Each Hilbert space of holomorphic functions on $mathbb C$ is a reproducing kernel Hilbert space?












0












$begingroup$


I would like to know is the following result true ?



"Each Hilbert space of holomorphic functions on $mathbb C$ is a reproducing kernel Hilbert space".



Where a Hilbert space of holomorphic functions on $mathbb C$ is a Hilbert space
$H subseteq O(mathbb C)$ such that the inclusion mapping $H hookrightarrow O(mathbb C)$ is continuous.



i.e., If $H subseteq O(mathbb C)$ is a Hilbert space of holomorphic functions, then the point evaluation map, $f mapsto f(z)$, is continuous for all $zin mathbb C$ .



If so, where can I find its proof?



Thank you in advance






complex-analysis functional-analysis hilbert-spaces reproducing-kernel-hilbert-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$.
    $endgroup$
    – gerw
    Dec 5 '18 at 10:48






  • 1




    $begingroup$
    Is the topology on $O(mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(mathbb{C})$.
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:43












  • $begingroup$
    I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=frac{1}{2pi i}int_C frac{f(zeta), dzeta}{zeta - z}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 5 '18 at 13:52










  • $begingroup$
    I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $Ksubset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones.
    $endgroup$
    – reuns
    Dec 5 '18 at 14:55








  • 1




    $begingroup$
    @GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$.
    $endgroup$
    – David C. Ullrich
    Dec 5 '18 at 15:51
















0












$begingroup$


I would like to know is the following result true ?



"Each Hilbert space of holomorphic functions on $mathbb C$ is a reproducing kernel Hilbert space".



Where a Hilbert space of holomorphic functions on $mathbb C$ is a Hilbert space
$H subseteq O(mathbb C)$ such that the inclusion mapping $H hookrightarrow O(mathbb C)$ is continuous.



i.e., If $H subseteq O(mathbb C)$ is a Hilbert space of holomorphic functions, then the point evaluation map, $f mapsto f(z)$, is continuous for all $zin mathbb C$ .



If so, where can I find its proof?



Thank you in advance






complex-analysis functional-analysis hilbert-spaces reproducing-kernel-hilbert-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$.
    $endgroup$
    – gerw
    Dec 5 '18 at 10:48






  • 1




    $begingroup$
    Is the topology on $O(mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(mathbb{C})$.
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:43












  • $begingroup$
    I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=frac{1}{2pi i}int_C frac{f(zeta), dzeta}{zeta - z}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 5 '18 at 13:52










  • $begingroup$
    I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $Ksubset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones.
    $endgroup$
    – reuns
    Dec 5 '18 at 14:55








  • 1




    $begingroup$
    @GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$.
    $endgroup$
    – David C. Ullrich
    Dec 5 '18 at 15:51














0












0








0





$begingroup$


I would like to know is the following result true ?



"Each Hilbert space of holomorphic functions on $mathbb C$ is a reproducing kernel Hilbert space".



Where a Hilbert space of holomorphic functions on $mathbb C$ is a Hilbert space
$H subseteq O(mathbb C)$ such that the inclusion mapping $H hookrightarrow O(mathbb C)$ is continuous.



i.e., If $H subseteq O(mathbb C)$ is a Hilbert space of holomorphic functions, then the point evaluation map, $f mapsto f(z)$, is continuous for all $zin mathbb C$ .



If so, where can I find its proof?



Thank you in advance






complex-analysis functional-analysis hilbert-spaces reproducing-kernel-hilbert-spaces







share|cite|improve this question











$endgroup$




I would like to know is the following result true ?



"Each Hilbert space of holomorphic functions on $mathbb C$ is a reproducing kernel Hilbert space".



Where a Hilbert space of holomorphic functions on $mathbb C$ is a Hilbert space
$H subseteq O(mathbb C)$ such that the inclusion mapping $H hookrightarrow O(mathbb C)$ is continuous.



i.e., If $H subseteq O(mathbb C)$ is a Hilbert space of holomorphic functions, then the point evaluation map, $f mapsto f(z)$, is continuous for all $zin mathbb C$ .



If so, where can I find its proof?



Thank you in advance






complex-analysis functional-analysis hilbert-spaces reproducing-kernel-hilbert-spaces




complex-analysis functional-analysis hilbert-spaces reproducing-kernel-hilbert-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 10:49







Z. Alfata

















asked Dec 5 '18 at 8:53









Z. AlfataZ. Alfata

895514




895514












  • $begingroup$
    @Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$.
    $endgroup$
    – gerw
    Dec 5 '18 at 10:48






  • 1




    $begingroup$
    Is the topology on $O(mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(mathbb{C})$.
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:43












  • $begingroup$
    I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=frac{1}{2pi i}int_C frac{f(zeta), dzeta}{zeta - z}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 5 '18 at 13:52










  • $begingroup$
    I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $Ksubset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones.
    $endgroup$
    – reuns
    Dec 5 '18 at 14:55








