If $f in H(Omega)$ holomorphic, then $|f(x) - f(y)| geq (1/2)|f'(c)||x-y|$
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I was reading Rudin's proof on the Open mapping theorem. He declare $Omega$ to be a region, which is connected and open.
Then he claims there is a neighbourhood of $c$ such that
$|f(x) - f(y)| geq (1/2)|f'(c)||x-y|.$ How is this obtained? Is it because being holomorphic yields the inequalities.
begin{align}
left |frac{ phi(x) - phi(y) }{x - y} - phi'(c) right | < (1/2)|phi'(c)| &iff -(1/2)|phi'(c)| <left| frac{ phi(x) - phi(y) }{x - y} right| - |phi'(c)| < (1/2)|phi'(c)| \
&iff (1/2)|phi'(c)||x-y| < left| phi(x) - phi(y) right|
end{align}
I am not sure how to fix the second $iff$ as it is only true in real analysis. Also I am not sure how to relax the last inequality to $leq$
complex-analysis inequality
asked Nov 16 at 3:54
Hawk
5,4211138102
add a comment |
up vote
1
down vote
favorite
I was reading Rudin's proof on the Open mapping theorem. He declare $Omega$ to be a region, which is connected and open.
Then he claims there is a neighbourhood of $c$ such that
$|f(x) - f(y)| geq (1/2)|f'(c)||x-y|.$ How is this obtained? Is it because being holomorphic yields the inequalities.
begin{align}
left |frac{ phi(x) - phi(y) }{x - y} - phi'(c) right | < (1/2)|phi'(c)| &iff -(1/2)|phi'(c)| <left| frac{ phi(x) - phi(y) }{x - y} right| - |phi'(c)| < (1/2)|phi'(c)| \
&iff (1/2)|phi'(c)||x-y| < left| phi(x) - phi(y) right|
end{align}
I am not sure how to fix the second $iff$ as it is only true in real analysis. Also I am not sure how to relax the last inequality to $leq$
complex-analysis inequality
asked Nov 16 at 3:54
Hawk
5,4211138102
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was reading Rudin's proof on the Open mapping theorem. He declare $Omega$ to be a region, which is connected and open.
Then he claims there is a neighbourhood of $c$ such that
$|f(x) - f(y)| geq (1/2)|f'(c)||x-y|.$ How is this obtained? Is it because being holomorphic yields the inequalities.
begin{align}
left |frac{ phi(x) - phi(y) }{x - y} - phi'(c) right | < (1/2)|phi'(c)| &iff -(1/2)|phi'(c)| <left| frac{ phi(x) - phi(y) }{x - y} right| - |phi'(c)| < (1/2)|phi'(c)| \
&iff (1/2)|phi'(c)||x-y| < left| phi(x) - phi(y) right|
end{align}
I am not sure how to fix the second $iff$ as it is only true in real analysis. Also I am not sure how to relax the last inequality to $leq$
complex-analysis inequality
asked Nov 16 at 3:54
Hawk
5,4211138102
I was reading Rudin's proof on the Open mapping theorem. He declare $Omega$ to be a region, which is connected and open.
Then he claims there is a neighbourhood of $c$ such that
$|f(x) - f(y)| geq (1/2)|f'(c)||x-y|.$ How is this obtained? Is it because being holomorphic yields the inequalities.
begin{align}
left |frac{ phi(x) - phi(y) }{x - y} - phi'(c) right | < (1/2)|phi'(c)| &iff -(1/2)|phi'(c)| <left| frac{ phi(x) - phi(y) }{x - y} right| - |phi'(c)| < (1/2)|phi'(c)| \
&iff (1/2)|phi'(c)||x-y| < left| phi(x) - phi(y) right|
end{align}
I am not sure how to fix the second $iff$ as it is only true in real analysis. Also I am not sure how to relax the last inequality to $leq$
complex-analysis inequality
complex-analysis inequality
asked Nov 16 at 3:54
Hawk
5,4211138102
asked Nov 16 at 3:54
Hawk
5,4211138102
asked Nov 16 at 3:54
Hawk
5,4211138102
asked Nov 16 at 3:54
Hawk
5,4211138102
asked Nov 16 at 3:54
Hawk
5,4211138102
5,4211138102
add a comment |
add a comment |
1 Answer
1
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up vote
2
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accepted
$|z_1-z_2| geq |z_2|-|z_1|$ for all complex numbers $z_1,z_2$. Hence $|frac {phi(x)-phi(y)} {x-y} -phi'(c)| geq |phi'(c)|- |frac {phi(x)-phi(y)} {x-y}|$ which gives $|frac {phi(x)-phi(y)} {x-y}| geq frac 1 2 |phi'(c)|$.
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$|z_1-z_2| geq |z_2|-|z_1|$ for all complex numbers $z_1,z_2$. Hence $|frac {phi(x)-phi(y)} {x-y} -phi'(c)| geq |phi'(c)|- |frac {phi(x)-phi(y)} {x-y}|$ which gives $|frac {phi(x)-phi(y)} {x-y}| geq frac 1 2 |phi'(c)|$.
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
add a comment |
up vote
2
down vote
accepted
$|z_1-z_2| geq |z_2|-|z_1|$ for all complex numbers $z_1,z_2$. Hence $|frac {phi(x)-phi(y)} {x-y} -phi'(c)| geq |phi'(c)|- |frac {phi(x)-phi(y)} {x-y}|$ which gives $|frac {phi(x)-phi(y)} {x-y}| geq frac 1 2 |phi'(c)|$.
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$|z_1-z_2| geq |z_2|-|z_1|$ for all complex numbers $z_1,z_2$. Hence $|frac {phi(x)-phi(y)} {x-y} -phi'(c)| geq |phi'(c)|- |frac {phi(x)-phi(y)} {x-y}|$ which gives $|frac {phi(x)-phi(y)} {x-y}| geq frac 1 2 |phi'(c)|$.
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
$|z_1-z_2| geq |z_2|-|z_1|$ for all complex numbers $z_1,z_2$. Hence $|frac {phi(x)-phi(y)} {x-y} -phi'(c)| geq |phi'(c)|- |frac {phi(x)-phi(y)} {x-y}|$ which gives $|frac {phi(x)-phi(y)} {x-y}| geq frac 1 2 |phi'(c)|$.
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
edited Nov 16 at 6:26
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
answered Nov 16 at 5:21
Kavi Rama Murthy
40k31750
40k31750
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
add a comment |
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
Well now I feel silly asking...
– Hawk
Nov 16 at 8:17
add a comment |
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