An entire function whose integral is bounded is identically zero












5












$begingroup$


Suppose $f$ has a power series at $0$ that converges in all of $mathbb{C}$ and $$int_{mathbb{C}} |f(x+iy)|dxdy$$



Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.



A hint is given: “Use polar coordinates to show $f(0)=0$”



Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why $f(0) = 0$ makes things any easier?
    $endgroup$
    – xyzzyz
    Jan 29 '18 at 0:03






  • 1




    $begingroup$
    You probably need to tell us what you do know.
    $endgroup$
    – user99914
    Jan 29 '18 at 0:07






  • 1




    $begingroup$
    @JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2
    $endgroup$
    – mysatellite
    Jan 29 '18 at 0:13












  • $begingroup$
    You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle?
    $endgroup$
    – anomaly
    Jan 29 '18 at 2:57










  • $begingroup$
    @anomaly Yes I know
    $endgroup$
    – mysatellite
    Jan 29 '18 at 3:00
















5












$begingroup$


Suppose $f$ has a power series at $0$ that converges in all of $mathbb{C}$ and $$int_{mathbb{C}} |f(x+iy)|dxdy$$



Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.



A hint is given: “Use polar coordinates to show $f(0)=0$”



Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why $f(0) = 0$ makes things any easier?
    $endgroup$
    – xyzzyz
    Jan 29 '18 at 0:03






  • 1




    $begingroup$
    You probably need to tell us what you do know.
    $endgroup$
    – user99914
    Jan 29 '18 at 0:07






  • 1




    $begingroup$
    @JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2
    $endgroup$
    – mysatellite
    Jan 29 '18 at 0:13












  • $begingroup$
    You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle?
    $endgroup$
    – anomaly
    Jan 29 '18 at 2:57










  • $begingroup$
    @anomaly Yes I know
    $endgroup$
    – mysatellite
    Jan 29 '18 at 3:00














5












5








5


4



$begingroup$


Suppose $f$ has a power series at $0$ that converges in all of $mathbb{C}$ and $$int_{mathbb{C}} |f(x+iy)|dxdy$$



Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.



A hint is given: “Use polar coordinates to show $f(0)=0$”



Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc






complex-analysis







share|cite|improve this question











$endgroup$




Suppose $f$ has a power series at $0$ that converges in all of $mathbb{C}$ and $$int_{mathbb{C}} |f(x+iy)|dxdy$$



Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.



A hint is given: “Use polar coordinates to show $f(0)=0$”



Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 '18 at 1:51







mysatellite

















asked Jan 28 '18 at 23:44









mysatellitemysatellite

2,14721231




2,14721231












  • $begingroup$
    Why $f(0) = 0$ makes things any easier?
    $endgroup$
    – xyzzyz
    Jan 29 '18 at 0:03






  • 1




    $begingroup$
    You probably need to tell us what you do know.
    $endgroup$
    – user99914
    Jan 29 '18 at 0:07






  • 1




    $begingroup$
    @JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2
    $endgroup$
    – mysatellite
    Jan 29 '18 at 0:13












  • $begingroup$
    You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle?
    $endgroup$
    – anomaly
    Jan 29 '18 at 2:57










  • $begingroup$
    @anomaly Yes I know
    $endgroup$
    – mysatellite
    Jan 29 '18 at 3:00


















  • $begingroup$
    Why $f(0) = 0$ makes things any easier?
    $endgroup$
    – xyzzyz
    Jan 29 '18 at 0:03






  • 1




    $begingroup$
    You probably need to tell us what you do know.
    $endgroup$
    – user99914
    Jan 29 '18 at 0:07






  • 1




    $begingroup$
    @JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2
    $endgroup$
    – mysatellite
    Jan 29 '18 at 0:13












  • $begingroup$
    You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle?
    $endgroup$
    – anomaly
    Jan 29 '18 at 2:57










  • $begingroup$
    @anomaly Yes I know
    $endgroup$
    – mysatellite
    Jan 29 '18 at 3:00
















$begingroup$
Why $f(0) = 0$ makes things any easier?
$endgroup$
– xyzzyz
Jan 29 '18 at 0:03




$begingroup$
Why $f(0) = 0$ makes things any easier?
$endgroup$
– xyzzyz
Jan 29 '18 at 0:03




1




1




$begingroup$
You probably need to tell us what you do know.
$endgroup$
– user99914
Jan 29 '18 at 0:07




$begingroup$
You probably need to tell us what you do know.
$endgroup$
– user99914
Jan 29 '18 at 0:07




1




1




$begingroup$
@JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2
$endgroup$
– mysatellite
Jan 29 '18 at 0:13






