me an example such that $f^{prime}(z_0)$ exist but $f^{primeprime}(z_0)$ does not exist?
$endgroup$
– user315531
Jan 23 at 7:16
add a comment |
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There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{infty}$ in that neighborhood.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 23:24