Is there a holomorphic function on open unit disc with this property? Announcing the arrival...
Is there a holomorphic function on open unit disc with this property?
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Can the unit complex 1-dimensional disc be embedded isometrically into complex euclidean space? Boundary behavior of a holomorphic function on $D$ ?Behavior of essential singularities in an 'open cone'Existence of a holomorphic function with specific caracteristicsHolomorphic Function on DiskContinuation to holomorphic functionAn estimation for holomorphic functions in the unit discClosed orbit for vector field $f(bar{z})$ where $f$ is holomorphic functionIs a holomorphic function on a subvariety of $mathbb C^n$ locally a restriction?extension of bounded holomorphic function on the disk
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Let $D={zin mathbb{C}mid |z|<1}$. Is there a holomorphic function $f:Dto mathbb{C}$ such that for every $nin mathbb{N} cup {0},;f^{(n)}$ has a continuous extension to $bar D$ but $f$ has no a holomorphic extension to any open neighborhood of $bar D$?
cv.complex-variables
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
$endgroup$
add a comment |
$begingroup$
Let $D={zin mathbb{C}mid |z|<1}$. Is there a holomorphic function $f:Dto mathbb{C}$ such that for every $nin mathbb{N} cup {0},;f^{(n)}$ has a continuous extension to $bar D$ but $f$ has no a holomorphic extension to any open neighborhood of $bar D$?
cv.complex-variables
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
$endgroup$
4
$begingroup$
Intuitively, you can construct such a function by using the Szegő kernel and any nonanalytic periodic function of period $=2pi$.
$endgroup$
– Daniele Tampieri
17 hours ago
add a comment |
$begingroup$
Let $D={zin mathbb{C}mid |z|<1}$. Is there a holomorphic function $f:Dto mathbb{C}$ such that for every $nin mathbb{N} cup {0},;f^{(n)}$ has a continuous extension to $bar D$ but $f$ has no a holomorphic extension to any open neighborhood of $bar D$?
cv.complex-variables
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
$endgroup$
Let $D={zin mathbb{C}mid |z|<1}$. Is there a holomorphic function $f:Dto mathbb{C}$ such that for every $nin mathbb{N} cup {0},;f^{(n)}$ has a continuous extension to $bar D$ but $f$ has no a holomorphic extension to any open neighborhood of $bar D$?
cv.complex-variables
cv.complex-variables
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
asked 17 hours ago
Ali TaghaviAli Taghavi
20252085
20252085
4
$begingroup$
Intuitively, you can construct such a function by using the Szegő kernel and any nonanalytic periodic function of period $=2pi$.
$endgroup$
– Daniele Tampieri
17 hours ago
add a comment |
4
$begingroup$
Intuitively, you can construct such a function by using the Szegő kernel and any nonanalytic periodic function of period $=2pi$.
$endgroup$
– Daniele Tampieri
17 hours ago
4
4
$begingroup$
Intuitively, you can construct such a function by using the Szegő kernel and any nonanalytic periodic function of period $=2pi$.
$endgroup$
– Daniele Tampieri
17 hours ago
$begingroup$
Intuitively, you can construct such a function by using the Szegő kernel and any nonanalytic periodic function of period $=2pi$.
$endgroup$
– Daniele Tampieri
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, for example
$$f(z)=sum_{n=1}^infty e^{-log^2n}z^{n^2}.$$
The radius of convergence is $1$ by Hadamard's test. The function has gaps (the density of the sequence $n^2$ among the integers is zero) which are sufficient to
apply Fabry's gap theorem (or Polya's gap theorem) to ensure that this $f$ does not have any analytic continuation to a larger region than the unit disk,
and all derivatives are continuous in the closed disk because their power series are absolutely and uniformly convergent
in the closed disk.
Refs. It is difficult to find a good exposition of Fabry's work in English,
one can read his original papers (in French) or the exposition of Bieberbach (in his book Analytische Fortsetzung, in German). Polya work is also in German. But for the Polya gap theorem
one can consult Koosis, The Logarithmic integral, vol. II.
