Sums of entire surjective functions












2












$begingroup$


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
    6 hours 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
    6 hours 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
    6 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    6 hours 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
    4 hours ago
















2












$begingroup$


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
    6 hours 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
    6 hours 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
    6 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    6 hours 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
    4 hours ago














2












2








2





$begingroup$


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









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




share|cite|improve this question








edited 4 hours ago







user137377













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asked 6 hours ago









user137377user137377

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142




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user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    6 hours 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
    6 hours 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
    6 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    6 hours 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
    4 hours 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
    6 hours 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
    6 hours 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
    6 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    6 hours 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
    4 hours 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
6 hours 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
6 hours 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
6 hours 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
6 hours 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
6 hours 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
6 hours ago




1




1




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




$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
6 hours 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
4 hours 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
4 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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 $1/2$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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 4 hours ago









          Noam D. ElkiesNoam D. Elkies

          56.4k11199282




          56.4k11199282























              2












              $begingroup$

              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 $1/2$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                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 $1/2$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 $1/2$.






                  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 $1/2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Alexandre EremenkoAlexandre Eremenko

                  50.7k6140258




                  50.7k6140258






















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