Mixing
Morera's theorem, Conway's proof
$begingroup$
I have a question about the proof of Morera's theorem as presented in Conway's text volume I.
Let $G$ be a region and let $ f:G to mathbb{C} $ be a continuous function.
Fix $ z_o$ in $G$. Then for any $z in G$, let $[z_o,z]$ denote a line segment from $ z_o$ to $z$.
Then,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq |f(z)-f(z_o)|$$
I want to ask how to establish that inequality, my attempt is this, I know that
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq frac{V([z_o,z])}{|z-z_o|}~ text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
where $ V([z_o,z]) $ is the length of $ [z_o,z]$.
So,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg| leq text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
I can't show that $$text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} leq |f(z)-f(z_o)| $$
just by the continuity of $f$.
I feel that I'm just missing some simple detail to show this. thank you.
complex-analysis
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
$endgroup$
|
show 1 more comment
$begingroup$
I have a question about the proof of Morera's theorem as presented in Conway's text volume I.
Let $G$ be a region and let $ f:G to mathbb{C} $ be a continuous function.
Fix $ z_o$ in $G$. Then for any $z in G$, let $[z_o,z]$ denote a line segment from $ z_o$ to $z$.
Then,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq |f(z)-f(z_o)|$$
I want to ask how to establish that inequality, my attempt is this, I know that
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq frac{V([z_o,z])}{|z-z_o|}~ text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
where $ V([z_o,z]) $ is the length of $ [z_o,z]$.
So,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg| leq text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
I can't show that $$text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} leq |f(z)-f(z_o)| $$
just by the continuity of $f$.
I feel that I'm just missing some simple detail to show this. thank you.
complex-analysis
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
$endgroup$
1
$begingroup$
Are you sure that you quoted the estimate from Conway's book correctly? What you need (in order to prove Morera) is to show that the left-hand side converges to zero for $ z to z_0$, and that follows from your estimate (with the supremum on the right-hand side) and the continuity of $f$.
$endgroup$
– Martin R
Jan 30 '16 at 9:21
$begingroup$
@MartinR thank you for the response. Yes, I check and I did quote it correctly. that's why it keeps bugging me about that. I know we can bound sup{...} to complete the proof. But I was puzzle about that inequality. Do you think this might be a typo?? I thought that before but I just want to post it on here to see if I miss anything.
$endgroup$
– Khoa ta
Jan 30 '16 at 22:24
1
$begingroup$
I don't see how that inequality can be obtained, but perhaps I am overlooking something. I don't have Conway's book to check it there.
$endgroup$
– Martin R
Jan 30 '16 at 23:33
$begingroup$
Let $z_0=0,,z=1$ and $f(w)=w(1-w)$. Thenbegin{align} &left|frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dwright| =int_0^1 x(1-x) dx=frac{1}{6},\ &|f(z)-f(z_o)|=0. end{align} The inequality is not correct.
$endgroup$
– ts375_zk26
Jan 30 '16 at 23:40
$begingroup$
@ts375_zk26 thank you for your response, it is clear now.
$endgroup$
– Khoa ta
Jan 31 '16 at 0:26
|
show 1 more comment
$begingroup$
I have a question about the proof of Morera's theorem as presented in Conway's text volume I.
Let $G$ be a region and let $ f:G to mathbb{C} $ be a continuous function.
Fix $ z_o$ in $G$. Then for any $z in G$, let $[z_o,z]$ denote a line segment from $ z_o$ to $z$.
Then,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq |f(z)-f(z_o)|$$
I want to ask how to establish that inequality, my attempt is this, I know that
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq frac{V([z_o,z])}{|z-z_o|}~ text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
where $ V([z_o,z]) $ is the length of $ [z_o,z]$.
So,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg| leq text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
I can't show that $$text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} leq |f(z)-f(z_o)| $$
just by the continuity of $f$.
I feel that I'm just missing some simple detail to show this. thank you.
complex-analysis
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
$endgroup$
I have a question about the proof of Morera's theorem as presented in Conway's text volume I.
Let $G$ be a region and let $ f:G to mathbb{C} $ be a continuous function.
Fix $ z_o$ in $G$. Then for any $z in G$, let $[z_o,z]$ denote a line segment from $ z_o$ to $z$.
Then,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq |f(z)-f(z_o)|$$
I want to ask how to establish that inequality, my attempt is this, I know that
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq frac{V([z_o,z])}{|z-z_o|}~ text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
where $ V([z_o,z]) $ is the length of $ [z_o,z]$.
