Subsequences of frames (Q2717572)

A sequence \((x_j)\) in a separable Hilbert space \(H\) is called a frame if there exist \(B\geq A>0\) such that \(A\|x\|^2\sum_j|\langle x, x_j\rangle|^2\leq B\|x\|^2\) for \(\forall x\in H\). \((B/A)^{1/2}\) is a constant of the frame. \((x_j)\) is tight if \(A= B=1\). \((x_j)\) and \((y_j)\) are equivalent if there exists an isomorphism \(T: [x_j]\to [y_j]\) such that \(Tx_j= y_j\) for \(\forall j\), where \([x_j]\): closed linear span of \((x_j)\). When \(c= \|T\|\cdot\|T^{-1}\|\), \((x_j)\) and \((y_j)\) are \(c\)-equivalent. The author proves the following Theorem:NEWLINENEWLINENEWLINE(1) Let \((x_j)\) be an \(n\)-dimensional frame with constant \(c\). We can find the function \(h: \mathbb{R}_+\to\mathbb{R}_+\) satisfying the following property: For \(\forall \varepsilon> 0\) there is a set \(\sigma\) of indices with the size \(|\sigma|> (1-\varepsilon)n\) such that \((x_j)_{j\in\sigma}\) is \(C\)-equivalent to an orthogonal basis, where \(C= h(\varepsilon)c\).NEWLINENEWLINENEWLINE(2) Given an \(\varepsilon> 0\), every \(\infty\)-dimensional frame has a subsequence \((1-\varepsilon)\)-equivalent to an orthogonal basis of \(\ell_2\).NEWLINENEWLINENEWLINE(3) There exists a tight frame in \(\ell_2\) which does not contain basis with brackets.