Characterizing Peano and Symmetric Derivatives and the GGR Conjecture’s Solution

Abstract

We provide three characterizations of the $n$th symmetric (Peano) derivative $f_{(n)}^{s}(x)$ in terms of symmetric generalized Riemann derivatives of a function $f$ at $x$ and a characterization of the $n$th Peano derivative $f_{(n)}(x)$ in terms of generalized Riemann derivatives of $f$ at $x$. The latter has been a conjecture by Ginchev, Guerragio, and Rocca since 1998.

A generalization of the GGR conjecture

“…, n − 1. The theorem has been recently proved in [5] and has been a conjecture by Ghinchev, Guerragio, and Rocca since 1998. We provide a new proof of this theorem, based on a generalization of it that produces numerous new sets of n-th Riemann smoothness conditions that can play the role of the above set in the GGR Theorem.…”

“…The proof of the GGR Theorem given in [5] has a part based on the theory of symmetric Peano and symmetric generalized Riemann derivatives, developed in that article, and a part based on a highly non-trivial combinatorial algorithm. The proof of the Generalized GGR Theorem, Theorem 0.1, is based entirely on analysis, by extending the notion of a generalized Riemann differentiation to the notion of a generalized Riemann smoothness.…”

“…The original GGR Conjecture was proved by Ginchev, Guerragio, and Rocca by hand for n = 1, 2, 3, 4 in [22] and with the help of a computer they proved it for n = 5, 6, 7, 8 in [24]. The conjecture has been recently proved for general n in [5] and is now a theorem.…”

“…The generalized Riemann derivatives were introduced by Denjoy in [14] in 1935, generalizing the Riemann derivatives R n f (c) = lim h→0 ∆ n (h)(f )/h n and the symmetric Riemann derivatives R s n f (c) = lim h→0 ∆ s n (h)(f )/h n , invented by Riemann in the mid 1800s; see [34]. Smoothness of order n was introduced by Marcinkiewicz and Zygmund in [29] in 1936, and has been recently investigated in [5].…”

“…. , n−1, was denoted in [5] as ∆ n,k and called the k-th backward shift of the n-th Riemann difference ∆ n = ∆ n,0 . In this way, the n-th symmetric Riemann difference is ∆ s n = ∆ n, n 2 .…”