Mixing
Find the element representing evaluation at a point in the Hilbert space of $L^2$ holomorphic functions on the disk
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The space $S$ of holomorphic functions on the disk $mathbbD$ which are also in $L^2(mathbbD)$ is a Hilbert space, with the inner product being $langle f, grangle = int f(z) overlineg(z) dA(z)$, where $dA$ denotes Lebesgue measure. The linear functional $L_z_0$, given by $fmapsto f(z_0)$, is a bounded linear functional on $S$, and thus there must exist some $g_z_0in S$ so that $L_z_0(f) = langle f, g_z_0rangle$ for all $f$. It turns out, this function is given by
$$g_z_0(w) = frac1(1-overlinez_0z)^2$$
I have seen this proven by representing $f$ and $g$ as power series, performing the multiplication, and working out the coefficients of the $g$ power series by brute force, which is fine. But I want to think that there is an easier way to do it, or at least a more conceptually satisfying way. For example, I would think to do it by first noting that for $z_0 = 0$, $g_z_0$ can be a constant function (I believe $1/pi$), and then getting the general case by composing with a Mobius transformation; indeed, the function $g_z_0$ is closely related to the derivative of such a Mobius transformation, but I have not been able to get such an approach to work. I realized that I don't really know any machinery for working with non-holomorphic functions (like $overlineg_z_0$), and area integrals of complex functions are also somewhat new to me.
In sum, Can anyone give me a conceptually satisfying way of approaching this problem, hopefully one that uses some classical complex analysis tricks like Cauchy's integral formula, Mobius transformations, and the like?
Or, Can you direct me to a resource which has the basic tools for solving this problem?
complex-analysis functional-analysis hilbert-spaces
asked Jul 16 at 20:28
Van Latimer
1177
add a comment |Â
up vote
2
down vote
favorite
The space $S$ of holomorphic functions on the disk $mathbbD$ which are also in $L^2(mathbbD)$ is a Hilbert space, with the inner product being $langle f, grangle = int f(z) overlineg(z) dA(z)$, where $dA$ denotes Lebesgue measure. The linear functional $L_z_0$, given by $fmapsto f(z_0)$, is a bounded linear functional on $S$, and thus there must exist some $g_z_0in S$ so that $L_z_0(f) = langle f, g_z_0rangle$ for all $f$. It turns out, this function is given by
$$g_z_0(w) = frac1(1-overlinez_0z)^2$$
I have seen this proven by representing $f$ and $g$ as power series, performing the multiplication, and working out the coefficients of the $g$ power series by brute force, which is fine. But I want to think that there is an easier way to do it, or at least a more conceptually satisfying way. For example, I would think to do it by first noting that for $z_0 = 0$, $g_z_0$ can be a constant function (I believe $1/pi$), and then getting the general case by composing with a Mobius transformation; indeed, the function $g_z_0$ is closely related to the derivative of such a Mobius transformation, but I have not been able to get such an approach to work. I realized that I don't really know any machinery for working with non-holomorphic functions (like $overlineg_z_0$), and area integrals of complex functions are also somewhat new to me.
In sum, Can anyone give me a conceptually satisfying way of approaching this problem, hopefully one that uses some classical complex analysis tricks like Cauchy's integral formula, Mobius transformations, and the like?
Or, Can you direct me to a resource which has the basic tools for solving this problem?
complex-analysis functional-analysis hilbert-spaces
asked Jul 16 at 20:28
Van Latimer
1177
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The space $S$ of holomorphic functions on the disk $mathbbD$ which are also in $L^2(mathbbD)$ is a Hilbert space, with the inner product being $langle f, grangle = int f(z) overlineg(z) dA(z)$, where $dA$ denotes Lebesgue measure. The linear functional $L_z_0$, given by $fmapsto f(z_0)$, is a bounded linear functional on $S$, and thus there must exist some $g_z_0in S$ so that $L_z_0(f) = langle f, g_z_0rangle$ for all $f$. It turns out, this function is given by
$$g_z_0(w) = frac1(1-overlinez_0z)^2$$
I have seen this proven by representing $f$ and $g$ as power series, performing the multiplication, and working out the coefficients of the $g$ power series by brute force, which is fine. But I want to think that there is an easier way to do it, or at least a more conceptually satisfying way. For example, I would think to do it by first noting that for $z_0 = 0$, $g_z_0$ can be a constant function (I believe $1/pi$), and then getting the general case by composing with a Mobius transformation; indeed, the function $g_z_0$ is closely related to the derivative of such a Mobius transformation, but I have not been able to get such an approach to work. I realized that I don't really know any machinery for working with non-holomorphic functions (like $overlineg_z_0$), and area integrals of complex functions are also somewhat new to me.
