• 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.