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Borel’s theorem, in its simplest form, states that for any sequence $(a_k)$ of real numbers, there is a function $f\in {\cal C}^{\infty}({\mathbb R},{\mathbb R})$ such that $f^{(k)}(0)=a_k$ for all $k$. I wonder if we can take $f$ to be increasing.

Obviously, this will not be possible if the sequence starts with some zeros and then a negative number. If, however, all the $a_k$ are $0$ or if the first nonzero $a_k$ is positive, there seems to be no reason why $f$ could not be increasing. Is this already known ?

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I was wondering how this theorem works for any sequence of complex numbers ? – Maman Apr 17 at 10:54
up vote 2 down vote accepted

In the form stated it won't be possible. E.g., if $a_1 = 0$ and $a_2 > 0$, then $f'$ will be increasing near $0$, so $f'(x)<f'(0)=0$ for (small) negative $x$, and $f$ will be decreasing in at least a small interval $(-\delta,0)$ for some $\delta>0$.

On the other hand, if $a_1>0$, you can get an increasing function as follows: Start with a function $f$ which has all the prescribed derivatives and has compact support (by multiplying an arbitrary solution with a $\mathcal{C}^\infty$ function with compact support which is $\equiv 1$ near $0$). Now it is enough to find a $\mathcal{C}^\infty$ function $g$ with $g^{(n)} (0) = 0$ for all $n$, and $g'(x)>0$ for $x \ne 0$. Then $f_t= f+tg$ will satisfy $f_t'(x) > 0$ for $x \ne 0$ if $t$ is sufficiently large, implying that $f_t$ is strictly increasing. The construction of $g$ is standard, e.g., $g(x) = e^{-1/x^2}$ for $x>0$, $g(0)=0$, and $g(-x) = -g(x)$ for $x>0$.

Similarly, if the smallest $k$ for which $a_k \ne 0$ is odd, and if $a_k > 0$, you can find an increasing function with the prescribed derivatives (basically by integrating the solution given above $k-1$ times.)

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