A note on finite dual frame pairs (Q2846746)

Let \(\mathbb{K}^d\) denote either one of the finite dimensional spaces \(\mathbb{R}^d\) or \(\mathbb{C}^d\). A sequence \(\{e_n\}_{n=1}^N\subset \mathbb{K}^d\) is said to be a frame for \(\mathbb{K}^d\), with frame bounds \(A,B>0\), provided that NEWLINE\[NEWLINEA\|x\|^2\leq \sum_{n=1}^N|\langle x,e_n\rangle|^2\leq B\|x\|^2,\qquad x\in \mathbb{K}^d.NEWLINE\]NEWLINE The frame \(\{e_n\}_{n=1}^N\) is said to be tight if one can take \(A=B\) in the above inequality.NEWLINENEWLINEFor any given frame \(\{e_n\}_{n=1}^N\), it is always possible to find at least one sequence \(\{f_n\}_{n=1}^N\subset \mathbb{K}^d\) such that NEWLINE\[NEWLINEx=\sum_{n=1}^N\langle x,e_n\rangle f_n=\sum_{n=1}^N\langle x,f_n\rangle e_n,\qquad x\in \mathbb{K}^d.NEWLINE\]NEWLINE This \(\{f_n\}_{n=1}^N\) is also a frame and the sequences \(\{e_n\}_{n=1}^N\) and \(\{f_n\}_{n=1}^N\) are called dual frames. A tight frame \(\{e_n\}_{n=1}^N\) has \(\{\frac{1}{A}e_n\}_{n=1}^N\) for (its canonical) dual frame. The following important result characterizes the possible norms of elements of a tight frame:NEWLINENEWLINEGiven a sequence \(\{a_n\}_{n=1}^N\subset [0,\infty)\) with \(N\geq d\), (1) there exists a tight frame \(\{e_n\}_{n=1}^N\) for \(\mathbb{K}^d\) such that \(a_n=\langle e_n,e_n\rangle\), \(n=1,\ldots,N\), if and only if (2) \(\max_{1\leq n\leq N}a_n\leq \frac{1}{d}\sum_{n=1}^Na_n\).NEWLINENEWLINEInequality (2) is known as the fundamental inequality of tight frames.NEWLINENEWLINEThe paper extends this result to a pair of dual frames \(\{e_n\}_{n=1}^N\) and \(\{f_n\}_{n=1}^N\) by characterizing the possible values of \(\langle e_n,f_n\rangle\). This is Theorem 3.1 of the paper, which states that given \(\{\alpha_n\}_{n=1}^N\subset\mathbb{K}\) with \(N>d\), the following statements are equivalent: {\parindent=0.6cm \begin{itemize}\item[(1)] There exist dual frames \(\{e_n\}_{n=1}^N\) and \(\{f_n\}_{n=1}^N\) for \(\mathbb{K}^d\) such that \(\alpha_n=\langle e_n,f_n\rangle\), \(1\leq n\leq N\). \item[(2)] There exists a tight frame \(\{g_n\}_{n=1}^N\) and a corresponding dual frame \(\{h_n\}_{n=1}^N\) for \(\mathbb{K}^d\) such that \(\alpha_n=\langle g_n,h_n\rangle\), \(1\leq n\leq N\). \item[(3)] \(d=\sum_{n=1}^N\alpha_n\). NEWLINENEWLINE\end{itemize}} The assumption \(N>d\) is due to the fact that for \(\{e_n\}_{n=1}^N\) to be a frame for \(\mathbb{K}^d\) it is necessary that \(N\geq d\), and if \(N=d\), any frame \(\{e_n\}_{n=1}^N\) is a basis and has a unique dual frame, namely, \(\{\|e_n\|^{-2}e_n\}_{n=1}^N\), in which case \(\alpha_n=1\) for all \(n\).NEWLINENEWLINEThe paper then studies to a good extent the frames \(\{e_n\}_{n=1}^N\) for \(\mathbb{K}^d\) having (what the authors name) the full range property, that is, those frames \(\{e_n\}_{n=1}^N\) such that for every \(\{\alpha_n\}_{n=1}^N\subset\mathbb{K}\) with \(\sum_{n=1}^N\alpha_n=d\), one can find a dual frame \(\{f_n\}_{n=1}^N\) satisfying that \(\alpha_n=\langle e_n,f_n\rangle\) for \(1\leq n\leq N\).