Constraint Qualification with Schauder Basis for Infinite Programming Problems

Abstract

We consider infinite programming problems with constraint sets defined by systems of infinite number of inequalities and equations given by continuously differentiable functions defined on Banach spaces. In the approach proposed here we represent these systems with the help of coefficients in a given Schauder basis. We prove Abadie constraint qualification under the new infinite-dimensional Relaxed Constant Rank Constraint Qualification Plus and we discuss the existence of Lagrange multipliers via Hurwicz set. The main tools are: Rank Theorem and Lyusternik–Graves theorem.

1 Introduction

Let E be a Banach space and $f_0,\ g_i:\ E \rightarrow \mathbb {R}$, $i\in I_0\cup I_1\subseteq \mathbb {N}$, $I_0\cap I_1=\emptyset$ are functions of class $C^1$. We consider the following infinite programming problem:

i.e., both index sets $I_0,I_1$ can be infinite.

This problem can be cast into a general framework by using Schauder basis. To this aim, we recall the following facts. Let F be a Banach space. Let $(b_{i})_{i\in \mathbb {N}}$ be a Schauder basis of F, i.e. for all $x\in F$ there is a unique sequence of scalars $(x_n)$, $n\in \mathbb {N}$ s.t. $x=\sum \limits _{n=1}^\infty x_n b_n$.

Proposition 1

([2, Theorem 1.1.3]) Let X be a Banach space and let $(v_n)_{n\in \mathbb {N}}$ be a Schauder basis of X. Then there is a sequence $(v_n^*)_{n\in \mathbb {N}} \in X^*$ such that

$$\begin{aligned} v_i^*( v_j) =\delta _{ij}= \left\{ \begin{array}{ll} 1 &{}\quad \text {if } i=j,\\ 0 &{}\quad \text {if } i\ne j \end{array}\right. \end{aligned}$$

(1)

and $x=\sum _{n\in \mathbb {N}} v_n^*(x)v_n$ for each $x \in X$. Such a pair $(v_n,v_n^*)$ is called biorthogonal system.

When F is finite-dimensional (${\text {dim}} F= n$) we assume that $(b_{i})_{i=1,\dots ,n}$ is a basis of F (see [14, Example (i), Chapter 6]). In the sequel, we use the term "basis" when referring to Schauder basis. Let us note that any separable Hilbert space has a basis.

Let $K\subset F$ be a cone defined as

$$\begin{aligned} K: = \left\{ y \in F \mid y= \sum _{i\in I_0\cup I_1} b_i^*(y) b_i, \ b_i^*(y)= 0,\ i\in I_0,\ b_i^*(y)\le 0, \ i\in I_1 \right\} , \end{aligned}$$

(2)

where $I_0\cup I_1\subseteq \mathbb {N}$, $I_0\cap I_1=\emptyset$. Cone K is closed, convex and has no interior points (see [15]). In the sequel, we refer to the cone K as a basis cone (see [16]).

Let $G:\ E\rightarrow F$ be a $C^1$ mapping, i.e. $G:\ E\rightarrow F$ is continuously Fréchet differentiable mapping. Consider the minimization problem,

where ${{\mathcal {F}}}:=\{x\in E\ \mid \ G(x)\in K\}$.

By taking $F=\ell _2$, $b_i=e_i$, $i\in \mathbb {N}$, where $(e_i)_{i\in \mathbb {N}}$, is the canonical basis of $\ell _2$, the problem (P) is of the form ($P_0$). Indeed, in this case, the set ${{\mathcal {F}}}$ takes the form

$$\begin{aligned} {{\mathcal {F}}}:=\{x\in E\ \mid \ g_i(x):=e_i^*(G(x))=0,\ i\in I_0,\ \ g_i(x):=e_i^*(G(x))\le 0,\ i\in I_1\}. \end{aligned}$$

In the case when F is finite-dimensional ($F = \mathbb {R}^\kappa$) and $b_i=e_i$, $i\in \{1,\dots ,\kappa \}$ is the canonical basis of the space F, cone K has a form

$$\begin{aligned} K= \{ y=(y_1,\dots ,y_\kappa ) \in F \mid y_i= 0,\ i\in I_0,\ y_i\le 0, \ i\in I_1 \}, \end{aligned}$$

(3)

where $I_0\cup I_1 = \{1,\dots ,\kappa \}$.

