If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is bounded then $f$ is constant function.











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Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.



Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you






complex-analysis







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  • Are you assuming that $f$ is entire?
    – Tito Eliatron
    Nov 21 at 22:32










  • @Tito Eliatron yes. I edited it.
    – ramanujan
    Nov 21 at 22:33












  • This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
    – John Douma
    Nov 21 at 22:42










  • @John Douma I used this
    – ramanujan
    Nov 21 at 22:45






  • 1




    Then I'd say you are good.
    – John Douma
    Nov 21 at 22:49















up vote
0
down vote

favorite












Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.



Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you






complex-analysis







share|cite|improve this question
























  • Are you assuming that $f$ is entire?
    – Tito Eliatron
    Nov 21 at 22:32










  • @Tito Eliatron yes. I edited it.
    – ramanujan
    Nov 21 at 22:33












  • This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
    – John Douma
    Nov 21 at 22:42










  • @John Douma I used this
    – ramanujan
    Nov 21 at 22:45






  • 1




    Then I'd say you are good.
    – John Douma
    Nov 21 at 22:49













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.



Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you






complex-analysis







share|cite|improve this question















Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.



Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 21 at 22:40

























asked Nov 21 at 22:30









ramanujan

675713




675713












  • Are you assuming that $f$ is entire?
    – Tito Eliatron
    Nov 21 at 22:32










  • @Tito Eliatron yes. I edited it.
    – ramanujan
    Nov 21 at 22:33












  • This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
    – John Douma
    Nov 21 at 22:42










  • @John Douma I used this
    – ramanujan
    Nov 21 at 22:45






  • 1




    Then I'd say you are good.
    – John Douma
    Nov 21 at 22:49


















  • Are you assuming that $f$ is entire?
    – Tito Eliatron
    Nov 21 at 22:32










  • @Tito Eliatron yes. I edited it.
    – ramanujan
    Nov 21 at 22:33












  • This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
    – John Douma
    Nov 21 at 22:42










  • @John Douma I used this
    – ramanujan
    Nov 21 at 22:45






  • 1




    Then I'd say you are good.
    – John Douma
    Nov 21 at 22:49
















Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32




Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32












@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 at 22:33






@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 at 22:33














This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 at 22:42




This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 at 22:42












@John Douma I used this
– ramanujan
Nov 21 at 22:45




@John Douma I used this
– ramanujan
Nov 21 at 22:45




1




1




Then I'd say you are good.
– John Douma
Nov 21 at 22:49




Then I'd say you are good.
– John Douma
Nov 21 at 22:49










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Another approach:



We have the identity
$$leftvert e^z right vert=e^{Re z}$$




Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.



We have
$$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$



Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$



which completes the proof.






share|cite|improve this answer





















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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Another approach:



    We have the identity
    $$leftvert e^z right vert=e^{Re z}$$




    Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.



    We have
    $$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$



    Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$



    which completes the proof.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Another approach:



      We have the identity
      $$leftvert e^z right vert=e^{Re z}$$




      Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.



      We have
      $$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$



      Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$



      which completes the proof.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Another approach:



        We have the identity
        $$leftvert e^z right vert=e^{Re z}$$




        Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.



        We have
        $$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$



        Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$



        which completes the proof.






        share|cite|improve this answer












        Another approach:



        We have the identity
        $$leftvert e^z right vert=e^{Re z}$$




        Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.



        We have
        $$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$



        Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$



        which completes the proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 1:55









        Szeto

        6,3092826




        6,3092826






























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