Prove that $T$ is a bounded operator on a disk algebra and prove the existence of a Borel measure on a...
$begingroup$
Let $A(D)$ be the space of holomorphic functions on the open unit disk $D$ and continuous on the closed disk $bar{D}$. Then $A(D)$ is a Banach space if we set $|f|=sup{|f(z)|:zinbar{D}}$. For $fin A(D)$ write
$$
f(z)=sum_{n=0}^{infty}a_nz^n.
$$
a) Let ${c_n}_{n=0}^{infty}$ be a sequence of complex numbers with the property that if $fin A(D)$ represented as above, and
$$
f^*(z)=sum_{n=0}^{infty}c_na_nz^n,
$$
then $f^*in A(D)$. Prove that there exists $C>0$ such that $|f^*|leq C|f|$ for all $fin A(D)$.
I am thinking that I need to prove $T:A(D)to A(D)$ with $T(f)=f^*$ is bounded (or continuous). Do you have any other ideas?
b) Let $omegain D$. Prove that there exists a Borel measure $mu_{omega}$ on $partial D={z:|z|=1}$ such that
$$
f(omega)=int_{partial D}f(t)dmu_{omega}(t),;fin A.
$$
I am thinking that we can use the Maximum Modulus principle, but is the measure $mu_{omega}$ unique? Could you please give me some ideas?
Thank you so much.
functional-analysis analysis banach-spaces holomorphic-functions maximum-principle
asked Dec 30 '18 at 13:30
RossRoss
1237
$endgroup$
add a comment |
$begingroup$
Let $A(D)$ be the space of holomorphic functions on the open unit disk $D$ and continuous on the closed disk $bar{D}$. Then $A(D)$ is a Banach space if we set $|f|=sup{|f(z)|:zinbar{D}}$. For $fin A(D)$ write
$$
f(z)=sum_{n=0}^{infty}a_nz^n.
$$
a) Let ${c_n}_{n=0}^{infty}$ be a sequence of complex numbers with the property that if $fin A(D)$ represented as above, and
$$
f^*(z)=sum_{n=0}^{infty}c_na_nz^n,
$$
then $f^*in A(D)$. Prove that there exists $C>0$ such that $|f^*|leq C|f|$ for all $fin A(D)$.
I am thinking that I need to prove $T:A(D)to A(D)$ with $T(f)=f^*$ is bounded (or continuous). Do you have any other ideas?
b) Let $omegain D$. Prove that there exists a Borel measure $mu_{omega}$ on $partial D={z:|z|=1}$ such that
$$
f(omega)=int_{partial D}f(t)dmu_{omega}(t),;fin A.
$$
I am thinking that we can use the Maximum Modulus principle, but is the measure $mu_{omega}$ unique? Could you please give me some ideas?
Thank you so much.
functional-analysis analysis banach-spaces holomorphic-functions maximum-principle
asked Dec 30 '18 at 13:30
RossRoss
1237
$endgroup$
1
$begingroup$
Hints; closed graph theorem for a), and Hahn-Banach theorem together with Riesz representation theorem for b) !
$endgroup$
– Bartosz Malman
Dec 30 '18 at 15:03
$begingroup$
@BartoszMalman Maybe the Beppo-Necky theorem is more efficient here?
$endgroup$
– TheOscillator
Dec 30 '18 at 15:12
add a comment |
$begingroup$
Let $A(D)$ be the space of holomorphic functions on the open unit disk $D$ and continuous on the closed disk $bar{D}$. Then $A(D)$ is a Banach space if we set $|f|=sup{|f(z)|:zinbar{D}}$. For $fin A(D)$ write
$$
f(z)=sum_{n=0}^{infty}a_nz^n.
$$
a) Let ${c_n}_{n=0}^{infty}$ be a sequence of complex numbers with the property that if $fin A(D)$ represented as above, and
$$
f^*(z)=sum_{n=0}^{infty}c_na_nz^n,
$$
then $f^*in A(D)$. Prove that there exists $C>0$ such that $|f^*|leq C|f|$ for all $fin A(D)$.
I am thinking that I need to prove $T:A(D)to A(D)$ with $T(f)=f^*$ is bounded (or continuous). Do you have any other ideas?
b) Let $omegain D$. Prove that there exists a Borel measure $mu_{omega}$ on $partial D={z:|z|=1}$ such that
$$
f(omega)=int_{partial D}f(t)dmu_{omega}(t),;fin A.
$$
I am thinking that we can use the Maximum Modulus principle, but is the measure $mu_{omega}$ unique? Could you please give me some ideas?
