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.










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 $f_n:mathbbCtomathbbC$. 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.










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.










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






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 $min_|f(z)|toinfty$ 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 4 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




















                      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

                      Magento 2 duplicate PHPSESSID cookie when using session_start() in custom php scriptMagento 2: User cant logged in into to account page, no error showing!Magento duplicate on subdomainGrabbing storeview from cookie (after using language selector)How do I run php custom script on magento2Magento 2: Include PHP script in headerSession lock after using Cm_RedisSessionscript php to update stockMagento set cookie popupMagento 2 session id cookie - where to find it?How to import Configurable product from csv with custom attributes using php scriptMagento 2 run custom PHP script

                      Can not update quote_id field of “quote_item” table magento 2Magento 2.1 - We can't remove the item. (Shopping Cart doesnt allow us to remove items before becomes empty)Add value for custom quote item attribute using REST apiREST API endpoint v1/carts/cartId/items always returns error messageCorrect way to save entries to databaseHow to remove all associated quote objects of a customer completelyMagento 2 - Save value from custom input field to quote_itemGet quote_item data using quote id and product id filter in Magento 2How to set additional data to quote_item table from controller in Magento 2?What is the purpose of additional_data column in quote_item table in magento2Set Custom Price to Quote item magento2 from controller

                      How to solve knockout JS error in Magento 2 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?(Magento2) knockout.js:3012 Uncaught ReferenceError: Unable to process bindingUnable to process binding Knockout.js magento 2Cannot read property `scopeLabel` of undefined on Product Detail PageCan't get Customer Data on frontend in Magento 2Magento2 Order Summary - unable to process bindingKO templates are not loading in Magento 2.1 applicationgetting knockout js error magento 2Product grid not load -— Unable to process binding Knockout.js magento 2Product form not loaded in magento2Uncaught ReferenceError: Unable to process binding “if: function()return (isShowLegend()) ” magento 2