Sufficient optimality conditions for problem (P) with F—finite dimensional and cone K, given by (3), are considered in [7]. In this paper, we concentrate on the case when F is infinite-dimensional and K is given by (2).

Remark 1

In Banach spaces F with a basis $(b_i)_{i\in \mathbb {N}}$, the basis cone, defined as

$$\begin{aligned} K_{\{b_i\}}:= \left\{ y \in F \mid y= \sum _{i\in \mathbb {N}} b_i^*(y) b_i, \ b_i^*(y)\le 0, \ i\in \mathbb {N} \right\} \end{aligned}$$

(4)

was considered in [15, 23]. For $I_0=\emptyset$ and $I_1=\mathbb {N}$, the basis cone $K_{\{b_i\}}$ coincides with K. In the space c of convergent sequences, the basis cone $K_{\{b_i\}}$ defined as (4) never coincide with the natural nonnegative cone, since the natural nonnegative cone has a nonvoid interior while the basis cone $K_{\{b_i\}}$ always has a void interior (see [16]).

Let us note that even in Hilbert spaces not all standard nonnegative cones can be represented in the form of (2) as can be seen from the following example.¹

Example 1

Let $F=L_2([0,1])$. Suppose that $f_k \ge 0$, $k=1,\dots$ is a basis of the nonnegative cone of $L_2([0,1])$. Observe that, due to the fact that $L_2^+([0,1])$ is generating, $f_k$, $k\in \mathbb {N}$, is also a basis of F. For each $k\in \mathbb {N}$ let $F_k$ be the support of $f_k$ and choose a subset $M_k \subset F_k$ of measure $<10^{-k}$, Then the set $M=\bigcup _{k\in \mathbb {N}} M_{k}$ has measure $<1/9$, hence the set $M^\prime =[0,1]\setminus M$ has measure greater than 0.

Let h(t) be the characteristic function of $M^\prime$, i.e. $h(t)=1$ for $t\in M^\prime$ and $h(t)=0$ for $t\notin M^\prime$. Obviously, $h(t)=0$ on M. On the other hand, $h(t)=\sum _{k\in \mathbb {N}} a_k f_k(t)$ with $a_k\ge 0$, $k\in \mathbb {N}$. Take any $k\in \mathbb {N}$ such that $a_k>0$. Then $h(t)>0$ on $F_k$ and in particular on $M_k$. However, $M_k\subset M$, where $h(t)=0$, a contradiction, i.e., such a basis does not exists.

Now, let us express the mapping appeared in the definition of problem (P) with the use of Schauder basis $b_i$, $i\in \mathbb {N}$. We have

$$\begin{aligned} G(x)=&\sum _{i\in I_0\cup I_1} b_i^*(G(x)) b_i=\sum _{i\in I_0\cup I_1} g_i(x) b_i, \\&\text {where } g_i(x):= b_i^* (G(x)), \ i\in I_0\cup I_1. \end{aligned}$$

By (2), we can rewrite set ${{\mathcal {F}}}$ in an equivalent way as follows

$$\begin{aligned} {{\mathcal {F}}}= \{ x \in E \mid g_i(x)=0,\ i \in I_0,\ g_i(x)\le 0,\ i\in I_1 \}. \end{aligned}$$

(5)

The aim of the present paper is to provide conditions ensuring Abadie CQ,

$$\begin{aligned} {{\mathcal {T}}}_{{{\mathcal {F}}}}(x_{0})=\varGamma _{{\mathcal {F}}}(x_{0}), \end{aligned}$$

where ${{\mathcal {T}}}_{{{\mathcal {F}}}}(x_{0})$ is the tangent cone to ${{\mathcal {F}}}$ at $x_0$ (see (6)) and $\varGamma _{{{\mathcal {F}}}}(x_{0})$ is the linearized cone to ${{\mathcal {F}}}$ at $x_0$ (see (7)). In the sequel we concentrate on the case $I_0\cup I_1=\mathbb {N}$.

Our main tool is Relaxed Constant Rank Constraint Qualification Plus introduced in Sect. 6. Other regularity conditions were recently proposed in [4, 6, 7, 11, 25].

