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Every power series is the Taylor series of some $C^infty$ function
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Do you have some reference to a proof of the so-called Borel theorem, i.e. every power series is the Taylor series of some $C^infty$ function?
real-analysis reference-request
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up vote
42
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Do you have some reference to a proof of the so-called Borel theorem, i.e. every power series is the Taylor series of some $C^infty$ function?
real-analysis reference-request
4
see exercise 1 on pages 16, 18-19 of C. Zuily 's Problems in Distribution and Partial Differential Equations, or Theorem 4.32 on page 191 of Karl Stromberg's An Introduction to Classical Real Analysis, or the Theorem on pages 50-51 William Donoghue's Distributions and Fourier Transform.
– Soarer
Sep 9 '11 at 7:48
6
@Peter, this is not quite related to generating functions I think.
– Soarer
Sep 9 '11 at 7:49
1
@Peter, Usually when I think about power series, I implicitly assume it's an actual series where series convergence etc is important, rather than formal power series. (Personally I reserve the term generating function for a meaningful sequence of numbers) In this case, convergence shouldn't be part of the condition so yes, you can call it a generating function probably.
– Soarer
Sep 9 '11 at 8:09
2
However, the remark "it is like asking for a proof that every sequence has a generating function" is wrong, because when you write down the power series like you said, it's very likely to have 0 radius of convergence.
– Soarer
Sep 9 '11 at 8:10
add a comment |Â
up vote
42
down vote
favorite
up vote
42
down vote
favorite
Do you have some reference to a proof of the so-called Borel theorem, i.e. every power series is the Taylor series of some $C^infty$ function?
real-analysis reference-request
Do you have some reference to a proof of the so-called Borel theorem, i.e. every power series is the Taylor series of some $C^infty$ function?
real-analysis reference-request
asked Sep 9 '11 at 7:38
user365
4
see exercise 1 on pages 16, 18-19 of C. Zuily 's Problems in Distribution and Partial Differential Equations, or Theorem 4.32 on page 191 of Karl Stromberg's An Introduction to Classical Real Analysis, or the Theorem on pages 50-51 William Donoghue's Distributions and Fourier Transform.
– Soarer
Sep 9 '11 at 7:48
6
@Peter, this is not quite related to generating functions I think.
– Soarer
Sep 9 '11 at 7:49
1
@Peter, Usually when I think about power series, I implicitly assume it's an actual series where series convergence etc is important, rather than formal power series. (Personally I reserve the term generating function for a meaningful sequence of numbers) In this case, convergence shouldn't be part of the condition so yes, you can call it a generating function probably.
– Soarer
Sep 9 '11 at 8:09
2
However, the remark "it is like asking for a proof that every sequence has a generating function" is wrong, because when you write down the power series like you said, it's very likely to have 0 radius of convergence.
– Soarer
Sep 9 '11 at 8:10
add a comment |Â
4
see exercise 1 on pages 16, 18-19 of C. Zuily 's Problems in Distribution and Partial Differential Equations, or Theorem 4.32 on page 191 of Karl Stromberg's An Introduction to Classical Real Analysis, or the Theorem on pages 50-51 William Donoghue's Distributions and Fourier Transform.
– Soarer
Sep 9 '11 at 7:48
6
@Peter, this is not quite related to generating functions I think.
– Soarer
Sep 9 '11 at 7:49
1
@Peter, Usually when I think about power series, I implicitly assume it's an actual series where series convergence etc is important, rather than formal power series. (Personally I reserve the term generating function for a meaningful sequence of numbers) In this case, convergence shouldn't be part of the condition so yes, you can call it a generating function probably.
– Soarer
Sep 9 '11 at 8:09
2
However, the remark "it is like asking for a proof that every sequence has a generating function" is wrong, because when you write down the power series like you said, it's very likely to have 0 radius of convergence.
– Soarer
Sep 9 '11 at 8:10
4
4
see exercise 1 on pages 16, 18-19 of C. Zuily 's Problems in Distribution and Partial Differential Equations, or Theorem 4.32 on page 191 of Karl Stromberg's An Introduction to Classical Real Analysis, or the Theorem on pages 50-51 William Donoghue's Distributions and Fourier Transform.
– Soarer
Sep 9 '11 at 7:48
see exercise 1 on pages 16, 18-19 of C. Zuily 's Problems in Distribution and Partial Differential Equations, or Theorem 4.32 on page 191 of Karl Stromberg's An Introduction to Classical Real Analysis, or the Theorem on pages 50-51 William Donoghue's Distributions and Fourier Transform.
– Soarer
Sep 9 '11 at 7:48
6
6
@Peter, this is not quite related to generating functions I think.
– Soarer
Sep 9 '11 at 7:49
@Peter, this is not quite related to generating functions I think.
– Soarer
Sep 9 '11 at 7:49
1
1
@Peter, Usually when I think about power series, I implicitly assume it's an actual series where series convergence etc is important, rather than formal power series. (Personally I reserve the term generating function for a meaningful sequence of numbers) In this case, convergence shouldn't be part of the condition so yes, you can call it a generating function probably.
