On a property of a uniformly bounded sequence of holomorphic functions on $D$












1












$begingroup$


Let $left{f_{n}：n=1,2,3ldotsright}$ a uniformly bounded sequence of holomorphic functions on a unit disk $D$.Suppose there exists a point $ain D$,so that $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$ for each $k=1,2,3...$ ($f^{left(kright)}_{n}$ is the $kth$ derivative of $f_{n}$). Show that $f_{n}rightarrow 0$ uniformaly on each compact subset of $D$.



My thought: since $left{f_{n}right}$ is uniformaly bounded,from montel theory it's a normal family which means that $left{f_{n}right}$ has a subsequence that convergents on each compact subsets of $D$. Additionally the limit of above subsequence is holomorphic on each compact subset. And also the limit function on compact subsets containing $a$ is constant because of identity principle and the condition $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$. Then I don't know how to approach to the result.






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please tell us what $D$ is.
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:57










  • $begingroup$
    @Hmm That's a different problem
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:58










  • $begingroup$
    Uniformly, not uniformaly.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jul 11 '16 at 7:23
















1












$begingroup$


Let $left{f_{n}：n=1,2,3ldotsright}$ a uniformly bounded sequence of holomorphic functions on a unit disk $D$.Suppose there exists a point $ain D$,so that $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$ for each $k=1,2,3...$ ($f^{left(kright)}_{n}$ is the $kth$ derivative of $f_{n}$). Show that $f_{n}rightarrow 0$ uniformaly on each compact subset of $D$.



My thought: since $left{f_{n}right}$ is uniformaly bounded,from montel theory it's a normal family which means that $left{f_{n}right}$ has a subsequence that convergents on each compact subsets of $D$. Additionally the limit of above subsequence is holomorphic on each compact subset. And also the limit function on compact subsets containing $a$ is constant because of identity principle and the condition $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$. Then I don't know how to approach to the result.






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please tell us what $D$ is.
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:57










  • $begingroup$
    @Hmm That's a different problem
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:58










  • $begingroup$
    Uniformly, not uniformaly.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jul 11 '16 at 7:23














1












1








1





$begingroup$


Let $left{f_{n}：n=1,2,3ldotsright}$ a uniformly bounded sequence of holomorphic functions on a unit disk $D$.Suppose there exists a point $ain D$,so that $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$ for each $k=1,2,3...$ ($f^{left(kright)}_{n}$ is the $kth$ derivative of $f_{n}$). Show that $f_{n}rightarrow 0$ uniformaly on each compact subset of $D$.



My thought: since $left{f_{n}right}$ is uniformaly bounded,from montel theory it's a normal family which means that $left{f_{n}right}$ has a subsequence that convergents on each compact subsets of $D$. Additionally the limit of above subsequence is holomorphic on each compact subset. And also the limit function on compact subsets containing $a$ is constant because of identity principle and the condition $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$. Then I don't know how to approach to the result.






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$




Let $left{f_{n}：n=1,2,3ldotsright}$ a uniformly bounded sequence of holomorphic functions on a unit disk $D$.Suppose there exists a point $ain D$,so that $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$ for each $k=1,2,3...$ ($f^{left(kright)}_{n}$ is the $kth$ derivative of $f_{n}$). Show that $f_{n}rightarrow 0$ uniformaly on each compact subset of $D$.



My thought: since $left{f_{n}right}$ is uniformaly bounded,from montel theory it's a normal family which means that $left{f_{n}right}$ has a subsequence that convergents on each compact subsets of $D$. Additionally the limit of above subsequence is holomorphic on each compact subset. And also the limit function on compact subsets containing $a$ is constant because of identity principle and the condition $lim_{nrightarrowinfty}f^{left(kright)}_{n}left(aright)=0$. Then I don't know how to approach to the result.






complex-analysis holomorphic-functions




complex-analysis holomorphic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 11 '16 at 8:35









Joe

7,29421230




7,29421230










asked Jul 11 '16 at 3:44









mikemike

90049




90049












  • $begingroup$
    Please tell us what $D$ is.
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:57










  • $begingroup$
    @Hmm That's a different problem
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:58










  • $begingroup$
    Uniformly, not uniformaly.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jul 11 '16 at 7:23


















  • $begingroup$
    Please tell us what $D$ is.
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:57










