Application of Maximum modulus theorem
Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $f$ is a constant function and find the value of $f$ in the domain $omega$. I tries to use maximum modulus theorem.$|f(z)|le sqrt{frac{pi}{2}}$ for all $z$ in the domain and hence $f=sqrt{frac{pi}{2}}$ for all $z$ in the domain.Is this valid?
Second attempt:
the function is bounded by 0 in the open disc of radius $frac{1}{pi}$. And then by maximum modulus theorem it is zero throughout the disk and since it is analytic after redefining then by unique of coefficient of power series, the result follows. Is this valid?
complex-analysis
asked Nov 26 at 8:09
user43529463
16118
|
show 1 more comment
Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $f$ is a constant function and find the value of $f$ in the domain $omega$. I tries to use maximum modulus theorem.$|f(z)|le sqrt{frac{pi}{2}}$ for all $z$ in the domain and hence $f=sqrt{frac{pi}{2}}$ for all $z$ in the domain.Is this valid?
Second attempt:
the function is bounded by 0 in the open disc of radius $frac{1}{pi}$. And then by maximum modulus theorem it is zero throughout the disk and since it is analytic after redefining then by unique of coefficient of power series, the result follows. Is this valid?
complex-analysis
asked Nov 26 at 8:09
user43529463
16118
Your answer is wrong because $f(frac 1 {npi})=0$ for all $n$.
– Kavi Rama Murthy
Nov 26 at 8:27
Can I put z=2/pi?
– user43529463
Nov 26 at 8:36
$frac 2 {pi}$ is inside the domain. But I don't understand how you are applying MMT.
– Kavi Rama Murthy
Nov 26 at 8:39
How you deduce it is equal to zero for all z?
– user43529463
Nov 26 at 8:40
I just try to plug in different value to find the maximum value of |f|.
– user43529463
Nov 26 at 8:41
|
show 1 more comment
Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $f$ is a constant function and find the value of $f$ in the domain $omega$. I tries to use maximum modulus theorem.$|f(z)|le sqrt{frac{pi}{2}}$ for all $z$ in the domain and hence $f=sqrt{frac{pi}{2}}$ for all $z$ in the domain.Is this valid?
Second attempt:
the function is bounded by 0 in the open disc of radius $frac{1}{pi}$. And then by maximum modulus theorem it is zero throughout the disk and since it is analytic after redefining then by unique of coefficient of power series, the result follows. Is this valid?
complex-analysis
asked Nov 26 at 8:09
user43529463
16118
Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $f$ is a constant function and find the value of $f$ in the domain $omega$. I tries to use maximum modulus theorem.$|f(z)|le sqrt{frac{pi}{2}}$ for all $z$ in the domain and hence $f=sqrt{frac{pi}{2}}$ for all $z$ in the domain.Is this valid?
Second attempt:
the function is bounded by 0 in the open disc of radius $frac{1}{pi}$. And then by maximum modulus theorem it is zero throughout the disk and since it is analytic after redefining then by unique of coefficient of power series, the result follows. Is this valid?
complex-analysis
complex-analysis
asked Nov 26 at 8:09
user43529463
16118
asked Nov 26 at 8:09
user43529463
16118
edited Nov 27 at 2:11
asked Nov 26 at 8:09
user43529463
16118
asked Nov 26 at 8:09
user43529463
16118
asked Nov 26 at 8:09
user43529463
16118
16118
Your answer is wrong because $f(frac 1 {npi})=0$ for all $n$.
– Kavi Rama Murthy
Nov 26 at 8:27
Can I put z=2/pi?
– user43529463
Nov 26 at 8:36
$frac 2 {pi}$ is inside the domain. But I don't understand how you are applying MMT.
– Kavi Rama Murthy
Nov 26 at 8:39
How you deduce it is equal to zero for all z?
– user43529463
Nov 26 at 8:40
I just try to plug in different value to find the maximum value of |f|.
– user43529463
Nov 26 at 8:41
|
show 1 more comment
Your answer is wrong because $f(frac 1 {npi})=0$ for all $n$.
– Kavi Rama Murthy
Nov 26 at 8:27
Can I put z=2/pi?
– user43529463
Nov 26 at 8:36
$frac 2 {pi}$ is inside the domain. But I don't understand how you are applying MMT.
– Kavi Rama Murthy
Nov 26 at 8:39
How you deduce it is equal to zero for all z?
– user43529463
Nov 26 at 8:40
I just try to plug in different value to find the maximum value of |f|.
– user43529463
Nov 26 at 8:41
Your answer is wrong because $f(frac 1 {npi})=0$ for all $n$.
– Kavi Rama Murthy
Nov 26 at 8:27
Your answer is wrong because $f(frac 1 {npi})=0$ for all $n$.
