The paper "On the constant in a transference inequality for the vector-valued Fourier transform" revisited

TL;DR

The paper provides a streamlined proof that, for every $r\geq 1$, the finite-sum function $f_r(x)=\sum_{m\in\mathbb{Z}} \left|\frac{\sin(\pi(x+m))}{\pi(x+m)}\right|^{2r}$ on $[0,1]$ attains a global minimum at $x=\tfrac{1}{2}$. Building on the $r=1$ identity $\sum_{m\in\mathbb{Z}} h(x+m)=1$ and Parseval via the Fourier kernel $g_x(t)=e^{2\pi i x t}$, the authors relate $f_r$ to convex sums of $s_m(x)=h(x+m)+h(x-(m+1))$. A central one-crossing lemma yields convex-dominance under mass transfers, with the crucial fact that $s_0(x)$ minimizes at $x=\tfrac{1}{2}$ and $s_m(x)$ ($m\geq 1$) maximizes there, enabling the comparison $\sum_{m\ge0} x_m^r \geq \sum_{m\ge0} y_m^r$ and thus $f_r(x)\geq f_r(\tfrac{1}{2})$. The result simplifies a known argument, providing a shorter, more direct route to the minimum-location claim and reinforcing the connection between transference-type constants and their one-dimensional minimizers.

Abstract

The standard proof of the equivalence of Fourier type on \(\mathbb R^d\) and on the torus \(\mathbb T^d\) is usually stated in terms of an implicit constant which can be expressed in terms of the global minimiser of the functions \[f_r(x)=\sum_{m\in\mathbb{Z}}\left|\frac{\sin(π(x+m))}{π(x+m)}\right|^{2r},\qquad x\in [0,1], \ r\ge 1.\] The aim of this note is to provide a short proof of a result of the authors which states that each \(f_r\) takes a global minimum at the point \(x = \frac12\).

Figures (1)

• Figure 1: A plot of $f_r$, where $r=1.02^{k}$ for $k=1,2,4,\dots,256$.

Theorems & Definitions (9)

• Proposition 1
• Lemma 1.1
• proof
• Lemma 1.2: One-crossing implies convex dominance
• proof
• Lemma 1.3
• Lemma 1.4
• proof
• proof : Proof of Proposition \ref{['prop']}