Is a de-zeroed Hardy function still Hardy?
Suppose a Hardy function $f(z)$ on the upper half complex plane or $fin H^{2+}$ (Chapter II, p.45 of Fulvio Ricci, Hardy Spaces in One Complex Variable) has a zero of order $m$ at $omega$ with $mathbf{Im}(omega)>0$. A Hardy $H^{2+}$ function $f(z=x+iy)$ on the upper half complex plane is a holomorphic function on the that plane where $f_y(x):=f(x+iy)in L^2(mathbf R)$ with norm $|f_y|_2$ with respect to $x$ for any given $y$, and $sup_y|f_y|_2<infty$. Is $displaystylefrac{f(z)-f(omega)}{(z-omega)^n}in H^{2+},, forall nle m$?
complex-analysis harmonic-analysis hardy-spaces
asked Oct 7 '18 at 23:18
Hans
4,94521032
|
show 7 more comments
Suppose a Hardy function $f(z)$ on the upper half complex plane or $fin H^{2+}$ (Chapter II, p.45 of Fulvio Ricci, Hardy Spaces in One Complex Variable) has a zero of order $m$ at $omega$ with $mathbf{Im}(omega)>0$. A Hardy $H^{2+}$ function $f(z=x+iy)$ on the upper half complex plane is a holomorphic function on the that plane where $f_y(x):=f(x+iy)in L^2(mathbf R)$ with norm $|f_y|_2$ with respect to $x$ for any given $y$, and $sup_y|f_y|_2<infty$. Is $displaystylefrac{f(z)-f(omega)}{(z-omega)^n}in H^{2+},, forall nle m$?
complex-analysis harmonic-analysis hardy-spaces
asked Oct 7 '18 at 23:18
Hans
4,94521032
can you define Hardy plz
– mathworker21
Oct 7 '18 at 23:27
@mathworker21: Just added a reference. Check the chapter on the half plane.
– Hans
Oct 7 '18 at 23:33
1
I saw definitions depending on $p$. I still don't know what "Hardy" means. I'd appreciate you either giving a definition in the problem or telling me what page of the PDF to look at so that I don't waste more time scrolling
– mathworker21
Oct 7 '18 at 23:47
@mathworker21: Chapter II, p.45, $p=2$.
– Hans
Oct 8 '18 at 0:15
1
@Hans: I don't see any definition of "Hardy" on that page. The only occurrence of the word "Hardy" on page 45 is in the chapter title.
– Henning Makholm
Oct 8 '18 at 0:48
|
show 7 more comments
Suppose a Hardy function $f(z)$ on the upper half complex plane or $fin H^{2+}$ (Chapter II, p.45 of Fulvio Ricci, Hardy Spaces in One Complex Variable) has a zero of order $m$ at $omega$ with $mathbf{Im}(omega)>0$. A Hardy $H^{2+}$ function $f(z=x+iy)$ on the upper half complex plane is a holomorphic function on the that plane where $f_y(x):=f(x+iy)in L^2(mathbf R)$ with norm $|f_y|_2$ with respect to $x$ for any given $y$, and $sup_y|f_y|_2<infty$. Is $displaystylefrac{f(z)-f(omega)}{(z-omega)^n}in H^{2+},, forall nle m$?
complex-analysis harmonic-analysis hardy-spaces
asked Oct 7 '18 at 23:18
Hans
4,94521032
Suppose a Hardy function $f(z)$ on the upper half complex plane or $fin H^{2+}$ (Chapter II, p.45 of Fulvio Ricci, Hardy Spaces in One Complex Variable) has a zero of order $m$ at $omega$ with $mathbf{Im}(omega)>0$. A Hardy $H^{2+}$ function $f(z=x+iy)$ on the upper half complex plane is a holomorphic function on the that plane where $f_y(x):=f(x+iy)in L^2(mathbf R)$ with norm $|f_y|_2$ with respect to $x$ for any given $y$, and $sup_y|f_y|_2<infty$. Is $displaystylefrac{f(z)-f(omega)}{(z-omega)^n}in H^{2+},, forall nle m$?
complex-analysis harmonic-analysis hardy-spaces
complex-analysis harmonic-analysis hardy-spaces
asked Oct 7 '18 at 23:18
Hans
4,94521032
asked Oct 7 '18 at 23:18
Hans
4,94521032
edited Nov 21 '18 at 7:35
asked Oct 7 '18 at 23:18
Hans
4,94521032
asked Oct 7 '18 at 23:18
Hans
4,94521032
asked Oct 7 '18 at 23:18
Hans
4,94521032
4,94521032
can you define Hardy plz
– mathworker21
Oct 7 '18 at 23:27
@mathworker21: Just added a reference. Check the chapter on the half plane.
