Decomposition into subspaces preconditioning: abstract framework


Operator preconditioning based on decomposition into subspaces has been developed in early 90’s in the works of Nepomnyaschikh, Matsokin, Oswald, Griebel, Dahmen, Kunoth, Rüde, Xu, and others, with inspiration from particular applications, e.g., to fictitious domains, additive Schwarz methods, multilevel methods etc. Our paper presents a revisited general additive splitting-based preconditioning scheme which is not connected to any particular preconditioning method. We primarily work with infinite-dimensional spaces. Motivated by the work of Faber, Manteuffel, and Parter published in 1990, we derive spectral and norm lower and upper bounds for the resulting preconditioned operator. The bounds depend on three pairs of constants which can be estimated independently in practice. We subsequently describe a nontrivial general relationship between the infinite-dimensional results and their finite-dimensional analogs valid for the Galerkin discretization. The presented abstract framework is universal and easily applicable to specific approaches, which is illustrated on several examples.

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The authors thank Radim Blaheta for pointing out the separate displacement method in elasticity. The authors thank Miroslav Bulíček, Vít Dolejší, Josef Málek, Endre Süli, and Jan Zeman for their careful reading the manuscript and for many helpful suggestions and comments.


The work was supported by the Grant Agency of the Czech Republic under the contract no. 17-04150J and by the Charles University, project GA UK no. 172915.

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In the Appendix, we give the proof of the following theorem.

Theorem 1

Let \({\mathcal A}:V\to V^\#\) be a linear, bounded, coercive, and self-adjoint operator. Using the standard definition of the operator norm, the boundedness constant \(C_{\mathcal A}\) and the coercivity constant \(c_{\mathcal A}\) can be expressed as

$$\begin{array}{@{}rcl@{}} C_{\mathcal A}&=&\Vert{\mathcal A}\Vert_{{\mathcal L}(V,V^\#)}= \sup\limits_{u\in V, \Vert u\Vert_{V}= 1}\langle {\mathcal A}u,u\rangle=M_{\mathcal A}, \end{array} $$
$$\begin{array}{@{}rcl@{}} c_{\mathcal A}&=&m_{\mathcal A}= \inf\limits_{v\in V, \Vert v\Vert_{V}= 1}\langle {\mathcal A}v,v\rangle= \frac{1}{\sup\limits_{f\in V^\#, \Vert f\Vert_{V^\#}= 1}\Vert {\mathcal A}^{-1}f\Vert_{V}}\\ &=&\left\{\Vert{\mathcal A}^{-1}\Vert_{{\mathcal L}(V^\#,V)}\right\}^{-1}. \end{array} $$


The equality (124) is well known. It follows from the following sequence of equalities

$$\begin{array}{@{}rcl@{}} C_{\mathcal A}&=&\Vert{\mathcal A}\Vert_{{\mathcal L}(V,V^\#)}=\Vert\tau{\mathcal A}\Vert_{{\mathcal L}(V,V)}= \sup\limits_{u\in V,\Vert u\Vert_{V}= 1}(\tau{\mathcal A}u,u)_{V}\\ &=& \sup\limits_{u\in V,\Vert u\Vert_{V}= 1}\langle{\mathcal A}u,u\rangle=M_{\mathcal A}. \end{array} $$

Here, we used the fact that for any self-adjoint operator S in a Hilbert space V

$$\begin{array}{@{}rcl@{}} \Vert S\Vert_{{\mathcal L}(V,V)}&=&\sup\limits_{z\in V, \Vert z\Vert_{V}= 1}\Vert Sz\Vert_{V}=\sup\limits_{z\in V, \Vert z\Vert_{V}= 1}(Sz,Sz)_{V}^{1/2}\\ &=& \sup\limits_{z\in V, \Vert z\Vert_{V}= 1}\vert(Sz,z)_{V}\vert; \end{array} $$

see [14, Theorem 4.10.1, p. 220], [24, Theorem 6.5.1]. The second statement (125) was published without proof in [41, Section 3.3]. Since \(c_{\mathcal A}=m_{\mathcal A}\), it remains to prove that

$$ m_{\mathcal A}=\frac{1}{\sup\limits_{f\in V^\#,\Vert f\Vert_{V^\#}= 1}\Vert{\mathcal A}^{-1}f\Vert_{V}}= \inf_{u\in V,\Vert u\Vert_{V}= 1}\Vert{\mathcal A}u\Vert_{V^\#}, $$

where the second equality results from the substitution \(f={\mathcal A}u/\Vert {\mathcal A}u\Vert _{V^\#}\), uV. Equivalently, it remains to prove that

$$ m_{\mathcal A}:=\inf_{u\in V,\Vert u\Vert_{V}= 1}(\tau{\mathcal A}u,u)_{V}=\inf_{u\in V,\Vert u\Vert_{V}= 1} \Vert \tau{\mathcal A}u\Vert_{V}. $$

Clearly \((\tau {\mathcal A}u,u)_{V}\le \Vert \tau {\mathcal A}u\Vert _{V} \Vert u\Vert _{V}\), therefore the inequality

$$m_{\mathcal A}\le \inf_{u\in V, \Vert u\Vert_{V}= 1}\Vert \tau{\mathcal A}u\Vert_{V} $$

is trivial. In order to prove the opposite inequality

$$m_{\mathcal A}\ge \inf_{u\in V, \Vert u\Vert_{V}= 1}\Vert \tau{\mathcal A}u\Vert_{V}, $$

we use the fact that \(m_{\mathcal A}\) belongs to the spectrum of \(\tau {\mathcal A}\) and therefore there exists a sequence \(\{v_{k}\}_{k = 1,2,\dots }\) in V, ∥vkV = 1, such that

$$ \lim_{k\to\infty}\Vert\tau{\mathcal A}v_{k}-m_{\mathcal A}v_{k}\Vert_{V}= 0; $$

see [24, Corollary 6.5.6]. We will finish the proof by contradiction. Assume that

$$m_{\mathcal A}<\inf_{u\in V,\Vert u\Vert_{V}= 1}\Vert \tau{\mathcal A}u\Vert_{V}-\triangle $$

for some △ > 0. Using the Cauchy-Schwarz inequality,

$$\begin{array}{@{}rcl@{}} \Vert\tau{\mathcal A}v_{k}-m_{\mathcal A}v_{k}{\Vert_{V}^{2}} &=& \Vert \tau{\mathcal A}v_{k}{\Vert_{V}^{2}}+m_{\mathcal A}^{2}-2m_{\mathcal A}(\tau{\mathcal A}v_{k},v_{k})_{V}\\ &\ge& \Vert \tau{\mathcal A}v_{k}{\Vert_{V}^{2}}+m_{\mathcal A}^{2}-2m_{\mathcal A}\Vert \tau{\mathcal A}\Vert_{V} = (\Vert\tau{\mathcal A}v_{k}\Vert_{V}-m_{\mathcal A})^{2}. \end{array} $$


$$\Vert\tau{\mathcal A}v_{k}-m_{\mathcal A}v_{k}{\Vert_{V}^{2}} \ge \triangle^{2}\quad \textrm{for all} k = 1,2,\dots, $$

which gives the contradiction with (130) and completes the proof. □

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Hrnčíř, J., Pultarová, I. & Strakoš, Z. Decomposition into subspaces preconditioning: abstract framework. Numer Algor 83, 57–98 (2020).

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  • Operator preconditioning
  • Decomposition into infinite-dimensional subspaces
  • Stable splitting
  • Norm and spectral equivalence of operators
  • Additive Schwarz methods
  • Multilevel methods
  • Separate displacement preconditioning