The novelty of our approach relies on the use of the Schauder bases and basic sequences in definitions of the main concepts, namely in the definitions of the CRC+ (Definition 9) and RCRCQ+ (Definition 10). Consequently, the assumptions of the main theorem (Theorem 6), of Proposition 9 on functional dependence and two auxiliary results (Propositions 7, 8) are expressed with the help of Schauder bases of some spaces generated by the derivative of the constraint map.

The proof of Theorem 6 is based on Rank Theorem 3 and Lyusternik–Graves Theorem (Theorem 5). Moreover, in Propositions 10 and 11 we prove the existence of Lagrange multipliers. To the best of our knowledge, the only result concerning the existence of Lagrange multipliers for which the proof is based on Rank Theorem is [9, Theorem 4.1].

Theorem 1

([9, Theorem 4.1]) Let $x_0$ be a local solution to problem (P) with $K=\{0\}$ and F be a Banach space. Let $x_0$ be a local solution to (P). Assume that the following conditions are fulfilled:

• (A) $f_0$ is Fréchet differentiable at $x_0$ and G is of class $C^1$ in a neighbourhood of $x_0$.

• (B) $E_2:= \ker DG(x_0)$ is topologically complemented in E, i.e., $E=E_1 \oplus E_2$, where $E_1$ is a closed subspace of E, $F_1:= DG(x_0)(E)$ is closed and topologically complemented in F, i.e., $F=F_1 \oplus F_2$, where $F_2$ is a closed subspace of F.

• (C) There exists a neighbourhood of $x_0$ such that for all x in this neighbourhood $DG(x)(E)\cap F_2= \{0\}$.

Then there exists $\lambda \in F^*$ such that $Df_0(x_0)=\lambda \circ DG(x_0)$.

Remark 2

By Proposition 1 of [8], condition (C) is equivalent to condition

$$\begin{aligned} DG(x)|_{E_1}:\ E_1 \rightarrow DG(x)(E)\quad \text {is an isomorphism.} \end{aligned}$$

In Proposition 7 of Sect. 4 we provide conditions under which we ensure that (B) holds. In Proposition 8 of Sect. 4 we provide conditions under which we ensure that (C) holds.

The organization of the paper is as follows. In Sect. 2 we recall basic concepts related to Schauder bases. In Sect. 3 we introduce Constant Rank Condition (CRC) and Constant Rank Condition Plus (CRC+). In Sect. 4, Conditions 4., 5., 6. of CRC+ allows us to prove Proposition 7 on the split of the space E. Condition CRC+ allows us to prove Proposition 8 on isomorphisms. Section 5 is devoted to the proof of Proposition 9 on functional dependence. In Sect. 6 we introduce Relaxed Constant Rank Constraint Qualification Plus, which is the main ingredient of main results of Sect. 7. In Sect. 7 we present main result Theorem 6, namely sufficient conditions under which Abadie condition holds for the investigated problem. Section 8 is devoted to the topic of existence of Lagrange multipliers to problem (P). The main result of this section is Proposition 10, in which an important assumption is weak*-closedness of the Hurwicz set ${{\mathcal {M}}}(x_0,0)$. In the closing Sect. 9 we discuss this assumption. We conclude the paper with graphical illustration of the main contribution.

2 Preliminaries

In this section, we recall concepts and facts used throughout the paper.

Definition 1

For a given, possibly nonconvex, set $Q\subset E$ and $x\in Q$ the tangent (Bouligand) cone to Q is defined as

$$\begin{aligned} {{\mathcal {T}}}_{Q}(x):=\{d\in E\ |\ \exists \ \{x_{k}\}\subset Q\,\ \{t_{k}\}\subset \mathbb {R}\,\ x_{k}\rightarrow x,\, t_{k}\downarrow 0,\ (x_{k}-x)/t_{k}\rightarrow d\}. \end{aligned}$$

(6)

The cone

$$\begin{aligned} \varGamma _{{\mathcal {F}}}(x):=\{d\in E\ \mid DG(x)d\in {{\mathcal {T}}}_{K}(G(x))\} \end{aligned}$$

(7)

is called the linearized cone to ${{\mathcal {F}}}$ at x.