– Hans
Oct 7 '18 at 23:33
1
I saw definitions depending on $p$. I still don't know what "Hardy" means. I'd appreciate you either giving a definition in the problem or telling me what page of the PDF to look at so that I don't waste more time scrolling
– mathworker21
Oct 7 '18 at 23:47
@mathworker21: Chapter II, p.45, $p=2$.
– Hans
Oct 8 '18 at 0:15
1
@Hans: I don't see any definition of "Hardy" on that page. The only occurrence of the word "Hardy" on page 45 is in the chapter title.
– Henning Makholm
Oct 8 '18 at 0:48
|
show 7 more comments
can you define Hardy plz
– mathworker21
Oct 7 '18 at 23:27
@mathworker21: Just added a reference. Check the chapter on the half plane.
– Hans
Oct 7 '18 at 23:33
1
I saw definitions depending on $p$. I still don't know what "Hardy" means. I'd appreciate you either giving a definition in the problem or telling me what page of the PDF to look at so that I don't waste more time scrolling
– mathworker21
Oct 7 '18 at 23:47
@mathworker21: Chapter II, p.45, $p=2$.
– Hans
Oct 8 '18 at 0:15
1
@Hans: I don't see any definition of "Hardy" on that page. The only occurrence of the word "Hardy" on page 45 is in the chapter title.
– Henning Makholm
Oct 8 '18 at 0:48
can you define Hardy plz
– mathworker21
Oct 7 '18 at 23:27
can you define Hardy plz
– mathworker21
Oct 7 '18 at 23:27
@mathworker21: Just added a reference. Check the chapter on the half plane.
– Hans
Oct 7 '18 at 23:33
@mathworker21: Just added a reference. Check the chapter on the half plane.
– Hans
Oct 7 '18 at 23:33
1
1
I saw definitions depending on $p$. I still don't know what "Hardy" means. I'd appreciate you either giving a definition in the problem or telling me what page of the PDF to look at so that I don't waste more time scrolling
– mathworker21
Oct 7 '18 at 23:47
I saw definitions depending on $p$. I still don't know what "Hardy" means. I'd appreciate you either giving a definition in the problem or telling me what page of the PDF to look at so that I don't waste more time scrolling
– mathworker21
Oct 7 '18 at 23:47
@mathworker21: Chapter II, p.45, $p=2$.
– Hans
Oct 8 '18 at 0:15
@mathworker21: Chapter II, p.45, $p=2$.
– Hans
Oct 8 '18 at 0:15
1
1
@Hans: I don't see any definition of "Hardy" on that page. The only occurrence of the word "Hardy" on page 45 is in the chapter title.
– Henning Makholm
Oct 8 '18 at 0:48
@Hans: I don't see any definition of "Hardy" on that page. The only occurrence of the word "Hardy" on page 45 is in the chapter title.
– Henning Makholm
Oct 8 '18 at 0:48
|
show 7 more comments
1 Answer
1
active
oldest
votes
Ironically, had I read further into Fulvio Ricci, Hardy Spaces in One Complex Variable, Chapter II, Theorem 5.4 provides the answer. This theorem also appears as Theorem 3.10 in Section 3, Chapter 20, of John B. Conway, Functions of One Complex Variable II, and Theorem 11.6 of Peter L. Duren, Theory of $H^p$ Spaces regarding the Blaschke product factorization.
Edit: Actually, it is not necessary to use the explicit factorization as referred to above.
The answer is affirmative
Proof: Consider $n=1$. Larger $nle m$ works the same.
$g(z):=frac{f(z)-f(omega)}{z-omega}$ is holomorphic on the closed upper half complex plane. Let $D(omega;R)$ be the closed disk centered at $omega$ and radius $0<R<mathbf{Im}(omega)$. $|g(z)|$ has a maximum and is bounded on the compact $D(omega,R)$.
For an arbitrary $yge0$, let $Omega_y:={x+iy: xinmathbf R}setminus D(omega;R)$. We have
$$|g(z)|le frac{|f(z)|}R+frac{|f(omega)|}{|z-omega|}, forall |z-omega|ge R.$$
$|z-omega|^2ge(R^2+x^2)$, where $x$ is a real number such that $x+x_0=mathbf{Re}(z-omega)$ and that $|x_0+imathbf{Im}(z-omega)|=R$.
By the Minkowski inequality
$$Big(int_{Omega_y} |g(x+iy)|^2 dxBig)^frac12 le frac1RBig(int_{Omega_y} |f(x+iy)|^2 dxBig)^frac12+|g(omega)|Big(2int_0^inftyfrac1{R^2+x^2}dxBig)^frac12$$
The first integral on the right hand side is uniformly bounded over $y$ by the definition of $H^{2+}$. $displaystyle 2int_0^inftyfrac1{R^2+x^2},dx=fracpi R$. We are done. $quadsquare$
answered Oct 8 '18 at 0:32
Hans
4,94521032
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
add a comment |
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Ironically, had I read further into Fulvio Ricci, Hardy Spaces in One Complex Variable, Chapter II, Theorem 5.4 provides the answer. This theorem also appears as Theorem 3.10 in Section 3, Chapter 20, of John B. Conway, Functions of One Complex Variable II, and Theorem 11.6 of Peter L. Duren, Theory of $H^p$ Spaces regarding the Blaschke product factorization.