So,
$$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg| leq text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} $$
I can't show that $$text{sup}{ ~|f(w)-f(z_o)| : w in [z_o,z] ~} leq |f(z)-f(z_o)| $$
just by the continuity of $f$.
I feel that I'm just missing some simple detail to show this. thank you.
complex-analysis
complex-analysis
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
asked Jan 30 '16 at 7:32
Khoa taKhoa ta
23719
23719
1
$begingroup$
Are you sure that you quoted the estimate from Conway's book correctly? What you need (in order to prove Morera) is to show that the left-hand side converges to zero for $ z to z_0$, and that follows from your estimate (with the supremum on the right-hand side) and the continuity of $f$.
$endgroup$
– Martin R
Jan 30 '16 at 9:21
$begingroup$
@MartinR thank you for the response. Yes, I check and I did quote it correctly. that's why it keeps bugging me about that. I know we can bound sup{...} to complete the proof. But I was puzzle about that inequality. Do you think this might be a typo?? I thought that before but I just want to post it on here to see if I miss anything.
$endgroup$
– Khoa ta
Jan 30 '16 at 22:24
1
$begingroup$
I don't see how that inequality can be obtained, but perhaps I am overlooking something. I don't have Conway's book to check it there.
$endgroup$
– Martin R
Jan 30 '16 at 23:33
$begingroup$
Let $z_0=0,,z=1$ and $f(w)=w(1-w)$. Thenbegin{align} &left|frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dwright| =int_0^1 x(1-x) dx=frac{1}{6},\ &|f(z)-f(z_o)|=0. end{align} The inequality is not correct.
$endgroup$
– ts375_zk26
Jan 30 '16 at 23:40
$begingroup$
@ts375_zk26 thank you for your response, it is clear now.
$endgroup$
– Khoa ta
Jan 31 '16 at 0:26
|
show 1 more comment
1
$begingroup$
Are you sure that you quoted the estimate from Conway's book correctly? What you need (in order to prove Morera) is to show that the left-hand side converges to zero for $ z to z_0$, and that follows from your estimate (with the supremum on the right-hand side) and the continuity of $f$.
$endgroup$
– Martin R
Jan 30 '16 at 9:21
$begingroup$
@MartinR thank you for the response. Yes, I check and I did quote it correctly. that's why it keeps bugging me about that. I know we can bound sup{...} to complete the proof. But I was puzzle about that inequality. Do you think this might be a typo?? I thought that before but I just want to post it on here to see if I miss anything.
$endgroup$
– Khoa ta
Jan 30 '16 at 22:24
1
$begingroup$
I don't see how that inequality can be obtained, but perhaps I am overlooking something. I don't have Conway's book to check it there.
$endgroup$
– Martin R
Jan 30 '16 at 23:33
$begingroup$
Let $z_0=0,,z=1$ and $f(w)=w(1-w)$. Thenbegin{align} &left|frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dwright| =int_0^1 x(1-x) dx=frac{1}{6},\ &|f(z)-f(z_o)|=0. end{align} The inequality is not correct.
$endgroup$
– ts375_zk26
Jan 30 '16 at 23:40
$begingroup$
@ts375_zk26 thank you for your response, it is clear now.
$endgroup$
– Khoa ta
Jan 31 '16 at 0:26
1
1
$begingroup$
Are you sure that you quoted the estimate from Conway's book correctly? What you need (in order to prove Morera) is to show that the left-hand side converges to zero for $ z to z_0$, and that follows from your estimate (with the supremum on the right-hand side) and the continuity of $f$.
$endgroup$
– Martin R
Jan 30 '16 at 9:21
$begingroup$
Are you sure that you quoted the estimate from Conway's book correctly? What you need (in order to prove Morera) is to show that the left-hand side converges to zero for $ z to z_0$, and that follows from your estimate (with the supremum on the right-hand side) and the continuity of $f$.
$endgroup$
– Martin R
Jan 30 '16 at 9:21
$begingroup$
@MartinR thank you for the response. Yes, I check and I did quote it correctly. that's why it keeps bugging me about that. I know we can bound sup{...} to complete the proof. But I was puzzle about that inequality. Do you think this might be a typo?? I thought that before but I just want to post it on here to see if I miss anything.
$endgroup$
– Khoa ta
Jan 30 '16 at 22:24
$begingroup$
@MartinR thank you for the response. Yes, I check and I did quote it correctly. that's why it keeps bugging me about that. I know we can bound sup{...} to complete the proof. But I was puzzle about that inequality. Do you think this might be a typo?? I thought that before but I just want to post it on here to see if I miss anything.
