论文标题
Gabor功能乘数在较高维度
Gabor Functional Multiplier in the Higher Dimensions
论文作者
论文摘要
对于两个给定的全额定晶格$ \ mathcal {l} = a \ Mathbb {z}^d $和$ \ nathcal {k} = b \ Mathbb {z}^d $ in $ \ mathbf {r}^d $ $ g（\ bf {t}）\在l^2（\ mathbf {r}^d）$中，称为parseval gabor frage Generator，如果$ \ sum _ {\ bf {\ bf {l}，\ bf {k} b \ bf {k}，\ bf {t} \ rangle}} g（\ bf {t} -a \ bf {l}）\ rangle |^2 = \ | f \ |^2 $持有任何$ f（\ bf {t}）\ in l^2（\ n l^2（\ math）众所周知，当且仅当$ | \ det（ab）| \ le 1 $时，Parseval Gabor框架发生器存在。 l^{\ infty}（\ mathbf {r}^d）$ in函数$ h \，如果具有$ hg $是$ l^2（\ mathbf {r}^d）的parseval gabor框架的属性，则称为功能性gabor帧乘数。可以猜想，且仅当函数的条件$ h \ in l^{\ infty}（\ mathbf {r}^d）$作为功能性GABOR帧乘数是$ H $必须是un -imodular和$ h（\ bf {x}）\ edrolline { -a \ bf {l}）\ + edline { \ forall \ \ bf {x} \ in \ Mathbf {r}^d $ {\ em a.e.}对于任何$ \ bf {l}，\ bf {k} \ in \ bf {k} \ in \ mathbb {z}^d $，$ \ bf {k}^d $，$ \ bf {k} {k} {k} {k} {k} \ not该猜想的一部分是正确的，可以很容易地证明，但是，只有在此日期仅在一维情况下证明了猜想的一部分。在本文中，我们证明了猜想的唯一一部分在两个维度的情况下。
For two given full-rank lattices $\mathcal{L}=A\mathbb{Z}^d$ and $\mathcal{K}=B\mathbb{Z}^d$ in $\mathbf{R}^d$, where $A$ and $B$ are nonsingular real $d\times d$ matrices, a function $g(\bf{t})\in L^2(\mathbf{R}^d)$ is called a Parseval Gabor frame generator if $\sum_{\bf{l},\bf{k}\in\mathbb{Z}^d}|\langle f, {e^{2πi\langle B\bf{k},\bf{t}\rangle}}g(\bf{t}-A\bf{l})\rangle|^2=\|f\|^2$ holds for any $f(\bf{t})\in L^2(\mathbf{R}^d)$. It is known that Parseval Gabor frame generators exist if and only if $|\det(AB)|\le 1$. A function $h\in L^{\infty}(\mathbf{R}^d)$ is called a functional Gabor frame multiplier if it has the property that $hg$ is a Parseval Gabor frame generator for $L^2(\mathbf{R}^d)$ whenever $g$ is. It is conjectured that an if and only if condition for a function $h\in L^{\infty}(\mathbf{R}^d)$ to be a functional Gabor frame multiplier is that $h$ must be unimodular and $h(\bf{x})\overline{h(\bf{x}-(B^T)^{-1}\bf{k})}=h(\bf{x}-A\bf{l})\overline{h(\bf{x}-A\bf{l}-(B^T)^{-1}\bf{k})},\ \forall\ \bf{x}\in \mathbf{R}^d$ {\em a.e.} for any $\bf{l},\bf{k}\in \mathbb{Z}^d$, $\bf{k}\not=\bf{0}$. The if part of this conjecture is true and can be proven easily, however the only if part of the conjecture has only been proven in the one dimensional case to this date. In this paper we prove that the only if part of the conjecture holds in the two dimensional case.
