Sums of entire surjective functions The Next CEO of Stack OverflowUniform 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
Sums of entire surjective functions
The Next CEO of Stack OverflowUniform 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
$begingroup$
Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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 Mar 22 at 21:10
user137377user137377
263
$endgroup$
add a comment |
$begingroup$
Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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 Mar 22 at 21:10
user137377user137377
263
$endgroup$
$begingroup$
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
Mar 22 at 21:14
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
Mar 22 at 21:18
$begingroup$
@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
Mar 22 at 21:21
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
Mar 22 at 21:31
$begingroup$
@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
Mar 22 at 23:17
add a comment |
$begingroup$
Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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 Mar 22 at 21:10
user137377user137377
263
$endgroup$
Suppose $(f_n)_n$ is a countable family of entire, surjective functions, each $f_n:mathbbCtomathbbC$. 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 Mar 22 at 21:10
user137377user137377
263
asked Mar 22 at 21:10
user137377user137377
263
edited Mar 22 at 23:22
user137377
asked Mar 22 at 21:10
user137377user137377
263
asked Mar 22 at 21:10
user137377user137377
263
asked Mar 22 at 21:10
user137377user137377
263
263
$begingroup$
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
Mar 22 at 21:14
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
Mar 22 at 21:18
$begingroup$
@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
Mar 22 at 21:21
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
Mar 22 at 21:31
$begingroup$
@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
Mar 22 at 23:17
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
Mar 22 at 21:14
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
Mar 22 at 21:18
$begingroup$
@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
Mar 22 at 21:21
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
Mar 22 at 21:31
$begingroup$
@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
Mar 22 at 23:17
$begingroup$
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
Mar 22 at 21:14
$begingroup$
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
Mar 22 at 21:14
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
Mar 22 at 21:18
$begingroup$
@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
Mar 22 at 21:18
$begingroup$
@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
Mar 22 at 21:21
$begingroup$
@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
Mar 22 at 21:21
1
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
Mar 22 at 21:31
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
$endgroup$
– Nik Weaver
Mar 22 at 21:31
$begingroup$
@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
Mar 22 at 23:17
$begingroup$
@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
Mar 22 at 23:17
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 Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
$endgroup$
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 $min_=r_k|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
for example, such that $n_k/ktoinfty$,
and consider the class of entire functions of the form
$$f(z)=sum_ninLambdac_nz^n.$$
It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.
Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
Proc. LMS, 1970, 21 525-539.
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
$endgroup$
1
$begingroup$
Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
$endgroup$
– Mateusz Kwaśnicki
Mar 23 at 8:25
1
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
$endgroup$
– Christian Remling
Mar 23 at 17:40
$begingroup$
@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
$endgroup$
– Alexandre Eremenko
Mar 23 at 20:52
$begingroup$
@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
$endgroup$
– Christian Remling
Mar 25 at 14:53
$begingroup$
@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
$endgroup$
– Alexandre Eremenko
Mar 25 at 18:46
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 Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
$endgroup$
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 Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
$endgroup$
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 Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
$endgroup$
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 Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
answered Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
answered Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
answered Mar 22 at 23:41
Noam D. ElkiesNoam D. Elkies
56.5k11199282
56.5k11199282
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 $min_=r_k|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
for example, such that $n_k/ktoinfty$,
and consider the class of entire functions of the form
$$f(z)=sum_ninLambdac_nz^n.$$
It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.
Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
Proc. LMS, 1970, 21 525-539.
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
$endgroup$
1
$begingroup$
Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
$endgroup$
– Mateusz Kwaśnicki
Mar 23 at 8:25
1
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
$endgroup$
– Christian Remling
Mar 23 at 17:40
$begingroup$
@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
$endgroup$
– Alexandre Eremenko
Mar 23 at 20:52
$begingroup$
@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
$endgroup$
– Christian Remling
Mar 25 at 14:53
$begingroup$
@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
$endgroup$
– Alexandre Eremenko
Mar 25 at 18:46
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 $min_=r_k|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
for example, such that $n_k/ktoinfty$,
and consider the class of entire functions of the form
$$f(z)=sum_ninLambdac_nz^n.$$
It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.
Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
Proc. LMS, 1970, 21 525-539.
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
$endgroup$
1
$begingroup$
Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
$endgroup$
– Mateusz Kwaśnicki
Mar 23 at 8:25
1
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
$endgroup$
– Christian Remling
Mar 23 at 17:40
$begingroup$
@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
$endgroup$
– Alexandre Eremenko
Mar 23 at 20:52
$begingroup$
@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
$endgroup$
– Christian Remling
Mar 25 at 14:53
$begingroup$
@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
$endgroup$
– Alexandre Eremenko
Mar 25 at 18:46
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 $min_=r_k|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
for example, such that $n_k/ktoinfty$,
and consider the class of entire functions of the form
$$f(z)=sum_ninLambdac_nz^n.$$
It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.
Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
Proc. LMS, 1970, 21 525-539.
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
$endgroup$
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 $min_=r_k|f(z)|toinfty$ as $kto infty.$
And of course linear combinations of functions of order less than $1/2$ are of order less
than $1/2$.
Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast,
for example, such that $n_k/ktoinfty$,
and consider the class of entire functions of the form
$$f(z)=sum_ninLambdac_nz^n.$$
It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.
Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
Proc. LMS, 1970, 21 525-539.
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
edited Mar 23 at 12:24
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
answered Mar 23 at 1:37
Alexandre EremenkoAlexandre Eremenko
50.9k6140260
50.9k6140260
1
$begingroup$
Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
$endgroup$
– Mateusz Kwaśnicki
Mar 23 at 8:25
1
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
$endgroup$
– Christian Remling
Mar 23 at 17:40
$begingroup$
@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
$endgroup$
– Alexandre Eremenko
Mar 23 at 20:52
$begingroup$
@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
$endgroup$
– Christian Remling
Mar 25 at 14:53
$begingroup$
@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
$endgroup$
– Alexandre Eremenko
Mar 25 at 18:46
add a comment |
1
$begingroup$
Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
$endgroup$
– Mateusz Kwaśnicki
Mar 23 at 8:25
1
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
$endgroup$
– Christian Remling
Mar 23 at 17:40
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@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
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– Alexandre Eremenko
Mar 23 at 20:52
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@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
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– Christian Remling
Mar 25 at 14:53
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@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
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– Alexandre Eremenko
Mar 25 at 18:46
1
1
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Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
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– Mateusz Kwaśnicki
Mar 23 at 8:25
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Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes.
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– Mateusz Kwaśnicki
Mar 23 at 8:25
1
1
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
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– Christian Remling
Mar 23 at 17:40
$begingroup$
Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-cequiv d$.
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– Christian Remling
Mar 23 at 17:40
$begingroup$
@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
$endgroup$
– Alexandre Eremenko
Mar 23 at 20:52
$begingroup$
@Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion.
$endgroup$
– Alexandre Eremenko
Mar 23 at 20:52
$begingroup$
@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
$endgroup$
– Christian Remling
Mar 25 at 14:53
$begingroup$
@AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter...
$endgroup$
– Christian Remling
Mar 25 at 14:53
$begingroup$
@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
$endgroup$
– Alexandre Eremenko
Mar 25 at 18:46
$begingroup$
@Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-)
$endgroup$
– Alexandre Eremenko
Mar 25 at 18:46
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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 $sum a_nf_n=0$...
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– M. Dus
Mar 22 at 21:14
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.
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– Mateusz Kwaśnicki
Mar 22 at 21:18
$begingroup$
@Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective.
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– Alexandre Eremenko
Mar 22 at 21:21
1
$begingroup$
Maybe $(f_n)$ denotes an infinite sequence of functions?
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– Nik Weaver
Mar 22 at 21:31
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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.
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– user137377
Mar 22 at 23:17