Thank you so much.
functional-analysis analysis banach-spaces holomorphic-functions maximum-principle
asked Dec 30 '18 at 13:30
RossRoss
1237
$endgroup$
Let $A(D)$ be the space of holomorphic functions on the open unit disk $D$ and continuous on the closed disk $bar{D}$. Then $A(D)$ is a Banach space if we set $|f|=sup{|f(z)|:zinbar{D}}$. For $fin A(D)$ write
$$
f(z)=sum_{n=0}^{infty}a_nz^n.
$$
a) Let ${c_n}_{n=0}^{infty}$ be a sequence of complex numbers with the property that if $fin A(D)$ represented as above, and
$$
f^*(z)=sum_{n=0}^{infty}c_na_nz^n,
$$
then $f^*in A(D)$. Prove that there exists $C>0$ such that $|f^*|leq C|f|$ for all $fin A(D)$.
I am thinking that I need to prove $T:A(D)to A(D)$ with $T(f)=f^*$ is bounded (or continuous). Do you have any other ideas?
b) Let $omegain D$. Prove that there exists a Borel measure $mu_{omega}$ on $partial D={z:|z|=1}$ such that
$$
f(omega)=int_{partial D}f(t)dmu_{omega}(t),;fin A.
$$
I am thinking that we can use the Maximum Modulus principle, but is the measure $mu_{omega}$ unique? Could you please give me some ideas?
Thank you so much.
functional-analysis analysis banach-spaces holomorphic-functions maximum-principle
functional-analysis analysis banach-spaces holomorphic-functions maximum-principle
asked Dec 30 '18 at 13:30
RossRoss
1237
asked Dec 30 '18 at 13:30
RossRoss
1237
asked Dec 30 '18 at 13:30
RossRoss
1237
asked Dec 30 '18 at 13:30
RossRoss
1237
asked Dec 30 '18 at 13:30
RossRoss
1237
1237
1
$begingroup$
Hints; closed graph theorem for a), and Hahn-Banach theorem together with Riesz representation theorem for b) !
$endgroup$
– Bartosz Malman
Dec 30 '18 at 15:03
$begingroup$
@BartoszMalman Maybe the Beppo-Necky theorem is more efficient here?
$endgroup$
– TheOscillator
Dec 30 '18 at 15:12
add a comment |
1
$begingroup$
Hints; closed graph theorem for a), and Hahn-Banach theorem together with Riesz representation theorem for b) !
$endgroup$
– Bartosz Malman
Dec 30 '18 at 15:03
$begingroup$
@BartoszMalman Maybe the Beppo-Necky theorem is more efficient here?
$endgroup$
– TheOscillator
Dec 30 '18 at 15:12
1
1
$begingroup$
Hints; closed graph theorem for a), and Hahn-Banach theorem together with Riesz representation theorem for b) !
$endgroup$
– Bartosz Malman
Dec 30 '18 at 15:03
$begingroup$
Hints; closed graph theorem for a), and Hahn-Banach theorem together with Riesz representation theorem for b) !
$endgroup$
– Bartosz Malman
Dec 30 '18 at 15:03
$begingroup$
@BartoszMalman Maybe the Beppo-Necky theorem is more efficient here?
$endgroup$
– TheOscillator
Dec 30 '18 at 15:12
$begingroup$
@BartoszMalman Maybe the Beppo-Necky theorem is more efficient here?
$endgroup$
– TheOscillator
Dec 30 '18 at 15:12
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
For a), use closed graph theorem to show $T$ is bounded. It suffices to show if $f_k to f$ and $Tf_k = f_k^* to g$ in $A(D)$, then $g= Tf = f^*$. It follows from Cauchy's integral formula
$$
a_n = frac{1}{2pi i}int_{|z|=r} frac{f(z)}{z^{n+1}}dz
$$ and uniform convergence of $(f_k)$ and $(f^*_k)$. (Hint: compare coefficients of power series expansions of $f^*$ and $g$.)
