Perturbation of the Moore–Penrose Metric generalized inverse with applications to the best approximate solution problem in L^p(Ω, μ)

ABSTRACT

Let $X=L^p(\mathrm{\Omega },\mu )$ ($1<p<\mathrm{\infty }$), let $T\in B\left(X\right)$ with closed range. In this paper, utilizing the gap between closed subspaces and the perturbation bounds of metric projections, we present some new perturbation results of the Moore–Penrose metric generalized inverse. As applications of our results, we also investigate the best approximate solution problem for the ill-posed operator equation Tx=y under some conditions. The main results have three parts, part one covers the null space preserving case, part two covers the range preserving case, and part three covers the general case. Examples in connection with the theoretical results will be also presented.

KEYWORDS:

• Metric generalized inverse
• perturbation
• best approximate solution

2010 MATHEMATICS SUBJECT CLASSIFICATION:

• Primary: 47A05
• Secondary: 46B20

Acknowledgments

The authors would like to thank the reviewers and editors for their valuable comments which lead to significant improvement of the paper. The first author would like to thank Prof. Yifeng Xue in East China Normal University for his constructive advices and kind support.

Disclosure statement

No potential conflict of interest was reported by the authors.

Notes

1 Here, we should indicate that, as we said above, in the literature, this type of generalized inverse was first introduced by Wang and his collaborators. But, Professor Nashed–The reviewer of our article [6], told us that the term ‘Moore–Penrose’ may be an inappropriate credit, since Moore never considered any metric properties and Penrose never considered any thing beyond matrices. Professor Nashed suggested us that the term ‘algebraic metric generalized inverse’ may be more appropriate, and we adopt his suggestion in that paper.

2 In fact, in order to calculate the metric projection ${\pi }_{\mathcal{R}\left(A\right)}$(The method of computing ${\pi }_{\mathcal{N}\left(A\right)}$ is exactly the same), by using the definition of p-norm, we only need to consider the minimum of the real function $f\left(t\right)=|x_1-2t{|}^p+|x_2-t{|}^p$. But note that, here, p is even, thus, we can consider the following function: $f\left(t\right)=(x_1-2t{)}^p+(x_2-t{)}^p$ By taking the derivative of $f\left(t\right)$ step by step, we can get $t=\left(1/(1+2^p)\right)(2^{p-1}{x}_1+x_2)$. In this case, we have $\left[\begin{array}{c}2t\\ t\end{array}\right]=\frac{1}{1+2^p}\left[\begin{array}{c}{2}^p{x}_1+2x_2\\ 2^{p-1}{x}_1+x_2\end{array}\right]=\frac{1}{1+2^p}\left[\begin{array}{cc}{2}^p& 2\\ 2^{p-1}& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right].$ Furthermore, $\frac{1}{1+2^p}\left[\begin{array}{cc}{2}^p& 2\\ 2^{p-1}& 1\end{array}\right]\left[\begin{array}{c}2t\\ t\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}2t\\ t\end{array}\right]=\left[\begin{array}{c}2t\\ t\end{array}\right].$ Note that $l_{{\mathbb{R}}^2}^p$ is reflexive and strictly convex Banach space, therefore, there exists a uniquely single-valued metric projection on each closed subspace of $l_{{\mathbb{R}}^2}^p$. Thus, we have ${\pi }_{\mathcal{R}\left(A\right)}={\pi }_{\mathcal{R}\left(\overline{A}\right)}=\frac{1}{1+2^p}\left[\begin{array}{cc}{2}^p& 2\\ 2^{p-1}& 1\end{array}\right].$

3 In order to calculate $\parallel A^M{\parallel }_p$(Similarly, we can calculate the p-norms of other operators, respectively), we only need to consider a conditional extremum problem, and then following the conventional Lagrange multiplier method. In fact, first note that p is even, and then using $F(x_1,x_2)\triangleq {∥\left[\begin{array}{cc}{2}^{p-1}& 1\\ 4^{p-1}& 2^{p-1}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]∥}_p^p=(1+2^{p^2-p})(2^{p-1}{x}_1+x_2),$ we can construct the Lagrange function $L(x_1,x_2,\lambda )=2^{p-1}{x}_1+x_2+\lambda (x_1^p+x_2^p-1)$, here $x_1^p+x_2^p=1$. By simple calculation, we can obtain that $F(x_1,x_2)$ has the maximum when $x_1=2x_2=2(1+2^p{)}^{-1/p}$. Thus, we get $\parallel A^M{\parallel }_p=\frac{2}{3}{\left(1+2^p\right)}^{-(2p^2-p+1)/p^2}(1+2^{p^2-p}{)}^{1/p}.$

Additional information

Funding

The work is partially supported by China Postdoctoral Science Foundation (grant no. 2015M582186), Key Research Project of the Education Department of Henan Province (grant nos. 17A630019, 18A110018), Henan Institute of Science and Technology Postdoctoral Science Foundation (grant no. 5201029470209).