Sums of entire surjective functions












4












$begingroup$


Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.






ca.classical-analysis-and-odes cv.complex-variables







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  • $begingroup$
    I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
    $endgroup$
    – M. Dus
    2 days ago






  • 1




    $begingroup$
    @M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    @Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
    $endgroup$
    – Alexandre Eremenko
    2 days ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    2 days ago










  • $begingroup$
    @AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
    $endgroup$
    – user137377
    2 days ago
















4












$begingroup$


Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.






ca.classical-analysis-and-odes cv.complex-variables







share|cite|improve this question









New contributor




user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
    $endgroup$
    – M. Dus
    2 days ago






  • 1




    $begingroup$
    @M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    @Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
    $endgroup$
    – Alexandre Eremenko
    2 days ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    2 days ago










  • $begingroup$
    @AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
    $endgroup$
    – user137377
    2 days ago














4












4








4





$begingroup$


Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.






ca.classical-analysis-and-odes cv.complex-variables







share|cite|improve this question









New contributor




user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.






ca.classical-analysis-and-odes cv.complex-variables




ca.classical-analysis-and-odes cv.complex-variables






share|cite|improve this question









New contributor




user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago







user137377













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asked 2 days ago









user137377user137377

263




263




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user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
    $endgroup$
    – M. Dus
    2 days ago






  • 1




    $begingroup$
    @M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    @Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
    $endgroup$
    – Alexandre Eremenko
    2 days ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    2 days ago










  • $begingroup$
    @AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
    $endgroup$
    – user137377
    2 days ago


















  • $begingroup$
    I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
    $endgroup$
    – M. Dus
    2 days ago






  • 1




    $begingroup$
    @M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    @Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
    $endgroup$
    – Alexandre Eremenko
    2 days ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    2 days ago










  • $begingroup$
    @AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
    $endgroup$
    – user137377
    2 days ago
















$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
2 days ago




$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
2 days ago




1




1




$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
2 days ago




$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
2 days ago












$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
2 days ago




$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
2 days ago




1




1




$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
2 days ago




$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
2 days ago












$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
2 days ago




$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
2 days ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z mapsto c + exp g(z)$). An explicit example is
$f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
of degree at most $1$; but $f$ is even, so must be constant,
from which it soon follows that $a_n=0$ for every $n$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)



    For example, all non-constant functions of order less than $1/2$ are surjective.
    This follows from an old theorem of Wiman that for such function $f$ there exists
    a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
    And of course linear combinations of functions of order less than $1/2$ are of order less
    than $1/2$.



    Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
    for example, such that $n_k/ktoinfty$,
    and consider the class of entire functions of the form
    $$f(z)=sum_{ninLambda}c_nz^n.$$
    It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.



    Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
    Proc. LMS, 1970, 21 525-539.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
      $endgroup$
      – Mateusz Kwaśnicki
      yesterday






    • 1




      $begingroup$
      Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
      $endgroup$
      – Christian Remling
      yesterday










    • $begingroup$
      @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
      $endgroup$
      – Alexandre Eremenko
      yesterday











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    4












    $begingroup$

    One expects there to be no such $a_n$ in general, because the
    "typical" entire functions is surjective (those that aren't are of the
    special form $z mapsto c + exp g(z)$). An explicit example is
    $f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
    is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
    of degree at most $1$; but $f$ is even, so must be constant,
    from which it soon follows that $a_n=0$ for every $n$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      One expects there to be no such $a_n$ in general, because the
      "typical" entire functions is surjective (those that aren't are of the
      special form $z mapsto c + exp g(z)$). An explicit example is
      $f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
      is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
      of degree at most $1$; but $f$ is even, so must be constant,
      from which it soon follows that $a_n=0$ for every $n$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        One expects there to be no such $a_n$ in general, because the
        "typical" entire functions is surjective (those that aren't are of the
        special form $z mapsto c + exp g(z)$). An explicit example is
        $f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
        is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
        of degree at most $1$; but $f$ is even, so must be constant,
        from which it soon follows that $a_n=0$ for every $n$.






        share|cite|improve this answer









        $endgroup$



        One expects there to be no such $a_n$ in general, because the
        "typical" entire functions is surjective (those that aren't are of the
        special form $z mapsto c + exp g(z)$). An explicit example is
        $f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
        is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
        of degree at most $1$; but $f$ is even, so must be constant,
        from which it soon follows that $a_n=0$ for every $n$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Noam D. ElkiesNoam D. Elkies

        56.4k11199282




        56.4k11199282























            3












            $begingroup$

            The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)



            For example, all non-constant functions of order less than $1/2$ are surjective.
            This follows from an old theorem of Wiman that for such function $f$ there exists
            a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
            And of course linear combinations of functions of order less than $1/2$ are of order less
            than $1/2$.



            Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
            for example, such that $n_k/ktoinfty$,
            and consider the class of entire functions of the form
            $$f(z)=sum_{ninLambda}c_nz^n.$$
            It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.



            Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
            Proc. LMS, 1970, 21 525-539.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
              $endgroup$
              – Mateusz Kwaśnicki
              yesterday






            • 1




              $begingroup$
              Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
              $endgroup$
              – Christian Remling
              yesterday










            • $begingroup$
              @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
              $endgroup$
              – Alexandre Eremenko
              yesterday
















            3












            $begingroup$

            The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)



            For example, all non-constant functions of order less than $1/2$ are surjective.
            This follows from an old theorem of Wiman that for such function $f$ there exists
            a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
            And of course linear combinations of functions of order less than $1/2$ are of order less
            than $1/2$.



            Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
            for example, such that $n_k/ktoinfty$,
            and consider the class of entire functions of the form
            $$f(z)=sum_{ninLambda}c_nz^n.$$
            It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.



            Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
            Proc. LMS, 1970, 21 525-539.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
              $endgroup$
              – Mateusz Kwaśnicki
              yesterday






            • 1




              $begingroup$
              Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
              $endgroup$
              – Christian Remling
              yesterday










            • $begingroup$
              @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
              $endgroup$
              – Alexandre Eremenko
              yesterday














            3












            3








            3





            $begingroup$

            The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)



            For example, all non-constant functions of order less than $1/2$ are surjective.
            This follows from an old theorem of Wiman that for such function $f$ there exists
            a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
            And of course linear combinations of functions of order less than $1/2$ are of order less
            than $1/2$.



            Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
            for example, such that $n_k/ktoinfty$,
            and consider the class of entire functions of the form
            $$f(z)=sum_{ninLambda}c_nz^n.$$
            It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.



            Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
            Proc. LMS, 1970, 21 525-539.






            share|cite|improve this answer











            $endgroup$



            The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)



            For example, all non-constant functions of order less than $1/2$ are surjective.
            This follows from an old theorem of Wiman that for such function $f$ there exists
            a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
            And of course linear combinations of functions of order less than $1/2$ are of order less
            than $1/2$.



            Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
            for example, such that $n_k/ktoinfty$,
            and consider the class of entire functions of the form
            $$f(z)=sum_{ninLambda}c_nz^n.$$
            It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.



            Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
            Proc. LMS, 1970, 21 525-539.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            Alexandre EremenkoAlexandre Eremenko

            50.7k6140258




            50.7k6140258








            • 1




              $begingroup$
              Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
              $endgroup$
              – Mateusz Kwaśnicki
              yesterday






            • 1




              $begingroup$
              Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
              $endgroup$
              – Christian Remling
              yesterday










            • $begingroup$
              @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
              $endgroup$
              – Alexandre Eremenko
              yesterday














            • 1




              $begingroup$
              Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
              $endgroup$
              – Mateusz Kwaśnicki
              yesterday






            • 1




              $begingroup$
              Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
              $endgroup$
              – Christian Remling
              yesterday










            • $begingroup$
              @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
              $endgroup$
              – Alexandre Eremenko
              yesterday








            1




            1




            $begingroup$
            Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
            $endgroup$
            – Mateusz Kwaśnicki
            yesterday




            $begingroup$
            Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
            $endgroup$
            – Mateusz Kwaśnicki
            yesterday




            1




            1




            $begingroup$
            Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
            $endgroup$
            – Christian Remling
            yesterday




            $begingroup$
            Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
            $endgroup$
            – Christian Remling
            yesterday












            $begingroup$
            @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
            $endgroup$
            – Alexandre Eremenko
            yesterday




            $begingroup$
            @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
            $endgroup$
            – Alexandre Eremenko
            yesterday










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