Prove that $T$ is a bounded operator on a disk algebra and prove the existence of a Borel measure on a...












5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














5












5








5





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $f in A(D)$.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056834%2fprove-that-t-is-a-bounded-operator-on-a-disk-algebra-and-prove-the-existence-o%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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 $f in A(D)$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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 $f in A(D)$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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 $f in A(D)$.






        share|cite|improve this answer











        $endgroup$



        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 $f in A(D)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 1 at 17:50

























        answered Jan 1 at 10:15









        SongSong

        18.5k21651




        18.5k21651






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056834%2fprove-that-t-is-a-bounded-operator-on-a-disk-algebra-and-prove-the-existence-o%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more