In sum, Can anyone give me a conceptually satisfying way of approaching this problem, hopefully one that uses some classical complex analysis tricks like Cauchy's integral formula, Mobius transformations, and the like?
Or, Can you direct me to a resource which has the basic tools for solving this problem?
complex-analysis functional-analysis hilbert-spaces
asked Jul 16 at 20:28
Van Latimer
1177
The space $S$ of holomorphic functions on the disk $mathbbD$ which are also in $L^2(mathbbD)$ is a Hilbert space, with the inner product being $langle f, grangle = int f(z) overlineg(z) dA(z)$, where $dA$ denotes Lebesgue measure. The linear functional $L_z_0$, given by $fmapsto f(z_0)$, is a bounded linear functional on $S$, and thus there must exist some $g_z_0in S$ so that $L_z_0(f) = langle f, g_z_0rangle$ for all $f$. It turns out, this function is given by
$$g_z_0(w) = frac1(1-overlinez_0z)^2$$
I have seen this proven by representing $f$ and $g$ as power series, performing the multiplication, and working out the coefficients of the $g$ power series by brute force, which is fine. But I want to think that there is an easier way to do it, or at least a more conceptually satisfying way. For example, I would think to do it by first noting that for $z_0 = 0$, $g_z_0$ can be a constant function (I believe $1/pi$), and then getting the general case by composing with a Mobius transformation; indeed, the function $g_z_0$ is closely related to the derivative of such a Mobius transformation, but I have not been able to get such an approach to work. I realized that I don't really know any machinery for working with non-holomorphic functions (like $overlineg_z_0$), and area integrals of complex functions are also somewhat new to me.
In sum, Can anyone give me a conceptually satisfying way of approaching this problem, hopefully one that uses some classical complex analysis tricks like Cauchy's integral formula, Mobius transformations, and the like?
Or, Can you direct me to a resource which has the basic tools for solving this problem?
complex-analysis functional-analysis hilbert-spaces
asked Jul 16 at 20:28
Van Latimer
1177
asked Jul 16 at 20:28
Van Latimer
1177
asked Jul 16 at 20:28
Van Latimer
1177
asked Jul 16 at 20:28
Van Latimer
1177
1177
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your Möbius transformation idea is good, but it needs one extra thing to work. We know that
$$
f(0) = frac1piiint_z f(z),dA(z)
tag1$$
Given $z_0$, apply (1) to $g = (fcirc phi^-1)(phi^-1)'$ where $phi(z)= (z-z_0)/(1-bar z_0z) $ is a Möbius transformation sending $z_0$ to $ 0$. (Here, $(phi^-1)'$ is that extra thing.) Since
$$phi'(z) = frac1-bar z_0 z + bar z_0(z-z_0) (1-bar z_0 z)^2 = frac^2 (1-bar z_0 z)^2 tag2$$
the change of variables $z=phi(w)$ yields
$$
f(z_0) (phi^-1)'(0) = g(0) = frac1piiint_z g(z),dA(z)
= frac1piiint_w fracf(w)phi'(w) |phi'(w)|^2,dA(w)
tag3$$
This simplifies to
$$
fracf(z_0)phi'(z_0) = frac1piiint_w f(w) overlinephi'(w) ,dA(w)
tag4$$
and, using (2), further simplifies to
$$
f(z_0) (1-|z_0|^2) = frac1piiint_w f(w) fracz_0(1-z_0 bar w)^2 ,dA(w)
tag6$$
which is the desired result
$$
f(z_0) = frac1piiint_w f(w) frac1(1-z_0 bar w)^2 ,dA(w)
tag7$$
What's the idea of putting the derivative factor $(phi^-1)'$ in the formula for $g$? Without it we'd end up with the Jacobian $|phi'|^2 = phi'overlinephi'$ as the factor of $f$ on the right hand side of (3). And we want an antiholomorphic weight there, so $phi'$ has to be canceled off.