$begingroup$
@JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2
$endgroup$
– mysatellite
Jan 29 '18 at 0:13














$begingroup$
You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle?
$endgroup$
– anomaly
Jan 29 '18 at 2:57




$begingroup$
You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle?
$endgroup$
– anomaly
Jan 29 '18 at 2:57












$begingroup$
@anomaly Yes I know
$endgroup$
– mysatellite
Jan 29 '18 at 3:00




$begingroup$
@anomaly Yes I know
$endgroup$
– mysatellite
Jan 29 '18 at 3:00










1 Answer
1






active

oldest

votes


















5












$begingroup$

Let $f(z)=sum_0 ^{infty} a_n z^{n}$ be the power series expansion. Write $z=re^{itheta}$ and integrate with respect to $theta$ from 0 to $2pi$. Integrating term by term is permitted because of uniform convergence. You get $2pi a_0= int_0 ^{2pi} f(re^{itheta}) dtheta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr dtheta =dxdy$ you will see that $|(int_0 R rdr) 2pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    The integral is of the absolute value of the function.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jan 29 '18 at 8:11










  • $begingroup$
    Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
    $endgroup$
    – Kavi Rama Murthy
    Jan 29 '18 at 22:34












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1 Answer
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active

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oldest

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5












$begingroup$

Let $f(z)=sum_0 ^{infty} a_n z^{n}$ be the power series expansion. Write $z=re^{itheta}$ and integrate with respect to $theta$ from 0 to $2pi$. Integrating term by term is permitted because of uniform convergence. You get $2pi a_0= int_0 ^{2pi} f(re^{itheta}) dtheta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr dtheta =dxdy$ you will see that $|(int_0 R rdr) 2pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    The integral is of the absolute value of the function.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jan 29 '18 at 8:11










  • $begingroup$
    Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
    $endgroup$
    – Kavi Rama Murthy
    Jan 29 '18 at 22:34
















5












$begingroup$

Let $f(z)=sum_0 ^{infty} a_n z^{n}$ be the power series expansion. Write $z=re^{itheta}$ and integrate with respect to $theta$ from 0 to $2pi$. Integrating term by term is permitted because of uniform convergence. You get $2pi a_0= int_0 ^{2pi} f(re^{itheta}) dtheta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr dtheta =dxdy$ you will see that $|(int_0 R rdr) 2pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    The integral is of the absolute value of the function.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jan 29 '18 at 8:11










  • $begingroup$
    Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
    $endgroup$
    – Kavi Rama Murthy
    Jan 29 '18 at 22:34














5












5








5





$begingroup$

Let $f(z)=sum_0 ^{infty} a_n z^{n}$ be the power series expansion. Write $z=re^{itheta}$ and integrate with respect to $theta$ from 0 to $2pi$. Integrating term by term is permitted because of uniform convergence. You get $2pi a_0= int_0 ^{2pi} f(re^{itheta}) dtheta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr dtheta =dxdy$ you will see that $|(int_0 R rdr) 2pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.






share|cite|improve this answer









$endgroup$



Let $f(z)=sum_0 ^{infty} a_n z^{n}$ be the power series expansion. Write $z=re^{itheta}$ and integrate with respect to $theta$ from 0 to $2pi$. Integrating term by term is permitted because of uniform convergence. You get $2pi a_0= int_0 ^{2pi} f(re^{itheta}) dtheta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr dtheta =dxdy$ you will see that $|(int_0 R rdr) 2pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 29 '18 at 7:34









Kavi Rama MurthyKavi Rama Murthy

73.9k53170




73.9k53170








  • 2




    $begingroup$
    The integral is of the absolute value of the function.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jan 29 '18 at 8:11










  • $begingroup$
    Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
    $endgroup$
    – Kavi Rama Murthy
    Jan 29 '18 at 22:34














  • 2




    $begingroup$
    The integral is of the absolute value of the function.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jan 29 '18 at 8:11










  • $begingroup$
    Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
    $endgroup$
    – Kavi Rama Murthy
    Jan 29 '18 at 22:34








2




2




$begingroup$
The integral is of the absolute value of the function.
$endgroup$
– Mariano Suárez-Álvarez
Jan 29 '18 at 8:11




$begingroup$
The integral is of the absolute value of the function.
$endgroup$
– Mariano Suárez-Álvarez
Jan 29 '18 at 8:11












$begingroup$
Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
$endgroup$
– Kavi Rama Murthy
Jan 29 '18 at 22:34




$begingroup$
Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value.
$endgroup$
– Kavi Rama Murthy
Jan 29 '18 at 22:34


















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