For b), let $B(mathbb{T})$ be a family of $gin C(mathbb{T})$ such that $g = f|_mathbb{T}$ for some $fin A(D)$. (Here, $mathbb{T}=partial D$.) Then, $B(mathbb{T})$ is a subspace of $ C(mathbb{T})$. Let
$$
phi_omega: gin B(mathbb{T}) mapsto f(omega)in mathbb{C}.
$$ Then, $phi_omega$ is a well-defined linear functional (since there is unique $fin A(D)$ such that $g = f|_mathbb{T}$) with $|phi_omega|_{B(mathbb{T})to mathbb{C}}leq 1$ by maximum modulus theorem. By Hahn-Banach theorem, there exists an extension $Phi_omega : C(mathbb{T}) to mathbb{C}$ of $phi_omega$ with $|Phi_omega|_{C(mathbb{T})to mathbb{C}}=|phi_omega|_{B(mathbb{T})to mathbb{C}}$. Now, by Riesz representation theorem, we know there exists a finite Borel measure $mu_omega$ on $mathbb{T}$ such that $$
Phi_omega(f|_mathbb{T})=f(omega) = int_mathbb{T} f|_mathbb{T}(t)dmu_omega(t)=int_mathbb{T} f(t)dmu_omega(t)
$$ for all $f in A(D)$.
answered Jan 1 at 10:15
SongSong
18.5k21651
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add a comment |
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$begingroup$
For a), use closed graph theorem to show $T$ is bounded. It suffices to show if $f_k to f$ and $Tf_k = f_k^* to g$ in $A(D)$, then $g= Tf = f^*$. It follows from Cauchy's integral formula
$$
a_n = frac{1}{2pi i}int_{|z|=r} frac{f(z)}{z^{n+1}}dz
$$ and uniform convergence of $(f_k)$ and $(f^*_k)$. (Hint: compare coefficients of power series expansions of $f^*$ and $g$.)
For b), let $B(mathbb{T})$ be a family of $gin C(mathbb{T})$ such that $g = f|_mathbb{T}$ for some $fin A(D)$. (Here, $mathbb{T}=partial D$.) Then, $B(mathbb{T})$ is a subspace of $ C(mathbb{T})$. Let
$$
phi_omega: gin B(mathbb{T}) mapsto f(omega)in mathbb{C}.
$$ Then, $phi_omega$ is a well-defined linear functional (since there is unique $fin A(D)$ such that $g = f|_mathbb{T}$) with $|phi_omega|_{B(mathbb{T})to mathbb{C}}leq 1$ by maximum modulus theorem. By Hahn-Banach theorem, there exists an extension $Phi_omega : C(mathbb{T}) to mathbb{C}$ of $phi_omega$ with $|Phi_omega|_{C(mathbb{T})to mathbb{C}}=|phi_omega|_{B(mathbb{T})to mathbb{C}}$. Now, by Riesz representation theorem, we know there exists a finite Borel measure $mu_omega$ on $mathbb{T}$ such that $$
Phi_omega(f|_mathbb{T})=f(omega) = int_mathbb{T} f|_mathbb{T}(t)dmu_omega(t)=int_mathbb{T} f(t)dmu_omega(t)
$$ for all $f in A(D)$.
answered Jan 1 at 10:15
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
For a), use closed graph theorem to show $T$ is bounded. It suffices to show if $f_k to f$ and $Tf_k = f_k^* to g$ in $A(D)$, then $g= Tf = f^*$. It follows from Cauchy's integral formula
$$
a_n = frac{1}{2pi i}int_{|z|=r} frac{f(z)}{z^{n+1}}dz
$$ and uniform convergence of $(f_k)$ and $(f^*_k)$. (Hint: compare coefficients of power series expansions of $f^*$ and $g$.)
For b), let $B(mathbb{T})$ be a family of $gin C(mathbb{T})$ such that $g = f|_mathbb{T}$ for some $fin A(D)$. (Here, $mathbb{T}=partial D$.) Then, $B(mathbb{T})$ is a subspace of $ C(mathbb{T})$. Let
$$
phi_omega: gin B(mathbb{T}) mapsto f(omega)in mathbb{C}.
$$ Then, $phi_omega$ is a well-defined linear functional (since there is unique $fin A(D)$ such that $g = f|_mathbb{T}$) with $|phi_omega|_{B(mathbb{T})to mathbb{C}}leq 1$ by maximum modulus theorem. By Hahn-Banach theorem, there exists an extension $Phi_omega : C(mathbb{T}) to mathbb{C}$ of $phi_omega$ with $|Phi_omega|_{C(mathbb{T})to mathbb{C}}=|phi_omega|_{B(mathbb{T})to mathbb{C}}$. Now, by Riesz representation theorem, we know there exists a finite Borel measure $mu_omega$ on $mathbb{T}$ such that $$
Phi_omega(f|_mathbb{T})=f(omega) = int_mathbb{T} f|_mathbb{T}(t)dmu_omega(t)=int_mathbb{T} f(t)dmu_omega(t)
$$ for all $f in A(D)$.
answered Jan 1 at 10:15
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
For a), use closed graph theorem to show $T$ is bounded. It suffices to show if $f_k to f$ and $Tf_k = f_k^* to g$ in $A(D)$, then $g= Tf = f^*$. It follows from Cauchy's integral formula
$$
a_n = frac{1}{2pi i}int_{|z|=r} frac{f(z)}{z^{n+1}}dz
$$ and uniform convergence of $(f_k)$ and $(f^*_k)$. (Hint: compare coefficients of power series expansions of $f^*$ and $g$.)
