## A problem in analysis

[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 $C[0,1]$ in place of the more clumsy $C([0,1])$ and $C(0,1)$ instead of $C((0,1))$. )

Theorem 1 (John Ma’s theorem) Suppose ${P_0, P_1, Q_0, Q_1\in C^2[-1,1]}$ are functions such that ${P_1, Q_1}$ are nowhere zero, and define for any (differentiable) ${p,q}$ on ${[-1,1]}$

$\displaystyle \begin{cases} P(x)=P(p,p')=P_0(x)p(x)+P_1(x)p'(x), \\ Q(x)=Q(q, q')=Q_0(x)q(x)+Q_1(x)q'(x). \end{cases}$

Then for any ${(\alpha, \beta)\in \mathbb{R}^2}$, there exists ${p, q\in C^2_c(-1,1)}$ such that

$\displaystyle \left(\int_{-1}^1 p(x)q(x)dx, \int_{-1}^1 P(x)Q(x)dx\right)=(\alpha, \beta).$

As a corollary, one can easily see that we can relax our conditions that $P_1, Q_1$ nowhere zero to the conditions that $P_1Q_1$ not identically zero.

For example, can we find such ${p, q}$ such that

$\displaystyle \left(\int_{-1}^1 pq, \int_{-1}^1 (p\sin x+x^3p') q'\right)$

to be any arbitrary point in ${\mathbb{R}^2}$? 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 ${q}$ is not required to be compactly supported in ${(-1,1)}$.

Lemma 2 Suppose ${P_0, P_1, Q_0, Q_1\in C^2[-1,1]}$ are functions such that ${P_1, Q_1}$ are nowhere zero, and define for any (differentiable) ${p,q}$ on ${[-1,1]}$

$\displaystyle \begin{cases} P(x)=P(p,p')=P_0(x)p(x)+P_1(x)p'(x), \\ Q(x)=Q(q, q')=Q_0(x)q(x)+Q_1(x)q'(x). \end{cases}$

Then for any ${(\alpha, \beta)\in \mathbb{R}^2}$, there exists ${p\in C^2_c(-1,1)}$, ${q\in C^2[-1,1]}$ such that

$\displaystyle \left(\int_{-1}^1 p(x)q(x)dx, \int_{-1}^1 P(x)Q(x)dx\right)=(\alpha, \beta).$

Proof: First note that ${P(p, p')}$ can be made exact up to an integrating factor, i.e.

$\displaystyle P=u^{-1}(uP_1p)'$

for some nowhere zero function ${u\in C^2[-1,1]}$. In fact, take ${u}$ to be the integrating factor

$\displaystyle u(x)=\exp\left(\int_{-1}^x \frac{P_0-P_1'}{P_1}\right)$

which is well-defined and nowhere zero. Then ${u'=\frac{P_0-P_1'}{P_1}u}$. Thus we have

$\displaystyle u^{-1}(u P_1 p)'=u^{-1}u'P_1p+P_1'p+P_1p'=(P_0-P_1')p+P_1'p+P_1p'=P_0p+P_1p'=P.$

Therefore for any ${p\in C_c^2(-1,1)}$ by integration by parts,

$\displaystyle \int_{-1}^1PQ=\int_{-1}^1 u^{-1}(u P_1 p)'Q=0-\int_{-1}^1 u P_1 p(u^{-1}Q)'=\int_{-1}^1 p L(x) \ \ \ \ \ (1)$

where ${L(x)=L(q,q',q''):=-(uP_1(u^{-1}Q)')}$. There is no need to explicitly write out all terms of ${L}$, except to note that

$\displaystyle L(q,q',q'')=L_2q''+L_1q'+L_0q$

for some continuous functions ${L_i=L_i(x)}$ with ${L_2}$ nowhere zero on ${[-1,1]}$.

Now, if ${\alpha\neq 0}$, ${\beta=k\alpha}$ for some ${k}$. We then solve for a nontrivial solution ${q\in C^2[-1,1]}$ of the second order linear homogeneous equation

$\displaystyle L(q, q', q'')=L_2q''+L_1q'+L_0q=kq. \ \ \ \ \ (2)$

(Such ${q}$ can be obtained by, for example, prescribing the initial values ${q(0)=q'(0)=1}$, but this is not very important here. ) As ${q}$ is nozero, we can find ${p\in C^2_c(-1,1)}$ such that ${\int_{-1}^1pq}$ to be ${\alpha}$ (first make this nonzero, then rescale ${p\rightarrow cp}$). But then by (1) and (2), we have

$\displaystyle \int_{-1}^1 PQ=k\int_{-1}^1pq=\beta.$

In the case where ${\alpha=0}$, we solve for a nontrivial solution ${q\in C^2[-1,1]}$ of the second order linear nonhomogeneous equation

