• Honam Mathematical Journal
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• v.36 no.1
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• pp.29-32
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• 2014

Given operators X and Y acting on a separable Hilbert space $\mathcal{H}$, an interpolating operator is a bounded operator A such that AX = Y. In this article, we investigate self-adjoint interpolation problems for operators in a tridiagonal algebra : Let $\mathcal{L}$ be a subspace lattice acting on a separable complex Hilbert space $\mathcal{H}$ and let X = ($x_{ij}$) and Y = ($y_{ij}$) be operators acting on $\mathcal{H}$. Then the following are equivalent: (1) There exists a self-adjoint operator A = ($a_{ij}$) in $Alg{\mathcal{L}}$ such that AX = Y. (2) There is a bounded real sequence {${\alpha}_n$} such that $y_{ij}={\alpha}_ix_{ij}$ for $i,j{\in}\mathbb{N}$.

• Proceedings of the Korean Society of Computational and Applied Mathematics Conference
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• 2003.09a
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• pp.4.1-4
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• 2003

Given operators X and Y acting on a Hilbert space H, an interpolating operator is a bounded operator A such that AX= Y. An interpolating operator for n-operators satisfies the equation AXi= Yi, for i = 1,2,...,n, In this article, we showed the following : Let H be a Hilbert space and let L be a subspace lattice on H. Let X and Y be operators acting on H. Assume that rangeX is dense in H. Then the following statements are equivalent : (1) There exists an operator A in AlgL such that AX = Y, A$\^$*/=A and every E in L reduces A. (2) sup｛(equation omitted) : n $\in$ N f$\sub$I/ $\in$ H and E$\sub$I/ $\in$ L｝<$\infty$ and = for all E in L and all f, g in H.

• Honam Mathematical Journal
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• v.30 no.4
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• pp.685-691
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• 2008

Given operators $X_i$ and $Y_i$ (i = 1, 2, ${\cdots}$, n) acting on a Hilbert space $\mathcal{H}$, an interpolating operator is a bounded operator A acting on $\mathcal{H}$ such that $AX_i$ = $Y_i$ for i= 1, 2, ${\cdots}$, n. In this article, if the range of $X_k$ is dense in H for a certain k in {1, 2, ${\cdots}$, n), then the following are equivalent: (1) There exists a self-adjoint operator A in $\mathcal{B}(\mathcal{H})$ stich that $AX_i$ = $Y_i$ for I = 1, 2, ${\cdots}$, n. (2) $sup\{{\frac{{\parallel}{\sum}^n_{i=1}Y_if_i{\parallel}}{{\parallel}{\sum}^n_{i=1}X_if_i{\parallel}}:f_i{\in}H}\}$ < ${\infty}$ and < $X_kf,Y_kg$ >=< $Y_kf,X_kg$> for all f, g in $\mathcal{H}$.

• Journal of applied mathematics & informatics
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• v.36 no.3_4
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• pp.237-244
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• 2018

Let ${\mathcal{H}}$ be an infinite dimensional separable Hilbert space with a fixed orthonormal base $\{e_1,e_2,{\cdots}\}$. Let L be the subspace lattice generated by the subspaces $\{[e_1],[e_1,e_2],[e_1,e_2,e_3],{\cdots}\}$ and let $Alg{\mathcal{L}}$ be the algebra of bounded operators which leave invariant all projections in ${\mathcal{L}}$. Let p and q be natural numbers (p < q). Let ${\mathcal{A}}$ be a linear manifold in $Alg{\mathcal{L}}$ such that $T_{(p,q)}=0$ for all T in ${\mathcal{A}}$. If ${\mathcal{A}}$ is a Lie ideal, then $T_{(p,p)}=T_{(p+1,p+1)}={\cdots}=T_{(q,q)}$ and $T_{(i,j)}=0$, $p{\eqslantless}i{\eqslantless}q$ and i < $j{\eqslantless}q$ for all T in ${\mathcal{A}}$.

• Journal of applied mathematics & informatics
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• v.19 no.1_2
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• pp.447-452
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• 2005

Given operators X and Y on a Hilbert space H, an interpolating operator is a bounded operator A such that AX = Y. In this article, we investigate compact interpolation problems for vectors in a tridiagonal algebra. Let L be a subspace lattice acting on a separable complex Hilbert space H and Alg L be a tridiagonal algebra. Let X = $(x_{ij})\;and\;Y\;=\;(y_{ij})$ be operators acting on H. Then the following are equivalent: (1) There exists a compact operator A = $(x_{ij})$ in AlgL such that AX = Y. (2) There is a sequence {$\alpha_n$} in $\mathbb{C}$ such that {$\alpha_n$} converges to zero and $$y_1\;_j=\alpha_1x_1\;_j+\alpha_2x_2\;_j\;y_{2k}\;_j=\alpha_{4k-1}x_{2k\;j}\;y_{2k+1\;j}=\alpha_{4k}x_{2k\;j}+\alpha_{4k+1}x_{2k+1\;j}+\alpha_{4k+2}x_{2k+2\;j\;for\;all\;k\;\epsilon\;\mathbb{N}$$.