– Soarer
Sep 9 '11 at 8:09
@Peter, Usually when I think about power series, I implicitly assume it's an actual series where series convergence etc is important, rather than formal power series. (Personally I reserve the term generating function for a meaningful sequence of numbers) In this case, convergence shouldn't be part of the condition so yes, you can call it a generating function probably.
– Soarer
Sep 9 '11 at 8:09
2
2
However, the remark "it is like asking for a proof that every sequence has a generating function" is wrong, because when you write down the power series like you said, it's very likely to have 0 radius of convergence.
– Soarer
Sep 9 '11 at 8:10
However, the remark "it is like asking for a proof that every sequence has a generating function" is wrong, because when you write down the power series like you said, it's very likely to have 0 radius of convergence.
– Soarer
Sep 9 '11 at 8:10
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
45
down vote
accepted
Borel's theorem states that given a sequence of real numbers $(a_n)_nin mathbb N$ there exists a $C^infty$ function $fin C^infty(mathbb R)$ such that
$frac f^(n)(0)n!=a_n $ , i.e. the Taylor series associated to $f$ is $Sigma a_nX^n$.
The function $f$ is never unique: you can always add to it a flat function, one all of whose derivatives at zero are zero, like the well-known Cauchy function $e^-1/x^2$ .
There is a huge caveat however: you can't go from the series to the function $f$ .
Firstly, the series might not be convergent at any $xneq 0in mathbb R $ ! An example is $Sigma a_n X^n=Sigma n^n X^n$ whose radius of convergence is zero.
Secondly, even if it does converge it might converge to the wrong function! For example if you start with Cauchy's function you get the zero Taylor series. It converges to zero, of course, but that is definitely not the Cauchy function you started with. So we should not read too much in Borel's theorem: it cannot force a non-analytic function to become analytic!
Borel's theorem is also valid in several variables. Given a sequence of $k$-tuples $(a_I)_Iin mathbb N^k$ of real numbers $a_I inmathbb R$, there exists a function $fin C^infty(mathbb R^k)$, again highly non-unique, whose derivatives satisfy
$frac partial^I f(0)I!=a_I $. [I have used multiindex notation with $I=(i_1,ldots,i_k)$, $I!=i_1!ldots i_k! ;etc.$]
There is a vast generalization due to Whitney of Borel's theorem. You can consider a closed subset $Zsubset mathbb R^k$ and continuous functions $phi_Iin C(Z) ; $ . Whitney gives necessary and sufficient growth and compatibility conditions on the $phi_I $ 's which will guarantee that there exists a $C^infty$ function $fin C^infty (U)$ defined on an open neighbourhood $U supset Z$ of $Z$ such that $frac partial^I f(0)I!=phi_I ; $. Borel' s theorem is then the case $Z=0$ .
Bibliography: Borel's theorem in several variables is proved in R.Narasimhan's book Analysis on Real and complex Manifolds, which also contains the precise statement of Whitney's theorem.
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
2
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
1
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
add a comment |Â
up vote
17
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Through this question, I was made aware of
ÃÂdám Besenyei. Peano's unnoticed proof of Borel's theorem, Amer. Math. Monthly 121 (2014), no. 1, 69–72.
In this short note, Besenyei presents a proof due to Peano of the theorem usually attributed to Borel. Peano's result first appeared in
Angelo Genocchi , Giuseppe Peano. Calculo differenziale e principii di calcolo integrale, Fratelli Bocca, Roma, 1884.
Note that Borel's result first appeared in his dissertation, published as
Émile Borel. Sur quelques points de la théorie des fonctions, Ann. Sci. l’École Norm. Sup. (3) 12 (1895), 9–55. MR1508908.
Peano's proof is short, and completely different from Borel's. Besenyei provides full details. I present a sketch:
Given a sequence $(c_n)_nge0$ of real numbers, we want a $C^infty$ function $f$ such that $f^(n)(0)=c_n$ for all $n$. Peano considers
$$ f(x)=sum_kge0fraca_k x^k1+b_kx^2, $$
for $(a_n)_nge0$ arbitrary, and $(b_n)_nge0$ a sequence of positive numbers, chosen so that $f$ is indeed $C^infty$ and can be differentiated term by term. Assuming that this is possible, one easily checks that $f^(n)(0)=a_n$ for $n=0,1$, and that if $nge2$, then
$$ fracf^(n)(0)n!=a_n+sum_j=1^lfloor n/2rfloor(-1)^ja_n-2jb_n-2j^j. $$
To see the latter, consider the power series expansion of $displaystyle fraca_k x^k1+b_kx^2$, valid for $|b_kx^2|<1$, and note that it implies that its $n$-th derivative at $0$ is either $0$ (if $n-k$ is odd), or
$$ n!(-1)^ja_n-2jb_n-2j^j, $$
if $n-k=2j$ for some $j$.
The point is that this recurrence allows us to define the $a_n$ (uniquely) in terms of the $b_n$ and the $c_n$, so that $f^(n)(0)=c_n$ for all $n$.