answered Jul 16 at 21:55
user357151
13.9k31140
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your Möbius transformation idea is good, but it needs one extra thing to work. We know that
$$
f(0) = frac1piiint_z f(z),dA(z)
tag1$$
Given $z_0$, apply (1) to $g = (fcirc phi^-1)(phi^-1)'$ where $phi(z)= (z-z_0)/(1-bar z_0z) $ is a Möbius transformation sending $z_0$ to $ 0$. (Here, $(phi^-1)'$ is that extra thing.) Since
$$phi'(z) = frac1-bar z_0 z + bar z_0(z-z_0) (1-bar z_0 z)^2 = frac^2 (1-bar z_0 z)^2 tag2$$
the change of variables $z=phi(w)$ yields
$$
f(z_0) (phi^-1)'(0) = g(0) = frac1piiint_z g(z),dA(z)
= frac1piiint_w fracf(w)phi'(w) |phi'(w)|^2,dA(w)
tag3$$
This simplifies to
$$
fracf(z_0)phi'(z_0) = frac1piiint_w f(w) overlinephi'(w) ,dA(w)
tag4$$
and, using (2), further simplifies to
$$
f(z_0) (1-|z_0|^2) = frac1piiint_w f(w) fracz_0(1-z_0 bar w)^2 ,dA(w)
tag6$$
which is the desired result
$$
f(z_0) = frac1piiint_w f(w) frac1(1-z_0 bar w)^2 ,dA(w)
tag7$$
What's the idea of putting the derivative factor $(phi^-1)'$ in the formula for $g$? Without it we'd end up with the Jacobian $|phi'|^2 = phi'overlinephi'$ as the factor of $f$ on the right hand side of (3). And we want an antiholomorphic weight there, so $phi'$ has to be canceled off.
answered Jul 16 at 21:55
user357151
13.9k31140
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
add a comment |Â
up vote
1
down vote
accepted
Your Möbius transformation idea is good, but it needs one extra thing to work. We know that
$$
f(0) = frac1piiint_z f(z),dA(z)
tag1$$
Given $z_0$, apply (1) to $g = (fcirc phi^-1)(phi^-1)'$ where $phi(z)= (z-z_0)/(1-bar z_0z) $ is a Möbius transformation sending $z_0$ to $ 0$. (Here, $(phi^-1)'$ is that extra thing.) Since
$$phi'(z) = frac1-bar z_0 z + bar z_0(z-z_0) (1-bar z_0 z)^2 = frac^2 (1-bar z_0 z)^2 tag2$$
the change of variables $z=phi(w)$ yields
$$
f(z_0) (phi^-1)'(0) = g(0) = frac1piiint_z g(z),dA(z)
= frac1piiint_w fracf(w)phi'(w) |phi'(w)|^2,dA(w)
tag3$$
This simplifies to
$$
fracf(z_0)phi'(z_0) = frac1piiint_w f(w) overlinephi'(w) ,dA(w)
tag4$$
and, using (2), further simplifies to
$$
f(z_0) (1-|z_0|^2) = frac1piiint_w f(w) fracz_0(1-z_0 bar w)^2 ,dA(w)
tag6$$
which is the desired result
$$
f(z_0) = frac1piiint_w f(w) frac1(1-z_0 bar w)^2 ,dA(w)
tag7$$
What's the idea of putting the derivative factor $(phi^-1)'$ in the formula for $g$? Without it we'd end up with the Jacobian $|phi'|^2 = phi'overlinephi'$ as the factor of $f$ on the right hand side of (3). And we want an antiholomorphic weight there, so $phi'$ has to be canceled off.