For b), let $B(mathbb{T})$ be a family of $gin C(mathbb{T})$ such that $g = f|_mathbb{T}$ for some $fin A(D)$. (Here, $mathbb{T}=partial D$.) Then, $B(mathbb{T})$ is a subspace of $ C(mathbb{T})$. Let
$$
phi_omega: gin B(mathbb{T}) mapsto f(omega)in mathbb{C}.
$$ Then, $phi_omega$ is a well-defined linear functional (since there is unique $fin A(D)$ such that $g = f|_mathbb{T}$) with $|phi_omega|_{B(mathbb{T})to mathbb{C}}leq 1$ by maximum modulus theorem. By Hahn-Banach theorem, there exists an extension $Phi_omega : C(mathbb{T}) to mathbb{C}$ of $phi_omega$ with $|Phi_omega|_{C(mathbb{T})to mathbb{C}}=|phi_omega|_{B(mathbb{T})to mathbb{C}}$. Now, by Riesz representation theorem, we know there exists a finite Borel measure $mu_omega$ on $mathbb{T}$ such that $$
Phi_omega(f|_mathbb{T})=f(omega) = int_mathbb{T} f|_mathbb{T}(t)dmu_omega(t)=int_mathbb{T} f(t)dmu_omega(t)
$$ for all $f in A(D)$.
answered Jan 1 at 10:15
SongSong
18.5k21651
$endgroup$
For a), use closed graph theorem to show $T$ is bounded. It suffices to show if $f_k to f$ and $Tf_k = f_k^* to g$ in $A(D)$, then $g= Tf = f^*$. It follows from Cauchy's integral formula
$$
a_n = frac{1}{2pi i}int_{|z|=r} frac{f(z)}{z^{n+1}}dz
$$ and uniform convergence of $(f_k)$ and $(f^*_k)$. (Hint: compare coefficients of power series expansions of $f^*$ and $g$.)
For b), let $B(mathbb{T})$ be a family of $gin C(mathbb{T})$ such that $g = f|_mathbb{T}$ for some $fin A(D)$. (Here, $mathbb{T}=partial D$.) Then, $B(mathbb{T})$ is a subspace of $ C(mathbb{T})$. Let
$$
phi_omega: gin B(mathbb{T}) mapsto f(omega)in mathbb{C}.
$$ Then, $phi_omega$ is a well-defined linear functional (since there is unique $fin A(D)$ such that $g = f|_mathbb{T}$) with $|phi_omega|_{B(mathbb{T})to mathbb{C}}leq 1$ by maximum modulus theorem. By Hahn-Banach theorem, there exists an extension $Phi_omega : C(mathbb{T}) to mathbb{C}$ of $phi_omega$ with $|Phi_omega|_{C(mathbb{T})to mathbb{C}}=|phi_omega|_{B(mathbb{T})to mathbb{C}}$. Now, by Riesz representation theorem, we know there exists a finite Borel measure $mu_omega$ on $mathbb{T}$ such that $$
Phi_omega(f|_mathbb{T})=f(omega) = int_mathbb{T} f|_mathbb{T}(t)dmu_omega(t)=int_mathbb{T} f(t)dmu_omega(t)
$$ for all $f in A(D)$.
answered Jan 1 at 10:15
SongSong
18.5k21651
edited Jan 1 at 17:50
answered Jan 1 at 10:15
SongSong
18.5k21651
answered Jan 1 at 10:15
SongSong
18.5k21651
answered Jan 1 at 10:15
SongSong
18.5k21651
18.5k21651
add a comment |
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$begingroup$
Hints; closed graph theorem for a), and Hahn-Banach theorem together with Riesz representation theorem for b) !
$endgroup$
– Bartosz Malman
Dec 30 '18 at 15:03
$begingroup$
@BartoszMalman Maybe the Beppo-Necky theorem is more efficient here?
$endgroup$
– TheOscillator
Dec 30 '18 at 15:12