Definition 2

For any $x\in {{\mathcal {F}}}$, where ${{\mathcal {F}}}$ is given by (5), I(x) denotes the set of active (inequality) indices of ${{\mathcal {F}}}$ at x,

$$\begin{aligned} I(x):=\{ i\in I_1 \mid g_i(x)=0 \}. \end{aligned}$$

In the example below we recall standard properties of tangent cone.

Example 2

Let $G:E\rightarrow \mathbb {R}^{n}$, $G(x)=(g_{1}(x),\dots ,g_{n}(x))$ and $K=\mathbb {R}_{-}^n=\{y=(y_{i})\in \mathbb {R}^{n}\ |\ y_{i}\le 0,\ i=1,\dots n\}$, i.e.

$$\begin{aligned} {{\mathcal {F}}}=\{x\in E\ | \ g_{i}(x)\le 0, \ i=1,\dots , n\}, \end{aligned}$$

where $g_{i}:E\rightarrow \mathbb {R}$, $i=1,\dots , n$. Let us calculate ${{\mathcal {T}}}_{K}(G(x_0))$, $x_0\in E$, $G(x_0)\in \text {cl}\, K=K$.

• 1. If $G(x_0)\in \text {int}\, \mathbb {R}_{-}^n$, then ${{\mathcal {T}}}_{K}(G(x_0))=\mathbb {R}^n.$

• 2. If $G(x_0)\in \text {bd}\, \mathbb {R}_{-}^n$, then $I(x_0):= \{i\in \{1,\dots ,n\}\ |\ g_{i}(x_0)=0\}$ is nonempty and ${{\mathcal {T}}}_{K}(G(x_0))=\mathbb {R}_{-}^n=\{y=(y_{i})\in \mathbb {R}^{n}\ |\ y_{i}\le 0,\ \ i\in I(x_0)\}$.

In the following proposition, we provide a characterization of tangent cone to K given by (2).

Proposition 2

For any $y_0\in K$, where K is given by (2), we have

$$\begin{aligned} {{\mathcal {T}}}_{K}(y_{0}) = \left\{ z\in F\ \mid \ b_i^*(z) \le 0,\ i\in I(y_{0}), \ b_i^*(z)=0,\ i\in I_0\right\} , \end{aligned}$$

(8)

where $I(y_{0}):=\{i\in I_1\ \mid \ b_i^*(y_0)= 0\}$.

Proof

Take any $\tilde{z}\in {{\mathcal {T}}}_{K}(y_{0})$. By the definition, there exist $r(t)\in F$ and $\varepsilon _0>0$ such that $r(t)/t\rightarrow 0^+$ and

$$\begin{aligned} y_{0}+t\tilde{z}+r(t)\in K \ \ \ \forall \ \ t\in [0,\varepsilon _{0}). \end{aligned}$$

(9)

By this, for all $t\in [0,\varepsilon _{0})$

$$\begin{aligned} \begin{aligned}&b_{i}^* (t\tilde{z})+ b_{i}^* ( r(t))\le 0\ \ \ \forall \ \ i\in I(y_{0}),\\&b_{i}^* ( t\tilde{z})+b_{i}^*( r(t))= 0\ \ \ \forall \ \ i\in I_0,\end{aligned} \end{aligned}$$

(10)

and consequently

$$\begin{aligned}&b_{i}^* ( \tilde{z}\rangle + b_{i}^* \left( \frac{r(t)}{t}\right) \le 0\ \ \ \forall \ \ i\in I(y_{0}),\\&b_{i}^* ( \tilde{z}\rangle + b_{i}^* \left( \frac{r(t)}{t}\right) = 0\ \ \ \forall \ \ i\in I_0. \end{aligned}$$

Since $r(t)/t\rightarrow 0$ as $t\rightarrow 0^+$, we obtain ${{\mathcal {T}}}_{K}(y_{0})\subset \{z\in F\ \mid \ b_{i}^* ( z)\le 0,\ i\in I(y_{0}),\ b_{i}^* ( z)=0,\ i\in I_0\}$.

