• East Asian mathematical journal
• /
• v.15 no.2
• /
• pp.313-324
• /
• 1999

What we will be concerned with is, first, the question of the condition about $\mathcal{L}$ that gives Alg$\mathcal{L}$ a von Neumann algebra, that is, the question of the condition about $\mathcal{L}$ that will give Alg$\mathcal{L}$ a self-adjoint algebra. Secondly, if Alg$\mathcal{L}$ is a von Neumann algebra, we want to find out what type it is.

• Honam Mathematical Journal
• /
• v.34 no.3
• /
• pp.409-422
• /
• 2012

Let $\mathcal{L}$ be a subspace lattice on a Hilbert space $\mathcal{H}$. And let X and Y be operators acting on a Hilbert space $\mathcal{H}$. Let $sp(x)=\{{\alpha}x\;:\;{\alpha}{\in}\mathcal{C}\}$ $x{\in}\mathcal{H}$. Assume that $\mathcal{H}=\overline{range\;X}{\oplus}sp(h)$ for some $h{\in}\mathcal{H}$ and < $h$, $E^{\bot}Xf$ >= 0 for each $f{\in}\mathcal{H}$ and $E{\in}\mathcal{L}$. Then there exists an operator A in Alg$\mathcal{L}$ such that AX = Y if and only if $sup\{\frac{{\parallel}E^{\bot}Yf{\parallel}}{{\parallel}E^{\bot}Yf{\parallel}}\;:\;f{\in}H,\;E{\in}\mathcal{L}\}$ = K < ${\infty}$. Moreover, if the necessary condition holds, then we may choose an operator A such that AX = Y and ${\parallel}||A{\parallel}=K$.

• Journal of applied mathematics & informatics
• /
• v.24 no.1_2
• /
• 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 $y_i=\alpha_ix_i\;for\;i{\in}\mathbb{N}$.

• Honam Mathematical Journal
• /
• v.39 no.1
• /
• pp.93-100
• /
• 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
• /
• v.31 no.2
• /
• pp.259-265
• /
• 2009

Let $\mathcal{L}$ be a subspace lattice on a Hilbert space $\mathcal{H}$. Let x and y be vectors in $\mathcal{H}$ and let $P_x$ be the projection onto sp(x). If $P_xE$ = $EP_x$ for each E ${\in}\;\mathcal{L}$, then the following are equivalent. (1) There exists an operator A in Alg$\mathcal{L}$ such that Ax = y, Af = 0 for all f in $sp(x)^{\perp}$ and A ${\geq}$ 0. (2) sup ${\frac{{\parallel}E^{\perp}y{\parallel}}{{\parallel}E^{\perp}x{\parallel}}:E{\in}\mathcal{L}}$ < ${\infty}$ < x, y > ${\geq}$ 0. Let X and Y be operators in $\mathcal{B}(\mathcal{H})$. Let P be the projection onto $\overline{rangeX}$. If PE = EP for each E ${\in}\;\mathcal{L}$, then the following are equivalent: (1) sup ${\frac{{\parallel}E^{\perp}Yf{\parallel}}{{\parallel}E^{\perp}Xf{\parallel}}:f{\in}\mathcal{H},E{\in}\mathcal{L}}$ < ${\infty}$ and < Xf, Yf > ${\geq}$ 0 for all f in H. (2) There exists a positive operator A in Alg$\mathcal{L}$ such that AX = Y.

• Bulletin of the Korean Mathematical Society
• /
• v.52 no.2
• /
• pp.351-361
• /
• 2015

We give examples of Lie ideals in a tridiagonal algebra $Alg\mathcal{L}_{\infty}$ and study some properties of Lie ideals in $Alg\mathcal{L}_{\infty}$. We also investigate relationships between Lie ideals in $Alg\mathcal{L}_{\infty}$. Let k be a fixed natural number. Let $\mathcal{A}$ be a linear manifold in $Alg\mathcal{L}_{\infty}$ such that $T_{(2k-1,2k)}=0$ for all $T{\in}\mathcal{A}$. Then $\mathcal{A}$ is a Lie ideal if and only if $T_{(2k-1,2k-1)}=T_{(2k,2k)}$ for all $T{\in}\mathcal{A}$.

