• Bulletin of the Korean Mathematical Society
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• v.41 no.2
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• pp.319-326
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• 2004

In this article, we investigate invertible interpolation problems in Alg(equation omitted) : Let(equation omitted) be a subspace lattice on a Hilbert space H and let X and Y be operators acting on H. When does there exist an invertible operator A in Alg(equation omitted) such that AX = Y?

• Journal of applied mathematics & informatics
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• v.12 no.1_2
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• pp.359-365
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• 2003

Given vectors x and y in a Hilbert space, an interpolating operator is a bounded operator T such that Tx = y. In this article, we investigate invertible interpolation problems in CSL-Algebra AlgL : Let L be a commutative subspace lattice on a Hilbert space H and x and y be vectors in H. When does there exist an invertible operator A in AlgL suth that An = ㅛ?

• The Pure and Applied Mathematics
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• v.14 no.3
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• pp.161-166
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• 2007

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 $AX_i=Y_i$, for i = 1,2,...,n. In this article, we showed the following: Let L, be a subspace lattice on a Hilbert space H and let X and Y be operators in B(H). Then the following are equivalent: (1) $$sup\{\frac{{\parallel}E^{\bot}Yf{\parallel}}{{\overline}{\parallel}E^{\bot}Xf{\parallel}}\;:\;f{\epsilon}H,\;E{\epsilon}L}\}\;<\;{\infty},\;sup\{\frac{{\parallel}Xf{\parallel}}{{\overline}{\parallel}Yf{\parallel}}\;:\;f{\epsilon}H\}\;<\;{\infty}$$ and $\bar{range\;X}=H=\bar{range\;Y}$. (2) There exists an invertible operator A in AlgL such that AX=Y.

• 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 $$k{\in}N$$.

• Honam Mathematical Journal
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• v.33 no.1
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• pp.115-120
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• 2011

Given vectors x and y in a separable complex Hilbert space $\cal{H}$, an interpolating operator is a bounded operator A such that Ax = y. We show the following : Let Alg$\cal{L}$ be a tridiagonal algebra on a separable complex Hilbert space H and let x = ($x_i$) and y = ($y_i$) be vectors in H. Then the following are equivalent: (1) There exists an invertible operator A = ($a_{kj}$) in Alg$\cal{L}$ such that Ax = y. (2) There exist bounded sequences $\{{\alpha}_n\}$ and $\{{{\beta}}_n\}$ in $\mathbb{C}$ such that for all $k\in\mathbb{N}$, ${\alpha}_{2k-1}\neq0,\;{\beta}_{2k-1}=\frac{1}{{\alpha}_{2k-1}},\;{\beta}_{2k}=\frac{\alpha_{2k}}{{\alpha}_{2k-1}\alpha_{2k+1}}$ and $$y_1={\alpha}_1x_1+{\alpha}_2x_2$$ $$y_{2k}={\alpha}_{4k-1}x_{2k}$$ $$y_{2k+1}={\alpha}_{4k}x_{2k}+{\alpha}_{4k+1}x_{2k+1}+{\alpha}_{4k+2}x_{2k+2}$$.