Now, to see the converse, take any $\tilde{z}\in \{z\in F\ \mid \ b_{i}^* (\tilde{z})\le 0,\ i\in I(y_{0}),\ b_{i}^* (\tilde{z})=0,\ i\in I_0\}$. We have

$$\begin{aligned} \tilde{z}=\lim _{k\rightarrow +\infty }z_k = \sum _{i=1}^k b_i^*(z)b_i. \end{aligned}$$

Since $z_k\in {{\mathcal {T}}}_{K}(y_{0})$, $k\in \mathbb {N}$ and ${{\mathcal {T}}}_{K}(y_{0})$ is closed, $\tilde{z}\in {{\mathcal {T}}}_{K}(y_{0})$. $\square$

Fact 1

Let $\bar{x} \in E$. Since G is assumed differentiable on E we have

$$\begin{aligned} 0&= \lim _{h\rightarrow 0} \frac{G(\bar{x}+h)-G(\bar{x})-DG(\bar{x})h}{\Vert h\Vert } \\&=\lim _{h\rightarrow 0}\frac{ \sum _{i\in \mathbb {N}} b_i^*(G(\bar{x}+h)-G(\bar{x})-DG(\bar{x})h)b_i}{\Vert h\Vert }\\&=\sum _{i\in \mathbb {N}} \lim _{h\rightarrow 0}\frac{ b_i^*(G(\bar{x}+h)-G(\bar{x})-DG(\bar{x})h)b_i}{\Vert h\Vert }\\&= \sum _{i\in \mathbb {N}} \lim _{h\rightarrow 0}\frac{ g_i(\bar{x}+h)-g_i(\bar{x})-Dg_i(\bar{x})h}{\Vert h\Vert } b_{i}. \end{aligned}$$

Hence, due to the uniqueness of the representation of elements of the space F in basis $(b_{i})_{i\in \mathbb {N}}$, the coefficients $g_{i}(\cdot )=b_i^{*}(g(\cdot ))$, $i\in \mathbb {N}$, are differentiable for all $x\in E$ and $b_i^*( DG(\bar{x})z)= Dg_i(x)z$ for any $z\in E$, $i\in \mathbb {N}$.

As a consequence we obtain the following proposition.

Proposition 3

Let $x_0\in {{\mathcal {F}}}$, where ${{\mathcal {F}}}$ is given by (5). Then

$$\begin{aligned} \begin{aligned} \varGamma _{{{\mathcal {F}}}}(x_{0})=&\left\{ d\in E \mid DG(x_{0}) d\in {{\mathcal {T}}}_{K}(G(x_{0}))\right\} \\ \subset&\{d\in E \mid Dg_{i}(x_{0}) d= 0,\ \ i\in I_0,\\ {}&Dg_{i}(x_{0}) d\le 0,\ \ i\in I(x_{0})\}, \end{aligned} \end{aligned}$$

(11)

where $I(x_{0}):=\{i\in I_1 \mid g_{i}(x_{0})=0\}$.

Proof

Take any $d \in \varGamma _{{\mathcal {F}}}(x_0)$. This means that

$$\begin{aligned} DG(x_0) d \in {{\mathcal {T}}}_{K}(G(x_0)). \end{aligned}$$

(12)

By Proposition 2,

$$\begin{aligned} {{\mathcal {T}}}_{K}(G(x_0))\subset \{z\in F\ \mid \ b_i^*(z) = 0,\ i\in I_0,\ b_i^*(z) \le 0,\ i\in I(G(x_0))\}. \end{aligned}$$

Hence, by (12),

$$\begin{aligned} b_i^*( DG(x_0) d )= 0, \ i \in I_0,\ b_i^*( DG(x_0) d )\le 0, \ i \in I(G(x_0)). \end{aligned}$$

Now, (11) follows from Fact 1. $\square$

2.1 Boundedly-Complete, Shrinking, Besselian and Hilbertian Bases

In this subsection we recall basic definitions and facts related selected types of bases in Banach spaces. These concepts will be extensively used in the sequel.

Definition 3

The closed subspace $E_1$ of the Banach space E is said to be split, or complemented, if there is a closed subspace $E_2 \subset E$ such that $E = E_1 \oplus E_2$.

Proposition 4

([1, Theorem 2.1.15]) If F is a Hilbert space and $F_1$ a closed subspace, then $F=F_1\oplus F_1^\perp$. Thus every closed subspace of a Hilbert space splits (see e.g. Definition 2.1.14 of [1]).