In order for the above to hold, one needs to ensure that $f$ so defined can indeed be differentiated term by term. For this, Besenyei checks that if $kge n+2$, then $(*)$
$$left|left(fraca_kx^k1+b_kx^2right)^(n)right|le(n+1)!fracb_k|x|^k-n-2$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then
$$ sum_kge n+2left|left(fraca_kx^k1+b_kx^2right)^(n)right| $$ is uniformly convergent on any finite interval, and the Weierstrass M-test allows us to differentiate termwise.
Finally, Besenyei proves $(*)$ in a straightforward fashion with estimates coming from Leibniz rule, after rewriting
$$fraca_k x^k1+b_kx^2=fraca_kb_kcdotfracx^k-12left(frac1x+frac1sqrtb_ki+frac1x-frac1sqrtb_kiright). $$
edited Apr 13 '17 at 12:20
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answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
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The statement of Borel's theorem that I know is that the map $C^infty(mathbbR,mathbbR) to mathbbR^mathbbN$ which sends a function $f$ to its sequence of derivatives at a point (say $0$), is surjective.
One reference, with proof, is to Kriegl and Michor's book A Convenient Setting for Global Analysis, section 15.4 (print version). (It used to be free online from the AMS bookstore, but I can no longer find a link to that version. Maybe I'm just not searching correctly.) There, it is stated as:
Borel's theorem. Suppose a Banach space $E$ has $C^infty_b$-bump functions. Then every formal power series with coefficients in $L^n_sym(E;F)$ for another Banach space $F$ is the Taylor-series of a smooth mapping $E to F$.
It is attributed there to Wells, Differentiable functions on Banach spaces with Lipschitz derivative in JDG 8, 1973, 135-152. Presumably, due to the name, Wells proved the version for Banach spaces. I would also presume that Wells had a reference to Borel's original version. There are also some following remarks on where it fails (mainly due to Colombeau).
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
1
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
45
down vote
accepted
Borel's theorem states that given a sequence of real numbers $(a_n)_nin mathbb N$ there exists a $C^infty$ function $fin C^infty(mathbb R)$ such that
$frac f^(n)(0)n!=a_n $ , i.e. the Taylor series associated to $f$ is $Sigma a_nX^n$.
The function $f$ is never unique: you can always add to it a flat function, one all of whose derivatives at zero are zero, like the well-known Cauchy function $e^-1/x^2$ .
There is a huge caveat however: you can't go from the series to the function $f$ .
Firstly, the series might not be convergent at any $xneq 0in mathbb R $ ! An example is $Sigma a_n X^n=Sigma n^n X^n$ whose radius of convergence is zero.
Secondly, even if it does converge it might converge to the wrong function! For example if you start with Cauchy's function you get the zero Taylor series. It converges to zero, of course, but that is definitely not the Cauchy function you started with. So we should not read too much in Borel's theorem: it cannot force a non-analytic function to become analytic!
Borel's theorem is also valid in several variables. Given a sequence of $k$-tuples $(a_I)_Iin mathbb N^k$ of real numbers $a_I inmathbb R$, there exists a function $fin C^infty(mathbb R^k)$, again highly non-unique, whose derivatives satisfy
$frac partial^I f(0)I!=a_I $. [I have used multiindex notation with $I=(i_1,ldots,i_k)$, $I!=i_1!ldots i_k! ;etc.$]
There is a vast generalization due to Whitney of Borel's theorem. You can consider a closed subset $Zsubset mathbb R^k$ and continuous functions $phi_Iin C(Z) ; $ . Whitney gives necessary and sufficient growth and compatibility conditions on the $phi_I $ 's which will guarantee that there exists a $C^infty$ function $fin C^infty (U)$ defined on an open neighbourhood $U supset Z$ of $Z$ such that $frac partial^I f(0)I!=phi_I ; $. Borel' s theorem is then the case $Z=0$ .
Bibliography: Borel's theorem in several variables is proved in R.Narasimhan's book Analysis on Real and complex Manifolds, which also contains the precise statement of Whitney's theorem.
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
2
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
1
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
add a comment |Â
up vote
45
down vote
accepted
Borel's theorem states that given a sequence of real numbers $(a_n)_nin mathbb N$ there exists a $C^infty$ function $fin C^infty(mathbb R)$ such that
$frac f^(n)(0)n!=a_n $ , i.e. the Taylor series associated to $f$ is $Sigma a_nX^n$.
The function $f$ is never unique: you can always add to it a flat function, one all of whose derivatives at zero are zero, like the well-known Cauchy function $e^-1/x^2$ .
There is a huge caveat however: you can't go from the series to the function $f$ .
Firstly, the series might not be convergent at any $xneq 0in mathbb R $ ! An example is $Sigma a_n X^n=Sigma n^n X^n$ whose radius of convergence is zero.
Secondly, even if it does converge it might converge to the wrong function! For example if you start with Cauchy's function you get the zero Taylor series. It converges to zero, of course, but that is definitely not the Cauchy function you started with. So we should not read too much in Borel's theorem: it cannot force a non-analytic function to become analytic!
