Proving non-negativity of Schur's interpolating function provided it holds for $t=1$ and $t\to∞$?

Question

I learnt that the function $x^{2t}(x-y)(x-z)+y^{2t}(y-z)(y-x)+z^{2t}(z-x)(z-y)$ is non-negative for all $x,y,z\in\mathbb{R}$.

I wonder what we can say about the case when the power is odd.

Let $f_t(x,y,z)=x^{2t-1}(x-y)(x-z)+y^{2t-1}(y-z)(y-x)+z^{2t-1}(z-x)(z-y)$.

Is the following true?

If

• $f_1(x,y,z)\ge0$ for all $x,y,z\in\mathbb{R}$

and

• $f_t(x,y,z)\ge0$ for all $x,y,z\in\mathbb{R}$ for sufficiently large $t$

then $f_t(x,y,z)\ge0$ for all $x,y,z\in\mathbb{R}$ for all $t\in\mathbb{N}^+$.

Answer 1

The statement is true. If $f_1(x,y,z)\ge0$ and $f_t(x,y,z)\ge0$ for sufficiently large $t$, then $f_t(x,y,z)\ge0$ for all $t\in\mathbb{N}^+$.

The expression for $f_t(x,y,z)$ is symmetric under any permutation of the variables $(x,y,z)$. This allows us, without loss of generality, to assume an ordering $x \ge y \ge z$ for our analysis.

Simple Cases

First, we address cases that can be resolved without invoking the full set of conditions.

• Degenerate Case: At least two variables are equal.

  If any two variables are equal, we may assume $x=y$. The expression simplifies to:

  $$f_t(x,x,z) = z^{2t-1}(z-x)^2$$

  The condition $f_1(x,x,z) \ge 0$ becomes $z(z-x)^2 \ge 0$. If $x \ne z$, this implies $z \ge 0$. If $z \ge 0$, then $z^{2t-1} \ge 0$ for all $t\in\mathbb{N}^+$, which ensures $f_t(x,x,z) \ge 0$. If $x=z$, then all variables are equal and $f_t=0$. In all degenerate subcases, the proposition holds.

For the remainder of the proof, we assume $x, y, z$ are distinct, so $x > y > z$.

• Case A: All variables are non-negative ($x > y > z \ge 0$).

  We can write $f_t(x,y,z) = (x-y)[x^{2t-1}(x-z) - y^{2t-1}(y-z)] + z^{2t-1}(x-z)(y-z)$. Every term in this sum is non-negative, so $f_t(x,y,z) \ge 0$ for all $t$. The conditions are met and the conclusion holds.

• Case B: All variables are negative ($0 > x > y > z$).

  Here, $f_t(x,y,z) < 0$ for all $t$. This can be seen by letting $x=-a, y=-b, z=-c$ with $0<a<b<c$, which yields $f_t(x,y,z) = -f_t(a,b,c)$, where $f_t(a,b,c)>0$ by Case A. Since $f_1(x,y,z)<0$, this case is ruled out by the problem's conditions.

The Main Case: Mixed Signs ($x>0>z$)

The only remaining possibility is that the variables are not all of the same sign. Given our ordering $x > y > z$, this means we must have $x>0$ and $z<0$.

Lemma: The "Large $t$" Condition Implies $x \ge |z|$

The asymptotic behavior of $f_t$ for large $t$ is determined by the term whose base has the largest absolute value. If we assume $|z|$ is the unique largest magnitude (i.e., $|z|>|x|\ge|y|$), then for large $t$, the sign of $f_t$ is determined by the sign of its $z$ term:

$$\text{sgn}(f_t) =\text{sgn}(z^{2t-1}(z-x)(z-y)) = \text{sgn}(z)\,\text{sgn}(z-x)\,\text{sgn}(z-y)$$

Since $z<0$, $z-x<0$, and $z-y<0$, the product of the signs is $(-)(-)(-) = (-)$. This implies $f_t < 0$ for large $t$, which contradicts the hypothesis. A similar analysis of all possibilities shows that the largest absolute value cannot uniquely belong to a negative variable. Given the ordering $x>0>z$, this forces the condition $x \ge |z|$.