  • $begingroup$
    @Hmm That's a different problem
    $endgroup$
    – zhw.
    Jul 11 '16 at 6:58










  • $begingroup$
    Uniformly, not uniformaly.
    $endgroup$
    – Mariano Suárez-Álvarez
    Jul 11 '16 at 7:23
















$begingroup$
Please tell us what $D$ is.
$endgroup$
– zhw.
Jul 11 '16 at 6:57




$begingroup$
Please tell us what $D$ is.
$endgroup$
– zhw.
Jul 11 '16 at 6:57












$begingroup$
@Hmm That's a different problem
$endgroup$
– zhw.
Jul 11 '16 at 6:58




$begingroup$
@Hmm That's a different problem
$endgroup$
– zhw.
Jul 11 '16 at 6:58












$begingroup$
Uniformly, not uniformaly.
$endgroup$
– Mariano Suárez-Álvarez
Jul 11 '16 at 7:23




$begingroup$
Uniformly, not uniformaly.
$endgroup$
– Mariano Suárez-Álvarez
Jul 11 '16 at 7:23










1 Answer
1






active

oldest

votes


















2












$begingroup$

I suppose $D={|z|<1}$.
Being our sequence ${f_n}_n$ of holomorphic functions on $D$, uniformly bounded, then by Montel theorem, admits some converging subsequence ${f_{n_nu}}_nu$: let's call $f$ the limit function.



Since $f_n^{(k)}(a)stackrel{nto+infty}{longrightarrow}0$ for all $kge1$, then this condition holds even for the subsequence $f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow}0$ for all $kge1$; but it's clear that $f_{n_nu}to f$ wrt compact subsets topology implies that
$f_{n_nu}^{(k)}to f^{(k)}$ for every $kge0$, thus
$$
f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow} f^{(k)}(a);;;;forall kge1
$$



from which we get that $f^{(k)}(a)=0$ for all $kge1$ and thus (expanding $f$ near $a$ in Taylor series first, and then applying identity principle) we get $fequiv f(a)$ on the whole $D$.



Now, this argument can be done for every converging subsequence of ${f_n}_n$, thus every converging subsequence of it, converges to the same function (the constant function $f(a)$), thus by the Montel converging criterion, we get that ${f_n}_n$ converges, and it converges necessarely to $f(a)$ (wrt the topology above).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
    $endgroup$
    – mike
    Jul 11 '16 at 9:11










  • $begingroup$
    @mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
    $endgroup$
    – Joe
    Jul 11 '16 at 12:23












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

I suppose $D={|z|<1}$.
Being our sequence ${f_n}_n$ of holomorphic functions on $D$, uniformly bounded, then by Montel theorem, admits some converging subsequence ${f_{n_nu}}_nu$: let's call $f$ the limit function.



Since $f_n^{(k)}(a)stackrel{nto+infty}{longrightarrow}0$ for all $kge1$, then this condition holds even for the subsequence $f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow}0$ for all $kge1$; but it's clear that $f_{n_nu}to f$ wrt compact subsets topology implies that
$f_{n_nu}^{(k)}to f^{(k)}$ for every $kge0$, thus
$$
f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow} f^{(k)}(a);;;;forall kge1
$$



from which we get that $f^{(k)}(a)=0$ for all $kge1$ and thus (expanding $f$ near $a$ in Taylor series first, and then applying identity principle) we get $fequiv f(a)$ on the whole $D$.



Now, this argument can be done for every converging subsequence of ${f_n}_n$, thus every converging subsequence of it, converges to the same function (the constant function $f(a)$), thus by the Montel converging criterion, we get that ${f_n}_n$ converges, and it converges necessarely to $f(a)$ (wrt the topology above).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
    $endgroup$
    – mike
    Jul 11 '16 at 9:11










  • $begingroup$
    @mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
    $endgroup$
    – Joe
    Jul 11 '16 at 12:23
















2












$begingroup$

I suppose $D={|z|<1}$.
Being our sequence ${f_n}_n$ of holomorphic functions on $D$, uniformly bounded, then by Montel theorem, admits some converging subsequence ${f_{n_nu}}_nu$: let's call $f$ the limit function.