• Communications of the Korean Mathematical Society
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• v.17 no.3
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• pp.487-493
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• 2002

Given operators X and Y acting on a Hilbert space H, an interpolating operator is a bounded operator A such that AX = Y. An interpolating operator for the n-operators satisfies the equation AX$\_$i/ : Y$\_$i/, for i = 1, 2 …, n. In this article, we obtained the following : Let X = (x$\_$ij/) and Y = (y$\_$ij/) be operators acting on H such that $\varkappa$$\_$ i$\sigma$ (i)/ 0 for all i. Then the following statements are equivalent. (1) There exists a unitary operator A in Alg(equation omitted) such that AX = Y and every E in (equation omitted) reduces A. (2) sup{(equation omitted)}<$\infty$ and (equation omitted) = 1 for all i = 1, 2, ….

• The Pure and Applied Mathematics
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• v.21 no.2
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• pp.113-120
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• 2014

In this paper, we first show that for any space X, there is a ${\sigma}$-complete Boolean subalgebra of $\mathcal{R}$(X) and that the subspace {${\alpha}{\mid}{\alpha}$ is a fixed ${\sigma}Z(X)^{\sharp}$-ultrafilter} of the Stone-space $S(Z({\Lambda}_X)^{\sharp})$ is the minimal basically disconnected cover of X. Using this, we will show that for any countably locally weakly Lindel$\ddot{o}$f space X, the set {$M{\mid}M$ is a ${\sigma}$-complete Boolean subalgebra of $\mathcal{R}$(X) containing $Z(X)^{\sharp}$ and $s_M^{-1}(X)$ is basically disconnected}, when partially ordered by inclusion, becomes a complete lattice.

• Honam Mathematical Journal
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• v.39 no.1
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• pp.93-100
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• 2017

Let $\mathcal{H}$ be an infinite dimensional separable Hilbert space with a fixed orthonormal base $\{e_1,e_2,{\cdots}\}$. Let $\mathcal{L}$ be the subspace lattice generated by the subspaces $\{[e_1],[e_1,e_2],[e_1,e_2,e_3],{\cdots}\}$ and let $Alg{\mathcal{L}}$ be the algebra of bounded operators which leave invariant all projections in $\mathcal{L}$. Let p and q be natural numbers($p{\leqslant}q$). Let $\mathcal{B}_{p,q}=\{T{\in}Alg\mathcal{L}{\mid}T_{(p,q)}=0\}$. Let $\mathcal{A}$ be a linear manifold in $Alg{\mathcal{L}}$ such that $\{0\}{\varsubsetneq}{\mathcal{A}}{\subset}{\mathcal{B}}_{p,q}$. If $\mathcal{A}$ is an ideal in $Alg{\mathcal{L}}$, then $T_{(i,j)}=0$, $p{\leqslant}i{\leqslant}q$ and $i{\leqslant}j{\leqslant}q$ for all T in $\mathcal{A}$.

• Honam Mathematical Journal
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• v.27 no.2
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• pp.243-250
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• 2005

Given operators X and Y acting on a separable Hilbert space ${\mathcal{H}}$, an interpolating operator is a bounded operator A such that AX = Y. We show the following: Let ${\mathcal{L}}$ be a subspace lattice acting on a separable complex Hilbert space ${\mathcal{H}}$. and let $X=(x_{ij})$ and $Y=(y_{ij})$ be operators acting on ${\mathcal{H}}$. Then the following are equivalent: (1) There exists an invertible operator $A=(a_{ij})$ in $Alg{\mathcal{L}}$ such that AX = Y. (2) There exist bounded sequences {${\alpha}_n$} and {${\beta}_n$} in ${\mathbb{C}}$ such that $${\alpha}_{2k-1}{\neq}0,\;{\beta}_{2k-1}=\frac{1}{{\alpha}_{2k-1}},\;{\beta}_{2k}=-\frac{{\alpha}_{2k}}{{\alpha}_{2k-1}{\alpha}_{2k+1}}$$ and $$y_{i1}={\alpha}_1x_{i1}+{\alpha}_2x_{i2}$$ $$y_{i\;2k}={\alpha}_{4k-1}x_{i\;2k}$$ $$y_{i\;2k+1}={\alpha}_{4k}x_{i\;2k}+{\alpha}_{4k+1}x_{i\;2k+1}+{\alpha}_{4k+2}x_{i\;2k+2}$$ for $$k{\in}N$$.

• The Pure and Applied Mathematics
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• v.11 no.2
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• pp.167-173
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• 2004

Given vectors x and y in a separable Hilbert space $\cal H$, an interpolating operator is a bounded operator A such that Ax = y. In this article, we investigate Hilbert-Schmidt interpolation problems for vectors in a tridiagonal algebra. We show the following: Let $\cal L$ be a subspace lattice acting on a separable complex Hilbert space $\cal H$ and let x = ($x_{i}$) and y = ($y_{i}$) be vectors in $\cal H$. Then the following are equivalent; (1) There exists a Hilbert-Schmidt operator A = ($a_{ij}$ in Alg$\cal L$ such that Ax = y. (2) There is a bounded sequence {$a_n$ in C such that ${\sum^{\infty}}_{n=1}\mid\alpha_n\mid^2 < \infty$ and $y_1 = \alpha_1x_1 + \alpha_2x_2$ ... $y_{2k} =\alpha_{4k-1}x_{2k}$ $y_{2k=1} = \alpha_{4kx2k} + \alpha_{4k+1}x_{2k+1} + \alpha_{4k+1}x_{2k+2}$ for K $\epsilon$ N.