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Cauchy estimate on angular domain
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This question is a (perhaps naive) 'simplification' of a result in a paper, so the answer could be negative.
Define the cone $Sigma(theta)$ for $thetain(0,pi/2]$,
$$Sigma(theta) = left z = x+iy : x>0, , $$
and define the norm $|f|_theta$ for functions $f$ analytic on $Sigma(theta)$ by
$$|f|_theta = sup_zin Sigma(theta)|f|$$
Let $Y_theta$ be the space of functions analytic on $Sigma(theta)$ with $|f|_theta < infty$. Also let $chi$ be a distinguished analytic function in $Y_pi/2$ with $chi(0) = 0$. It would seem then, that the following inequality is true: Let $fin Y_theta$. Then for any $theta'<theta$,
$$ |chi f'|_theta' leq C fracftheta-theta' $$
where the constant $C$ can depend on $chi$. How is this proven, and how does $chi$ help?
Remarks
- (basic Cauchy Estimate) Let $B(r)$ denote the ball around $0$ of radius $r$, and let $|f|_r$ denote $sup_zin B(r) |f(z)|$. Lets say $fin X_r$ if $f:B(r)to mathbb C$ is analytic on its domain, with $|f|_r<infty$. From the usual Cauchy formula $f'(z) = frac12À iint_partial D fracf(w) dw(w-z)^2$ it is not hard to prove that for any $fin X_r$, with any $r'<r$,
$$ |f'|_r' leq C fracfr-r'.$$
complex-analysis inequality pde cauchy-integral-formula
asked Jul 30 at 17:07
Calvin Khor
8,00411132
add a comment |Â
up vote
0
down vote
favorite
This question is a (perhaps naive) 'simplification' of a result in a paper, so the answer could be negative.
Define the cone $Sigma(theta)$ for $thetain(0,pi/2]$,
$$Sigma(theta) = left z = x+iy : x>0, , $$
and define the norm $|f|_theta$ for functions $f$ analytic on $Sigma(theta)$ by
$$|f|_theta = sup_zin Sigma(theta)|f|$$
Let $Y_theta$ be the space of functions analytic on $Sigma(theta)$ with $|f|_theta < infty$. Also let $chi$ be a distinguished analytic function in $Y_pi/2$ with $chi(0) = 0$. It would seem then, that the following inequality is true: Let $fin Y_theta$. Then for any $theta'<theta$,
$$ |chi f'|_theta' leq C fracftheta-theta' $$
where the constant $C$ can depend on $chi$. How is this proven, and how does $chi$ help?
Remarks
- (basic Cauchy Estimate) Let $B(r)$ denote the ball around $0$ of radius $r$, and let $|f|_r$ denote $sup_zin B(r) |f(z)|$. Lets say $fin X_r$ if $f:B(r)to mathbb C$ is analytic on its domain, with $|f|_r<infty$. From the usual Cauchy formula $f'(z) = frac12À iint_partial D fracf(w) dw(w-z)^2$ it is not hard to prove that for any $fin X_r$, with any $r'<r$,
$$ |f'|_r' leq C fracfr-r'.$$
complex-analysis inequality pde cauchy-integral-formula
asked Jul 30 at 17:07
Calvin Khor
8,00411132
Why is $chi(0)$ defined?
– zhw.
Jul 30 at 17:16
@zhw. I want to assume it continuously extends to 0 with value 0. In fact the paper im reading uses $chi= min(|z|,1)$? Which is weird because they then try to estimate $fracd^ndz^n (chi f')$, so the above is me trying to make sense of this. Hope this helps
– Calvin Khor
Jul 30 at 17:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is a (perhaps naive) 'simplification' of a result in a paper, so the answer could be negative.
