Refined conjecture for the Minkowski question mark function convolution

Let ?(x)?(x) denote the Minkowski question mark function, extended by ?(x)=0?(x)=0 for x0x\leq 0 and ?(x)=1?(x)=1 for x1x\geq 1. For β[0,2]{1}\beta\in[0,2]\setminus\{1\} and NNN\in\mathbb{N}, define the corresponding discrete convolution sum by

i=0N1[?(i+1N)?(iN)][?(βiN)?(βi+1N)].\sum_{i=0}^{N-1}\left[?\left(\frac{i+1}{N}\right)-?\left(\frac{i}{N}\right)\right]\left[?\left(\beta-\frac{i}{N}\right)-?\left(\beta-\frac{i+1}{N}\right)\right].

Refined conjecture. For every β[0,2]{1}\beta\in[0,2]\setminus\{1\}, there exists a constant C(β)0C(\beta)\geq 0 such that

i=0N1[?(i+1N)?(iN)][?(βiN)?(βi+1N)]C(β)N,N.\sum_{i=0}^{N-1}\left[?\left(\frac{i+1}{N}\right)-?\left(\frac{i}{N}\right)\right]\left[?\left(\beta-\frac{i}{N}\right)-?\left(\beta-\frac{i+1}{N}\right)\right]\sim\frac{C(\beta)}{N},\qquad N\to\infty.

Moreover, C(β)C(\beta) is discontinuous only at β=1\beta=1, where both one-sided limits are infinite, and this singularity is integrable. This conjecture refines the analysis needed to establish decay at infinity of the Fourier transform associated with the Minkowski measure, with the case k=2k=2 corresponding to the convolution of two copies of the formal derivative of the question mark function. The asymptotic estimate and the asserted regularity of C(β)C(\beta) remain unproved in the supplied text.

Sources & referencesView supporting material

Primary source

Giedrius Alkauskas, “Fourier-Stieltjes coefficients of the Minkowski question mark function”, arXiv:1008.4014 (2012).

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