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
$endgroup$
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, for example
$$f(z)=sum_{n=1}^infty e^{-log^2n}z^{n^2}.$$
The radius of convergence is $1$ by Hadamard's test. The function has gaps (the density of the sequence $n^2$ among the integers is zero) which are sufficient to
apply Fabry's gap theorem (or Polya's gap theorem) to ensure that this $f$ does not have any analytic continuation to a larger region than the unit disk,
and all derivatives are continuous in the closed disk because their power series are absolutely and uniformly convergent
in the closed disk.
Refs. It is difficult to find a good exposition of Fabry's work in English,
one can read his original papers (in French) or the exposition of Bieberbach (in his book Analytische Fortsetzung, in German). Polya work is also in German. But for the Polya gap theorem
one can consult Koosis, The Logarithmic integral, vol. II.
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
$endgroup$
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
add a comment |
$begingroup$
Yes, for example
$$f(z)=sum_{n=1}^infty e^{-log^2n}z^{n^2}.$$
The radius of convergence is $1$ by Hadamard's test. The function has gaps (the density of the sequence $n^2$ among the integers is zero) which are sufficient to
apply Fabry's gap theorem (or Polya's gap theorem) to ensure that this $f$ does not have any analytic continuation to a larger region than the unit disk,
and all derivatives are continuous in the closed disk because their power series are absolutely and uniformly convergent
in the closed disk.
Refs. It is difficult to find a good exposition of Fabry's work in English,
one can read his original papers (in French) or the exposition of Bieberbach (in his book Analytische Fortsetzung, in German). Polya work is also in German. But for the Polya gap theorem
one can consult Koosis, The Logarithmic integral, vol. II.
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
$endgroup$
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
add a comment |
$begingroup$
Yes, for example
$$f(z)=sum_{n=1}^infty e^{-log^2n}z^{n^2}.$$
The radius of convergence is $1$ by Hadamard's test. The function has gaps (the density of the sequence $n^2$ among the integers is zero) which are sufficient to
apply Fabry's gap theorem (or Polya's gap theorem) to ensure that this $f$ does not have any analytic continuation to a larger region than the unit disk,
and all derivatives are continuous in the closed disk because their power series are absolutely and uniformly convergent
in the closed disk.
Refs. It is difficult to find a good exposition of Fabry's work in English,
one can read his original papers (in French) or the exposition of Bieberbach (in his book Analytische Fortsetzung, in German). Polya work is also in German. But for the Polya gap theorem
one can consult Koosis, The Logarithmic integral, vol. II.
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
$endgroup$
Yes, for example
$$f(z)=sum_{n=1}^infty e^{-log^2n}z^{n^2}.$$
The radius of convergence is $1$ by Hadamard's test. The function has gaps (the density of the sequence $n^2$ among the integers is zero) which are sufficient to
apply Fabry's gap theorem (or Polya's gap theorem) to ensure that this $f$ does not have any analytic continuation to a larger region than the unit disk,
and all derivatives are continuous in the closed disk because their power series are absolutely and uniformly convergent
in the closed disk.
Refs. It is difficult to find a good exposition of Fabry's work in English,
one can read his original papers (in French) or the exposition of Bieberbach (in his book Analytische Fortsetzung, in German). Polya work is also in German. But for the Polya gap theorem
one can consult Koosis, The Logarithmic integral, vol. II.
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
edited 11 hours ago
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
answered 15 hours ago
Alexandre EremenkoAlexandre Eremenko
51.5k6142262
51.5k6142262
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
add a comment |
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
$begingroup$
If you use $f(z)=sum_{n=1}^infty e^{-nlog n}z^{2^n}$ it should still have the properties required since $sum_{n=1}^infty {e^{-nlog n}2^{kn}} < infty$ for all $k ge 0$ and $({e^{-nlog n}})^{frac{1}{2^n}} to 1$ and now the easier Hadamard gap theorem which appears in many complex analysis (and related subjects) texts applies
$endgroup$
– Conrad
7 hours ago
add a comment |
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4
$begingroup$
Intuitively, you can construct such a function by using the Szegő kernel and any nonanalytic periodic function of period $=2pi$.
$endgroup$
– Daniele Tampieri
17 hours ago