Borel's theorem is also valid in several variables. Given a sequence of $k$-tuples $(a_I)_Iin mathbb N^k$ of real numbers $a_I inmathbb R$, there exists a function $fin C^infty(mathbb R^k)$, again highly non-unique, whose derivatives satisfy
$frac partial^I f(0)I!=a_I $. [I have used multiindex notation with $I=(i_1,ldots,i_k)$, $I!=i_1!ldots i_k! ;etc.$]
There is a vast generalization due to Whitney of Borel's theorem. You can consider a closed subset $Zsubset mathbb R^k$ and continuous functions $phi_Iin C(Z) ; $ . Whitney gives necessary and sufficient growth and compatibility conditions on the $phi_I $ 's which will guarantee that there exists a $C^infty$ function $fin C^infty (U)$ defined on an open neighbourhood $U supset Z$ of $Z$ such that $frac partial^I f(0)I!=phi_I ; $. Borel' s theorem is then the case $Z=0$ .
Bibliography: Borel's theorem in several variables is proved in R.Narasimhan's book Analysis on Real and complex Manifolds, which also contains the precise statement of Whitney's theorem.
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
2
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
1
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
add a comment |Â
up vote
45
down vote
accepted
up vote
45
down vote
accepted
Borel's theorem states that given a sequence of real numbers $(a_n)_nin mathbb N$ there exists a $C^infty$ function $fin C^infty(mathbb R)$ such that
$frac f^(n)(0)n!=a_n $ , i.e. the Taylor series associated to $f$ is $Sigma a_nX^n$.
The function $f$ is never unique: you can always add to it a flat function, one all of whose derivatives at zero are zero, like the well-known Cauchy function $e^-1/x^2$ .
There is a huge caveat however: you can't go from the series to the function $f$ .
Firstly, the series might not be convergent at any $xneq 0in mathbb R $ ! An example is $Sigma a_n X^n=Sigma n^n X^n$ whose radius of convergence is zero.
Secondly, even if it does converge it might converge to the wrong function! For example if you start with Cauchy's function you get the zero Taylor series. It converges to zero, of course, but that is definitely not the Cauchy function you started with. So we should not read too much in Borel's theorem: it cannot force a non-analytic function to become analytic!
Borel's theorem is also valid in several variables. Given a sequence of $k$-tuples $(a_I)_Iin mathbb N^k$ of real numbers $a_I inmathbb R$, there exists a function $fin C^infty(mathbb R^k)$, again highly non-unique, whose derivatives satisfy
$frac partial^I f(0)I!=a_I $. [I have used multiindex notation with $I=(i_1,ldots,i_k)$, $I!=i_1!ldots i_k! ;etc.$]
There is a vast generalization due to Whitney of Borel's theorem. You can consider a closed subset $Zsubset mathbb R^k$ and continuous functions $phi_Iin C(Z) ; $ . Whitney gives necessary and sufficient growth and compatibility conditions on the $phi_I $ 's which will guarantee that there exists a $C^infty$ function $fin C^infty (U)$ defined on an open neighbourhood $U supset Z$ of $Z$ such that $frac partial^I f(0)I!=phi_I ; $. Borel' s theorem is then the case $Z=0$ .
Bibliography: Borel's theorem in several variables is proved in R.Narasimhan's book Analysis on Real and complex Manifolds, which also contains the precise statement of Whitney's theorem.
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
Borel's theorem states that given a sequence of real numbers $(a_n)_nin mathbb N$ there exists a $C^infty$ function $fin C^infty(mathbb R)$ such that
$frac f^(n)(0)n!=a_n $ , i.e. the Taylor series associated to $f$ is $Sigma a_nX^n$.
The function $f$ is never unique: you can always add to it a flat function, one all of whose derivatives at zero are zero, like the well-known Cauchy function $e^-1/x^2$ .
There is a huge caveat however: you can't go from the series to the function $f$ .
Firstly, the series might not be convergent at any $xneq 0in mathbb R $ ! An example is $Sigma a_n X^n=Sigma n^n X^n$ whose radius of convergence is zero.
Secondly, even if it does converge it might converge to the wrong function! For example if you start with Cauchy's function you get the zero Taylor series. It converges to zero, of course, but that is definitely not the Cauchy function you started with. So we should not read too much in Borel's theorem: it cannot force a non-analytic function to become analytic!
Borel's theorem is also valid in several variables. Given a sequence of $k$-tuples $(a_I)_Iin mathbb N^k$ of real numbers $a_I inmathbb R$, there exists a function $fin C^infty(mathbb R^k)$, again highly non-unique, whose derivatives satisfy
$frac partial^I f(0)I!=a_I $. [I have used multiindex notation with $I=(i_1,ldots,i_k)$, $I!=i_1!ldots i_k! ;etc.$]
There is a vast generalization due to Whitney of Borel's theorem. You can consider a closed subset $Zsubset mathbb R^k$ and continuous functions $phi_Iin C(Z) ; $ . Whitney gives necessary and sufficient growth and compatibility conditions on the $phi_I $ 's which will guarantee that there exists a $C^infty$ function $fin C^infty (U)$ defined on an open neighbourhood $U supset Z$ of $Z$ such that $frac partial^I f(0)I!=phi_I ; $. Borel' s theorem is then the case $Z=0$ .