Edit: Actually, it is not necessary to use the explicit factorization as referred to above.
The answer is affirmative
Proof: Consider $n=1$. Larger $nle m$ works the same.
$g(z):=frac{f(z)-f(omega)}{z-omega}$ is holomorphic on the closed upper half complex plane. Let $D(omega;R)$ be the closed disk centered at $omega$ and radius $0<R<mathbf{Im}(omega)$. $|g(z)|$ has a maximum and is bounded on the compact $D(omega,R)$.
For an arbitrary $yge0$, let $Omega_y:={x+iy: xinmathbf R}setminus D(omega;R)$. We have
$$|g(z)|le frac{|f(z)|}R+frac{|f(omega)|}{|z-omega|}, forall |z-omega|ge R.$$
$|z-omega|^2ge(R^2+x^2)$, where $x$ is a real number such that $x+x_0=mathbf{Re}(z-omega)$ and that $|x_0+imathbf{Im}(z-omega)|=R$.
By the Minkowski inequality
$$Big(int_{Omega_y} |g(x+iy)|^2 dxBig)^frac12 le frac1RBig(int_{Omega_y} |f(x+iy)|^2 dxBig)^frac12+|g(omega)|Big(2int_0^inftyfrac1{R^2+x^2}dxBig)^frac12$$
The first integral on the right hand side is uniformly bounded over $y$ by the definition of $H^{2+}$. $displaystyle 2int_0^inftyfrac1{R^2+x^2},dx=fracpi R$. We are done. $quadsquare$
answered Oct 8 '18 at 0:32
Hans
4,94521032
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
add a comment |
Ironically, had I read further into Fulvio Ricci, Hardy Spaces in One Complex Variable, Chapter II, Theorem 5.4 provides the answer. This theorem also appears as Theorem 3.10 in Section 3, Chapter 20, of John B. Conway, Functions of One Complex Variable II, and Theorem 11.6 of Peter L. Duren, Theory of $H^p$ Spaces regarding the Blaschke product factorization.
Edit: Actually, it is not necessary to use the explicit factorization as referred to above.
The answer is affirmative
Proof: Consider $n=1$. Larger $nle m$ works the same.
$g(z):=frac{f(z)-f(omega)}{z-omega}$ is holomorphic on the closed upper half complex plane. Let $D(omega;R)$ be the closed disk centered at $omega$ and radius $0<R<mathbf{Im}(omega)$. $|g(z)|$ has a maximum and is bounded on the compact $D(omega,R)$.
For an arbitrary $yge0$, let $Omega_y:={x+iy: xinmathbf R}setminus D(omega;R)$. We have
$$|g(z)|le frac{|f(z)|}R+frac{|f(omega)|}{|z-omega|}, forall |z-omega|ge R.$$
$|z-omega|^2ge(R^2+x^2)$, where $x$ is a real number such that $x+x_0=mathbf{Re}(z-omega)$ and that $|x_0+imathbf{Im}(z-omega)|=R$.
By the Minkowski inequality
$$Big(int_{Omega_y} |g(x+iy)|^2 dxBig)^frac12 le frac1RBig(int_{Omega_y} |f(x+iy)|^2 dxBig)^frac12+|g(omega)|Big(2int_0^inftyfrac1{R^2+x^2}dxBig)^frac12$$
The first integral on the right hand side is uniformly bounded over $y$ by the definition of $H^{2+}$. $displaystyle 2int_0^inftyfrac1{R^2+x^2},dx=fracpi R$. We are done. $quadsquare$
answered Oct 8 '18 at 0:32
Hans
4,94521032
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
add a comment |
Ironically, had I read further into Fulvio Ricci, Hardy Spaces in One Complex Variable, Chapter II, Theorem 5.4 provides the answer. This theorem also appears as Theorem 3.10 in Section 3, Chapter 20, of John B. Conway, Functions of One Complex Variable II, and Theorem 11.6 of Peter L. Duren, Theory of $H^p$ Spaces regarding the Blaschke product factorization.
Edit: Actually, it is not necessary to use the explicit factorization as referred to above.
The answer is affirmative
Proof: Consider $n=1$. Larger $nle m$ works the same.
$g(z):=frac{f(z)-f(omega)}{z-omega}$ is holomorphic on the closed upper half complex plane. Let $D(omega;R)$ be the closed disk centered at $omega$ and radius $0<R<mathbf{Im}(omega)$. $|g(z)|$ has a maximum and is bounded on the compact $D(omega,R)$.
