# Every power series is the Taylor series of some $C^{\infty}$ function

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?

• 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
• @Peter, this is not quite related to generating functions I think. – Soarer Sep 9 '11 at 7:49
• @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
• 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

Borel's theorem states that given a sequence of real numbers $(a_n)_{n\in \mathbb N}$ there exists a $C^\infty$ function $f\in C^\infty(\mathbb R)$ 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 $x\neq 0\in \mathbb R$ ! 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 $(a_I)_{I\in \mathbb N^k}$ of real numbers $a_I \in\mathbb R$, there exists a function $f\in C^\infty(\mathbb R^k)$, 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 $Z\subset \mathbb R^k$ and continuous functions $\phi_I\in C(Z) \;$ . Whitney gives necessary and sufficient growth and compatibility conditions on the $\phi_I$ 's which will guarantee that there exists a $C^\infty$ function $f\in C^\infty (U)$ 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.

• Incidentally, there's a nice functional analytic reason why there's no splitting map. If there were, we would have a continuous linear injection $\mathbb{R}^{\mathbb{N}} \to C^0(\mathbb{R},\mathbb{R})$ but any continuous linear map $\mathbb{R}^{\mathbb{N}} \to V$ to a normable space has to factor through one of the projections $\mathbb{R}^{\mathbb{N}} \to \mathbb{R}^n$. (NB, I'm sure that you know this, I'm putting it for the benefit of anyone else reading.) – Andrew Stacey 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 $(a_n)_{n\in\mathbb{N}}$" in your first sentence, do you assume there exists a function $a(n)$ that can generate the $a_n$ for $n\in\mathbb{N}$. What if $a_n$ were "random"? What if I chose $a_n$ at random? – Pixel Jan 29 '14 at 9:57
• Dear @pbs: I'm not sure I understand your question but you can choose any arbitrary sequence $(a_n)_{n\in \mathbb N}$ 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

Through this question, I was made aware of

Ádám 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 $(c_n)_{n\ge0}$ of real numbers, we want a $C^\infty$ function $f$ such that $f^{(n)}(0)=c_n$ for all $n$. Peano considers $$f(x)=\sum_{k\ge0}\frac{a_k x^k}{1+b_kx^2},$$ for $(a_n)_{n\ge0}$ arbitrary, and $(b_n)_{n\ge0}$ 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 $n\ge2$, then $$\frac{f^{(n)}(0)}{n!}=a_n+\sum_{j=1}^{\lfloor n/2\rfloor}(-1)^ja_{n-2j}{b_{n-2j}}^{j}.$$ To see the latter, consider the power series expansion of $\displaystyle \frac{a_k x^k}{1+b_kx^2}$, 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 $$n!(-1)^ja_{n-2j}{b_{n-2j}}^{j},$$ 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 $k\ge n+2$, then $(*)$ $$\left|\left(\frac{a_kx^k}{1+b_kx^2}\right)^{(n)}\right|\le(n+1)!\frac{|a_k|k!}{b_k}|x|^{k-n-2}$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then $$\sum_{k\ge n+2}\left|\left(\frac{a_kx^k}{1+b_kx^2}\right)^{(n)}\right|$$ 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 $$\frac{a_k x^k}{1+b_kx^2}=\frac{a_k}{b_k}\cdot\frac{x^{k-1}}2\left(\frac1{x+\frac1{\sqrt{b_k}}i}+\frac1{x-\frac1{\sqrt{b_k}}i}\right).$$

The statement of Borel's theorem that I know is that the map $C^\infty(\mathbb{R},\mathbb{R}) \to \mathbb{R}^{\mathbb{N}}$ 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).

• "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. isn't 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. – Andrew Stacey Sep 9 '11 at 12:20
• I misread, then. Sorry about that. – J. M. isn't 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! – Andrew Stacey Sep 9 '11 at 12:29
• 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