Prove that there do not exist real polynomials \(a,b,c,d\) such that \[ a(x)<b(x)<c(x)<d(x)\quad(0<x<1) \] and \[ b(x)<d(x)<a(x)<c(x)\quad(-1<x<0). \]
Solution
1. Subtracting \(a\) from all four polynomials, it is enough to ruleout polynomials \(B,C,D\) satisfying\[ 0<B(x)<C(x)<D(x) \]for \(0<x<1\) and\[ B(x)<D(x)<0<C(x) \]for \(-1<x<0\).
Set \[ B=b-a, \qquad C=c-a, \qquad D=d-a. \] Subtracting the same real number \(a(x)\) from every entry in each ordering preserves all strict inequalities and replaces \(a\) by the zero polynomial.
2. The three normalized polynomials satisfy\[ B(0)=C(0)=D(0)=0. \]
Letting \(x\to0^+\) in \[ 0<B(x)<C(x)<D(x) \] and using continuity gives \[ 0\le B(0)\le C(0)\le D(0). \] Letting \(x\to0^-\) in \[ B(x)<D(x)<0<C(x) \] gives \[ B(0)\le D(0)\le0\le C(0). \] Thus \[ 0\le B(0)\le D(0)\le0, \] so \(B(0)=D(0)=0\). The first chain then forces \(C(0)=0\) as well.
3. Write the lowest nonzero terms as\[ B(x)=\beta x^i+O(x^{i+1}), \qquad C(x)=\gamma x^j+O(x^{j+1}), \qquad D(x)=\delta x^k+O(x^{k+1}), \]where\[ \beta,\gamma,\delta>0. \]
The strict inequalities in step 1 show that none of \(B,C,D\) is the zero polynomial. Step 2 shows that each has positive vanishing order at \(0\), so the displayed lowest terms exist.
1. Since \(x^i,x^j,x^k\) are then positive, their firstnonzero coefficients must satisfy\[ \beta>0,\qquad\gamma>0,\qquad\delta>0. \]
4. The positive-side ordering implies\[ i\ge j\ge k. \]
If \(i<j\), then as \(x\to0^+\), \[ \frac{B(x)}{C(x)} \sim \frac{\beta}{\gamma}x^{i-j} \longrightarrow \infty, \] contradicting \(0<B(x)<C(x)\). Hence \(i\ge j\).
Likewise, if \(j<k\), then \[ \frac{C(x)}{D(x)} \longrightarrow \infty \] as \(x\to0^+\), contradicting \(C(x)<D(x)\). Thus \(j\ge k\).
5. The negative-side signs imply that \(i\) and \(k\) are odd while\(j\) is even.
For sufficiently small negative \(x\), step 1 gives \[ B(x)<0,\qquad D(x)<0,\qquad C(x)>0. \] The leading coefficients in step 3 are all positive. Therefore the sign of each polynomial near \(0\) on the negative side is the sign of \(x\) raised to its vanishing order. Hence \(i\) and \(k\) must be odd and \(j\) must be even.
6. Consequently,\[ i>j>k. \]
By step 4, \[ i\ge j\ge k. \] By step 5, \(i\) and \(j\) have opposite parity, so they cannot be equal; hence \(i>j\). Similarly \(j\) and \(k\) have opposite parity, so \(j>k\).
7. For all sufficiently small negative \(x\),\[ B(x)>D(x), \]contradicting the required ordering.
By steps 3 and 6, \[ \frac{B(x)}{D(x)} \sim \frac{\beta}{\delta}x^{i-k}. \] The integer \(i-k\) is positive and even, so as \(x\to0^-\), \[ \frac{B(x)}{D(x)} \longrightarrow0^+. \] Thus for all sufficiently small negative \(x\), \[ 0<\frac{B(x)}{D(x)}<1. \] Step 1 also gives \(D(x)<0\). Multiplying the last inequality by the negative number \(D(x)\) reverses the order and yields \[ B(x)>D(x). \] This contradicts the required inequality \(B(x)<D(x)\).
8. Therefore no such four polynomials exist.
If the original polynomials existed, step 1 would produce normalized polynomials \(B,C,D\), but step 7 shows those cannot exist.
9. Q.E.D.
Step 8 is the required conclusion.