Lacunary function doesn't converge anywhere on the boundary
$begingroup$
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
asked Dec 5 '18 at 19:37
AlpAlp
112
$endgroup$
add a comment |
$begingroup$
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
asked Dec 5 '18 at 19:37
AlpAlp
112
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 5 '18 at 19:37
AlpAlp
112
$endgroup$
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
asked Dec 5 '18 at 19:37
AlpAlp
112
asked Dec 5 '18 at 19:37
AlpAlp
112
edited Dec 5 '18 at 21:41
Alp
asked Dec 5 '18 at 19:37
AlpAlp
112
asked Dec 5 '18 at 19:37
AlpAlp
112
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
$endgroup$
$begingroup$
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
|
show 5 more comments
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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.
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
$endgroup$
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.
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
answered Dec 5 '18 at 19:54
saulspatzsaulspatz
16.1k31331
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
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$begingroup$
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