– Kavi Rama Murthy
Nov 26 at 8:27
Can I put z=2/pi?
– user43529463
Nov 26 at 8:36
Can I put z=2/pi?
– user43529463
Nov 26 at 8:36
$frac 2 {pi}$ is inside the domain. But I don't understand how you are applying MMT.
– Kavi Rama Murthy
Nov 26 at 8:39
$frac 2 {pi}$ is inside the domain. But I don't understand how you are applying MMT.
– Kavi Rama Murthy
Nov 26 at 8:39
How you deduce it is equal to zero for all z?
– user43529463
Nov 26 at 8:40
How you deduce it is equal to zero for all z?
– user43529463
Nov 26 at 8:40
I just try to plug in different value to find the maximum value of |f|.
– user43529463
Nov 26 at 8:41
I just try to plug in different value to find the maximum value of |f|.
– user43529463
Nov 26 at 8:41
|
show 1 more comment
1 Answer
1
active
oldest
votes
Let $g(z)=zf(z)$. Then $|g(z)| leq sqrt {|z|}$ so $g$ has a removable singularity at $0$ and it becomes analytic in the unit disk if define $g(0)$ to be $0$. Hence $frac {g(z)} z$ has a removable singularity at $0$ which means $f$ has a removable singularity at $0$. Now, since $f (frac 1 {npi}) =0$ for all $n$ the zeros of $f$ have a limit point so $f$ must be identically $0$.
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
|
show 3 more comments
Your 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
Let $g(z)=zf(z)$. Then $|g(z)| leq sqrt {|z|}$ so $g$ has a removable singularity at $0$ and it becomes analytic in the unit disk if define $g(0)$ to be $0$. Hence $frac {g(z)} z$ has a removable singularity at $0$ which means $f$ has a removable singularity at $0$. Now, since $f (frac 1 {npi}) =0$ for all $n$ the zeros of $f$ have a limit point so $f$ must be identically $0$.
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
|
show 3 more comments
Let $g(z)=zf(z)$. Then $|g(z)| leq sqrt {|z|}$ so $g$ has a removable singularity at $0$ and it becomes analytic in the unit disk if define $g(0)$ to be $0$. Hence $frac {g(z)} z$ has a removable singularity at $0$ which means $f$ has a removable singularity at $0$. Now, since $f (frac 1 {npi}) =0$ for all $n$ the zeros of $f$ have a limit point so $f$ must be identically $0$.
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
|
show 3 more comments
Let $g(z)=zf(z)$. Then $|g(z)| leq sqrt {|z|}$ so $g$ has a removable singularity at $0$ and it becomes analytic in the unit disk if define $g(0)$ to be $0$. Hence $frac {g(z)} z$ has a removable singularity at $0$ which means $f$ has a removable singularity at $0$. Now, since $f (frac 1 {npi}) =0$ for all $n$ the zeros of $f$ have a limit point so $f$ must be identically $0$.
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
Let $g(z)=zf(z)$. Then $|g(z)| leq sqrt {|z|}$ so $g$ has a removable singularity at $0$ and it becomes analytic in the unit disk if define $g(0)$ to be $0$. Hence $frac {g(z)} z$ has a removable singularity at $0$ which means $f$ has a removable singularity at $0$. Now, since $f (frac 1 {npi}) =0$ for all $n$ the zeros of $f$ have a limit point so $f$ must be identically $0$.
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
answered Nov 26 at 8:49
Kavi Rama Murthy
49.1k31854
49.1k31854
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
|
show 3 more comments
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
So the problem cannot be solved using maximum modulus theorem?
– user43529463
Nov 26 at 8:55
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
As far as I can see MMT theorem is of no use here. @kenkenb
– Kavi Rama Murthy
Nov 26 at 8:58
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
Could you explain the last sentence?
– user43529463
Nov 26 at 9:26
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
What happens to 2/pi?
– user43529463
Nov 26 at 9:27
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
At $frac 2 {pi}$ we only know that $|f(frac 2 {pi})|leq sqrt {frac {pi} 2}$. This is not very useful.
– Kavi Rama Murthy
Nov 26 at 9:29
|
show 3 more comments
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Your answer is wrong because $f(frac 1 {npi})=0$ for all $n$.
– Kavi Rama Murthy
Nov 26 at 8:27
Can I put z=2/pi?
– user43529463
Nov 26 at 8:36
$frac 2 {pi}$ is inside the domain. But I don't understand how you are applying MMT.
– Kavi Rama Murthy
Nov 26 at 8:39
How you deduce it is equal to zero for all z?
– user43529463
Nov 26 at 8:40
I just try to plug in different value to find the maximum value of |f|.
– user43529463
Nov 26 at 8:41