# Comptes Rendus
Mathématique

J. Marshall Ash, Stefan Catoiu and Hajrudin Fejzić

A new proof of the GGR conjecture

Volume 361 (2023), p. 349-353

https://doi.org/10.5802/crmath.413

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This article is licensed under the
CREATIVE COMMONS ATTRIBUTION 4.0 INTERNATIONAL LICENSE.
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Functional analysis / Analyse fonctionnelle

# A new proof of the GGR conjecture

J. Marshall Ash<sup>a</sup>, Stefan Catoiu<sup>*</sup>,<sup>a</sup> and Hajrudin Fejzić<sup>b</sup>

<sup>a</sup> Department of Mathematics, DePaul University, Chicago, IL 60614

<sup>b</sup> Department of Mathematics, California State University, San Bernardino, CA 92407

E-mails: mash@depaul.edu (J. M. Ash), scatoiu@depaul.edu (S. Catoiu), hfejzic@csusb.edu (H. Fejzić)

**Abstract.** For each positive integer $n$, function $f$, and point $x$, the 1998 conjecture by Ginchev, Guerragio, and Rocca states that the existence of the $n$th Peano derivative $f_{(n)}(x)$ is equivalent to the existence of all $n(n+1)/2$ generalized Riemann derivatives,

$$D_{k,-j} f(x) = \lim_{h \to 0} \frac{1}{h^n} \sum_{i=0}^{k} (-1)^i \binom{k}{i} f(x + (k - i - j)h),$$

for $j, k$ with $0 \le j < k \le n$. A version of it for $n \ge 2$ replaces all $-j$ with $j$ and eliminates all $j = k - 1$. Both the GGR conjecture and its version were recently proved by the authors using non-inductive proofs based on highly non-trivial combinatorial algorithms. This article provides a second, inductive, algebraic proof to each of these theorems, based on a reduction to (Laurent) polynomials.

**Mathematical subject classification (2010).** 26A24, 13F20, 15A03, 26A27.

**Funding.** This paper is in final form and no version of it will be submitted for publication elsewhere.

Manuscript received 10 June 2021, revised 30 April 2022, accepted 24 August 2022.

Arguably, the most important application of higher order derivatives is Taylor's theorem, asserting that an $n$ times differentiable function $f$ at a point $c$ is approximated near $c$ by a polynomial $p$ with error $f(c+h) - p(c+h) = o(h^n)$ as $h \to 0$. It is also well known that Taylor's theorem provides only a sufficient condition for this approximation to happen, and all functions $f$ with this property are said to be $n$ times Peano differentiable at $c$.

In 1998, Ginchev, Guerragio, and Rocca (GGR) conjectured the following result:

**Conjecture (GGR Conjecture).** When $n \ge 2$, the following two conditions,

(i) $f$ is $n-1$ times Peano differentiable at $c$ and

(ii) $\lim_{h \to 0} \frac{1}{h^n} \sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f(c + (j-k)h)$ exists for all $k$ with $0 \le k \le n-1$, are sufficient to make $f$ an $n$ times differentiable function at $c$.

They proved the theorem by hand for $n = 2, 3, 4$ in [7], and with the use of a computer they proved it for $n = 5, 6, 7, 8$ in [8], leaving the rest as a conjecture. The GGR conjecture was recently proved in [2] and is now a theorem. The result proved in [2] is slightly stronger for $n$ odd, by eliminating the second condition for $k = 0$. A variant of the conjecture, where the bounds for $k$

<footnote>* Corresponding author.</footnote>
are replaced by $-(n-2) \le k \le 0$, is proved in [4]. Article [6] sheds some light towards extending the GGR conjecture for $n=1$.

