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hbghlyj
Posted at 2022-6-21 05:58:32
Every power series is the Taylor series of some $C^∞$ function
ncatlab.org/nlab/show/Borel%27s+theorem
math.stackexchange.com/questions/63050/every- … function/63062#63062
abesenyei.web.elte.hu/publications/borel.pdf
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 $e^{-1/x^2}$.
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
$$\frac{a_k x^k}{1+b_kx^2}=\frac{a_k}{b_k}\cdot\frac{x^{k-1}}2\left(\frac1{x+\frac1{\sqrt{b_k}}i}+\frac1{x-\frac1{\sqrt{b_k}}i}\right). $$ |
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战巡
Posted at 2018-10-2 08:11:37
