• 제목/요약/키워드: invertible interpolation problem

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INVERTIBLE INTERPOLATION PROBLEMS IN CSL-ALGEBRA ALGL

  • Jo, Young-Soo;Kang, Joo-Ho
    • Journal of applied mathematics & informatics
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    • 제12권1_2호
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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?

INVERTIBLE INTERPOLATION ON AX = Y IN ALGL

  • Kang, Joo-Ho
    • 한국수학교육학회지시리즈B:순수및응용수학
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    • 제14권3호
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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.

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