Sums of entire surjective functionsUniform Convergence of piecewise Continuous Uniformly Convergent...
Sums of entire surjective functions
Uniform Convergence of piecewise Continuous Uniformly Convergent Functionsentire functions of one complex variable with prescribed value and order.The boundedness of an entire function along the imaginary axisEntire composite square roots of functions of finite orderCan an entire function have every root function?Surjective entire functionsInterchanging sums and integrals in a specific instanceExponential type of a product of entire functionsUnconditionally convergent series in some functional spacesCoefficients of entire functions with specified zero set
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Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.
ca.classical-analysis-and-odes cv.complex-variables
asked 7 hours ago
user137377user137377
192
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add a comment |
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Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.
ca.classical-analysis-and-odes cv.complex-variables
asked 7 hours ago
user137377user137377
192
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
7 hours ago
1
$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
7 hours ago
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@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
7 hours ago
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 hours ago
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@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
5 hours ago
add a comment |
$begingroup$
Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.
ca.classical-analysis-and-odes cv.complex-variables
asked 7 hours ago
user137377user137377
192
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbb{C}tomathbb{C}$. Can one always find complex scalars $(a_n)_n$, not all zero, such that $sum_{n=1}^{infty} a_n f_n$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $(f_n)_n$ are not polynomials.
ca.classical-analysis-and-odes cv.complex-variables
ca.classical-analysis-and-odes cv.complex-variables
asked 7 hours ago
user137377user137377
192
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
user137377user137377
192
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
user137377
asked 7 hours ago
user137377user137377
192
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
user137377user137377
192
asked 7 hours ago
user137377user137377
192
192
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user137377 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
7 hours ago
1
$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
7 hours ago
$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
7 hours ago
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 hours ago
$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
5 hours ago
add a comment |
$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
7 hours ago
1
$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
7 hours ago
$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
7 hours ago
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 hours ago
$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
5 hours ago
$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
7 hours ago
$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
7 hours ago
1
1
$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
7 hours ago
$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
7 hours ago
$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
7 hours ago
$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
7 hours ago
1
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 hours ago
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 hours ago
$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
5 hours ago
$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z mapsto c + exp g(z)$). An explicit example is
$f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
of degree at most $1$; but $f$ is even, so must be constant,
from which it soon follows that $a_n=0$ for every $n$.
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
$endgroup$
add a comment |
$begingroup$
The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)
For example, all non-constant functions of order less than $1/2$ are surjective.
This follows from an old theorem of Wiman that for such function $f$ there exists
a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z mapsto c + exp g(z)$). An explicit example is
$f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
of degree at most $1$; but $f$ is even, so must be constant,
from which it soon follows that $a_n=0$ for every $n$.
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
$endgroup$
add a comment |
$begingroup$
One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z mapsto c + exp g(z)$). An explicit example is
$f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
of degree at most $1$; but $f$ is even, so must be constant,
from which it soon follows that $a_n=0$ for every $n$.
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
$endgroup$
add a comment |
$begingroup$
One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z mapsto c + exp g(z)$). An explicit example is
$f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
of degree at most $1$; but $f$ is even, so must be constant,
from which it soon follows that $a_n=0$ for every $n$.
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
$endgroup$
One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z mapsto c + exp g(z)$). An explicit example is
$f_n(z) = cos z/n$: any convergent linear combination $f = sum_n a_n f_n$
is of order $1$, so if $f$ is not surjective then $g$ is a polynomial
of degree at most $1$; but $f$ is even, so must be constant,
from which it soon follows that $a_n=0$ for every $n$.
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
answered 4 hours ago
Noam D. ElkiesNoam D. Elkies
56.4k11199282
56.4k11199282
add a comment |
add a comment |
$begingroup$
The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)
For example, all non-constant functions of order less than $1/2$ are surjective.
This follows from an old theorem of Wiman that for such function $f$ there exists
a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
add a comment |
$begingroup$
The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)
For example, all non-constant functions of order less than $1/2$ are surjective.
This follows from an old theorem of Wiman that for such function $f$ there exists
a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
add a comment |
$begingroup$
The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)
For example, all non-constant functions of order less than $1/2$ are surjective.
This follows from an old theorem of Wiman that for such function $f$ there exists
a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)
For example, all non-constant functions of order less than $1/2$ are surjective.
This follows from an old theorem of Wiman that for such function $f$ there exists
a sequence $r_ktoinfty$ such that $min_{|z|=r_k}|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
answered 2 hours ago
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
50.7k6140258
add a comment |
add a comment |
user137377 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $ngeq 3$, will yield $sum a_nf_n=0$...
$endgroup$
– M. Dus
7 hours ago
1
$begingroup$
@M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant.
$endgroup$
– Mateusz Kwaśnicki
7 hours ago
$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
$endgroup$
– Alexandre Eremenko
7 hours ago
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
7 hours ago
$begingroup$
@AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family.
$endgroup$
– user137377
5 hours ago