  • 1




    $begingroup$
    @GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$.
    $endgroup$
    – David C. Ullrich
    Dec 5 '18 at 15:51


















  • $begingroup$
    @Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$.
    $endgroup$
    – gerw
    Dec 5 '18 at 10:48






  • 1




    $begingroup$
    Is the topology on $O(mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(mathbb{C})$.
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:43












  • $begingroup$
    I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=frac{1}{2pi i}int_C frac{f(zeta), dzeta}{zeta - z}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 5 '18 at 13:52










  • $begingroup$
    I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $Ksubset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones.
    $endgroup$
    – reuns
    Dec 5 '18 at 14:55








  • 1




    $begingroup$
    @GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$.
    $endgroup$
    – David C. Ullrich
    Dec 5 '18 at 15:51
















$begingroup$
@Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$.
$endgroup$
– gerw
Dec 5 '18 at 10:48




$begingroup$
@Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$.
$endgroup$
– gerw
Dec 5 '18 at 10:48




1




1




$begingroup$
Is the topology on $O(mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(mathbb{C})$.
$endgroup$
– MaoWao
Dec 5 '18 at 13:43






$begingroup$
Is the topology on $O(mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(mathbb{C})$.
$endgroup$
– MaoWao
Dec 5 '18 at 13:43














$begingroup$
I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=frac{1}{2pi i}int_C frac{f(zeta), dzeta}{zeta - z}.$$
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 13:52




$begingroup$
I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=frac{1}{2pi i}int_C frac{f(zeta), dzeta}{zeta - z}.$$
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 13:52












$begingroup$
I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $Ksubset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones.
$endgroup$
– reuns
Dec 5 '18 at 14:55






$begingroup$
I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $Ksubset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones.
$endgroup$
– reuns
Dec 5 '18 at 14:55






1




1




$begingroup$
@GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$.
$endgroup$
– David C. Ullrich
Dec 5 '18 at 15:51




$begingroup$
@GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$.
$endgroup$
– David C. Ullrich
Dec 5 '18 at 15:51










1 Answer
1






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oldest

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$begingroup$

If the inclusion $i: H to O(mathbb{C})$ is continuous, then of course the evaluations are continuous, since they are continuous on $O(mathbb{C})$ so that $f(z) = (if)(z)$ is a composition of two continuous maps. I'm assuming here that $O(mathbb{C})$ is supplied with the usual topology of uniform convergence on compacta.



A perhaps more interesting case happens if the continuity of this inclusion is dropped. Without this assumption, the continuity of evaluations certainly should not be true, although an explicit counterexample might be difficult to construct. If you allow the use of the axiom of choice, then the following construction I think will work.



Let $E$ be the vector space of all entire functions. It should not be difficult to show by example that a linearly independent sequence of functions $f_n in E$ exists such that $|f_n(0)| to infty$ as $n to infty$. By the axiom of choice, we can complete this sequence to a vector space basis ${f_alpha}_{alpha in A}$ for $E$.



There certainly exists a Hilbert space $H$ with a a vector space basis of the same cardinality as $A$, say this basis is ${h_alpha}_{alpha in A}$. By scaling, we can assume that $|h_alpha|_{H} = 1$. Now, define the norm on $E$ in the following way: $$|sum_{i=1}^n c_if_{alpha_i}|_{E} = |sum_{i=1}^n c_ih_{alpha_i}|_{H}.$$ This makes $E$ into a Hilbert space of holomorphic functions on $mathbb{C}$. Certainly $|f_n|_E = 1$, yet $|f_n(0)| to infty$ by construction, so the evaluation at $z = 0$ is discontinuous.






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    $begingroup$

    If the inclusion $i: H to O(mathbb{C})$ is continuous, then of course the evaluations are continuous, since they are continuous on $O(mathbb{C})$ so that $f(z) = (if)(z)$ is a composition of two continuous maps. I'm assuming here that $O(mathbb{C})$ is supplied with the usual topology of uniform convergence on compacta.



    A perhaps more interesting case happens if the continuity of this inclusion is dropped. Without this assumption, the continuity of evaluations certainly should not be true, although an explicit counterexample might be difficult to construct. If you allow the use of the axiom of choice, then the following construction I think will work.



    Let $E$ be the vector space of all entire functions. It should not be difficult to show by example that a linearly independent sequence of functions $f_n in E$ exists such that $|f_n(0)| to infty$ as $n to infty$. By the axiom of choice, we can complete this sequence to a vector space basis ${f_alpha}_{alpha in A}$ for $E$.



    There certainly exists a Hilbert space $H$ with a a vector space basis of the same cardinality as $A$, say this basis is ${h_alpha}_{alpha in A}$. By scaling, we can assume that $|h_alpha|_{H} = 1$. Now, define the norm on $E$ in the following way: $$|sum_{i=1}^n c_if_{alpha_i}|_{E} = |sum_{i=1}^n c_ih_{alpha_i}|_{H}.$$ This makes $E$ into a Hilbert space of holomorphic functions on $mathbb{C}$. Certainly $|f_n|_E = 1$, yet $|f_n(0)| to infty$ by construction, so the evaluation at $z = 0$ is discontinuous.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      If the inclusion $i: H to O(mathbb{C})$ is continuous, then of course the evaluations are continuous, since they are continuous on $O(mathbb{C})$ so that $f(z) = (if)(z)$ is a composition of two continuous maps. I'm assuming here that $O(mathbb{C})$ is supplied with the usual topology of uniform convergence on compacta.