Definition 4

([2, Definition 3.2.8]) Let X be a Banach space. A sequence $(v_n)_{n\in \mathbb {N}}$ in X is boundedly-complete if whenever $(a_n)_{n\in \mathbb {N}}$ is a sequence of scalars such that

$$\begin{aligned} \sup _{N\in \mathbb {N}} \left\| \sum _{n=1}^N a_n v_n \right\| < +\infty \end{aligned}$$

then the series $\sum _{n\in \mathbb {N}}a_n v_n$ converges.

Remark 3

Let $(v_n)_{n\in \mathbb {N}}$ in X be boundedly-complete. Then every subsequence $(v_{n_k})_{k \in \mathbb {N}}$ is boundedly-complete. Indeed, suppose that $(v_n)_{n\in \mathbb {N}}$ is boundedly-complete and $(a_{k})$, $k\in \mathbb {N}$, is such that $\sup _{N\in \mathbb {N}} \Vert \sum _{k=1}^N a_{k} v_{n_k} \Vert < \infty$. Let

$$\begin{aligned} \bar{a}_k=\left\{ \begin{array}{ll} a_{k}, &{}\quad k\in (n_l)_{l\in \mathbb {N}},\\ 0, &{}\quad \text {otherwise}. \end{array} \right. \end{aligned}$$

Then

$$\begin{aligned}&\sup _{N\in \mathbb {N}} \left\| \sum _{k=1}^N a_{k} v_{n_k} \right\|< \infty \iff \sup _{N\in \mathbb {N}} \left\| \sum _{n=1}^N \bar{a}_n v_n\right\| < \infty \\&\quad \implies \sum _{n\in \mathbb {N}} \bar{a}_n v_n \text { converges} \iff \sum _{k\in \mathbb {N}}a_{k} v_{n_k} \text { converges}. \end{aligned}$$

Definition 5

([2, Definition 3.2.5]) A basis $(v_n)_{n\in \mathbb {N}}$ of a Banach space X is shirnking if the sequence of its bioorthogonal functionals $(v_n^*)_{n\in \mathbb {N}}$ is a basis of $X^*$, i.e., $\overline{{\text {span}}}(v_n^*,\ n\in \mathbb {N})=X^*$.

Proposition 5

([2, Theorem 3.2.10]) Let $(v_n, v_n^*)$ be a biorthogonal system in a Banach space X. The following are equivalent:

1.:

$(v_n)_{n\in \mathbb {N}}$ is boundedly-complete,

2.:

$(v_n^*)_{n\in \mathbb {N}}$ is shrinking basis for $\overline{{\text {span}}}(v_n^{*},\ n\in \mathbb {N})$,

3.:

the canonical map ${\text {eval}}_X:\ X\rightarrow \overline{{\text {span}}}(v_n^*,\ n\in \mathbb {N})^*$ defined by ${\text {eval}}_X (x)(h)=h(x)$ for all $x\in X$ and $h\in \overline{{\text {span}}}(v_n^*,\ n\in \mathbb {N})$, is an isomorphism.

Remark 4

(Corollary 3.2.11 of [2]) Every Schauder basis of a Hilbert space is boundedly-complete. $c_0$ has no boundedly-complete basis.

Let us recall James Theorem from 1951.

Theorem 2

(James Theorem, see [2, Theorem 3.2.13]) Let X be a Banach space. If X has a basis $(v_n)_{n\in \mathbb {N}}$ then X is reflexive if and only if $(v_n)_{n\in \mathbb {N}}$ is both boundedly-complete and shrinking.

Definition 6

([26, Definition 11.1]]) We say that basis $(c_i)_{i\in \mathbb {N}}$ in a real Banach space is

• 1. Besselian if

  $$\begin{aligned} \sum _{i\in \mathbb {N}} \alpha _i c_i \quad \text {converges} \implies \sum _{i\in \mathbb {N}} (\alpha _i)^2<+\infty . \end{aligned}$$

• 2. Hilbertian if

  $$\begin{aligned} \sum _{i\in \mathbb {N}} (\alpha _i)^2<+\infty \implies \sum _{i\in \mathbb {N}} \alpha _i c_i \quad \text {converges}, \end{aligned}$$

  i.e., for every $\alpha _i \in \mathbb {R}$, $i\in \mathbb {N}$, with $\sum _{i\in \mathbb {N}} (\alpha _i)^2<+\infty$ there exists an (obviously unique) x such that

  $$\begin{aligned} c_i^*(x)=\alpha _i, \quad i\in \mathbb {N}. \end{aligned}$$

Remark 5

The natural basis of $\ell _2$ is Besselian. Not all bases in Hilbert spaces are Besselian or Hilbertian. For $L_2[-\pi ,\pi ]$ see [26, Example 11.2].