The original statement of the GGR conjecture is actually an equivalent version of the above, by the principle of mathematical induction: If $D_{n,k}$ denote the above limits, Ginchev, Guerragio, and Rocca conjectured that the $\binom{n+1}{2}$ limits $D_{m,k}$ for $0 \le k < m \le n$ will be enough for $f$ to be $n$ times Peano differentiable at $c$. The limit $D_{n,0} = D_{n,0}f(c)$ is called the $n^{\text{th}}$ Riemann derivative of $f$ at $c$, and $D_{n,n/2}f(c)$ is the $n^{\text{th}}$ symmetric Riemann derivative of $f$ at $c$. Both of these derivatives were invented by Riemann in the mid 1800s, see [11]. The Peano derivatives were invented by Peano in [10] in 1892, and then developed greatly by de la Vallée Poussin in [12]. For this reason, they are often referred to as de la Vallée Poussin derivatives.

The purpose of this note is to provide a new, simple proof to both the GGR conjecture and its variant.

* * *

Let $R_n(h)$ be the difference defined recursively by $R_1(h) = f(c+h) - f(c)$, and $R_n(h) = R_{n-1}(2h) - 2^{n-1}R_{n-1}(h)$ for $n \ge 2$. Closed form formulas for generalizations of these differences are deduced in [3]; they involve the Gaussian or $q$-binomial coefficients, so they are $q$-analogues of the Riemann differences. Other $q$-analogues of Riemann differences are found in [1] and [5].

We will use the following 1936 result of Marcinkiewicz and Zygmund in [9].

**Theorem 1.** Suppose $f$ is $n-1$ times Peano differentiable at $c$. If $\lim_{h \to 0} R_n(h)/h^n$ exists, then $f$ is $n$ times Peano differentiable at $c$.

If we denote $\Delta_k(h)$ as the difference $\sum_{j=0}^n (-1)^{n-j} \binom{n}{j} f(c+(j-k)h)$ in the GGR limit condition, then this condition can be concisely written as $\lim_{h \to 0} \Delta_k(h)/h^n$ exists. It is also obvious that if all GGR limit conditions are met then the following linear combination

$$ \lim_{h \to 0} \frac{\sum_k c_k \Delta_k(s_k h)}{h^n} \text{ exists, where } c_k, s_k \text{ are arbitrary real constants.} $$

Hence, if we can show that $R(h)$ is a linear combination $\sum_k c_k \Delta_k(s_k h)$, then the GGR conjecture follows from Theorem 1.

That this is indeed the case will follow from the analogous result for polynomials, via the linear isomorphism, $\Delta(h) := \sum c_j f(c+b_j h) \mapsto d(t) := \sum c_j t^{b_j}$, from the $\mathbb{R}$-space of all differences of $f$ at $c$ and $h$ with integer nodes (the $b_j$), to the $\mathbb{R}$-space $\mathbb{R}[t, t^{-1}]$ of all Laurent polynomials in indeterminate $t$ with real coefficients. In this way, (1) if

$$ \Delta_k(h) = \sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f(c + (j-k)h), $$

then

$$ d_k(t) = \sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} t^{j-k} = t^{-k} (t-1)^n; $$

(2) the polynomial corresponding to $\Delta_k(sh)$ is $d_k(t^s)$; and (3) if $r_n(t)$ is the polynomial that corresponds to $R_n(h)$ under this linear isomorphism, then its recursive definition is $r_1(t) = t-1$, and $r_n(t) = r_{n-1}(t^2) - 2^{n-1}r_{n-1}(t)$ for $n \ge 2$.

Based on these properties of the above linear isomorphism, we will be done by showing that the following result is true.

**Theorem 2.** There are constants $c_k$ and $s_k$ such that $r_n(t) = \sum c_k d_k(t^{s_k})$.

Before proceeding with the proof of Theorem 2, we need to make a clarification. Our solution to the theorem has the numbers $s_k$ non-negative integers instead of real numbers, so we can think of $s_k$ as $s$, with $s \ge 0$. Then the $c_k$ are viewed as $c_{k,s}$, for a more precisely indexed sum
$\sum c_k d_k(t^{s_k}) = \sum_{s=0}^{\infty} \sum_k c_{k,s} d_k(t^s) = \sum_{s=0}^{\infty} \sum_k c_{k,s} t^{-sk} (t^s - 1)^n$, where the ranges for $k$ will be clarified later, since these are different for different cases in the proof of the theorem. Summarizing, in order to prove Theorem 2, it suffices to show that

$$r_n(t) \in V_n := \operatorname{span} \{t^{-sk} (t^s - 1)^n \mid k = (0, 1, \dots, n-1, s=1, 2, \dots)\},$$

where (0) means that the value 0 is taken only for $n$ even.

