• Kyungpook Mathematical Journal
• /
• 제45권2호
• /
• pp.293-299
• /
• 2005

Let ${\cal{L}}$ be a commutative subspace lattice on a Hilbert space ${\cal{H}}$ and X and Y be operators on ${\cal{H}}$. Let $${\cal{M}}_X=\{{\sum}{\limits_{i=1}^n}E_{i}Xf_{i}:n{\in}{\mathbb{N}},f_{i}{\in}{\cal{H}}\;and\;E_{i}{\in}{\cal{L}}\}$$ and $${\cal{M}}_Y=\{{\sum}{\limits_{i=1}^n}E_{i}Yf_{i}:n{\in}{\mathbb{N}},f_{i}{\in}{\cal{H}}\;and\;E_{i}{\in}{\cal{L}}\}.$$ Then the following are equivalent. (i) There is an operator A in $Alg{\cal{L}}$ such that AX = Y, Ag = 0 for all g in ${\overline{{\cal{M}}_X}}^{\bot},A^*A=AA^*$ and every E in ${\cal{L}}$ reduces A. (ii) ${\sup}\;\{K(E, f)\;:\;n\;{\in}\;{\mathbb{N}},f_i\;{\in}\;{\cal{H}}\;and\;E_i\;{\in}\;{\cal{L}}\}\;<\;\infty,\;{\overline{{\cal{M}}_Y}}\;{\subset}\;{\overline{{\cal{M}}_X}}$and there is an operator T acting on ${\cal{H}}$ such that ${\langle}EX\;f,Tg{\rangle}={\langle}EY\;f,Xg{\rangle}$ and ${\langle}ET\;f,Tg{\rangle}={\langle}EY\;f,Yg{\rangle}$ for all f, g in ${\cal{H}}$ and E in ${\cal{L}}$, where $K(E,\;f)\;=\;{\parallel}{\sum{\array}{n\\i=1}}\;E_{i}Y\;f_{i}{\parallel}/{\parallel}{\sum{\array}{n\\i=1}}\;E_{i}Xf_{i}{\parallel}$.

• Journal of applied mathematics & informatics
• /
• 제15권1_2호
• /
• pp.503-510
• /
• 2004

Given vectors x and y in a Hilbert space, an interpolating operator is a bounded operator T such that Tx = y. An interpolating operator for n vectors satisfies the equation $Tx_i\;=\;y_i,\;for\;i\;=\;1,\;2,\;\cdots,\;n$. In this paper the following is proved: Let H be a Hilbert space and L be a commutative subspace lattice on H. Let H and y be vectors in H. Let $M_x\;=\;\{{\sum{n}{i=1}}\;{\alpha}_iE_ix\;:\;n\;{\in}\;N,\;{\alpha}_i\;{\in}\;{\mathbb{C}}\;and\;E_i\;{\in}\;L\}\;and\;M_y\;=\;\{{\sum{n}{i=1}}\;{\alpha}_iE_iy\;:\;n\;{\in}\;N,\;{\alpha}_i\;{\in}\;{\mathbb{C}}\;and\;E_i\;{\in}\;L\}. Then the following are equivalent. (1) There exists an operator A in AlgL such that Ax = y, Af = 0 for all f in ${\overline{M_x}}^{\bot}$, AE = EA for all $E\;{\in}\;L\;and\;A^{*}\;=\;A$. (2) $sup\;\{\frac{{\parallel}{{\Sigma}_{i=1}}^{n}\;{\alpha}_iE_iy{\parallel}}{{\parallel}{{\Sigma}_{i=1}}^{n}\;{\alpha}_iE_iy{\parallel}}\;:\;n\;{\in}\;N,\;{\alpha}_i\;{\in}\;{\mathbb{C}}\;and\;E_i\;{\in}\;L\}\;<\;{\infty},\;{\overline{M_u}}\;{\subset}{\overline{M_x}}$ and < Ex, y >=< Ey, x > for all E in L.

• 호남수학학술지
• /
• 제27권4호
• /
• pp.649-654
• /
• 2005

Given operators X and Y acting on a separable complex Hilbert space ${\mathcal{H}}$, an interpolating operator is a bounded operator A such that AX = Y. We show the following: Let $Alg{\mathcal{L}}$ 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 unitary operator $A=(a_{ij})$ in $Alg{\mathcal{L}}$ such that AX = Y. (2) There is a bounded sequence {${\alpha}_n$} in ${\mathbb{C}}$ such that ${\mid}{\alpha}_j{\mid}=1$ and $y_{ij}={\alpha}_jx_{ij}$ for $j{\in}{\mathbb{N}}$.