Remark 6

([27, Corollary 4.6]) The space $L_1[0,1]$ has a Besselian basis.

We say that sequence $(b_i)_{i\in \mathbb {N}}\in F$ is a basic sequence if $(b_i)_{i\in \mathbb {N}}$ is Schauder basis of $\overline{{\text {span}}}( b_i, i\in \mathbb {N} )$, where $\overline{{\text {span}}}$ denotes the closure of the span and the closure is taken in the strong topology of the space.

Lemma 1

([17, Lemma 3.1]) The basis $(c_i)_{i\in \mathbb {N}}$ of Banach space X is Hilbertian (Besselian) if and only if the basic sequence $(c_i^*)_{i\in \mathbb {N}}$ in $X^*$ is Besselian (Hilbertian).

Proposition 6

([26, Proposition 3.1]) Let $\{v_n\}$ be a sequence in Banach space X, and assume that $v_n\ne 0$ for every n.

Define $Y:=\{ (c_n)\mid \sum _{n} c_n v_n\quad \text {converges in } X\}$ and the set

$$\begin{aligned} \Vert (c_n)\Vert _Y:= \sup _{N} \left\| \sum _{n=1}^N c_n v_n \right\| . \end{aligned}$$

(13)

Then the following hold:

• 1. Y is a Banach space.

• 2. If $\{v_n\}$ is a basis for X, then Y is topologically isomorphic to X via the mapping $(c_n)\mapsto \sum _{n} c_n v_n$.

For any infinite subset $J=\{ j_1,j_2,\dots \}\subset \mathbb {N}$ let us denote

$$\begin{aligned} Y(J)=\left\{ (c_i)_{i\in J}\mid \sum _{i \in J} c_i v_i\quad \text {converges in } X\right\} . \end{aligned}$$

(14)

with the associated norm

$$\begin{aligned} \Vert (c_i)\Vert _{Y(J)}:= \sup _{N} \left\| \sum _{i=1}^N c_{j_i} v_i \right\| \end{aligned}$$

(15)

and $\ell _2(J):=\{ (c_j)_{j\in J} \mid \sum _{i\in J} (c_j)^2 < +\infty \}$ with the associated norm

$$\begin{aligned} \Vert (c_j)_{j\in J} \Vert _{\ell _2(J)}= \sum _{i\in J} (c_j)^2. \end{aligned}$$

Remark 7

Observe that if X is a Hilbert space with the inner product $\langle \cdot \ |\ \cdot \rangle$, the norm $\Vert x\Vert =\sqrt{\langle x\ |\ x\rangle }$ and the orthonormal basis $(x_{i})_{i\in J}$, then

$$\begin{aligned} \Vert (c_{i})\Vert _{Y(J)}\begin{array}{l} =\sup _{N}\sqrt{\left\langle \sum _{i=1}^N c_{j_i} x_i\ |\ \sum _{i=1}^N c_{j_i} x_i\right\rangle }\\ =\sup _{N}\sqrt{\sum _{i=1}^N\sum _{k=1}^N \left\langle c_{j_i} x_i\ |\ c_{j_k} x_k \right\rangle }\\ =\sup _{N}\sqrt{\sum _{i=1}^N (c_{j_i})^{2}}=\sqrt{\sup _{N}\sum _{i=1}^N (c_{j_i})^{2}} \\ =\sqrt{\sum _{i\in J} (c_{i})^{2}}=\Vert (c_{i})_{i\in J}\Vert _{\ell _{2}(J)}. \end{array} \end{aligned}$$

By James Theorem (Theorem 2), $(x_{i})_{i\in J}$ is boundedly-complete in X. In view of this, when $(x_{i})_{i\in J}$ is an orthonormal, then $Y(J)=\ell _{2}(J):=\overline{{\text {span}}}(e_{i},\ i\in J)$, where $e_{i}^{j}=0$ if $j=i$ and 0 otherwise, for all $i,j\in J$.

