• Honam Mathematical Journal
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• v.32 no.2
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• pp.255-260
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• 2010

Given vectors x and y in a separable complex Hilbert space $\mathcal{H}$, 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. We show the following : Let Alg$\mathcal{L}$ be a tridiagonal algebra on a separable complex Hilbert space $\mathcal{H}$ and let x = $(x_i)$ and y = $(y_i)$ be vectors in H. Then the following are equivalent: (1) There exists a compact operator A = $(a_{ij})$ in Alg$\mathcal{L}$ such that Ax = y. (2) There is a sequence ${{\alpha}_n}$ in $\mathbb{C}$ such that ${{\alpha}_n}$ converges to zero and for all k ${\in}$ $\mathbb{N}$, $y_1 = {\alpha}_1x_1 + {\alpha}_2x_2$ $y_{2k} = {\alpha}_{4k-1}x_{2k}$ $y_{2k+1}={\alpha}_{4k}x_{2k}+{\alpha}_{4k+1}x_{2k+1}+{\alpha}_{4k+2}+x_{2k+2}$.

• 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 ${\mathcal{A}}$.

• Honam Mathematical Journal
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• v.26 no.4
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• pp.379-389
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• 2004

Given operators X and Y acting on a Hilbert space ${\mathcal{H}}$, 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 $i=1,2,{\cdots},n$. In this article, we obtained the following : Let ${\mathcal{H}}$ be a Hilbert space and let ${\mathcal{L}}$ be a commutative subspace lattice on ${\mathcal{H}}$. Let X and Y be operators acting on ${\mathcal{H}}$. Then the following statements are equivalent. (1) There exists an operator A in $Alg{\mathcal{L}}$ such that AX = Y, A is positive and every E in ${\mathcal{L}}$ reduces A. (2) sup ${\frac{{\parallel}{\sum}^n_{i=1}\;E_iY\;f_i{\parallel}}{{\parallel}{\sum}^n_{i=1}\;E_iX\;f_i{\parallel}}}:n{\in}{\mathbb{N}},\;E_i{\in}{\mathcal{L}}$ and $f_i{\in}{\mathcal{H}}<{\infty}$ and <${\sum}^n_{i=1}\;E_iY\;f_i$, ${\sum}^n_{i=1}\;E_iX\;f_i>\;{\geq}0$, $n{\in}{\mathbb{N}}$, $E_i{\in}{\mathcal{L}}$ and $f_i{\in}H$.

• 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 $i,j{\in}\mathbb{N}$.

• The Pure and Applied Mathematics
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• v.11 no.1
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• pp.29-36
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• 2004

Given vectors x and ｙ in a Hilbert space, an interpolating operator is a bounded operator T such that Tx=y. In this paper the following is proved: Let $\cal{L}$ be a subspace lattice on a Hilbert space $\cal{H}$. Let x and ｙ be vectors in $\cal{H}$ and let $P_x$, be the projection onto sp(x). If $P_xE=EP_x$ for each $ E \in \cal{L}$ then the following are equivalent. (1) There exists an operator A in Alg(equation omitted) such that Ax=y, Af = 0 for all f in ($sp(x)^\perp$) and $A=-A^\ast$. (2) (equation omitted)

• 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 $$k{\in}N$$.

• Journal of applied mathematics & informatics
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• v.32 no.3_4
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• pp.441-446
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• 2014

In this paper the following is proved: Let $\mathcal{L}$ be a subspace lattice on a Hilbert space $\mathcal{H}$ and X and Y be operators acting on $\mathcal{H}$. Then there exists a compact operator A in $Alg\mathcal{L}$ such that AX = Y if and only if ${\sup}\{\frac{{\parallel}E^{\perp}Yf{\parallel}}{{\parallel}E^{\perp}Xf{\parallel}}\;:\;f{\in}\mathcal{H},\;E{\in}\mathcal{L}\}$ = K < ${\infty}$ and Y is compact. Moreover, if the necessary condition holds, then we may choose an operator A such that AX = Y and ${\parallel}A{\parallel}=K$.