$\displaystyle L(q, q', q'')=L_2q''+L_1q'+L_0q=1. \ \ \ \ \ (3)$

By prescribing the initial values ${q(0)=q'(0)=1}$, such ${q}$ must be non-constant. In particular, there exists ${[-\varepsilon, \varepsilon]\subset(-1,1)}$ such that ${1/q}$ is well-defined and strictly decreasing on ${[-\varepsilon, \varepsilon]}$. We then take nonzero odd function ${\widetilde p\in C^2_c(-\varepsilon,\varepsilon)}$ such that ${ \widetilde p\geq 0 }$ for ${x\leq 0}$. Now we take ${p=\widetilde p/q}$ on ${[-\varepsilon, \varepsilon]}$ which can then be extended trivially on ${[-1,1]}$. We thus have

$\displaystyle \int_{-1}^1pq=\int_{-\varepsilon}^\varepsilon\widetilde p=0.$

Also by (1), (3),

$\displaystyle \int_{-1}^1PQ=\int_{-1}^1p=\int_{-\varepsilon}^\varepsilon \widetilde p/q>0.$

as ${\widetilde p}$ is odd and ${1/q}$ is strictly decreasing. By rescaling ${q\rightarrow cq}$, we have

$\displaystyle \left(\int_{-1}^1 pq, \int_{-1}^1 PQ\right)=(0, \beta).$

This completes the proof. $\Box$

Proof (Proof of John Ma’s theorem): Fix ${(\alpha, \beta)}$. Let ${P_0, P_1, Q_0, Q_1}$ are given as in Theorem 1. We then apply Lemma 2, but with ${P_0(x)}$ (in Lemma) replaced by

$\displaystyle \widetilde P_0(x)=P_0(\frac{x}{2}) , \quad x\in [-1,1].$

Similarly, we define

$\displaystyle \widetilde P_1(x)=2P_1(\frac{x}{2}),\widetilde Q_0(x)=Q_0(\frac{x}{2}),\widetilde Q_1(x)=2Q_1(\frac{x}{2}), \quad x\in [-1,1].$

Then by the lemma, there exists ${\widetilde p\in C^2_c(-1,1)}$, ${\widetilde q\in C^2[-1,1]}$ such that for

$\displaystyle \begin{cases} \widetilde P(x)=\widetilde P_0(x)\widetilde p(x)+\widetilde P_1(x)\widetilde p'(x), \\ \widetilde Q(x)=\widetilde Q_0(x)\widetilde q(x)+\widetilde Q_1(x)\widetilde q'(x), \end{cases}$

$\displaystyle \left( \int_{-1}^1 \widetilde p(x)\widetilde q(x)dx, \int_{-1}^1 \widetilde P(x)\widetilde Q(x)dx\right)=(2\alpha, 2 \beta).$

We now extend ${\widetilde p}$ trivially to ${[-2,2]}$ and extend ${\widetilde q}$ to ${[-2,2]}$ such that ${\widetilde q\in C^2_c(-2,2)}$. (By definition a ${C^1}$ function on a compact set is one which can be extended to be a ${C^1}$ function on a slightly bigger open set. ) We claim that the required ${p,q}$ is then

$\displaystyle p(x)=\widetilde p(2x), q(x)=\widetilde q(2x) ,\quad x\in [-1,1].$

In fact, as ${\widetilde p(y)}$ is supported in ${(-1,1)}$,

$\displaystyle \begin{array}{rcl} \int_{-1}^1 p(x)q(x)dx=\int_{-1}^1\tilde p(2x)\tilde q(2x)dx=\frac{1}{2}\int_{-2}^2\tilde p(y)\tilde q(y)dy& = & \frac{1}{2}\int_{-1}^1\tilde p(y)\tilde q(y)dy \\ & = & \alpha \end{array}$

and

$\displaystyle \begin{array}{rcl} \int_{-1}^1 P(x)Q(x)dx &=&\int_{-1}^1\left[P_0(x)p(x)+P_1(x)p'(x)\right]\left[Q_0(x)q(x)+Q_1(x)q'(x)\right]dx\\ &=&\int_{-1}^1\left[P_0(x)\widetilde p(2x)+2P_1(x)\widetilde p'(2x)\right]\left[Q_0(x)\widetilde q(2x)+2Q_1(x)\widetilde q'(2x)\right]dx\\ &=&\frac{1}{2}\int_{-2}^2\left[P_0(\frac{y}{2})\widetilde p(y)+2P_1(\frac{y}{2})\widetilde p'(y)\right]\left[Q_0(\frac{y}{2})\widetilde q(y)+2Q_1(\frac{y}{2})\widetilde q'(y)\right]dy\\ &=&\frac{1}{2}\int_{-1}^1\left[P_0(\frac{y}{2})\widetilde p(y)+2P_1(\frac{y}{2})\widetilde p'(y)\right]\left[Q_0(\frac{y}{2})\widetilde q(y)+2Q_1(\frac{y}{2})\widetilde q'(y)\right]dy \\ &=&\frac{1}{2}\int_{-1}^1\left[\widetilde P_0(y)\widetilde p(y)+\widetilde P_1(y)\widetilde p'(y)\right]\left[\widetilde Q_0(y)\widetilde q(y)+ \widetilde Q_1(y)\widetilde q'(y)\right]dy\\ &=&\frac{1}{2}\int_{-1}^1\widetilde P(y)\widetilde Q(y)dy\\ &=&\beta. \end{array}$

This completes the proof of John Ma’s theorem. $\Box$