answered Jul 16 at 21:55
user357151
13.9k31140
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your Möbius transformation idea is good, but it needs one extra thing to work. We know that
$$
f(0) = frac1piiint_z f(z),dA(z)
tag1$$
Given $z_0$, apply (1) to $g = (fcirc phi^-1)(phi^-1)'$ where $phi(z)= (z-z_0)/(1-bar z_0z) $ is a Möbius transformation sending $z_0$ to $ 0$. (Here, $(phi^-1)'$ is that extra thing.) Since
$$phi'(z) = frac1-bar z_0 z + bar z_0(z-z_0) (1-bar z_0 z)^2 = frac^2 (1-bar z_0 z)^2 tag2$$
the change of variables $z=phi(w)$ yields
$$
f(z_0) (phi^-1)'(0) = g(0) = frac1piiint_z g(z),dA(z)
= frac1piiint_w fracf(w)phi'(w) |phi'(w)|^2,dA(w)
tag3$$
This simplifies to
$$
fracf(z_0)phi'(z_0) = frac1piiint_w f(w) overlinephi'(w) ,dA(w)
tag4$$
and, using (2), further simplifies to
$$
f(z_0) (1-|z_0|^2) = frac1piiint_w f(w) fracz_0(1-z_0 bar w)^2 ,dA(w)
tag6$$
which is the desired result
$$
f(z_0) = frac1piiint_w f(w) frac1(1-z_0 bar w)^2 ,dA(w)
tag7$$
What's the idea of putting the derivative factor $(phi^-1)'$ in the formula for $g$? Without it we'd end up with the Jacobian $|phi'|^2 = phi'overlinephi'$ as the factor of $f$ on the right hand side of (3). And we want an antiholomorphic weight there, so $phi'$ has to be canceled off.
answered Jul 16 at 21:55
user357151
13.9k31140
Your Möbius transformation idea is good, but it needs one extra thing to work. We know that
$$
f(0) = frac1piiint_z f(z),dA(z)
tag1$$
Given $z_0$, apply (1) to $g = (fcirc phi^-1)(phi^-1)'$ where $phi(z)= (z-z_0)/(1-bar z_0z) $ is a Möbius transformation sending $z_0$ to $ 0$. (Here, $(phi^-1)'$ is that extra thing.) Since
$$phi'(z) = frac1-bar z_0 z + bar z_0(z-z_0) (1-bar z_0 z)^2 = frac^2 (1-bar z_0 z)^2 tag2$$
the change of variables $z=phi(w)$ yields
$$
f(z_0) (phi^-1)'(0) = g(0) = frac1piiint_z g(z),dA(z)
= frac1piiint_w fracf(w)phi'(w) |phi'(w)|^2,dA(w)
tag3$$
This simplifies to
$$
fracf(z_0)phi'(z_0) = frac1piiint_w f(w) overlinephi'(w) ,dA(w)
tag4$$
and, using (2), further simplifies to
$$
f(z_0) (1-|z_0|^2) = frac1piiint_w f(w) fracz_0(1-z_0 bar w)^2 ,dA(w)
tag6$$
which is the desired result
$$
f(z_0) = frac1piiint_w f(w) frac1(1-z_0 bar w)^2 ,dA(w)
tag7$$
What's the idea of putting the derivative factor $(phi^-1)'$ in the formula for $g$? Without it we'd end up with the Jacobian $|phi'|^2 = phi'overlinephi'$ as the factor of $f$ on the right hand side of (3). And we want an antiholomorphic weight there, so $phi'$ has to be canceled off.
answered Jul 16 at 21:55
user357151
13.9k31140
answered Jul 16 at 21:55
user357151
13.9k31140
answered Jul 16 at 21:55
user357151
13.9k31140
answered Jul 16 at 21:55
user357151
13.9k31140
13.9k31140
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
add a comment |Â
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
This is beautiful, thank you. That’s exactly what I needed.
– Van Latimer
Jul 16 at 22:58
add a comment |Â
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