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I'm trying to prove the Borel Lemma, which is:

For every series $a_0,a_1,a_2,\dots$ in $\mathbb{C}$ exists $f \in C^{\infty}(\mathbb{R})$ such as $$ f^{(k)}(0) = a_k $$ for every $k \in \mathbb{N}_0$.

My idea: I would set $$f(x) = \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} x^k,$$ so if the series $(a_k)_{k \in \mathbb{N}_0}$ converges this $f$ would do it. But I don't know anything about the series, so it could diverge. To avoid this, I would think about to cut the sum by multiplying with a cut-off-function $\chi \in C^{\infty}(\mathbb{C})$ so that $\chi(x) = 0$ for every $|x| > R, R \in \mathbb{C}$. But I don't know how to continue..

I would be thankful for any kind of help.

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Using cut-off to produce functions $f_n$ such that $f_n(x)=x^n$ for $|x|<r_n$ and $f_n(x)=0$ for $|x|>r_{n-1}$. Then $$f(x)=\sum_{k=0}^\infty \frac{a_k}{k!}f_k(x)$$ should do the trick, provided $r_n\to0$ fast enough (how fast do you need for the sum to be "friendly"?)

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I'm not sure, if I'm done, but I have some thoughts to share and discuss. For $f_k$ I chose $f_k(x) = x^k \cdot \chi_k(x)$ with $$\chi_k(x) = \begin{cases} 1,& |x| \leq r_k\\ 0,& |x| \geq r_k +1\\ \in C^{\infty},& \text{ else} \end{cases} $$ with a series $(r_k)_{k \in \mathbb{N}_0} \subset \mathbb{C}$ with $r_k \rightarrow 0.$ –  gisma Jan 19 '13 at 12:20
    
So then $$\chi_k^{(m)}(x) = \begin{cases} 0,& |x| \leq r_k \text{ or } |x| \geq r_k +1\\ \in C^{\infty},& \text{ else} \end{cases} .$$ –  gisma Jan 19 '13 at 12:21
    
Now I observe the $m-$th derivative: $$\begin{align} f^{(m)}(x) =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} (x^k \cdot \chi_k(x))^{(m)}\\ =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} \left( \sum\limits_{\beta=0}^{m}{m \choose \beta}\ (x^k)^{(\beta)} \cdot \chi_k(x)^{(m-\beta)}\right)\\ =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} \left( (x^k)^{(m)}\cdot \chi_k(x)+\sum\limits_{\beta=1}^{m}{m \choose \beta}\ (x^k)^{(\beta)} \cdot \chi_k(x)^{(m-\beta)}\right) \end{align} $$ –  gisma Jan 19 '13 at 12:21
    
where $$(x^k)^{(m)} = \begin{cases} \frac{k!}{(m-1)!}\ x^{k-m}&,\ m<k\\ k!&,\ m=k\\ 0&,\ m>k \end{cases} $$ It should be $\chi_k(x) \leq c_k, c_k \in \mathbb{C}$ and $\chi_k^{(m)}(x)$ cutting off (for a proper $(r_k)$) the trouble making $(x^k)^{(\beta)}$ for $|x| >1$, hence $f \in C^{\infty}(\mathbb{C})$ even if $(a_k)$ is divergent, but only fo a proper series $(r_k)$ letting $\chi_k^{(m-\beta)}(x)$ cut off those $a_k$ making trouble. For $a_k$ convergent this is obvious. –  gisma Jan 19 '13 at 12:22
    
Now we should ensure that $f$ is doing the trick: $f^{(k)}(0) = a_k$. Because of $\chi_k(0) = 1$ and $(x^k)^{(m)}\Big|_{x=0} \neq 0$ only for $m=k$ it follows that $$ f^{(k)}(0) = \frac{a_k}{k!} \left( k! \right) = a_k.$$ If there are no serious misleadings in my thoughts the only thing that misses is to find the proper $(r_k)$ ensuring that $\chi_k$ is cutting off right, even if $(a_k)$ is divergent. This $(r_k)$ should be directly related with $(a_k)$. –  gisma Jan 19 '13 at 12:22
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