Bibliography: Borel's theorem in several variables is proved in R.Narasimhan's book Analysis on Real and complex Manifolds, which also contains the precise statement of Whitney's theorem.
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
edited Sep 9 '11 at 10:50
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
answered Sep 9 '11 at 9:06
Georges Elencwajg
116k7174315
116k7174315
2
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
1
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
add a comment |Â
2
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
1
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
2
2
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $mathbbR^mathbbN to C^0(mathbbR,mathbbR)$ but any continuous linear map $mathbbR^mathbbN to V$ to a normable space has to factor through one of the projections $mathbbR^mathbbN to mathbbR^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.)
– Loop Space
Sep 9 '11 at 12:22
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
Thank you for your comment, Andrew.
– Georges Elencwajg
Sep 9 '11 at 12:34
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
When you say "given a sequence of real numbers $(a_n)_ninmathbbN$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $ninmathbbN$. What if $a_n$ were "random"? What if I chose $a_n$ at random?
– Antinous
Jan 29 '14 at 9:57
1
1
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_nin mathbb N$ of real numbers in Borel's theorem, as "random" as you like or on the contrary given by any "rule" you care to make up.
– Georges Elencwajg
Jan 29 '14 at 10:33
add a comment |Â
up vote
17
down vote
Through this question, I was made aware of
ÃÂdám Besenyei. Peano's unnoticed proof of Borel's theorem, Amer. Math. Monthly 121 (2014), no. 1, 69–72.
In this short note, Besenyei presents a proof due to Peano of the theorem usually attributed to Borel. Peano's result first appeared in
Angelo Genocchi , Giuseppe Peano. Calculo differenziale e principii di calcolo integrale, Fratelli Bocca, Roma, 1884.
Note that Borel's result first appeared in his dissertation, published as
Émile Borel. Sur quelques points de la théorie des fonctions, Ann. Sci. l’École Norm. Sup. (3) 12 (1895), 9–55. MR1508908.
Peano's proof is short, and completely different from Borel's. Besenyei provides full details. I present a sketch:
Given a sequence $(c_n)_nge0$ of real numbers, we want a $C^infty$ function $f$ such that $f^(n)(0)=c_n$ for all $n$. Peano considers
$$ f(x)=sum_kge0fraca_k x^k1+b_kx^2, $$
for $(a_n)_nge0$ arbitrary, and $(b_n)_nge0$ a sequence of positive numbers, chosen so that $f$ is indeed $C^infty$ and can be differentiated term by term. Assuming that this is possible, one easily checks that $f^(n)(0)=a_n$ for $n=0,1$, and that if $nge2$, then
$$ fracf^(n)(0)n!=a_n+sum_j=1^lfloor n/2rfloor(-1)^ja_n-2jb_n-2j^j. $$
To see the latter, consider the power series expansion of $displaystyle fraca_k x^k1+b_kx^2$, valid for $|b_kx^2|<1$, and note that it implies that its $n$-th derivative at $0$ is either $0$ (if $n-k$ is odd), or
$$ n!(-1)^ja_n-2jb_n-2j^j, $$
if $n-k=2j$ for some $j$.
The point is that this recurrence allows us to define the $a_n$ (uniquely) in terms of the $b_n$ and the $c_n$, so that $f^(n)(0)=c_n$ for all $n$.
In order for the above to hold, one needs to ensure that $f$ so defined can indeed be differentiated term by term. For this, Besenyei checks that if $kge n+2$, then $(*)$
$$left|left(fraca_kx^k1+b_kx^2right)^(n)right|le(n+1)!fracb_k|x|^k-n-2$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then
$$ sum_kge n+2left|left(fraca_kx^k1+b_kx^2right)^(n)right| $$ is uniformly convergent on any finite interval, and the Weierstrass M-test allows us to differentiate termwise.
Finally, Besenyei proves $(*)$ in a straightforward fashion with estimates coming from Leibniz rule, after rewriting
$$fraca_k x^k1+b_kx^2=fraca_kb_kcdotfracx^k-12left(frac1x+frac1sqrtb_ki+frac1x-frac1sqrtb_kiright). $$
edited Apr 13 '17 at 12:20
Community♦
1
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
add a comment |Â
up vote
17
down vote
Through this question, I was made aware of
ÃÂdám Besenyei. Peano's unnoticed proof of Borel's theorem, Amer. Math. Monthly 121 (2014), no. 1, 69–72.
In this short note, Besenyei presents a proof due to Peano of the theorem usually attributed to Borel. Peano's result first appeared in
Angelo Genocchi , Giuseppe Peano. Calculo differenziale e principii di calcolo integrale, Fratelli Bocca, Roma, 1884.
