An entire function whose integral is bounded is identically zero
$begingroup$
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
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
$endgroup$
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
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
edited Jan 29 '18 at 1:51
mysatellite
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
asked Jan 28 '18 at 23:44
mysatellitemysatellite
2,14721231
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
1 Answer
1
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votes
$begingroup$
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.
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
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
add a comment |
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$begingroup$
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.
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
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.
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
answered Jan 29 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
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
add a comment |
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
add a comment |
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$begingroup$
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