Define the cone $Sigma(theta)$ for $thetain(0,pi/2]$,
$$Sigma(theta) = left z = x+iy : x>0, , $$
and define the norm $|f|_theta$ for functions $f$ analytic on $Sigma(theta)$ by
$$|f|_theta = sup_zin Sigma(theta)|f|$$
Let $Y_theta$ be the space of functions analytic on $Sigma(theta)$ with $|f|_theta < infty$. Also let $chi$ be a distinguished analytic function in $Y_pi/2$ with $chi(0) = 0$. It would seem then, that the following inequality is true: Let $fin Y_theta$. Then for any $theta'<theta$,
$$ |chi f'|_theta' leq C fracftheta-theta' $$
where the constant $C$ can depend on $chi$. How is this proven, and how does $chi$ help?
Remarks
- (basic Cauchy Estimate) Let $B(r)$ denote the ball around $0$ of radius $r$, and let $|f|_r$ denote $sup_zin B(r) |f(z)|$. Lets say $fin X_r$ if $f:B(r)to mathbb C$ is analytic on its domain, with $|f|_r<infty$. From the usual Cauchy formula $f'(z) = frac12À iint_partial D fracf(w) dw(w-z)^2$ it is not hard to prove that for any $fin X_r$, with any $r'<r$,
$$ |f'|_r' leq C fracfr-r'.$$
complex-analysis inequality pde cauchy-integral-formula
asked Jul 30 at 17:07
Calvin Khor
8,00411132
This question is a (perhaps naive) 'simplification' of a result in a paper, so the answer could be negative.
Define the cone $Sigma(theta)$ for $thetain(0,pi/2]$,
$$Sigma(theta) = left z = x+iy : x>0, , $$
and define the norm $|f|_theta$ for functions $f$ analytic on $Sigma(theta)$ by
$$|f|_theta = sup_zin Sigma(theta)|f|$$
Let $Y_theta$ be the space of functions analytic on $Sigma(theta)$ with $|f|_theta < infty$. Also let $chi$ be a distinguished analytic function in $Y_pi/2$ with $chi(0) = 0$. It would seem then, that the following inequality is true: Let $fin Y_theta$. Then for any $theta'<theta$,
$$ |chi f'|_theta' leq C fracftheta-theta' $$
where the constant $C$ can depend on $chi$. How is this proven, and how does $chi$ help?
Remarks
- (basic Cauchy Estimate) Let $B(r)$ denote the ball around $0$ of radius $r$, and let $|f|_r$ denote $sup_zin B(r) |f(z)|$. Lets say $fin X_r$ if $f:B(r)to mathbb C$ is analytic on its domain, with $|f|_r<infty$. From the usual Cauchy formula $f'(z) = frac12À iint_partial D fracf(w) dw(w-z)^2$ it is not hard to prove that for any $fin X_r$, with any $r'<r$,
$$ |f'|_r' leq C fracfr-r'.$$
complex-analysis inequality pde cauchy-integral-formula
asked Jul 30 at 17:07
Calvin Khor
8,00411132
asked Jul 30 at 17:07
Calvin Khor
8,00411132
asked Jul 30 at 17:07
Calvin Khor
8,00411132
asked Jul 30 at 17:07
Calvin Khor
8,00411132
8,00411132
Why is $chi(0)$ defined?
– zhw.
Jul 30 at 17:16
@zhw. I want to assume it continuously extends to 0 with value 0. In fact the paper im reading uses $chi= min(|z|,1)$? Which is weird because they then try to estimate $fracd^ndz^n (chi f')$, so the above is me trying to make sense of this. Hope this helps
– Calvin Khor
Jul 30 at 17:27
add a comment |Â
Why is $chi(0)$ defined?
– zhw.
Jul 30 at 17:16
@zhw. I want to assume it continuously extends to 0 with value 0. In fact the paper im reading uses $chi= min(|z|,1)$? Which is weird because they then try to estimate $fracd^ndz^n (chi f')$, so the above is me trying to make sense of this. Hope this helps
– Calvin Khor
Jul 30 at 17:27
Why is $chi(0)$ defined?
– zhw.
Jul 30 at 17:16
Why is $chi(0)$ defined?
– zhw.