* * *

The proof of Theorem 2 is much shorter for the variant of the GGR conjecture than it is for the conjecture itself. For this reason, we deal with the variant first.

**Proof of Theorem 2 (Variant Case)**

In this case, by replacing the index of summation $k$ with $-k$, the range $-(n-2) \le k \le 0$ becomes
$0 \le k \le n-2$. In this way, $V_n$ becomes

$$V_n = \operatorname{span} \{t^{sk} (t^s - 1)^n \mid k = 0, \dots, n-2; s = 1, 2, \dots\}.$$

The following lemma provides a new set of generators for the space $V_n$.

**Lemma 3.** $V_n = \operatorname{span}\{(t^s - 1)^{n+k} \mid k = 0, 1, \dots, n-2, s = 1, 2, \dots\}$.

**Proof.** It suffices to show the following equality of subspaces:

$$\operatorname{span} \{t^k (t-1)^n \mid k = 0, 1, \dots, n-2\} = \operatorname{span} \{(t-1)^{n+k} \mid k = 0, 1, \dots, n-2\}.$$

Indeed, this is the result of multiplying by $(t-1)^n$ both sides of the obvious equation

$$\operatorname{span}\{1, t, t^2, \dots, t^{n-2}\} = \operatorname{span}\{1, t-1, (t-1)^2, \dots, (t-1)^{n-2}\}. \quad \square$$

We are now ready to proceed with the proof of Theorem 2 in its variant case.

**Proof of Theorem 2 (Variant Case).** Induct on $n$. When $n=2$, $r_2 = (t-1)^2$ is clearly in $V_2$. Suppose $r_n \in V_n$, for some $n, n \ge 2$, and prove the same property for $n+1$. By Lemma 3, $r_n$ is a linear combination of polynomials of the form $(t^s - 1)^{n+k}$, where $s$ is a positive integer and $k=0, 1, \dots, n-2$. By the recursion, $r_{n+1}(t) = r_n(t^2) - 2^n r_n(t)$ will be a linear combination of polynomials

$$(t^{2s}-1)^{n+k}-2^n(t^s-1)^{n+k}, \text{ for various } k \text{ and } s.$$

By Lemma 3, these polynomials belong to $V_{n+1}$ in all cases, except for $k=0$, when

$$(t^{2s}-1)^n-2^n(t^s-1)^n = (t^s-1)^n((t^s+1)^n-2^n) = (t^s-1)^{n+1}p(t^s),$$

where $p$ is a polynomial in $t$ of degree $n-1$, so that $(t^s - 1)^{n+1}p(t^s)$ belongs to the subspace $\operatorname{span}\{(t^s - 1)^{n+1}, (t^s - 1)^{n+2}, \dots, (t^s - 1)^{2n}\}$ of $V_{n+1}$. $\square$

**Proof of Theorem 2 (GGR Case)**

The GGR case in the proof of Theorem 2 is similar to the variant case. The proof of the inductive step is now split further into two subcases, $n$ odd and $n$ even. In both cases, following the refined result of the GGR Theorem from [2],

$$V_n = \operatorname{span} \{t^{-sk} (t^s - 1)^n \mid k = (0), 1, \dots, n-1; s = 1, 2, \dots\},$$

where (0) means that 0 is taken for $n$ even, and not taken for $n$ odd. More explicitly, this is

