on Lorentz Conjecture and best uniform approximation

Overview

This post develops Saff and Varga’s affirmative results for Lorentz’s common-origin conjecture: does requiring every best uniform polynomial approximant to vanish at zero force the function to be odd? Their theorem answers yes for polynomials and a specified class of entire-function profiles. (Saff & Varga, 1980) The continuous counterexample constructed in LorentzLean is discussed in on self-similar counterexamples to Lorentz conjectures, which explains how self-similarity preserves the approximation conditions while breaking oddness. The analytic results below remain valid.

🏷️ 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 $n$.

If $f$ is odd, reflection sends a best approximant $p(x)$ to the equally good approximant $-p(-x)$. Uniqueness therefore makes $p_n^{\ast }(x;f)$ odd, so $p_n^{\ast }(0;f)=0$ for every $n$. Lorentz proposed the converse in work of the late 1970s, as recorded by Saff and Varga. (Saff & Varga, 1980)

Historical formulation: 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:

$$\begin{array}{rl}{E}_n(x)& :={ϵ}_n(x)+{ϵ}_n(-x)\\ & =f(x)+f(-x)\\ & -[p_n^{\ast }(x)+p_n^{\ast }(-x)].\end{array}$$

By construction, $E_n$ is even. Write $f_e(x)=\frac{1}{2}(f(x)+f(-x))$ for the even part of $f$. The proposed conclusion is precisely $f_e=0$.

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 $E_n$ has at least $n+1$ zeros in $[-1,1]$, counted with multiplicity at most two. Every zero counted twice in this count is an alternation point. This does not say that every zero at an alternation point must be counted twice.

Proof Sketch

By the Chebyshev alternation theorem, there are $n+2$ points ${\xi }_j^{(n)}$ with residual signs ${\sigma }_j=\lambda (-1{)}^j$, where $\lambda \in \{-1,1\}$. Since $|{ϵ}_n(-{\xi }_j^{(n)})|\le \Vert f-p_n^{\ast }{\Vert }_{\infty }$, the even error satisfies

$$\begin{array}{rl}& {\sigma }_j{E}_n({\xi }_j^{(n)})\\ & =\Vert f-p_n^{\ast }{\Vert }_{\infty }\\ & +{\sigma }_j{ϵ}_n(-{\xi }_j^{(n)})\ge 0.\end{array}$$

These weak sign inequalities give a zero in each interval between successive alternation points. If one interior node must serve both adjacent intervals without another zero available, the neighboring signs agree and differentiability makes that zero stationary, allowing it to be counted twice. This gives at least $n+1$ zeros with the stated multiplicity convention.

Lemma 2 (Lorentz Assumption Effect)

Assume $[f(x)+f(-x)]\in C^1[-1,1]$ and $p_{2k+1}^{\ast }(0;f)=0$ for every $k\ge 0$. Then $E_{2k+1}$ has at least $2k+4$ zeros in $[-1,1]$, counted with multiplicity at most two.

Proof Detail

Uniform convergence of the best approximants gives $f(0)=0$, hence $E_{2k+1}(0)=0$. Its derivative at zero also vanishes by evenness. If the approximation error is positive, the origin cannot be an alternation point: the original residual ${ϵ}_{2k+1}$ is zero there. Lemma 1 therefore counts the origin at most once, and its stationary zero supplies an additional count. The total is at least $2k+3$; pairing nonzero roots by evenness, and counting the origin twice, raises it to at least $2k+4$. If the approximation error is zero, $E_{2k+1}$ is identically zero and the conclusion is immediate.

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.

Saff–Varga analytic rigidity

Suppose $p_{2k+1}^{\ast }(0;f)=0$ for every $k\ge 0$. Define $F$ on $[0,1]$ by $F(x^2)=f(x)+f(-x)$. If $F$ extends to an entire function with

$$\begin{array}{c}\underset{r\to \infty }{\mathrm{limsup}}\frac{\log M_F(r)}{r}<\frac{\pi }{2},\\ M_F(r)=\underset{|z|=r}{\max }|F(z)|,\end{array}$$

then $f$ is odd. The case $F=0$ is immediate. In particular, the conclusion holds for polynomial $f$.