• Honam Mathematical Journal
• /
• v.31 no.3
• /
• pp.421-428
• /
• 2009

Given operators X and Y acting on a Hilbert space $\mathcal{H}$, an interpolating operator is a bounded operator A such that AX = Y. In this paper, we showed the following : Let $\mathcal{L}$ be a subspace lattice acting on a Hilbert space $\mathcal{H}$ and let $X_i$ and $Y_i$ be operators in B($\mathcal{H}$) for i = 1, 2, ${\cdots}$. Let $P_i$ be the projection onto $\overline{rangeX_i}$ for all i = 1, 2, ${\cdots}$. If $P_kE$ = $EP_k$ for some k in $\mathbb{N}$ and all E in $\mathcal{L}$, then the following are equivalent: (1) $sup\;\{{\frac{{\parallel}E^{\perp}({\sum}^n_{i=1}Y_if_i){\parallel}}{{\parallel}E^{\perp}({\sum}^n_{i=1}Y_if_i){\parallel}}:f{\in}H,n{\in}{\mathbb{N}},E{\in}\mathcal{L}}\}$ < ${\infty}$ range $\overline{rangeY_k}\;=\;\overline{rangeX_k}\;=\;\mathcal{H}$, and < $X_kf,\;X_kg$ >=< $Y_kf,\;Y_kg$ > for some k in $\mathbb{N}$ and for all f and g in $\mathcal{H}$. (2) There exists an operator A in Alg$\mathcal{L}$ such that $AX_i$ = $Y_i$ for i = 1, 2, ${\cdots}$ and AA$^*$ = I = A$^*$A.

• Honam Mathematical Journal
• /
• v.30 no.2
• /
• pp.329-334
• /
• 2008

Given operators X and Y acting on a Hilbert space $\mathcal{H}$, an interpolating operator is a bounded operator A such that AX = Y. In this article, the following is proved: Let $\mathcal{L}$ be a subspace lattice on $\mathcal{H}$ and let X and Y be operators acting on a Hilbert space H. Let P be the projection onto the $\overline{rangeX}$. If PE = EP for each E ${\in}$ $\mathcal{L}$, then the following are equivalent: (1) sup ${{\frac{{\parallel}E^{\perp}Yf{\parallel}}{{\parallel}E^{\perp}Xf{\parallel}}}:f{\in}\mathcal{H},\;E{\in}\mathcal{L}}$ < ${\infty},\;\overline{rangeY}\;{\subset}\;\overline{rangeX}$, and there is a bounded operator T acting on $\mathcal{H}$ such that < Xf, Tg >=< Yf, Xg >, < Tf, Tg >=< Yf, Yg > for all f and gin $\mathcal{H}$ and $T^*h$ = 0 for h ${\in}\;{\overline{rangeX}}^{\perp}$. (2) There is a normal operator A in AlgL such that AX = Y and Ag = 0 for all g in range ${\overline{rangeX}}^{\perp}$.

• Honam Mathematical Journal
• /
• v.27 no.4
• /
• 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 $j{\in}{\mathbb{N}}$.

• Honam Mathematical Journal
• /
• v.36 no.4
• /
• pp.907-911
• /
• 2014

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 $\mathcal{H}$ and let $x=(x_i)$ and $y=(y_i)$ be vectors in $\mathcal{H}$. Then the following are equivalent: (1) There exists a unitary operator $A=(a_{ij})$ in $Alg{\mathcal{L}}$ such that Ax = y. (2) There is a bounded sequence $\{{\alpha}_i\}$ in $\mathbb{C}$ such that ${\mid}{\alpha}_i{\mid}=1$ and $y_i={\alpha}_ix_i$ for $i{\in}\mathbb{N}$.