[Updated: 17-5-2011 for a natural corollary. ] Last week, when I was giving a seminar I encountered a problem in analysis (which I was unable to solve at that time). After discussing with John Ma, we have come up with the following: (we write in place of the more clumsy and instead of . )

Theorem 1 (John Ma’s theorem)Suppose are functions such that are nowhere zero, and define for any (differentiable) onThen for any , there exists such that

As a corollary, one can easily see that we can relax our conditions that nowhere zero to the conditions that not identically zero.

For example, can we find such such that

to be any arbitrary point in ? This theorem tells us that the answer is yes.

We first prove the following main lemma which is nearly identical to Theorem 1 except that is not required to be compactly supported in .

Lemma 2Suppose are functions such that are nowhere zero, and define for any (differentiable) onThen for any , there exists , such that

*Proof:* First note that can be made exact up to an integrating factor, i.e.

for some nowhere zero function . In fact, take to be the integrating factor

which is well-defined and nowhere zero. Then . Thus we have

Therefore for any by integration by parts,

where . There is no need to explicitly write out all terms of , except to note that

for some continuous functions with nowhere zero on .

Now, if , for some . We then solve for a nontrivial solution of the second order linear homogeneous equation

(Such can be obtained by, for example, prescribing the initial values , but this is not very important here. ) As is nozero, we can find such that to be (first make this nonzero, then rescale ). But then by (1) and (2), we have

In the case where , we solve for a nontrivial solution of the second order linear nonhomogeneous equation

By prescribing the initial values , such must be non-constant. In particular, there exists such that is well-defined and strictly decreasing on . We then take nonzero odd function such that for . Now we take on which can then be extended trivially on . We thus have

as is odd and is strictly decreasing. By rescaling , we have

This completes the proof.

*Proof (Proof of John Ma’s theorem):* Fix . Let are given as in Theorem 1. We then apply Lemma 2, but with (in Lemma) replaced by

Similarly, we define

Then by the lemma, there exists , such that for

We now extend trivially to and extend to such that . (By definition a function on a compact set is one which can be extended to be a function on a slightly bigger open set. ) We claim that the required is then

In fact, as is supported in ,

and

This completes the proof of John Ma’s theorem.