$$V_n = \begin{cases} \operatorname{span}\{(t-1)^n, t^{-1}(t-1)^n, \dots, t^{-(n-1)}(t-1)^n, \dots\}, & \text{$n$ even,} \\ \operatorname{span}\{t^{-1}(t-1)^n, t^{-2}(t-1)^n, \dots, t^{-(n-1)}(t-1)^n, \dots\}, & \text{$n$ odd,} \end{cases}$$
where the last dots in both cases mean that the generating set also includes the previously listed generators evaluated at $t^s$, for all $s$ at least 2. Let $W_n$ be the subspace of $V_n$ spanned by all generators with $s=1$. Then $W_n$ has the expression

$$W_n = \begin{cases} \operatorname{span}\{(t-1)^n, t^{-1}(t-1)^n, \dots, t^{-(n-1)}(t-1)^n\}, & n \text{ even,} \\ \operatorname{span}\{t^{-1}(t-1)^n, t^{-2}(t-1)^n, \dots, t^{-(n-1)}(t-1)^n\}, & n \text{ odd.} \end{cases}$$

The following lemma provides new sets of generators for $W_n$ in both parity cases.

**Lemma 4.** *With the above notation,*

$$W_n = \begin{cases} \operatorname{span}\{t^{-n/2}(t-1)^n, t^{-1}(t-1)^{n+1}, \dots, t^{-(n-1)}(t-1)^{n+1}\}, & n \text{ even,} \\ \operatorname{span}\{t^{-(n-1)/2}(t-1)^n, t^{-2}(t-1)^{n+1}, \dots, t^{-(n-1)}(t-1)^{n+1}\}, & n \text{ odd.} \end{cases}$$

**Proof.** When $n$ is even, the result follows from $t^{-n/2}(t-1)^n$ being one of the generators in the definition of $W_n$, and $t^{-k}(t-1)^{n+1} = t^{-(k-1)}(t-1)^n - t^{-k}(t-1)^n$, for each $k=1, \dots, n-1$. The case when $n$ is odd has a similar proof. $\square$

We are now ready to prove Theorem 2 in the GGR case.

**Proof of Theorem 2 (Case GGR).** Induct on $n$. When $n=1$, $r_1 = t-1 \in V_1$. We assume that $r_n \in V_n$ and prove that $r_{n+1} \in V_{n+1}$ in two possible cases:

**Case 1.** When $n$ is even,

$$V_{n+1} = \operatorname{span}\{t^{-1}(t-1)^{n+1}, t^{-2}(t-1)^{n+1}, \dots, t^{-n}(t-1)^{n+1}, \dots\}.$$

The inductive hypothesis and Lemma 4 imply that $r_n(x)$ is a linear combination of

$t^{-n/2}(t-1)^n$ and $t^{-k}(t-1)^{n+1}$, for $k=1, \dots, n-1$,

and their evaluations at $t^s$, for $s$ at least 2. Then $r_{n+1}(t) = r_n(t^2) - 2^n r_n(t)$ is a linear combination of two kinds of polynomials and their evaluations at $t^s$, for $s \ge 2$. The first kind of polynomial has the form $t^{-n}(t^2-1)^n - 2^n t^{-n/2}(t-1)^n$

$$= t^{-n}(t-1)^n((t+1)^n - 2^n t^{n/2}) = t^{-n}(t-1)^{n+1}p(t),$$

where $p(t)$ is a polynomial degree $n-1$, hence the whole expression lives inside of

$$(t-1)^{n+1} \operatorname{span}\{t^{-1}, t^{-2}, \dots, t^{-n}\},$$

a subspace of $V_{n+1}$. The polynomials of the second kind are polynomials of the form

$$t^{-2k}(t^2-1)^{n+1} - 2^n t^{-k}(t-1)^{n+1}, \text{ for } k=1, \dots, n-1.$$

Their second term is a scalar multiple of a generator of $V_{n+1}$, while their first term is an $(s=2)$-dilation of the same generator, so all polynomials of the second kind also belong to $V_{n+1}$. We conclude that $r_{n+1} \in V_{n+1}$, as needed.

