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The solution by iteration of a composed K-positive definite operator equation in a Banach space. (English) Zbl 1228.47040

The article deals with linear equations of the type \(Lu = f\) in a Banach space \(E\), where \(L = A + B\), \(A\) is a \(K\)-positive definite (or an asymptotically \(K\)-positive definite) operator, and \(B\) is a linear operator. In the article, an operator \(A\) is called \(K\) positive definite if \(\langle Au,jKu \rangle \geq \alpha\|Ku\|^2\), \(u \in D(A)\), where \(K\) is an operator defined on a dense subspace \(E_0\) of \(E\) and \(K\) has a bounded inverse on \(R(T)\); an operator \(A\) is called asymptotically \(K\) positive definite if \(\langle K^{n-1}Au,j(K^n) \rangle \geq ck_n\|K^nu\|^2\), \(u \in D(A)\), where \(c > 0\), \(k_n \geq 1\), \(k_n \to 1\) (in these definitions, \(j(\cdot) \in J(\cdot)\), \(J\) is the normalized duality mapping for \(E\)). The main results are concerned with conditions under which the iterations \(x_{n+1} = x_n + r_n\), \(r_n = K^{-1}f - K^{-1}Ar_n\), converge strongly to the unique solution of \(Ax = f\) (the original equation \(Lu = f\) is not considered in the correspondent Theorems 3.3 and 3.5).
Reviewer’s remarks. All these results seem to be erroneous; in order to verify this fact, it is sufficient to cite some sentences from the article: Since \(A\) is continuously invertible, the operator \(T = A^{-1}\) is completely continuous. Hence \(T\) is locally lipschitzian and accretive. It follows that (3.3) (i.e., \(u + Tu = g\)) has a unique solution. Further, the author considers an example when \(B = A\) and thus \(B^{-1}A = I\)! The definition of an asymptotically \(K\) positive definite operator is also senseless.

MSC:

47B99 Special classes of linear operators
47B44 Linear accretive operators, dissipative operators, etc.
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References:

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