$endgroup$
– Khoa ta
Jan 30 '16 at 22:24
1
1
$begingroup$
I don't see how that inequality can be obtained, but perhaps I am overlooking something. I don't have Conway's book to check it there.
$endgroup$
– Martin R
Jan 30 '16 at 23:33
$begingroup$
I don't see how that inequality can be obtained, but perhaps I am overlooking something. I don't have Conway's book to check it there.
$endgroup$
– Martin R
Jan 30 '16 at 23:33
$begingroup$
Let $z_0=0,,z=1$ and $f(w)=w(1-w)$. Thenbegin{align} &left|frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dwright| =int_0^1 x(1-x) dx=frac{1}{6},\ &|f(z)-f(z_o)|=0. end{align} The inequality is not correct.
$endgroup$
– ts375_zk26
Jan 30 '16 at 23:40
$begingroup$
Let $z_0=0,,z=1$ and $f(w)=w(1-w)$. Thenbegin{align} &left|frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dwright| =int_0^1 x(1-x) dx=frac{1}{6},\ &|f(z)-f(z_o)|=0. end{align} The inequality is not correct.
$endgroup$
– ts375_zk26
Jan 30 '16 at 23:40
$begingroup$
@ts375_zk26 thank you for your response, it is clear now.
$endgroup$
– Khoa ta
Jan 31 '16 at 0:26
$begingroup$
@ts375_zk26 thank you for your response, it is clear now.
$endgroup$
– Khoa ta
Jan 31 '16 at 0:26
|
show 1 more comment
1 Answer
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$begingroup$
He meant to use the Cauchy ML-inequality and then take the limit I believe.
$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq max_{z in [z,z0]} |f(z)-f(z_o)| $
and then take the lim as z approaches z0 to get the result.
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
$endgroup$
add a comment |
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$begingroup$
He meant to use the Cauchy ML-inequality and then take the limit I believe.
$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq max_{z in [z,z0]} |f(z)-f(z_o)| $
and then take the lim as z approaches z0 to get the result.
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
$endgroup$
add a comment |
$begingroup$
He meant to use the Cauchy ML-inequality and then take the limit I believe.
$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq max_{z in [z,z0]} |f(z)-f(z_o)| $
and then take the lim as z approaches z0 to get the result.
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
$endgroup$
add a comment |
$begingroup$
He meant to use the Cauchy ML-inequality and then take the limit I believe.
$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq max_{z in [z,z0]} |f(z)-f(z_o)| $
and then take the lim as z approaches z0 to get the result.
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
$endgroup$
He meant to use the Cauchy ML-inequality and then take the limit I believe.
$bigg| frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dw bigg|leq max_{z in [z,z0]} |f(z)-f(z_o)| $
and then take the lim as z approaches z0 to get the result.
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
answered Jan 1 at 22:27
David WarrenDavid Warren
586313
586313
add a comment |
add a comment |
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$begingroup$
Are you sure that you quoted the estimate from Conway's book correctly? What you need (in order to prove Morera) is to show that the left-hand side converges to zero for $ z to z_0$, and that follows from your estimate (with the supremum on the right-hand side) and the continuity of $f$.
$endgroup$
– Martin R
Jan 30 '16 at 9:21
$begingroup$
@MartinR thank you for the response. Yes, I check and I did quote it correctly. that's why it keeps bugging me about that. I know we can bound sup{...} to complete the proof. But I was puzzle about that inequality. Do you think this might be a typo?? I thought that before but I just want to post it on here to see if I miss anything.
$endgroup$
– Khoa ta
Jan 30 '16 at 22:24
1
$begingroup$
I don't see how that inequality can be obtained, but perhaps I am overlooking something. I don't have Conway's book to check it there.
$endgroup$
– Martin R
Jan 30 '16 at 23:33
$begingroup$
Let $z_0=0,,z=1$ and $f(w)=w(1-w)$. Thenbegin{align} &left|frac{1}{z-z_o} int_{[z_o,z]} [f(w)-f(z_o)] dwright| =int_0^1 x(1-x) dx=frac{1}{6},\ &|f(z)-f(z_o)|=0. end{align} The inequality is not correct.
$endgroup$
– ts375_zk26
Jan 30 '16 at 23:40
$begingroup$
@ts375_zk26 thank you for your response, it is clear now.
$endgroup$
– Khoa ta
Jan 31 '16 at 0:26