Sums of entire surjective functionsUniform Convergence of piecewise Continuous Uniformly Convergent Functionsentire functions of one complex variable with prescribed value and order.The boundedness of an entire function along the imaginary axisEntire composite square roots of functions of finite orderCan an entire function have every root function?Surjective entire functionsInterchanging sums and integrals in a specific instanceExponential type of a product of entire functionsUnconditionally convergent series in some functional spacesCoefficients of entire functions with specified zero set

Sums of entire surjective functions


Uniform Convergence of piecewise Continuous Uniformly Convergent Functionsentire functions of one complex variable with prescribed value and order.The boundedness of an entire function along the imaginary axisEntire composite square roots of functions of finite orderCan an entire function have every root function?Surjective entire functionsInterchanging sums and integrals in a specific instanceExponential type of a product of entire functionsUnconditionally convergent series in some functional spacesCoefficients of entire functions with specified zero set













3












$begingroup$


Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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.
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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
    7 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
    7 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
    7 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    7 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
    5 hours ago















3












$begingroup$


Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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
    7 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
    7 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
    7 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    7 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
    5 hours ago













3












3








3





$begingroup$


Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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 $f_n:mathbbCtomathbbC$. 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 5 hours ago







user137377













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.









asked 7 hours ago









user137377user137377

192




192




New contributor




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






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
    7 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
    7 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
    7 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    7 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
    5 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
    7 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
    7 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
    7 hours ago






  • 1




    $begingroup$
    Maybe $(f_n)$ denotes an infinite sequence of functions?
    $endgroup$
    – Nik Weaver
    7 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
    5 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
7 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
7 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
7 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
7 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
7 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
7 hours ago




1




1




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




$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 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
5 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
5 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

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 $min_|f(z)|toinfty$ 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









      3












      $begingroup$

      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












        $begingroup$

        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












          3








          3





          $begingroup$

          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 5 hours 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 $min_|f(z)|toinfty$ 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









              $endgroup$

















                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 $min_|f(z)|toinfty$ 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









                $endgroup$















                  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 $min_|f(z)|toinfty$ 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 $min_|f(z)|toinfty$ 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 3 hours ago









                  Alexandre EremenkoAlexandre Eremenko

                  50.7k6140258




                  50.7k6140258




















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