Note that Borel's result first appeared in his dissertation, published as
Émile Borel. Sur quelques points de la théorie des fonctions, Ann. Sci. l’École Norm. Sup. (3) 12 (1895), 9–55. MR1508908.
Peano's proof is short, and completely different from Borel's. Besenyei provides full details. I present a sketch:
Given a sequence $(c_n)_nge0$ of real numbers, we want a $C^infty$ function $f$ such that $f^(n)(0)=c_n$ for all $n$. Peano considers
$$ f(x)=sum_kge0fraca_k x^k1+b_kx^2, $$
for $(a_n)_nge0$ arbitrary, and $(b_n)_nge0$ a sequence of positive numbers, chosen so that $f$ is indeed $C^infty$ and can be differentiated term by term. Assuming that this is possible, one easily checks that $f^(n)(0)=a_n$ for $n=0,1$, and that if $nge2$, then
$$ fracf^(n)(0)n!=a_n+sum_j=1^lfloor n/2rfloor(-1)^ja_n-2jb_n-2j^j. $$
To see the latter, consider the power series expansion of $displaystyle fraca_k x^k1+b_kx^2$, valid for $|b_kx^2|<1$, and note that it implies that its $n$-th derivative at $0$ is either $0$ (if $n-k$ is odd), or
$$ n!(-1)^ja_n-2jb_n-2j^j, $$
if $n-k=2j$ for some $j$.
The point is that this recurrence allows us to define the $a_n$ (uniquely) in terms of the $b_n$ and the $c_n$, so that $f^(n)(0)=c_n$ for all $n$.
In order for the above to hold, one needs to ensure that $f$ so defined can indeed be differentiated term by term. For this, Besenyei checks that if $kge n+2$, then $(*)$
$$left|left(fraca_kx^k1+b_kx^2right)^(n)right|le(n+1)!fracb_k|x|^k-n-2$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then
$$ sum_kge n+2left|left(fraca_kx^k1+b_kx^2right)^(n)right| $$ is uniformly convergent on any finite interval, and the Weierstrass M-test allows us to differentiate termwise.
Finally, Besenyei proves $(*)$ in a straightforward fashion with estimates coming from Leibniz rule, after rewriting
$$fraca_k x^k1+b_kx^2=fraca_kb_kcdotfracx^k-12left(frac1x+frac1sqrtb_ki+frac1x-frac1sqrtb_kiright). $$
edited Apr 13 '17 at 12:20
Community♦
1
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
add a comment |Â
up vote
17
down vote
up vote
17
down vote
Through this question, I was made aware of
ÃÂdám Besenyei. Peano's unnoticed proof of Borel's theorem, Amer. Math. Monthly 121 (2014), no. 1, 69–72.
In this short note, Besenyei presents a proof due to Peano of the theorem usually attributed to Borel. Peano's result first appeared in
Angelo Genocchi , Giuseppe Peano. Calculo differenziale e principii di calcolo integrale, Fratelli Bocca, Roma, 1884.
Note that Borel's result first appeared in his dissertation, published as
Émile Borel. Sur quelques points de la théorie des fonctions, Ann. Sci. l’École Norm. Sup. (3) 12 (1895), 9–55. MR1508908.
Peano's proof is short, and completely different from Borel's. Besenyei provides full details. I present a sketch:
Given a sequence $(c_n)_nge0$ of real numbers, we want a $C^infty$ function $f$ such that $f^(n)(0)=c_n$ for all $n$. Peano considers
$$ f(x)=sum_kge0fraca_k x^k1+b_kx^2, $$
for $(a_n)_nge0$ arbitrary, and $(b_n)_nge0$ a sequence of positive numbers, chosen so that $f$ is indeed $C^infty$ and can be differentiated term by term. Assuming that this is possible, one easily checks that $f^(n)(0)=a_n$ for $n=0,1$, and that if $nge2$, then
$$ fracf^(n)(0)n!=a_n+sum_j=1^lfloor n/2rfloor(-1)^ja_n-2jb_n-2j^j. $$
To see the latter, consider the power series expansion of $displaystyle fraca_k x^k1+b_kx^2$, valid for $|b_kx^2|<1$, and note that it implies that its $n$-th derivative at $0$ is either $0$ (if $n-k$ is odd), or
$$ n!(-1)^ja_n-2jb_n-2j^j, $$
if $n-k=2j$ for some $j$.
The point is that this recurrence allows us to define the $a_n$ (uniquely) in terms of the $b_n$ and the $c_n$, so that $f^(n)(0)=c_n$ for all $n$.
In order for the above to hold, one needs to ensure that $f$ so defined can indeed be differentiated term by term. For this, Besenyei checks that if $kge n+2$, then $(*)$
$$left|left(fraca_kx^k1+b_kx^2right)^(n)right|le(n+1)!fracb_k|x|^k-n-2$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then
$$ sum_kge n+2left|left(fraca_kx^k1+b_kx^2right)^(n)right| $$ is uniformly convergent on any finite interval, and the Weierstrass M-test allows us to differentiate termwise.
