Indefinite Abstract Splines with a Quadratic Constraint

Abstract

We study an extension to Krein spaces of the abstract interpolating spline problem in Hilbert spaces, introduced by M. Atteia. This is a quadratically constrained quadratic programming problem, where the objective function is not convex, while the equality constraint is sign indefinite. We characterize the existence of solutions and, if there are any, we describe the set of solutions as the union of a family of affine manifolds parallel to a fixed subspace, which depend on the original data.

Appendix: Terminology and Notations Related to Krein Spaces

In the following, we present the standard notation and some basic results on indefinite inner product spaces and, in particular, on Krein spaces. For a complete exposition on the subject (and the proofs of the results below) see, for example, [33,34,35,36,37].

An indefinite inner product space $(\mathcal {F}, \left[ \,\cdot , \cdot \, \right] )$ is a (complex) vector space $\mathcal {F}$ endowed with a Hermitian sesquilinear form $\left[ \,\cdot , \cdot \, \right] : \mathcal {F}\times \mathcal {F} {\rightarrow } {\mathbb {C}}$.

Two vectors $x,y\in \mathcal {F}$ are orthogonal, denoted by $x{[\bot ]}y$, if $\left[ \,x , y\, \right] =0$.

If $\mathcal {S}$ is a subset of an indefinite inner product space $\mathcal {F}$, the orthogonal companion to $\mathcal {S}$ is defined by

$$\begin{aligned} \mathcal {S}^{{[\bot ]}}=\{ x\in \mathcal {F} : \left[ \,x , s\, \right] =0 \; \text {for every} \,\,s\in \mathcal {S}\}, \end{aligned}$$

and it is always a subspace of $\mathcal {F}$.

Definition A.1

An indefinite inner product space $(\mathcal {H}, \left[ \,\cdot , \cdot \, \right] )$ is a Krein space, if it can be decomposed as a direct (orthogonal) sum of a Hilbert space and an anti-Hilbert space, i.e. there exist subspaces $\mathcal {H}_\pm$ of $\mathcal {H}$ such that $(\mathcal {H}_+, \left[ \,\cdot , \cdot \, \right] )$ and $(\mathcal {H}_-, -\left[ \,\cdot , \cdot \, \right] )$ are Hilbert spaces,

$$\begin{aligned} \mathcal {H}=\mathcal {H}_+ \dotplus \mathcal {H}_-, \end{aligned}$$

(17)

and $\mathcal {H}_+$ is orthogonal to $\mathcal {H}_-$ with respect to the indefinite inner product. Sometimes we use the notation $\left[ \,\cdot , \cdot \, \right] _\mathcal {H}$ instead of $\left[ \,\cdot , \cdot \, \right]$ to emphasize the Krein space considered.

A pair of subspaces $\mathcal {H}_\pm$ as in (17) is called a fundamental decomposition of $\mathcal {H}$. Given a Krein space $\mathcal {H}$ and a fundamental decomposition $\mathcal {H}=\mathcal {H}_+\dotplus \mathcal {H}_-$, the direct sum of the Hilbert spaces $(\mathcal {H}_+, \left[ \,\cdot , \cdot \, \right] )$ and $(\mathcal {H}_-, -\left[ \,\cdot , \cdot \, \right] )$ is denoted by $(\mathcal {H},\left\langle \,\cdot , \cdot \, \right\rangle )$.

If $\mathcal {H}=\mathcal {H}_+ \dotplus \mathcal {H}_-$ and $\mathcal {H}=\mathcal {H}'_+ \dotplus \mathcal {H}'_-$ are two different fundamental decompositions of $\mathcal {H}$, then the corresponding associated inner products $\left\langle \,\cdot , \cdot \, \right\rangle$ and $\left\langle \,\cdot , \cdot \, \right\rangle '$ turn out to be equivalent on $\mathcal {H}$. Therefore, the norm topology on $\mathcal {H}$ does not depend on the chosen fundamental decomposition.

If $\mathcal {H}$ and $\mathcal {K}$ are Krein spaces, $\mathcal {L}(\mathcal {H}, \mathcal {K})$ stands for the vector space of linear transformations which are bounded with respect to any of the associated Hilbert spaces.

Given $T\in \mathcal {L}(\mathcal {H},\mathcal {K})$, the adjoint operator of T is the unique operator $T^\#\in \mathcal {L}(\mathcal {K}, \mathcal {H})$ such that

$$\begin{aligned} \left[ \,Tx , y\, \right] _\mathcal {K}=[x,T^\#y]_\mathcal {H},\quad \text {for every} \,\,x\in \mathcal {H}, y\in \mathcal {K}. \end{aligned}$$

We frequently use that if $T\in \mathcal {L}(\mathcal {H},\mathcal {K})$ and $\mathcal {M}$ is a closed subspace of $\mathcal {K}$, then

$$\begin{aligned} T^\#(\mathcal {M})^{{[\bot ]}}= T^{-1}(\mathcal {M}^{{[\bot ]}}). \end{aligned}$$

A vector $x\in \mathcal {F}$ is positive, negative, or neutral, if $\left[ \,x , x\, \right] >0$, $\left[ \,x , x\, \right] <0$, or $\left[ \,x , x\, \right] =0$, respectively. A set $\mathcal {M}$ of $\mathcal {F}$ is positive (negative) if x is positive (negative) for every $x\in \mathcal {M}$, $x\ne 0$; and it is non-negative (non-positive) if $\left[ \,x , x\, \right] \ge 0$ ($\left[ \,x , x\, \right] \le 0$) for every $x\in \mathcal {M}$.

A subspace $\mathcal {M}$ of a Krein space is uniformly positive if there exists $\alpha >0$ such that

$$\begin{aligned} \left[ \,x , x\, \right] \ge \alpha \Vert x\Vert ^2\quad \text { for every}\,\, x\in \mathcal {M}. \end{aligned}$$

Uniformly negative subspaces are defined in a similar way.

A subspace $\mathcal {M}$ of a Krein space $\mathcal {H}$ is regular if $\mathcal {M}+\mathcal {M}^{[\bot ]}=\mathcal {H}$, or equivalently, if there exists a projection $Q\in \mathcal {L}(\mathcal {H})$ onto $\mathcal {M}$ such that $Q^\#=Q$. Regular subspaces are closed.

The following proposition shows that closed uniformly definite subspaces are regular subspaces (see [34, Chapter 1, §7]).

Proposition A.1

Let $\mathcal {M}$ be a subspace of a Krein space $\mathcal {H}$. Then, $\mathcal {M}$ is closed and uniformly positive (negative) if and only if $\mathcal {M}$ is regular and non-negative (non-positive).