• 호남수학학술지
• /
• 제32권2호
• /
• pp.255-260
• /
• 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
• /
• 제24권1_2호
• /
• pp.535-539
• /
• 2007

Given operators X and Y acting on a separable complex Hilbert space $\mathcal{H}$, an interpolating operator is a bounded operator A such that AX=Y. In this article, we show the following: Let $Alg\mathcal{L}$ be a tridiagonal algebra on a separable complex Hilbert space $\mathcal{H}$ and let $X=(x_{ij})\;and\;Y=(y_{ij})$ be operators in $\mathcal{H}$. Then the following are equivalent: (1) There exists a normal operator $A=(a_{ij})\;in\;Alg\mathcal{L}$ such that AX=Y. (2) There is a bounded sequence $\{\alpha_n\}\;in\;\mathbb{C}$ such that $y_{ij}=\alpha_jx_{ij}\;for\;i,\;j\;{\in}\;\mathbb{N}$. 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. 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 $\mathcal{H}$. Then the following are equivalent: (1) There exists a normal operator $A=(a_{ij})\;in\;Alg\mathcal{L}$ such that Ax=y. (2) There is a bounded sequence $\{\alpha_n\}$ in $\mathbb{C}$ such that $y_i=\alpha_ix_i\;for\;i{\in}\mathbb{N}$.

• 한국수학교육학회지시리즈B:순수및응용수학
• /
• 제11권2호
• /
• pp.167-173
• /
• 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.

• 한국수학교육학회지시리즈B:순수및응용수학
• /
• 제15권4호
• /
• pp.401-406
• /
• 2008

Given operators X and Y acting on a separable complex Hilbert space H, an interpolating operator is a bounded operator A such that AX=Y. In this article, we investigate Hilbert-Schmidt interpolation problems for operators in a tridiagonal algebra and we get the following: Let ${\pounds}$ be a subspace lattice acting on a separable complex Hilbert space H and let X=$(x_{ij})$ and Y=$(y_{ij})$ be operators acting on H. Then the following are equivalent: (1) There exists a Hilbert-Schmidt operator $A=(a_{ij})$ in Alg${\pounds}$ such that AX=Y. (2) There is a bounded sequence $\{{\alpha}_n\}$ in $\mathbb{C}$ such that ${\sum}_{n=1}^{\infty}|{\alpha}_n|^2<{\infty}$ and $$y1_i={\alpha}_1x_{1i}+{\alpha}_2x_{2i}$$ $$y2k_i={\alpha}_{4k-1}x_2k_i$$ $$y{2k+1}_i={\alpha}_{4k}x_{2k}_i+{\alpha}_{4k+1}x_{2k+1}_i+{\alpha}_{4k+2}x_{2k+2}_i\;for\;all\;i,\;k\;\mathbb{N}$$.

• 대한수학회보
• /
• 제39권3호
• /
• pp.423-430
• /
• 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 n-operators satisfies the equation $AX_{}i$ = $Y_{i}$ for i/ = 1,2,…, n. In this article, we obtained the following : Let X ＝ ($x_{i\sigma(i)}$ and Y ＝ ($y_{ij}$ be operators in B(H) such that $X_{i\sigma(i)}\neq\;0$ for all i. Then the following statements are equivalent. (1) There exists an operator A in Alg L such that AX = Y, every E in L reduces A and A is a self-adjoint operator. (2) sup ${\frac{\parallel{\sum^n}_{i=1}E_iYf_i\parallel}{\parallel{\sum^n}_{i=1}E_iXf_i\parallel}n\;\epsilon\;N,E_i\;\epsilon\;L and f_i\;\epsilon\;H}$ < $\infty$ and $x_{i,\sigma(i)}y_{i,\sigma(i)}$ is real for all i = 1,2, ....

• 호남수학학술지
• /
• 제27권2호
• /
• pp.243-250
• /
• 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 $$k{\in}N$$.

• 호남수학학술지
• /
• 제33권1호
• /
• pp.115-120
• /
• 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 $$y_1={\alpha}_1x_1+{\alpha}_2x_2$$ $$y_{2k}={\alpha}_{4k-1}x_{2k}$$ $$y_{2k+1}={\alpha}_{4k}x_{2k}+{\alpha}_{4k+1}x_{2k+1}+{\alpha}_{4k+2}x_{2k+2}$$.