Finally, Besenyei proves $(*)$ in a straightforward fashion with estimates coming from Leibniz rule, after rewriting
$$fraca_k x^k1+b_kx^2=fraca_kb_kcdotfracx^k-12left(frac1x+frac1sqrtb_ki+frac1x-frac1sqrtb_kiright). $$
edited Apr 13 '17 at 12:20
Community♦
1
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
Through this question, I was made aware of
ÃÂdám Besenyei. Peano's unnoticed proof of Borel's theorem, Amer. Math. Monthly 121 (2014), no. 1, 69–72.
In this short note, Besenyei presents a proof due to Peano of the theorem usually attributed to Borel. Peano's result first appeared in
Angelo Genocchi , Giuseppe Peano. Calculo differenziale e principii di calcolo integrale, Fratelli Bocca, Roma, 1884.
Note that Borel's result first appeared in his dissertation, published as
Émile Borel. Sur quelques points de la théorie des fonctions, Ann. Sci. l’École Norm. Sup. (3) 12 (1895), 9–55. MR1508908.
Peano's proof is short, and completely different from Borel's. Besenyei provides full details. I present a sketch:
Given a sequence $(c_n)_nge0$ of real numbers, we want a $C^infty$ function $f$ such that $f^(n)(0)=c_n$ for all $n$. Peano considers
$$ f(x)=sum_kge0fraca_k x^k1+b_kx^2, $$
for $(a_n)_nge0$ arbitrary, and $(b_n)_nge0$ a sequence of positive numbers, chosen so that $f$ is indeed $C^infty$ and can be differentiated term by term. Assuming that this is possible, one easily checks that $f^(n)(0)=a_n$ for $n=0,1$, and that if $nge2$, then
$$ fracf^(n)(0)n!=a_n+sum_j=1^lfloor n/2rfloor(-1)^ja_n-2jb_n-2j^j. $$
To see the latter, consider the power series expansion of $displaystyle fraca_k x^k1+b_kx^2$, valid for $|b_kx^2|<1$, and note that it implies that its $n$-th derivative at $0$ is either $0$ (if $n-k$ is odd), or
$$ n!(-1)^ja_n-2jb_n-2j^j, $$
if $n-k=2j$ for some $j$.
The point is that this recurrence allows us to define the $a_n$ (uniquely) in terms of the $b_n$ and the $c_n$, so that $f^(n)(0)=c_n$ for all $n$.
In order for the above to hold, one needs to ensure that $f$ so defined can indeed be differentiated term by term. For this, Besenyei checks that if $kge n+2$, then $(*)$
$$left|left(fraca_kx^k1+b_kx^2right)^(n)right|le(n+1)!fracb_k|x|^k-n-2$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then
$$ sum_kge n+2left|left(fraca_kx^k1+b_kx^2right)^(n)right| $$ is uniformly convergent on any finite interval, and the Weierstrass M-test allows us to differentiate termwise.
Finally, Besenyei proves $(*)$ in a straightforward fashion with estimates coming from Leibniz rule, after rewriting
$$fraca_k x^k1+b_kx^2=fraca_kb_kcdotfracx^k-12left(frac1x+frac1sqrtb_ki+frac1x-frac1sqrtb_kiright). $$
edited Apr 13 '17 at 12:20
Community♦
1
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
answered May 10 '13 at 19:52
Andrés E. Caicedo
63.2k7151236
63.2k7151236
add a comment |Â
add a comment |Â
up vote
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down vote
The statement of Borel's theorem that I know is that the map $C^infty(mathbbR,mathbbR) to mathbbR^mathbbN$ which sends a function $f$ to its sequence of derivatives at a point (say $0$), is surjective.
One reference, with proof, is to Kriegl and Michor's book A Convenient Setting for Global Analysis, section 15.4 (print version). (It used to be free online from the AMS bookstore, but I can no longer find a link to that version. Maybe I'm just not searching correctly.) There, it is stated as:
Borel's theorem. Suppose a Banach space $E$ has $C^infty_b$-bump functions. Then every formal power series with coefficients in $L^n_sym(E;F)$ for another Banach space $F$ is the Taylor-series of a smooth mapping $E to F$.
It is attributed there to Wells, Differentiable functions on Banach spaces with Lipschitz derivative in JDG 8, 1973, 135-152. Presumably, due to the name, Wells proved the version for Banach spaces. I would also presume that Wells had a reference to Borel's original version. There are also some following remarks on where it fails (mainly due to Colombeau).
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
1
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
add a comment |Â
up vote
10
down vote
The statement of Borel's theorem that I know is that the map $C^infty(mathbbR,mathbbR) to mathbbR^mathbbN$ which sends a function $f$ to its sequence of derivatives at a point (say $0$), is surjective.
One reference, with proof, is to Kriegl and Michor's book A Convenient Setting for Global Analysis, section 15.4 (print version). (It used to be free online from the AMS bookstore, but I can no longer find a link to that version. Maybe I'm just not searching correctly.) There, it is stated as:
Borel's theorem. Suppose a Banach space $E$ has $C^infty_b$-bump functions. Then every formal power series with coefficients in $L^n_sym(E;F)$ for another Banach space $F$ is the Taylor-series of a smooth mapping $E to F$.
