on Lorentz Conjecture and best uniform approximation

Overview

This post examines the first of several deep conjectures posed by G. G. Lorentz regarding the symmetry of best uniform approximations. Specifically, we investigate whether the requirement that all minimax polynomials of a continuous function vanish at a single point (the origin) is sufficient to force the function itself to be odd. While partially resolved for polynomials and certain analytic classes by Saff and Varga (Saff & Varga, 1980), the general conjecture remains a formidable open problem in approximation theory.

🏷️ The Conjecture

Let $f\in C[-1,1]$ be a real-valued continuous function. For each $n\ge 0$, let $p_n^{\ast }(x;f)$ denote the unique best uniform approximation to $f$ from the space ${\mathcal{P}}_n$ of polynomials of degree at most $n$.

It is a well-known elementary result that if $f$ is an odd function, then $p_n^{\ast }(x;f)$ is also an odd function for every $n$, which implies $p_n^{\ast }(0;f)=0$ for all $n\ge 0$. In the late 1970s, G. G. Lorentz conjectured that the converse is also true [@lorentz1977approximation].

Lorentz Conjecture 1

If $p_n^{\ast }(0;f)=0$ for all $n=0,1,2,\dots$, then $f$ must be an odd function.

🏷️ Rigorous Analysis: The Even Error Function

To analyze the conjecture, Saff and Varga (Saff & Varga, 1980) introduced the even error function $E_n(x)$. For any $f\in C[-1,1]$, let ${ϵ}_n(x)=f(x)-p_n^{\ast }(x)$. We define:

$$E_n(x):={ϵ}_n(x)+{ϵ}_n(-x)=[f(x)+f(-x)]-[p_n^{\ast }(x)+p_n^{\ast }(-x)]$$

By construction, $E_n(x)$ is an even function. Note that $f_e(x)=\frac{1}{2}(f(x)+f(-x))$ is the even part of $f$. The conjecture is equivalent to showing that $p_n^{\ast }(0)=0$ for all $n$ implies $f_e(x)=0$.

1. Zeros of the Even Error

The authors established two critical lemmas regarding the number of zeros of $E_n(x)$ based on the Chebyshev alternation property.

Lemma 1 (Existence of Zeros)

Let $f\in C[-1,1]$ satisfy $[f(x)+f(-x)]\in C^1[-1,1]$. Then, for each $n\ge 0$, $E_n$ has at least $n+1$ zeros in $[-1,1]$, where each zero is counted with multiplicity at most 2. Specifically, any zero $x_0$ of $E_n$ that is also an alternation point ${\xi }_j^{(n)}$ is counted with multiplicity 2.

Proof Sketch

By the Chebyshev Equioscillation Theorem, there exist $n+2$ alternation points $\{{\xi }_j^{(n)}\}$ where the error ${ϵ}_n({\xi }_j^{(n)})$ attains its maximum magnitude with alternating signs. Since $|{ϵ}_n(-{\xi }_j^{(n)})|\le \Vert f-p_n^{\ast }{\Vert }_{\infty }$, the even error satisfies:

$$\lambda (-1{)}^j{E}_n({\xi }_j^{(n)})=\Vert f-p_n^{\ast }{\Vert }_{\infty }+\lambda (-1{)}^j{ϵ}_n(-{\xi }_j^{(n)})\ge 0$$

This sign-alternation property, combined with the continuity of $E_n$, forces the existence of zeros between consecutive alternation points. Evenness and the boundary behavior ensure the total count reaches at least $n+1$.

Lemma 2 (Lorentz Assumption Effect)

If $p_{2k+1}^{\ast }(0;f)=0$ for all $k\ge 0$, then $E_{2k+1}(x)$ has at least $2k+4$ zeros in $[-1,1]$ (counting multiplicities).

Proof Detail

The assumption $p_{2k+1}^{\ast }(0)=0$ implies $f(0)=0$ (by convergence) and thus $E_{2k+1}(0)=0$. Since $E_{2k+1}$ is even and differentiable, the origin $x=0$ is a zero of order at least 2. In Lemma 1, this double zero at the origin is only guaranteed to be counted once because it cannot be an alternation point (since $E_{2k+1}({\xi }_j)\ne 0$). Consequently, we gain one additional zero from the multiplicity at $x=0$, and evenness further constrains the total count to at least $2k+4$.

2. Proof for Analytic Functions

The main result of (Saff & Varga, 1980) establishes the conjecture for functions whose even part grows slowly in the complex plane.

