Application of Maximum modulus theorem












0














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







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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


















0














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







share|cite|improve this question
























  • 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
















0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 2:11

























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




















  • 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












1 Answer
1






active

oldest

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2














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 $0$.






share|cite|improve this answer





















  • 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











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1 Answer
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1 Answer
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active

oldest

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active

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2














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 $0$.






share|cite|improve this answer





















  • 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
















2














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 $0$.






share|cite|improve this answer





















  • 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














2












2








2






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 $0$.






share|cite|improve this answer












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 $0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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