Since $f_n^{(k)}(a)stackrel{nto+infty}{longrightarrow}0$ for all $kge1$, then this condition holds even for the subsequence $f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow}0$ for all $kge1$; but it's clear that $f_{n_nu}to f$ wrt compact subsets topology implies that
$f_{n_nu}^{(k)}to f^{(k)}$ for every $kge0$, thus
$$
f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow} f^{(k)}(a);;;;forall kge1
$$



from which we get that $f^{(k)}(a)=0$ for all $kge1$ and thus (expanding $f$ near $a$ in Taylor series first, and then applying identity principle) we get $fequiv f(a)$ on the whole $D$.



Now, this argument can be done for every converging subsequence of ${f_n}_n$, thus every converging subsequence of it, converges to the same function (the constant function $f(a)$), thus by the Montel converging criterion, we get that ${f_n}_n$ converges, and it converges necessarely to $f(a)$ (wrt the topology above).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
    $endgroup$
    – mike
    Jul 11 '16 at 9:11










  • $begingroup$
    @mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
    $endgroup$
    – Joe
    Jul 11 '16 at 12:23














2












2








2





$begingroup$

I suppose $D={|z|<1}$.
Being our sequence ${f_n}_n$ of holomorphic functions on $D$, uniformly bounded, then by Montel theorem, admits some converging subsequence ${f_{n_nu}}_nu$: let's call $f$ the limit function.



Since $f_n^{(k)}(a)stackrel{nto+infty}{longrightarrow}0$ for all $kge1$, then this condition holds even for the subsequence $f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow}0$ for all $kge1$; but it's clear that $f_{n_nu}to f$ wrt compact subsets topology implies that
$f_{n_nu}^{(k)}to f^{(k)}$ for every $kge0$, thus
$$
f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow} f^{(k)}(a);;;;forall kge1
$$



from which we get that $f^{(k)}(a)=0$ for all $kge1$ and thus (expanding $f$ near $a$ in Taylor series first, and then applying identity principle) we get $fequiv f(a)$ on the whole $D$.



Now, this argument can be done for every converging subsequence of ${f_n}_n$, thus every converging subsequence of it, converges to the same function (the constant function $f(a)$), thus by the Montel converging criterion, we get that ${f_n}_n$ converges, and it converges necessarely to $f(a)$ (wrt the topology above).






share|cite|improve this answer











$endgroup$



I suppose $D={|z|<1}$.
Being our sequence ${f_n}_n$ of holomorphic functions on $D$, uniformly bounded, then by Montel theorem, admits some converging subsequence ${f_{n_nu}}_nu$: let's call $f$ the limit function.



Since $f_n^{(k)}(a)stackrel{nto+infty}{longrightarrow}0$ for all $kge1$, then this condition holds even for the subsequence $f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow}0$ for all $kge1$; but it's clear that $f_{n_nu}to f$ wrt compact subsets topology implies that
$f_{n_nu}^{(k)}to f^{(k)}$ for every $kge0$, thus
$$
f_{n_nu}^{(k)}(a)stackrel{nuto+infty}{longrightarrow} f^{(k)}(a);;;;forall kge1
$$



from which we get that $f^{(k)}(a)=0$ for all $kge1$ and thus (expanding $f$ near $a$ in Taylor series first, and then applying identity principle) we get $fequiv f(a)$ on the whole $D$.



Now, this argument can be done for every converging subsequence of ${f_n}_n$, thus every converging subsequence of it, converges to the same function (the constant function $f(a)$), thus by the Montel converging criterion, we get that ${f_n}_n$ converges, and it converges necessarely to $f(a)$ (wrt the topology above).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jul 11 '16 at 12:22

























answered Jul 11 '16 at 8:34









JoeJoe

7,29421230




7,29421230












  • $begingroup$
    Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
    $endgroup$
    – mike
    Jul 11 '16 at 9:11










  • $begingroup$
    @mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
    $endgroup$
    – Joe
    Jul 11 '16 at 12:23


















  • $begingroup$
    Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
    $endgroup$
    – mike
    Jul 11 '16 at 9:11










  • $begingroup$
    @mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
    $endgroup$
    – Joe
    Jul 11 '16 at 12:23
















$begingroup$
Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
$endgroup$
– mike
Jul 11 '16 at 9:11




$begingroup$
Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk.
$endgroup$
– mike
Jul 11 '16 at 9:11












$begingroup$
@mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
$endgroup$
– Joe
Jul 11 '16 at 12:23




$begingroup$
@mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-)
$endgroup$
– Joe
Jul 11 '16 at 12:23


















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