Correction to: Indefinite Abstract Splines with a Quadratic Constraint

1 Correction to: Journal of Optimization Theory and Applications (2020) 186:209–225https://doi.org/10.1007/s10957-020-01692-z

2 Introduction

In [1] the statement of Lemma 4.1 is incorrect. In fact, if the dimension of $\mathcal {H}$ is infinite it is always possible to find a sequence in $\mathcal {C}_{V}$ which converges weakly to a vector not contained in $\mathcal {C}_{V}$, see Proposition 2.1 below. We apologize for this mistake. If the dimension of $\mathcal {H}$ is finite, Lemma 4.1 is not necessary to prove Theorem 4.1. In this erratum, we provide the correct statement for Theorem 4.1 for the finite dimensional case, as well as its proof.

3 Corrected Result

Let us start by proving that $\mathcal {C}_{V}$ is not necessarily weakly closed if $\mathcal {H}$ is infinite dimensional. Recall that $(\mathcal {H}, \left\langle \,\cdot , \cdot \, \right\rangle )$ is a separable Hilbert space, $(\mathcal {E}, \left[ \,\cdot , \cdot \, \right] )$ is a Krein space and $V\in \mathcal {L}(\mathcal {H},\mathcal {E})$ is surjective. If $\dim N(V)^\bot$ is finite the problem can be treated as if $\mathcal {H}$ were finite dimensional, simply considering $N(V)^\bot$ instead of $\mathcal {H}$.

Proposition 2.1

If $\dim N(V)^\bot$ is infinite then $\mathcal {C}_{V}$ is not weakly closed.

Proof

Assume that $\dim N(V)^\bot$ is infinite. Then, $\mathcal {E}$ is also infinite dimensional (and separable) because V is surjective. Let $\mathcal {E}=\mathcal {E}_+[\dotplus ]\mathcal {E}_-$ be a fundamental decomposition of $\mathcal {E}$. Hence, $(\mathcal {E}_+, \left[ \,\cdot , \cdot \, \right] )$ and $(\mathcal {E}_-, -\left[ \,\cdot , \cdot \, \right] )$ are Hilbert spaces. Without loss of generality, we can assume that $\dim \mathcal {E}_+$ is infinite.

Let us consider an orthonormal basis ${(e^+_n)}_{n\ge 1}$ of $\mathcal {E}_+$. As a consequence of Bessel’s inequality we have that $e_n^+\xrightarrow []{w} 0$, i.e.

$$\begin{aligned} \left[ \,e_n^+ , z\, \right] \rightarrow 0 \qquad \text {for every }z\in \mathcal {E}^+. \end{aligned}$$

For each $n\ge 1$ there exists a unique $x_n\in N(V)^\bot$ such that $Vx_n=e_n^+$. Also, choose $e^{-}\in \mathcal {E}_-$ such that $\left[ \,e^- , e^-\, \right] =-1$ and the unique $x_0\in N(V)^\bot$ such that $Vx_0=e^-$. Then, define

$$\begin{aligned} y_n= x_n + x_0, \qquad n\ge 1. \end{aligned}$$

Below we show that the sequence ${(y_n)}_{n\ge 1}$ is contained in $\mathcal {C}_{V}$ and it weakly converges to $x_0\notin \mathcal {C}_{V}$. On the one hand, if $n\ge 1$,

$$\begin{aligned} \left[ \,Vy_n , Vy_n\, \right] =\left[ \,e_n^+ + e^- , e_n^+ + e^-\, \right] =\left[ \,e_n^+ , e_n^+\, \right] + \left[ \,e^- , e^-\, \right] =1-1=0, \end{aligned}$$

i.e. $y_n\in \mathcal {C}_{V}$. On the other hand, given $x\in \mathcal {H}$,

$$\begin{aligned} \big |\left\langle \,y_n-x_0 , x\, \right\rangle \big |&=\big |\left\langle \,x_n , x\, \right\rangle \big |=\big |\left\langle \,V^\dag e_n^+ , x\, \right\rangle \big |= \big |\left[ \,e_n^+ , (V^\dag )^\# x\, \right] \big |\xrightarrow [n\rightarrow \infty ]{} 0. \end{aligned}$$

Therefore, $y_n\xrightarrow []{w} x_0$ and $x_0\notin \mathcal {C}_{V}$ since $\left[ \,Vx_0 , Vx_0\, \right] =\left[ \,e^- , e^-\, \right] =-1\ne 0$. $\square$

Now we give the correct statement and proof of [1, Theorem 4.1], which establishes sufficient conditions for the existence of indefinite interpolating splines for every $z_0\in \mathcal {E}$ in the finite dimensional setting.

Theorem 2.1

[1, Theorem 4.1] Suppose that $\mathcal {H}$ is a finite dimensional space. If R(L) is a uniformly positive subspace of $(\mathcal {K}\times \mathcal {E}, \left[ \,\cdot , \cdot \, \right] _\rho )$ for some $\rho \ne 0$ then $\mathcal {S}_{z_0}\ne \varnothing$ for every $z_0\in \mathcal {E}$.

Proof

In order to prove the theorem, we apply [1, Proposition 4.2]. To this end, we first show that $T^\#Tx\in R(L^\#L)$ for every $x\in \mathcal {H}$.

By [1, Proposition A.1], R(L) is a regular subspace. Then, for every $(y,z)\in \mathcal {K}\times \mathcal {E}$ there exists (a unique) $x\in \mathcal {H}$ such that $Lx - (y,z)\in R(L)^{{[\bot ]}}$, or equivalently, $L^\#Lx=L^\#(y,z)$. Since T and V are surjective, for each $(y,z)\in \mathcal {K}\times \mathcal {E}$ there exist $u,w\in \mathcal {H}$ such that $y=Tu$ and $z=Vw$. Therefore, there exists $x\in \mathcal {H}$ such that

$$\begin{aligned} T^\#Tu + \rho V^\#Vw =L^\#(Tu, Vw)=(T^\#T + \rho V^\#V)x, \end{aligned}$$

and consequently $R(L^\#L)=R(T^\#T+\rho V^\#V)=R(T^\#T)+R(V^\#V)$. Thus, $R(T^\#T)\subseteq R(L^\#L)$.

Given $z_0\in \mathcal {E}$, let $x_0,u_0\in \mathcal {H}$ be such that $Vx_0=z_0$ and $T^\#Tx_0=L^\#Lu_0$. Since $\mathcal {H}$ is a finite dimensional space, $(R(L^\#L),\left( \,\cdot , \cdot \, \right) )$ is a Hilbert space and $\mathcal {C}_{V}\cap \mathcal {B}_L$ is compact. Then ${{\,\mathrm{d}\,}}(u_0,\mathcal {C}_{V}\cap \mathcal {B}_L)$ is attained. In this case, [1, Proposition 4.2] ensures that $\mathcal {S}_{z_0}\ne \varnothing$. $\square$