It is attributed there to Wells, Differentiable functions on Banach spaces with Lipschitz derivative in JDG 8, 1973, 135-152. Presumably, due to the name, Wells proved the version for Banach spaces. I would also presume that Wells had a reference to Borel's original version. There are also some following remarks on where it fails (mainly due to Colombeau).
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
1
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
add a comment |Â
up vote
10
down vote
up vote
10
down vote
The statement of Borel's theorem that I know is that the map $C^infty(mathbbR,mathbbR) to mathbbR^mathbbN$ which sends a function $f$ to its sequence of derivatives at a point (say $0$), is surjective.
One reference, with proof, is to Kriegl and Michor's book A Convenient Setting for Global Analysis, section 15.4 (print version). (It used to be free online from the AMS bookstore, but I can no longer find a link to that version. Maybe I'm just not searching correctly.) There, it is stated as:
Borel's theorem. Suppose a Banach space $E$ has $C^infty_b$-bump functions. Then every formal power series with coefficients in $L^n_sym(E;F)$ for another Banach space $F$ is the Taylor-series of a smooth mapping $E to F$.
It is attributed there to Wells, Differentiable functions on Banach spaces with Lipschitz derivative in JDG 8, 1973, 135-152. Presumably, due to the name, Wells proved the version for Banach spaces. I would also presume that Wells had a reference to Borel's original version. There are also some following remarks on where it fails (mainly due to Colombeau).
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
The statement of Borel's theorem that I know is that the map $C^infty(mathbbR,mathbbR) to mathbbR^mathbbN$ which sends a function $f$ to its sequence of derivatives at a point (say $0$), is surjective.
One reference, with proof, is to Kriegl and Michor's book A Convenient Setting for Global Analysis, section 15.4 (print version). (It used to be free online from the AMS bookstore, but I can no longer find a link to that version. Maybe I'm just not searching correctly.) There, it is stated as:
Borel's theorem. Suppose a Banach space $E$ has $C^infty_b$-bump functions. Then every formal power series with coefficients in $L^n_sym(E;F)$ for another Banach space $F$ is the Taylor-series of a smooth mapping $E to F$.
It is attributed there to Wells, Differentiable functions on Banach spaces with Lipschitz derivative in JDG 8, 1973, 135-152. Presumably, due to the name, Wells proved the version for Banach spaces. I would also presume that Wells had a reference to Borel's original version. There are also some following remarks on where it fails (mainly due to Colombeau).
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
answered Sep 9 '11 at 8:31
Loop Space
2,55011220
2,55011220
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
1
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
add a comment |Â
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
1
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
"but I can no longer find a link to that version" - It's gone, it looks. There's evidence in the WayBack Machine that it was once free, but it's no longer in the current AMS site. Oh well.
– J. M. is not a mathematician
Sep 9 '11 at 11:15
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
@J.M. My "Maybe I'm just not searching correctly" was a veiled hint that by putting the title into a search engine you'll get a copy - but without actually saying so. The reason I'm now actually saying so is that the first link you get from doing that is to Michor's home page where he has a copy for download. I strongly suspect that this is due to a recent reorganisation of the AMS website and not actually a policy change.
– Loop Space
Sep 9 '11 at 12:20
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
I misread, then. Sorry about that.
– J. M. is not a mathematician
Sep 9 '11 at 12:25
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
@J.M. No, you didn't. It (the free download) truly has gone from the AMS website. But I happen to know that they've done some major reorganisation of their website and so it might well be just because of that. As I can't be sure, at first I was reticent about posting an actual link to a PDF copy as it might be illegal. But as Michor has a link on his homepage, I don't feel that I'm on such shaky ground pointing this out. My apologies for obfuscation!
– Loop Space
Sep 9 '11 at 12:29
1
1
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
The book can be downloaded at P. Michor's website mat.univie.ac.at/~michor/apbookh-ams.pdf
– Zoran Skoda
Apr 2 '15 at 14:42
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4
see exercise 1 on pages 16, 18-19 of C. Zuily 's Problems in Distribution and Partial Differential Equations, or Theorem 4.32 on page 191 of Karl Stromberg's An Introduction to Classical Real Analysis, or the Theorem on pages 50-51 William Donoghue's Distributions and Fourier Transform.
– Soarer
Sep 9 '11 at 7:48
6
@Peter, this is not quite related to generating functions I think.
– Soarer
Sep 9 '11 at 7:49
1
@Peter, Usually when I think about power series, I implicitly assume it's an actual series where series convergence etc is important, rather than formal power series. (Personally I reserve the term generating function for a meaningful sequence of numbers) In this case, convergence shouldn't be part of the condition so yes, you can call it a generating function probably.
– Soarer
Sep 9 '11 at 8:09
2
However, the remark "it is like asking for a proof that every sequence has a generating function" is wrong, because when you write down the power series like you said, it's very likely to have 0 radius of convergence.
– Soarer
Sep 9 '11 at 8:10