Example of function $fin C^{infty}$ but not holomorphic
$begingroup$
I know that holomorphic function are infinitely differentiable .
I think converse not true .I am searching for counterexample. But I did not get .
Please can anyone suggest me how to find such example.
any help will be appreciated.
complex-analysis holomorphic-functions
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
$endgroup$
add a comment |
$begingroup$
I know that holomorphic function are infinitely differentiable .
I think converse not true .I am searching for counterexample. But I did not get .
Please can anyone suggest me how to find such example.
any help will be appreciated.
complex-analysis holomorphic-functions
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
$endgroup$
$begingroup$
I think $zmapsto overline{z}$ is an easy counter. In terms of real it is just $(x,y)mapsto(x,-y)$
$endgroup$
– PSG
Dec 5 '18 at 7:20
$begingroup$
The trick is that $f(z)=bar{z}$ is not complex differentiable, failing to satisfy the Cauchy-Riemann equations
$endgroup$
– obscurans
Dec 5 '18 at 7:21
add a comment |
$begingroup$
I know that holomorphic function are infinitely differentiable .
I think converse not true .I am searching for counterexample. But I did not get .
Please can anyone suggest me how to find such example.
any help will be appreciated.
complex-analysis holomorphic-functions
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
$endgroup$
I know that holomorphic function are infinitely differentiable .
I think converse not true .I am searching for counterexample. But I did not get .
Please can anyone suggest me how to find such example.
any help will be appreciated.
complex-analysis holomorphic-functions
complex-analysis holomorphic-functions
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
asked Dec 5 '18 at 7:09
MathLoverMathLover
49310
49310
$begingroup$
I think $zmapsto overline{z}$ is an easy counter. In terms of real it is just $(x,y)mapsto(x,-y)$
$endgroup$
– PSG
Dec 5 '18 at 7:20
$begingroup$
The trick is that $f(z)=bar{z}$ is not complex differentiable, failing to satisfy the Cauchy-Riemann equations
$endgroup$
– obscurans
Dec 5 '18 at 7:21
add a comment |
$begingroup$
I think $zmapsto overline{z}$ is an easy counter. In terms of real it is just $(x,y)mapsto(x,-y)$
$endgroup$
– PSG
Dec 5 '18 at 7:20
$begingroup$
The trick is that $f(z)=bar{z}$ is not complex differentiable, failing to satisfy the Cauchy-Riemann equations
$endgroup$
– obscurans
Dec 5 '18 at 7:21
$begingroup$
I think $zmapsto overline{z}$ is an easy counter. In terms of real it is just $(x,y)mapsto(x,-y)$
$endgroup$
– PSG
Dec 5 '18 at 7:20
$begingroup$
I think $zmapsto overline{z}$ is an easy counter. In terms of real it is just $(x,y)mapsto(x,-y)$
$endgroup$
– PSG
Dec 5 '18 at 7:20
$begingroup$
The trick is that $f(z)=bar{z}$ is not complex differentiable, failing to satisfy the Cauchy-Riemann equations
$endgroup$
– obscurans
Dec 5 '18 at 7:21
$begingroup$
The trick is that $f(z)=bar{z}$ is not complex differentiable, failing to satisfy the Cauchy-Riemann equations
$endgroup$
– obscurans
Dec 5 '18 at 7:21
add a comment |
1 Answer
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Complex differentiable (even once) implies analytic implies infinitely differentiable.
For (real) infinitely differentiable but not analytic, standard example is $e^{-frac{1}{x^2}}$.
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Complex differentiable (even once) implies analytic implies infinitely differentiable.
For (real) infinitely differentiable but not analytic, standard example is $e^{-frac{1}{x^2}}$.
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
$endgroup$
add a comment |
$begingroup$
Complex differentiable (even once) implies analytic implies infinitely differentiable.
For (real) infinitely differentiable but not analytic, standard example is $e^{-frac{1}{x^2}}$.
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
$endgroup$
add a comment |
$begingroup$
Complex differentiable (even once) implies analytic implies infinitely differentiable.
For (real) infinitely differentiable but not analytic, standard example is $e^{-frac{1}{x^2}}$.
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
$endgroup$
Complex differentiable (even once) implies analytic implies infinitely differentiable.
For (real) infinitely differentiable but not analytic, standard example is $e^{-frac{1}{x^2}}$.
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
answered Dec 5 '18 at 7:12
obscuransobscurans
898311
898311
add a comment |
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$begingroup$
I think $zmapsto overline{z}$ is an easy counter. In terms of real it is just $(x,y)mapsto(x,-y)$
$endgroup$
– PSG
Dec 5 '18 at 7:20
$begingroup$
The trick is that $f(z)=bar{z}$ is not complex differentiable, failing to satisfy the Cauchy-Riemann equations
$endgroup$
– obscurans
Dec 5 '18 at 7:21