Lacunary function doesn't converge anywhere on the boundary












2












$begingroup$


Consider the Lacunary function
$$f(z)=sum _{n=0}^{infty}z^{2^{n}}=z+z^{2}+z^{4}+z^{8}+cdots $$



which is analytic over the interior of the open unit disk. It is easy to show that $f(z)$ doesn't have limits at $2^n$-th roots of unity (see wikipedia), which form a dense subset of unit circle. Wikipedia claims that




hence by continuous extension every point on the unit circle must be a singularity of $f$.




Why does this hold? Suppose $f$ has a limit at one point of the unit circle, does it necessarily have limits everywhere in some neighborhood of it, and thus contradicts to the fact that the there is a dense subset of unit circle where $f$ doesn't have limits? (generally this is not sure, see Riemann Function)



I don't see how the linked post solve my question. I am asking about a very specific power series.






real-analysis complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Possible duplicate of Convergence power series in boundary
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 19:57










  • $begingroup$
    Protip: If you are going to suggest someone to check wikipedia or any online resourse, provide the link in your post. square brackets for title and parenthesis for url.
    $endgroup$
    – Mefitico
    Dec 5 '18 at 20:27
















2












$begingroup$


Consider the Lacunary function
$$f(z)=sum _{n=0}^{infty}z^{2^{n}}=z+z^{2}+z^{4}+z^{8}+cdots $$



which is analytic over the interior of the open unit disk. It is easy to show that $f(z)$ doesn't have limits at $2^n$-th roots of unity (see wikipedia), which form a dense subset of unit circle. Wikipedia claims that




hence by continuous extension every point on the unit circle must be a singularity of $f$.




Why does this hold? Suppose $f$ has a limit at one point of the unit circle, does it necessarily have limits everywhere in some neighborhood of it, and thus contradicts to the fact that the there is a dense subset of unit circle where $f$ doesn't have limits? (generally this is not sure, see Riemann Function)



I don't see how the linked post solve my question. I am asking about a very specific power series.






real-analysis complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Possible duplicate of Convergence power series in boundary
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 19:57










  • $begingroup$
    Protip: If you are going to suggest someone to check wikipedia or any online resourse, provide the link in your post. square brackets for title and parenthesis for url.
    $endgroup$
    – Mefitico
    Dec 5 '18 at 20:27














2












2








2





$begingroup$


Consider the Lacunary function
$$f(z)=sum _{n=0}^{infty}z^{2^{n}}=z+z^{2}+z^{4}+z^{8}+cdots $$



which is analytic over the interior of the open unit disk. It is easy to show that $f(z)$ doesn't have limits at $2^n$-th roots of unity (see wikipedia), which form a dense subset of unit circle. Wikipedia claims that




hence by continuous extension every point on the unit circle must be a singularity of $f$.




Why does this hold? Suppose $f$ has a limit at one point of the unit circle, does it necessarily have limits everywhere in some neighborhood of it, and thus contradicts to the fact that the there is a dense subset of unit circle where $f$ doesn't have limits? (generally this is not sure, see Riemann Function)



I don't see how the linked post solve my question. I am asking about a very specific power series.






real-analysis complex-analysis







share|cite|improve this question











$endgroup$




Consider the Lacunary function
$$f(z)=sum _{n=0}^{infty}z^{2^{n}}=z+z^{2}+z^{4}+z^{8}+cdots $$



which is analytic over the interior of the open unit disk. It is easy to show that $f(z)$ doesn't have limits at $2^n$-th roots of unity (see wikipedia), which form a dense subset of unit circle. Wikipedia claims that




hence by continuous extension every point on the unit circle must be a singularity of $f$.




Why does this hold? Suppose $f$ has a limit at one point of the unit circle, does it necessarily have limits everywhere in some neighborhood of it, and thus contradicts to the fact that the there is a dense subset of unit circle where $f$ doesn't have limits? (generally this is not sure, see Riemann Function)



I don't see how the linked post solve my question. I am asking about a very specific power series.






real-analysis complex-analysis




real-analysis complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 21:41







Alp

















asked Dec 5 '18 at 19:37









AlpAlp

112




112








  • 1




    $begingroup$
    Possible duplicate of Convergence power series in boundary
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 19:57










  • $begingroup$
    Protip: If you are going to suggest someone to check wikipedia or any online resourse, provide the link in your post. square brackets for title and parenthesis for url.
    $endgroup$
    – Mefitico
    Dec 5 '18 at 20:27














  • 1




    $begingroup$
    Possible duplicate of Convergence power series in boundary
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 19:57










  • $begingroup$
    Protip: If you are going to suggest someone to check wikipedia or any online resourse, provide the link in your post. square brackets for title and parenthesis for url.
    $endgroup$
    – Mefitico
    Dec 5 '18 at 20:27








1




1




$begingroup$
Possible duplicate of Convergence power series in boundary
$endgroup$
– Eric Towers
Dec 5 '18 at 19:57




$begingroup$
Possible duplicate of Convergence power series in boundary
$endgroup$
– Eric Towers
Dec 5 '18 at 19:57












$begingroup$
Protip: If you are going to suggest someone to check wikipedia or any online resourse, provide the link in your post. square brackets for title and parenthesis for url.
$endgroup$
– Mefitico
Dec 5 '18 at 20:27




