Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question

1 Answer 1

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.