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 generalized Rolle’s theorem gives ${\beta }_{k+1}\in [0,1]$ with $R_k^{(k+1)}({\beta }_{k+1})=0$. Since $\mathrm{deg}{s}_k\le k$, this means $F^{(k+1)}({\beta }_{k+1})=0$. Also $F(0)=0$, so setting ${\beta }_0=0$ supplies a zero for every derivative order.

Set $G(z)=F((z+1)/2)$. Its exponential type is less than $\pi /4$, and each derivative has a zero in $[-1,1]$. Schoenberg’s theorem in this normalization forces $G=0$. (Schoenberg, 1936) Thus $F=0$ and $f$ is odd. The length of the zero-containing interval matters: one cannot replace $[0,1]$ by an arbitrary bounded interval while keeping the same type threshold.

🏷️ Structural Consequences

1. Exclusion of Even Functions

Remark 4 of Saff and Varga excludes nonzero even functions without an analytic hypothesis, using a lemma about differences of successive best approximants.

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, its best approximants would be even. In the variable $t=x^2$, the polynomials $s_k(t)=p_{2k+1}^{\ast }(\sqrt{t})+p_{2k+1}^{\ast }(-\sqrt{t})$ would be best degree-$k$ approximants to $F(t)=2f(\sqrt{t})$ on $[0,1]$. They all vanish at the endpoint $t=0$. A lemma of Lorentz used by Saff and Varga forces consecutive best approximants sharing this endpoint value to agree. Starting from $s_0=0$, all $s_k$ vanish, and uniform convergence gives $F=0$, a contradiction.

2. Application to Zolotareff Polynomials

The polynomial case also gives an origin constraint for Zolotareff residuals.

Zolotareff Polynomials

Let $m\ge 1$, $\sigma \ne 0$, and let $q$ be the best degree-$(2m-1)$ approximant to $x^{2m+1}+\sigma x^{2m}$. Define $Z_{2m+1}(x;\sigma )=x^{2m+1}+\sigma x^{2m}-q(x)$. Its residual alternation makes zero its best approximant at every degree at most $2m-1$. At every degree at least $2m+1$, its best approximant is itself. If $Z_{2m+1}(0;\sigma )=0$, all its odd-degree approximants would therefore vanish at zero. The analytic rigidity theorem would make it odd, contradicting its nonzero $x^{2m}$ coefficient. Hence $Z_{2m+1}(0;\sigma )\ne 0$.

🏷️ Dependence on Every Odd-Degree Condition

Remark 3 shows that no individual odd-degree condition can be dropped from the analytic rigidity theorem. For a prescribed $m\ge 0$, set

$$\begin{array}{c}{\alpha }_m=-\cos \left(\frac{\pi }{2(m+1)}\right),\\ Q_{m+1}(t)=T_{m+1}((1-{\alpha }_m)t+{\alpha }_m),\\ f_m(x)=Q_{m+1}(x^2).\end{array}$$

Then $f_m$ is a nonzero even polynomial of degree $2m+2$, with $f_m(0)=0$. Its $2m+1$ alternating extrema make zero its best approximant at degrees at most $2m-1$. At degrees at least $2m+2$, its best approximant is itself. Consequently,

$$\begin{array}{c}{p}_{2k+1}^{\ast }(0;f_m)=0\\ \text{for every }k\ne m.\end{array}$$

although $f_m$ is not odd. This is a sharpness statement about the list of degree conditions. It does not establish sharpness of the exponential-type threshold.

For comparison, the even function $\cos (\pi x/2)$ does not satisfy the common-origin hypothesis: its best degree-one approximant is the constant $1/2$, since its range is $[0,1]$ and symmetry forces that approximant to be constant. Its profile $2\cos (\pi \sqrt{z}/2)$ is entire of exponential type zero, not $\pi /2$.

See Also

• on self-similar counterexamples to Lorentz conjectures gives continuous counterexamples to the common-origin and normalized plateau implications, with the finite certificates and analytic persistence arguments formalized in LorentzLean.
• on best uniform approximation from subspaces studies a different Lorentz conjecture about optimal monomial spans. It is not Conjecture 2 in Saff and Varga’s symmetry paper, which concerns common roots away from zero.

📚 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.