$begingroup$
Protip: If you are going to suggest someone to check wikipedia or any online resourse, provide the link in your post. square brackets for title and parenthesis for url.
$endgroup$
– Mefitico
Dec 5 '18 at 20:27










1 Answer
1






active

oldest

votes


















1












$begingroup$

I believe they are saying that the unit circle is a natural boundary for the function. If $f$ were analytic at any point on the unit circle, it would be analytic in a small disk about that point, but the disk would necessarily contain an $n$th root of unity, contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay... but do limits exist there at all? I know analytic continuation will fail.
    $endgroup$
    – Alp
    Dec 5 '18 at 19:55












  • $begingroup$
    Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
    $endgroup$
    – Daniele Tampieri
    Dec 5 '18 at 19:59








  • 1




    $begingroup$
    That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:00










  • $begingroup$
    @DanieleTampieri How does one see this?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    Some interesting remarks on Sierpinski's paper are given here
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:45











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1 Answer
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1 Answer
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$begingroup$

I believe they are saying that the unit circle is a natural boundary for the function. If $f$ were analytic at any point on the unit circle, it would be analytic in a small disk about that point, but the disk would necessarily contain an $n$th root of unity, contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay... but do limits exist there at all? I know analytic continuation will fail.
    $endgroup$
    – Alp
    Dec 5 '18 at 19:55












  • $begingroup$
    Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
    $endgroup$
    – Daniele Tampieri
    Dec 5 '18 at 19:59








  • 1




    $begingroup$
    That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:00










  • $begingroup$
    @DanieleTampieri How does one see this?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    Some interesting remarks on Sierpinski's paper are given here
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:45
















1












$begingroup$

I believe they are saying that the unit circle is a natural boundary for the function. If $f$ were analytic at any point on the unit circle, it would be analytic in a small disk about that point, but the disk would necessarily contain an $n$th root of unity, contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay... but do limits exist there at all? I know analytic continuation will fail.
    $endgroup$
    – Alp
    Dec 5 '18 at 19:55












  • $begingroup$
    Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
    $endgroup$
    – Daniele Tampieri
    Dec 5 '18 at 19:59








  • 1




    $begingroup$
    That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:00










  • $begingroup$
    @DanieleTampieri How does one see this?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    Some interesting remarks on Sierpinski's paper are given here
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:45














1












1








1





$begingroup$

I believe they are saying that the unit circle is a natural boundary for the function. If $f$ were analytic at any point on the unit circle, it would be analytic in a small disk about that point, but the disk would necessarily contain an $n$th root of unity, contradiction.






share|cite|improve this answer









$endgroup$



I believe they are saying that the unit circle is a natural boundary for the function. If $f$ were analytic at any point on the unit circle, it would be analytic in a small disk about that point, but the disk would necessarily contain an $n$th root of unity, contradiction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 19:54









saulspatzsaulspatz

16.1k31331




16.1k31331












  • $begingroup$
    Okay... but do limits exist there at all? I know analytic continuation will fail.
    $endgroup$
    – Alp
    Dec 5 '18 at 19:55












  • $begingroup$
    Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
    $endgroup$
    – Daniele Tampieri
    Dec 5 '18 at 19:59








  • 1




    $begingroup$
    That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:00










  • $begingroup$
    @DanieleTampieri How does one see this?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    Some interesting remarks on Sierpinski's paper are given here
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:45


















  • $begingroup$
    Okay... but do limits exist there at all? I know analytic continuation will fail.
    $endgroup$
    – Alp
    Dec 5 '18 at 19:55












  • $begingroup$
    Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
    $endgroup$
    – Daniele Tampieri
    Dec 5 '18 at 19:59








  • 1




    $begingroup$
    That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:00










  • $begingroup$
    @DanieleTampieri How does one see this?
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    Some interesting remarks on Sierpinski's paper are given here
    $endgroup$
    – saulspatz
    Dec 5 '18 at 20:45
















$begingroup$
Okay... but do limits exist there at all? I know analytic continuation will fail.
$endgroup$
– Alp
Dec 5 '18 at 19:55






$begingroup$
Okay... but do limits exist there at all? I know analytic continuation will fail.
$endgroup$
– Alp
Dec 5 '18 at 19:55














$begingroup$
Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:59






$begingroup$
Even if the disk is a natural boundary for $f$, in general $f(zeta)$ can exist for a given $zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle.
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:59






1




1




$begingroup$
That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
$endgroup$
– saulspatz
Dec 5 '18 at 20:00




$begingroup$
That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know.
$endgroup$
– saulspatz
Dec 5 '18 at 20:00












$begingroup$
@DanieleTampieri How does one see this?
$endgroup$
– saulspatz
Dec 5 '18 at 20:01




$begingroup$
@DanieleTampieri How does one see this?
$endgroup$
– saulspatz
Dec 5 '18 at 20:01




1




1




$begingroup$
Some interesting remarks on Sierpinski's paper are given here
$endgroup$
– saulspatz
Dec 5 '18 at 20:45




$begingroup$
Some interesting remarks on Sierpinski's paper are given here
$endgroup$
– saulspatz
Dec 5 '18 at 20:45


















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