Jul 30 at 17:16
@zhw. I want to assume it continuously extends to 0 with value 0. In fact the paper im reading uses $chi= min(|z|,1)$? Which is weird because they then try to estimate $fracd^ndz^n (chi f')$, so the above is me trying to make sense of this. Hope this helps
– Calvin Khor
Jul 30 at 17:27
@zhw. I want to assume it continuously extends to 0 with value 0. In fact the paper im reading uses $chi= min(|z|,1)$? Which is weird because they then try to estimate $fracd^ndz^n (chi f')$, so the above is me trying to make sense of this. Hope this helps
– Calvin Khor
Jul 30 at 17:27
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
If you didn't include the factor $chi,$ the result would imply $f'$ is bounded in $Sigma (theta').$ There are certainly counterexamples to that. But if you take, say, $chi(z) = z,$ then $chi$ tamps down the possible growth of $f'$ near $0$ and we have a chance.
Here's a sketch with $chi(z) = z:$ Show that if $re^it in Sigma (theta'),$ then
$$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$$
Here $c$ is a constant that depends only on $theta, theta'.$ Now we can apply Cauchy's estimates directly:
$$|chi(re^it)f'(re^it)| = r|f'(re^it)| le rfracfd(re^it, partial Sigma (theta)).$$
Apply $(1)$ to finish.
answered Jul 30 at 17:49
zhw.
65.2k42769
1
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
add a comment |Â
up vote
1
down vote
Expanding on the accepted answer.
If you didn't include the factor $Ç$,
the result would imply $f′$
is bounded in $Σ(θ′)$.There are certainly counterexamples to that.
One family of counterexamples is $z^s e^-z$, $s∈(0,1)$, where the branch cut of $z^s$ is chosen to lie outside the sector. The derivatives near the origin are like $1/|z|^1-s$ (c.f. the bound $|zf'(z)|_theta'<|f|_theta$)
Show that if $re^it∈Σ(theta′)$,
then
$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$
The distance $d(re^it, partial Sigma (theta))$ is obtained by dropping a perpendicular to the closest ray of $partial Sigma (theta)$. Therefore, it is half a chord on a circle of radius $r$, across a sector of angle $2(theta-t)$,
hence, $d(re^it, partial Sigma (theta)) = fracr2sin(theta-t)$. For the range $alphain(theta-theta',theta] subset (0,pi/2)$,
$$sin alpha > fracsin(theta) thetaalpha > frac2pi alpha,$$
so setting $t = theta - alpha$, we see that
$$d(re^it, partial Sigma (theta)) ge fracrpi (theta-t) ge fracrpi (theta-theta') .$$
answered Jul 31 at 8:39
Calvin Khor
8,00411132
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you didn't include the factor $chi,$ the result would imply $f'$ is bounded in $Sigma (theta').$ There are certainly counterexamples to that. But if you take, say, $chi(z) = z,$ then $chi$ tamps down the possible growth of $f'$ near $0$ and we have a chance.
Here's a sketch with $chi(z) = z:$ Show that if $re^it in Sigma (theta'),$ then
$$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$$
Here $c$ is a constant that depends only on $theta, theta'.$ Now we can apply Cauchy's estimates directly:
$$|chi(re^it)f'(re^it)| = r|f'(re^it)| le rfracfd(re^it, partial Sigma (theta)).$$
Apply $(1)$ to finish.
answered Jul 30 at 17:49
zhw.
65.2k42769
1
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
add a comment |Â
up vote
1
down vote
accepted
If you didn't include the factor $chi,$ the result would imply $f'$ is bounded in $Sigma (theta').$ There are certainly counterexamples to that. But if you take, say, $chi(z) = z,$ then $chi$ tamps down the possible growth of $f'$ near $0$ and we have a chance.
Here's a sketch with $chi(z) = z:$ Show that if $re^it in Sigma (theta'),$ then
$$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$$
Here $c$ is a constant that depends only on $theta, theta'.$ Now we can apply Cauchy's estimates directly:
$$|chi(re^it)f'(re^it)| = r|f'(re^it)| le rfracfd(re^it, partial Sigma (theta)).$$
Apply $(1)$ to finish.
answered Jul 30 at 17:49
zhw.