**Case 2.** When $n$ is odd,

$$V_{n+1} = \operatorname{span}\{(t-1)^{n+1}, t^{-1}(t-1)^{n+1}, \dots, t^{-n}(t-1)^{n+1}, \dots\}.$$

The inductive hypothesis and Lemma 4 imply that $r_n(x)$ is a linear combination of

$t^{-(n-1)/2}(t-1)^n$ and $t^{-k}(t-1)^{n+1}$, for $k=2, \dots, n-1$,

and their evaluations at $t^s$, for $s$ at least 2. Then $r_{n+1}(t) = r_n(t^2) - 2^n r_n(t)$ is a linear combination of two kinds of polynomials and their evaluations at $t^s$, for $s \ge 2$. The first kind of polynomial is of the form $t^{-(n-1)}(t^2-1)^n - 2^n t^{-(n-1)/2}(t-1)^n$

$$= t^{-(n-1)}(t-1)^n((t+1)^n - 2^n t^{(n-1)/2}) = t^{-(n-1)}(t-1)^{n+1}p(t),$$

where $p(t)$ is a polynomial degree $n-1$, hence the above expression lives inside of

$$(t-1)^{n+1} \operatorname{span}\{1, t^{-1}, t^{-2}, \dots, t^{-(n-1)}\},$$
a subspace of $V_{n+1}$. The polynomials of the second kind are polynomials of the form

$$t^{-2k} (t^2 - 1)^{n+1} - 2^n t^{-k} (t-1)^{n+1}, \text{ for } k=2, \dots, n-1.$$

Their second term is a scalar multiple of a generator of $V_{n+1}$, while their first term is an $(s=2)$-dilation of the same generator, so all polynomials of the second kind also belong to $V_{n+1}$. We conclude that $r_{n+1} \in V_{n+1}$ in this case as well. $\square$

## References

[1] J. M. Ash, S. Catoiu, "Quantum symmetric $L^p$ derivatives", *Trans. Am. Math. Soc.* **360** (2008), no. 2, p. 959-987.

[2] J. M. Ash, S. Catoiu, "Characterizing Peano and symmetric derivatives and the GGR conjecture's solution", *Int. Math. Res. Not.*, IMRN (2022), no. 10, p. 7893-7921.

[3] J. M. Ash, S. Catoiu, H. Fejzić, "Gaussian Riemann derivatives", https://arxiv.org/abs/2211.09209, to appear in Israel J. Math., 23 pp., online first, 2022.

[4] ——, "Two pointwise characterizations of the Peano derivative", https://arxiv.org/abs/2209.04088, preprint, 2022.

[5] J. M. Ash, S. Catoiu, R. Ríos-Collantes-de Terán, "On the $n^{th}$ quantum derivative", *J. Lond. Math. Soc.* **66** (2002), no. 1, p. 114-130.

[6] S. Catoiu, "A differentiability criterion for continuous functions", *Monatsh. Math.* **197** (2022), no. 2, p. 285-291.

[7] I. Ginchev, A. Guerraggio, M. Rocca, "Equivalence of Peano and Riemann derivatives", in *Generalized convexity and optimization for economic and financial decisions (Verona, 1998)*, Pitagora Editrice, Bologna, 1998, p. 169-178.

[8] I. Ginchev, M. Rocca, "On Peano and Riemann derivatives", *Rend. Circ. Mat. Palermo* **49** (2000), no. 3, p. 463-480.

[9] J. Marcinkiewicz, A. Zygmund, "On the differentiability of functions and summability of trigonometric series", *Fundam. Math.* **26** (1936), p. 1-43.

[10] G. Peano, "Sulla formula di Taylor", *Atti Acad. Sci. Torino* **27** (1892), p. 40-46.

[11] B. Riemann, *Über die Darstellung einer Funktion durch eine trigonometrische Reihe*, Ges. Werke, 2. Aufl., Leipzig, Dieterichschen Buchhandlung, 1892 (1867), 227-271 pages.

[12] C.-J. de la Vallée Poussin, "Sur l'approximation des fonctions d'une variable réelle et de leurs dérivées par les pôlynomes et les suites limitées de Fourier", *Belg. Bull. Sciences* **1908** (1908), p. 193-254.