How to prove that $f(z)$ is bounded if we know that $limlimits_{|z|to infty}frac{f(z)}{z}= 0$?












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Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.

How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$






complex-analysis functional-analysis complex-numbers







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$endgroup$








  • 2




    $begingroup$
    Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:27










  • $begingroup$
    Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
    $endgroup$
    – Story123
    Dec 23 '18 at 21:04


















-1












$begingroup$


Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.

How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$






complex-analysis functional-analysis complex-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:27










  • $begingroup$
    Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
    $endgroup$
    – Story123
    Dec 23 '18 at 21:04
















-1












-1








-1





$begingroup$


Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.

How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$






complex-analysis functional-analysis complex-numbers







share|cite|improve this question











$endgroup$




Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.

How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$






complex-analysis functional-analysis complex-numbers




complex-analysis functional-analysis complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 21:44









user376343

3,8483829




3,8483829










asked Dec 23 '18 at 19:22









Idan AvivIdan Aviv

1014




1014








  • 2




    $begingroup$
    Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:27










  • $begingroup$
    Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
    $endgroup$
    – Story123
    Dec 23 '18 at 21:04
















  • 2




    $begingroup$
    Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 19:27










  • $begingroup$
    Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
    $endgroup$
    – Story123
    Dec 23 '18 at 21:04










2




2




$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27




$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27












$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04






$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04












2 Answers
2






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3












$begingroup$

If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 19:38










  • $begingroup$
    Typo. I meant $n>1$. I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 19:54










  • $begingroup$
    I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 20:17










  • $begingroup$
    1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 21:18










  • $begingroup$
    One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 22:12



















1












$begingroup$

Hint If $f(z)$ is entire, so is
$$g(z)=frac{f(z)-f(0)}{z}$$



Next, it is easy to see that
$$lim_{z to infty}g(z)=0$$
which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.



Finally,
$$0= lim_{z to infty} g(z)= C$$






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    3












    $begingroup$

    If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 19:38










    • $begingroup$
      Typo. I meant $n>1$. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 19:54










    • $begingroup$
      I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 20:17










    • $begingroup$
      1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 21:18










    • $begingroup$
      One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 22:12
















    3












    $begingroup$

    If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 19:38










    • $begingroup$
      Typo. I meant $n>1$. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 19:54










    • $begingroup$
      I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 20:17










    • $begingroup$
      1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 21:18










    • $begingroup$
      One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 22:12














    3












    3








    3





    $begingroup$

    If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.






    share|cite|improve this answer











    $endgroup$



    If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 19:53

























    answered Dec 23 '18 at 19:32









    José Carlos SantosJosé Carlos Santos

    164k22131234




    164k22131234












    • $begingroup$
      Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 19:38










    • $begingroup$
      Typo. I meant $n>1$. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 19:54










    • $begingroup$
      I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 20:17










    • $begingroup$
      1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 21:18










    • $begingroup$
      One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 22:12


















    • $begingroup$
      Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 19:38










    • $begingroup$
      Typo. I meant $n>1$. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 19:54










    • $begingroup$
      I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 20:17










    • $begingroup$
      1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 21:18










    • $begingroup$
      One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
      $endgroup$
      – Idan Aviv
      Dec 23 '18 at 22:12
















    $begingroup$
    Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 19:38




    $begingroup$
    Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 19:38












    $begingroup$
    Typo. I meant $n>1$. I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 19:54




    $begingroup$
    Typo. I meant $n>1$. I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 19:54












    $begingroup$
    I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 20:17




    $begingroup$
    I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 20:17












    $begingroup$
    1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 21:18




    $begingroup$
    1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 21:18












    $begingroup$
    One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 22:12




    $begingroup$
    One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
    $endgroup$
    – Idan Aviv
    Dec 23 '18 at 22:12











    1












    $begingroup$

    Hint If $f(z)$ is entire, so is
    $$g(z)=frac{f(z)-f(0)}{z}$$



    Next, it is easy to see that
    $$lim_{z to infty}g(z)=0$$
    which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.



    Finally,
    $$0= lim_{z to infty} g(z)= C$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint If $f(z)$ is entire, so is
      $$g(z)=frac{f(z)-f(0)}{z}$$



      Next, it is easy to see that
      $$lim_{z to infty}g(z)=0$$
      which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.



      Finally,
      $$0= lim_{z to infty} g(z)= C$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint If $f(z)$ is entire, so is
        $$g(z)=frac{f(z)-f(0)}{z}$$



        Next, it is easy to see that
        $$lim_{z to infty}g(z)=0$$
        which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.



        Finally,
        $$0= lim_{z to infty} g(z)= C$$






        share|cite|improve this answer









        $endgroup$



        Hint If $f(z)$ is entire, so is
        $$g(z)=frac{f(z)-f(0)}{z}$$



        Next, it is easy to see that
        $$lim_{z to infty}g(z)=0$$
        which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.



        Finally,
        $$0= lim_{z to infty} g(z)= C$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 21:55









        N. S.N. S.

        104k7114209




        104k7114209






























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