      A perhaps more interesting case happens if the continuity of this inclusion is dropped. Without this assumption, the continuity of evaluations certainly should not be true, although an explicit counterexample might be difficult to construct. If you allow the use of the axiom of choice, then the following construction I think will work.



      Let $E$ be the vector space of all entire functions. It should not be difficult to show by example that a linearly independent sequence of functions $f_n in E$ exists such that $|f_n(0)| to infty$ as $n to infty$. By the axiom of choice, we can complete this sequence to a vector space basis ${f_alpha}_{alpha in A}$ for $E$.



      There certainly exists a Hilbert space $H$ with a a vector space basis of the same cardinality as $A$, say this basis is ${h_alpha}_{alpha in A}$. By scaling, we can assume that $|h_alpha|_{H} = 1$. Now, define the norm on $E$ in the following way: $$|sum_{i=1}^n c_if_{alpha_i}|_{E} = |sum_{i=1}^n c_ih_{alpha_i}|_{H}.$$ This makes $E$ into a Hilbert space of holomorphic functions on $mathbb{C}$. Certainly $|f_n|_E = 1$, yet $|f_n(0)| to infty$ by construction, so the evaluation at $z = 0$ is discontinuous.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        If the inclusion $i: H to O(mathbb{C})$ is continuous, then of course the evaluations are continuous, since they are continuous on $O(mathbb{C})$ so that $f(z) = (if)(z)$ is a composition of two continuous maps. I'm assuming here that $O(mathbb{C})$ is supplied with the usual topology of uniform convergence on compacta.



        A perhaps more interesting case happens if the continuity of this inclusion is dropped. Without this assumption, the continuity of evaluations certainly should not be true, although an explicit counterexample might be difficult to construct. If you allow the use of the axiom of choice, then the following construction I think will work.



        Let $E$ be the vector space of all entire functions. It should not be difficult to show by example that a linearly independent sequence of functions $f_n in E$ exists such that $|f_n(0)| to infty$ as $n to infty$. By the axiom of choice, we can complete this sequence to a vector space basis ${f_alpha}_{alpha in A}$ for $E$.



        There certainly exists a Hilbert space $H$ with a a vector space basis of the same cardinality as $A$, say this basis is ${h_alpha}_{alpha in A}$. By scaling, we can assume that $|h_alpha|_{H} = 1$. Now, define the norm on $E$ in the following way: $$|sum_{i=1}^n c_if_{alpha_i}|_{E} = |sum_{i=1}^n c_ih_{alpha_i}|_{H}.$$ This makes $E$ into a Hilbert space of holomorphic functions on $mathbb{C}$. Certainly $|f_n|_E = 1$, yet $|f_n(0)| to infty$ by construction, so the evaluation at $z = 0$ is discontinuous.






        share|cite|improve this answer











        $endgroup$



        If the inclusion $i: H to O(mathbb{C})$ is continuous, then of course the evaluations are continuous, since they are continuous on $O(mathbb{C})$ so that $f(z) = (if)(z)$ is a composition of two continuous maps. I'm assuming here that $O(mathbb{C})$ is supplied with the usual topology of uniform convergence on compacta.



        A perhaps more interesting case happens if the continuity of this inclusion is dropped. Without this assumption, the continuity of evaluations certainly should not be true, although an explicit counterexample might be difficult to construct. If you allow the use of the axiom of choice, then the following construction I think will work.



        Let $E$ be the vector space of all entire functions. It should not be difficult to show by example that a linearly independent sequence of functions $f_n in E$ exists such that $|f_n(0)| to infty$ as $n to infty$. By the axiom of choice, we can complete this sequence to a vector space basis ${f_alpha}_{alpha in A}$ for $E$.



        There certainly exists a Hilbert space $H$ with a a vector space basis of the same cardinality as $A$, say this basis is ${h_alpha}_{alpha in A}$. By scaling, we can assume that $|h_alpha|_{H} = 1$. Now, define the norm on $E$ in the following way: $$|sum_{i=1}^n c_if_{alpha_i}|_{E} = |sum_{i=1}^n c_ih_{alpha_i}|_{H}.$$ This makes $E$ into a Hilbert space of holomorphic functions on $mathbb{C}$. Certainly $|f_n|_E = 1$, yet $|f_n(0)| to infty$ by construction, so the evaluation at $z = 0$ is discontinuous.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 '18 at 17:27

























        answered Dec 5 '18 at 13:45









        Bartosz MalmanBartosz Malman

        8161620




        8161620






























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            Residenzschloss Arolsen

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            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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