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

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