Proof: Proposition 1 (Partial Validity)

The proof relies on reducing the symmetry problem to a zero-counting argument for entire functions. First, we perform a symmetric reduction by defining $F(x^2)=f(x)+f(-x)$ and $s_k(x^2)=p_{2k+1}^{\ast }(x)+p_{2k+1}^{\ast }(-x)$. Since the minimax polynomial $p_{2k+1}^{\ast }$ is of degree at most $2k+1$, its even part $s_k(t)$ is necessarily a polynomial in $t=x^2$ of degree at most $k$.

Let $R_k(t):=F(t)-s_k(t)$ denote the error in the $t$-domain for $t\in [0,1]$. By Lemma 2, the even error function $E_{2k+1}(x)$ possesses at least $2k+4$ zeros in $[-1,1]$. Invoking the evenness of $E_{2k+1}$ and the fact that $E_{2k+1}(0)=0$, we deduce that $R_k(t)$ must have at least $k+2$ zeros in the interval $[0,1]$, where the zero at the origin is counted with multiplicity 1 in this domain.

Applying the generalized Rolle’s Theorem to $R_k(t)$, we conclude that the $(k+1)$-th derivative $R_k^{(k+1)}(t)$ must vanish at some point ${\beta }_{k+1}\in (0,1)$. Since $s_k(t)$ is a polynomial of degree at most $k$, its $(k+1)$-th derivative vanishes identically, implying $F^{(k+1)}({\beta }_{k+1})=0$. By including the boundary condition $F(0)=0$ (and setting ${\beta }_0=0$), we obtain a sequence of zeros $\{F^{(j)}({\beta }_j){\}}_{j=0}^{\infty }$ with ${\beta }_j\in [0,1]$.

Finally, we invoke Schoenberg’s Theorem (Schoenberg, 1936): if $F(z)$ is an entire function of exponential type $\tau <\pi /2$ and each of its derivatives vanishes at some point in a bounded interval, then $F(z)$ must be identically zero. Consequently, $F(x^2)=2f_e(x)=0$, which forces $f$ to be an odd function. ▮

🏷️ Structural Consequences

1. Exclusion of Even Functions

A direct consequence of the zero-counting logic (Remark 4 in the paper) is that a non-zero even function can never satisfy the Lorentz condition.

Result: Non-Evenness

If $f\in C[-1,1]$ is not identically zero and satisfies $p_n^{\ast }(0;f)=0$ for all $n\ge 0$, then $f$ cannot be an even function.

Proof Sketch: If $f$ were even, the best approximation $p_n^{\ast }$ would also be even. But then $p_n^{\ast }(0)=0$ for all $n$ implies (via a lemma of Lorentz) that $p_{n+1}^{\ast }-p_n$ would not have enough zeros in $(-1,1)$ to satisfy the standard monotonicity properties unless $p_n\equiv 0$, forcing $f\equiv 0$.

2. Application to Zolotareff Polynomials

The result provides an immediate proof that Zolotareff polynomials (which are non-odd) cannot vanish at the origin if they are optimal approximations.

Zolotareff Polynomials

For any $\sigma \ne 0$, the odd-degree Zolotareff polynomial $Z_{2m+1}(x;\sigma )=x^{2m+1}+\sigma x^{2m}-q(x)$ satisfies $Z_{2m+1}(0;\sigma )\ne 0$. If it were zero, it would satisfy the Lorentz condition for all $k\ge m$ (and also for $k<m$ since the approximation is zero), forcing the polynomial to be odd, which contradicts the $\sigma x^{2m}$ term.

🏷️ Sharpness and Counterexamples

The assumption on the exponential type $\tau <\pi /2$ is sharp. Saff and Varga demonstrated this using the function:

$$f(x)=\cos \left(\frac{\pi x}{2}\right)$$

This function is even, yet it satisfies $p_{2k+1}^{\ast }(0)=0$ for all $k\ge 0$. Here, $F(z)=2\cos (\frac{\pi \sqrt{z}}{2})$ has exponential type exactly $\pi /2$.

Furthermore, Remark 3 in the paper shows that the condition $p_{2k+1}^{\ast }(0)=0$ must hold for every $k$. If even a single degree $m$ is omitted, one can construct an even polynomial $f_m(x)$ that satisfies the condition for all $k\ne m$.

🔗 See Also

• on best uniform approximation from subspaces --- Explores the 2nd conjecture by G G Lorentz regarding optimal monomial spans.
• on eigenvalue estimate of kernel --- The ETP property is the fundamental reason for the non-oscillation of kernel eigenfunctions.

📚 References

🐻  Saff, E.B. & Varga, R.S. 1980. Remarks on some conjectures of G. G. Lorentz. Journal of Approximation Theory 30(1), 29–36.

🐻  Schoenberg, I.J. 1936. On the zeros of successive derivatives of integral functions. Transactions of the American Mathematical Society 40(1), 12–23.