65.2k42769
1
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you didn't include the factor $chi,$ the result would imply $f'$ is bounded in $Sigma (theta').$ There are certainly counterexamples to that. But if you take, say, $chi(z) = z,$ then $chi$ tamps down the possible growth of $f'$ near $0$ and we have a chance.
Here's a sketch with $chi(z) = z:$ Show that if $re^it in Sigma (theta'),$ then
$$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$$
Here $c$ is a constant that depends only on $theta, theta'.$ Now we can apply Cauchy's estimates directly:
$$|chi(re^it)f'(re^it)| = r|f'(re^it)| le rfracfd(re^it, partial Sigma (theta)).$$
Apply $(1)$ to finish.
answered Jul 30 at 17:49
zhw.
65.2k42769
If you didn't include the factor $chi,$ the result would imply $f'$ is bounded in $Sigma (theta').$ There are certainly counterexamples to that. But if you take, say, $chi(z) = z,$ then $chi$ tamps down the possible growth of $f'$ near $0$ and we have a chance.
Here's a sketch with $chi(z) = z:$ Show that if $re^it in Sigma (theta'),$ then
$$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$$
Here $c$ is a constant that depends only on $theta, theta'.$ Now we can apply Cauchy's estimates directly:
$$|chi(re^it)f'(re^it)| = r|f'(re^it)| le rfracfd(re^it, partial Sigma (theta)).$$
Apply $(1)$ to finish.
answered Jul 30 at 17:49
zhw.
65.2k42769
edited Jul 30 at 18:42
answered Jul 30 at 17:49
zhw.
65.2k42769
answered Jul 30 at 17:49
zhw.
65.2k42769
answered Jul 30 at 17:49
zhw.
65.2k42769
65.2k42769
1
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
add a comment |Â
1
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
1
1
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
Oh, it seems I don't need to assume $chi$ is bounded? Thank you very much, it seems I now have much more to think about. So it continues...
– Calvin Khor
Jul 30 at 18:10
add a comment |Â
up vote
1
down vote
Expanding on the accepted answer.
If you didn't include the factor $Ç$,
the result would imply $f′$
is bounded in $Σ(θ′)$.There are certainly counterexamples to that.
One family of counterexamples is $z^s e^-z$, $s∈(0,1)$, where the branch cut of $z^s$ is chosen to lie outside the sector. The derivatives near the origin are like $1/|z|^1-s$ (c.f. the bound $|zf'(z)|_theta'<|f|_theta$)
Show that if $re^it∈Σ(theta′)$,
then
$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$
The distance $d(re^it, partial Sigma (theta))$ is obtained by dropping a perpendicular to the closest ray of $partial Sigma (theta)$. Therefore, it is half a chord on a circle of radius $r$, across a sector of angle $2(theta-t)$,
hence, $d(re^it, partial Sigma (theta)) = fracr2sin(theta-t)$. For the range $alphain(theta-theta',theta] subset (0,pi/2)$,
$$sin alpha > fracsin(theta) thetaalpha > frac2pi alpha,$$
so setting $t = theta - alpha$, we see that
$$d(re^it, partial Sigma (theta)) ge fracrpi (theta-t) ge fracrpi (theta-theta') .$$
answered Jul 31 at 8:39
Calvin Khor
8,00411132
add a comment |Â
up vote
1
down vote
Expanding on the accepted answer.
If you didn't include the factor $Ç$,
the result would imply $f′$
is bounded in $Σ(θ′)$.There are certainly counterexamples to that.
