Sums of entire surjective functions












3












$begingroup$


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
    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 $f_n:mathbb{C}tomathbb{C}$. 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
    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 $f_n:mathbb{C}tomathbb{C}$. 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 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.
Check out our Code of Conduct.





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 $min_{|z|=r_k}|f(z)|toinfty$ 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









    $endgroup$













      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: "504"
      };
      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
      });


      }
      });






      user137377 is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326094%2fsums-of-entire-surjective-functions%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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 $min_{|z|=r_k}|f(z)|toinfty$ 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









              $endgroup$


















                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 $1/2$.






                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 $min_{|z|=r_k}|f(z)|toinfty$ 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 3 hours ago









                  Alexandre EremenkoAlexandre Eremenko

                  50.7k6140258




                  50.7k6140258






















                      user137377 is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      user137377 is a new contributor. Be nice, and check out our Code of Conduct.













                      user137377 is a new contributor. Be nice, and check out our Code of Conduct.












                      user137377 is a new contributor. Be nice, and check out our Code of Conduct.
















                      Thanks for contributing an answer to MathOverflow!


                      • 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%2fmathoverflow.net%2fquestions%2f326094%2fsums-of-entire-surjective-functions%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

                      A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                      Image
                      0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{...
                      Read more

                      Calculate evaluation metrics using cross_val_predict sklearn

                      Image
                      1 In the sklearn.model_selection.cross_val_predict page it is stated: Generate cross-validated estimates for each input data point. It is not appropriate to pass these predictions into an evaluation metric. Can someone explain what does it mean? If this gives estimate of Y (y prediction) for every Y (true Y), why can't I calculate metrics such as RMSE or coefficient of determination using these results? python scikit-learn cross-validation share | improve this question edited Nov 28 '18 at 17:52 desertnaut 20.3k 7 43 79 ...
                      Read more

                      Insert data from modal to MySQL (multiple modal on website)

                      Image
                      0 I have problem with inserting data to MySQL from modal. My modal: <a href="#" class="badge badge-pill badge-success">6 komentarzy</a> <a href="#" class="badge badge-pill badge-danger">brak komentarzy</a> <a data-toggle="modal" href="#add_desk_comm_{$desk_['desk_id']}" data-target="#add_desk_comm_{$desk_['desk_id']}" class="ediiit">(dodaj)</a> <div class="modal fade" id="add_desk_comm_{$desk_['desk_id']}" tabindex="-1" role="dialog" aria-labelledby="edit_printer" aria-hidden="true"> <div class="modal-dialog" role="document"> ...
                      Read more