For an arbitrary $yge0$, let $Omega_y:={x+iy: xinmathbf R}setminus D(omega;R)$. We have
$$|g(z)|le frac{|f(z)|}R+frac{|f(omega)|}{|z-omega|}, forall |z-omega|ge R.$$
$|z-omega|^2ge(R^2+x^2)$, where $x$ is a real number such that $x+x_0=mathbf{Re}(z-omega)$ and that $|x_0+imathbf{Im}(z-omega)|=R$.
By the Minkowski inequality
$$Big(int_{Omega_y} |g(x+iy)|^2 dxBig)^frac12 le frac1RBig(int_{Omega_y} |f(x+iy)|^2 dxBig)^frac12+|g(omega)|Big(2int_0^inftyfrac1{R^2+x^2}dxBig)^frac12$$
The first integral on the right hand side is uniformly bounded over $y$ by the definition of $H^{2+}$. $displaystyle 2int_0^inftyfrac1{R^2+x^2},dx=fracpi R$. We are done. $quadsquare$
answered Oct 8 '18 at 0:32
Hans
4,94521032
Ironically, had I read further into Fulvio Ricci, Hardy Spaces in One Complex Variable, Chapter II, Theorem 5.4 provides the answer. This theorem also appears as Theorem 3.10 in Section 3, Chapter 20, of John B. Conway, Functions of One Complex Variable II, and Theorem 11.6 of Peter L. Duren, Theory of $H^p$ Spaces regarding the Blaschke product factorization.
Edit: Actually, it is not necessary to use the explicit factorization as referred to above.
The answer is affirmative
Proof: Consider $n=1$. Larger $nle m$ works the same.
$g(z):=frac{f(z)-f(omega)}{z-omega}$ is holomorphic on the closed upper half complex plane. Let $D(omega;R)$ be the closed disk centered at $omega$ and radius $0<R<mathbf{Im}(omega)$. $|g(z)|$ has a maximum and is bounded on the compact $D(omega,R)$.
For an arbitrary $yge0$, let $Omega_y:={x+iy: xinmathbf R}setminus D(omega;R)$. We have
$$|g(z)|le frac{|f(z)|}R+frac{|f(omega)|}{|z-omega|}, forall |z-omega|ge R.$$
$|z-omega|^2ge(R^2+x^2)$, where $x$ is a real number such that $x+x_0=mathbf{Re}(z-omega)$ and that $|x_0+imathbf{Im}(z-omega)|=R$.
By the Minkowski inequality
$$Big(int_{Omega_y} |g(x+iy)|^2 dxBig)^frac12 le frac1RBig(int_{Omega_y} |f(x+iy)|^2 dxBig)^frac12+|g(omega)|Big(2int_0^inftyfrac1{R^2+x^2}dxBig)^frac12$$
The first integral on the right hand side is uniformly bounded over $y$ by the definition of $H^{2+}$. $displaystyle 2int_0^inftyfrac1{R^2+x^2},dx=fracpi R$. We are done. $quadsquare$
answered Oct 8 '18 at 0:32
Hans
4,94521032
edited Nov 21 '18 at 7:26
answered Oct 8 '18 at 0:32
Hans
4,94521032
answered Oct 8 '18 at 0:32
Hans
4,94521032
answered Oct 8 '18 at 0:32
Hans
4,94521032
4,94521032
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
add a comment |
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
I don't quite see the irony here. Is it ironic that the very text you're reading contains information about the topics it presents?
– John Hughes
Oct 8 '18 at 3:00
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: Yes, indeed. I read a bit further into the paper I used as a reference in the question after I posted the question and found the factorization theorem I was looking for. However, now I think the theorem is unnecessary to prove the affirmative statement. What is your opinion?
– Hans
Oct 8 '18 at 3:07
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
@JohnHughes: I just posted my proof. Please review.
– Hans
Oct 8 '18 at 4:49
add a comment |
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can you define Hardy plz
– mathworker21
Oct 7 '18 at 23:27
@mathworker21: Just added a reference. Check the chapter on the half plane.
– Hans
Oct 7 '18 at 23:33
1
I saw definitions depending on $p$. I still don't know what "Hardy" means. I'd appreciate you either giving a definition in the problem or telling me what page of the PDF to look at so that I don't waste more time scrolling
– mathworker21
Oct 7 '18 at 23:47
@mathworker21: Chapter II, p.45, $p=2$.
– Hans
Oct 8 '18 at 0:15
1
@Hans: I don't see any definition of "Hardy" on that page. The only occurrence of the word "Hardy" on page 45 is in the chapter title.
– Henning Makholm
Oct 8 '18 at 0:48