One family of counterexamples is $z^s e^-z$, $s∈(0,1)$, where the branch cut of $z^s$ is chosen to lie outside the sector. The derivatives near the origin are like $1/|z|^1-s$ (c.f. the bound $|zf'(z)|_theta'<|f|_theta$)
Show that if $re^it∈Σ(theta′)$,
then
$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$
The distance $d(re^it, partial Sigma (theta))$ is obtained by dropping a perpendicular to the closest ray of $partial Sigma (theta)$. Therefore, it is half a chord on a circle of radius $r$, across a sector of angle $2(theta-t)$,
hence, $d(re^it, partial Sigma (theta)) = fracr2sin(theta-t)$. For the range $alphain(theta-theta',theta] subset (0,pi/2)$,
$$sin alpha > fracsin(theta) thetaalpha > frac2pi alpha,$$
so setting $t = theta - alpha$, we see that
$$d(re^it, partial Sigma (theta)) ge fracrpi (theta-t) ge fracrpi (theta-theta') .$$
answered Jul 31 at 8:39
Calvin Khor
8,00411132
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Expanding on the accepted answer.
If you didn't include the factor $Ç$,
the result would imply $f′$
is bounded in $Σ(θ′)$.There are certainly counterexamples to that.
One family of counterexamples is $z^s e^-z$, $s∈(0,1)$, where the branch cut of $z^s$ is chosen to lie outside the sector. The derivatives near the origin are like $1/|z|^1-s$ (c.f. the bound $|zf'(z)|_theta'<|f|_theta$)
Show that if $re^it∈Σ(theta′)$,
then
$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$
The distance $d(re^it, partial Sigma (theta))$ is obtained by dropping a perpendicular to the closest ray of $partial Sigma (theta)$. Therefore, it is half a chord on a circle of radius $r$, across a sector of angle $2(theta-t)$,
hence, $d(re^it, partial Sigma (theta)) = fracr2sin(theta-t)$. For the range $alphain(theta-theta',theta] subset (0,pi/2)$,
$$sin alpha > fracsin(theta) thetaalpha > frac2pi alpha,$$
so setting $t = theta - alpha$, we see that
$$d(re^it, partial Sigma (theta)) ge fracrpi (theta-t) ge fracrpi (theta-theta') .$$
answered Jul 31 at 8:39
Calvin Khor
8,00411132
Expanding on the accepted answer.
If you didn't include the factor $Ç$,
the result would imply $f′$
is bounded in $Σ(θ′)$.There are certainly counterexamples to that.
One family of counterexamples is $z^s e^-z$, $s∈(0,1)$, where the branch cut of $z^s$ is chosen to lie outside the sector. The derivatives near the origin are like $1/|z|^1-s$ (c.f. the bound $|zf'(z)|_theta'<|f|_theta$)
Show that if $re^it∈Σ(theta′)$,
then
$tag 1 d(re^it, partial Sigma (theta)) ge cr(theta - theta').$
The distance $d(re^it, partial Sigma (theta))$ is obtained by dropping a perpendicular to the closest ray of $partial Sigma (theta)$. Therefore, it is half a chord on a circle of radius $r$, across a sector of angle $2(theta-t)$,
hence, $d(re^it, partial Sigma (theta)) = fracr2sin(theta-t)$. For the range $alphain(theta-theta',theta] subset (0,pi/2)$,
$$sin alpha > fracsin(theta) thetaalpha > frac2pi alpha,$$
so setting $t = theta - alpha$, we see that
$$d(re^it, partial Sigma (theta)) ge fracrpi (theta-t) ge fracrpi (theta-theta') .$$
answered Jul 31 at 8:39
Calvin Khor
8,00411132
edited Jul 31 at 9:00
answered Jul 31 at 8:39
Calvin Khor
8,00411132
answered Jul 31 at 8:39
Calvin Khor
8,00411132
answered Jul 31 at 8:39
Calvin Khor
8,00411132
8,00411132
add a comment |Â
add a comment |Â
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Why is $chi(0)$ defined?
– zhw.
Jul 30 at 17:16
@zhw. I want to assume it continuously extends to 0 with value 0. In fact the paper im reading uses $chi= min(|z|,1)$? Which is weird because they then try to estimate $fracd^ndz^n (chi f')$, so the above is me trying to make sense of this. Hope this helps
– Calvin Khor
Jul 30 at 17:27