Main Proof for the Mixed-Sign Case

We now prove that if $x>0>z$ and $x \ge |z|$, the condition $f_1 \ge 0$ implies $f_t \ge 0$ for all $t$. We consider two subcases based on the sign of $y$.

• Subcase 1: $y \ge 0$ (so $x>y\ge 0 > z$ and $x\ge|z|$).

  Let $z=-a$ where $a>0$. We have $x \ge a$. We are given $f_1 \ge 0$ and need to show $f_t \ge 0$. Let $F(t)=f_t$. We show that $G(t) = F(t)/a^{2t-1}$ is non-decreasing for $t \ge 1$.

  $$G(t) = (x-y)\left[ \left(\frac{x}{a}\right)^{2t-1}(x+a) - \left(\frac{y}{a}\right)^{2t-1}(y+a) \right] - (x+a)(y+a)$$

  This is equivalent to showing that $K(t) = (\frac{x}{a})^{2t-1}(x+a) - (\frac{y}{a})^{2t-1}(y+a)$ is non-decreasing. Let's analyze $K(t+1) - K(t)$:

  $$K(t+1) - K(t) = (x+a)\left(\frac{x}{a}\right)^{2t-1}\left(\left(\frac{x}{a}\right)^2-1\right) - (y+a)\left(\frac{y}{a}\right)^{2t-1}\left(\left(\frac{y}{a}\right)^2-1\right)$$

  If $y \le a$, the term $((y/a)^2-1)$ is $\le 0$, which makes the entire second part of the difference non-negative. Since $x \ge a$, the first part is also non-negative. The difference is $\ge 0$.

  If $y > a$, so $x>y>a>0$, the difference is positive if $\left(\frac{x}{y}\right)^{2t-1} > \frac{(y-a)(y+a)^2}{(x-a)(x+a)^2}$. This is true because the LHS is $>1$ while the RHS is $<1$.

  Thus $K(t)$ is non-decreasing, so $G(t)$ is non-decreasing. Since $f_1 \ge 0 \implies G(1) \ge 0$, we have $G(t) \ge G(1) \ge 0$ for all $t$, which means $f_t \ge 0$.

• Subcase 2: $y < 0$ (so $x>0>y>z$ and $x\ge|z|$).

  Let $y=-b$ and $z=-c$, so we have $x>0$, $c>b>0$, and the lemma gives $x \ge c$. We rearrange the original expression for $f_t$:

  $$f_t(x,y,z) = (x-z)[x^{2t-1}(x-y) + z^{2t-1}(y-z)] - y^{2t-1}(x-y)(y-z)$$

  Substituting $y=-b$ and $z=-c$:

  $$f_t(x,y,z) = (x+c)[x^{2t-1}(x+b) - c^{2t-1}(c-b)] + b^{2t-1}(x+b)(c-b)$$

  We can show that this expression is a sum of positive terms.

  • The final term $b^{2t-1}(x+b)(c-b)$ is positive because $x,b,c$ are positive and $c>b$.

  • The term $(x+c)$ is positive.

  • For the bracketed term, $[x^{2t-1}(x+b) - c^{2t-1}(c-b)]$, we compare the two parts.

    • From the lemma, we have $x \ge c$, so $x^{2t-1} \ge c^{2t-1}$.

    • We also have $x>0$ and $c>b>0$, which implies $x+b > c-b$ (since $x\ge c$).

    • Combining these, $x^{2t-1}(x+b) > c^{2t-1}(c-b)$, so the bracketed term is strictly positive.

  Since all parts are positive, $f_t(x,y,z) > 0$ for all $t \in \mathbb{N}^+$. The conditions are met and the conclusion holds trivially.