3 Constant Rank Condition

In this section we introduce the Constant Rank Condition (CRC)² for a possibly infinite family of functions defined on a Banach space via Schauder basis. For other forms of CRC which do not refer to Schauder basis, see e.g. [9].

Let E be a Banach space and F a Hilbert space with basis $(b_{i})_{i\in \mathbb {N}}$. Consider $f:E\rightarrow \overline{{\text {span}}}(b_{i},\ i\in \mathbb {N})$, i.e., $f(x)=\sum _{i\in \mathbb {N}}f_{i}(x)b_{i}$, where $f_{i}=b_{i}^{*}(f(x)):E\rightarrow \mathbb {R}$, $i\in \mathbb {N}$ are continuous functionals.

Definition 7

Let $(f_{i})_{i\in J_1}:\ E \rightarrow Y(J_1)$, $J_1\subset \mathbb {N}$ be of class $C^1$. We say that the Constant Rank Condition (CRC in short) holds for $(f_{i})_{i\in J_1}$ at $x_0\in E$ if there exist a neighbourhood $V(x_0)$ and a subset $J_2\subset J_1$ such that

1.:

$(Df_i(x))_{i\in J_2}$ forms a Schauder basis for $\overline{{\text {span}}}(D f_{i}(x), i \in J_1)$ for all $x\in V(x_0)$,

2.:

for any $x\in V(x_0)$, there exists a topological isomorphism (linear)

$$\begin{aligned} z_x: \overline{{\text {span}}}( D f_{i}(x), i \in J_2) \rightarrow \overline{{\text {span}}}(D f_{i}(x_0), i \in J_2), \end{aligned}$$

(16)

such that $z_x(D f_{i}(x))= D f_{i}(x_0)$, $i \in J_2$.

We interpret $z_x$ as an isomorphism of functionals, i.e., $z_x(Df_i(x)(\cdot ))=Df_i(x_0)(\cdot )$, $x\in V(x_0)$, $i\in J_2$, hence, by (16), $(Df_i(x_0))_{i\in J_2}(E)$ and $(Df_i(x))_{i\in J_2}(E)$ are isomorphic for all $x\in V(x_{0})$.

Remark 8

Let us note that, for $(f_{i})_{i\in J_1}$, $J_1\subset \mathbb {N}$, where $J_1$ is finite, the condition

$$\begin{aligned} \text {rank}\,\{ Df_i(x_0), i\in J_1 \} = \text {rank}\,\{ Df_i(x), i\in J_1 \}, \quad x\in V(x_0), \end{aligned}$$

is equivalent to the existence of isomorphism $z_x$ given in Definition 7 for $(f_{i})_{i\in J_1}$ at $x_0$ (see [7, Definition 2.1]).

Definition 8

([2, Definition 1.3.1]) Two bases (or basic sequences) $(u_n)_{n\in \mathbb {N}}$ and $(v_n)_{n\in \mathbb {N}}$ in the respective real Banach spaces X and Y are said to be equivalent, if whenever we take a sequence of scalars $(a_n)_{n\in \mathbb {N}}$, then $\sum _{n\in \mathbb {N}}a_n u_n$ converges if and only if $\sum _{n\in \mathbb {N}}a_n v_n$ converges.

Let us recall the following fact.

Fact 2

([2, Theorem 1.3.2]) Let $(u_i)_{i\in J}$, $J\subset \mathbb {N}$, be a basic sequence in a Banach space X and let $(v_i)_{i\in J}$ be a basic sequence in a Banach space Y. The following are equivalent:

• $(u_i)_{i\in J}$ is a basic sequence equivalent to $(v_i)_{i\in J},$

• There is an isomorphism T of $\overline{{\text {span}}}(u_i )_{i\in J}$ onto $\overline{{\text {span}}}(v_i)_{i\in J}$ s.t. $T(u_i)=v_i$ for every $i\in J$.

In other words, condition 2. of CRC means that a basic sequence $(Df_i(x))_{i\in J_2}$ is equivalent to a basic sequence $(Df_i(x_0))_{i\in J_2}$ in $E^{*}$ for all x from some neighbourhood $V(x_0)$.

In the sequel, e.g. in Definition 10, we will use CRC with some additional conditions, which motivates the following definition.

Definition 9

Let $(f_{i})_{i\in J_1}:\ E \rightarrow Y(J_1)$, $J_1\subset \mathbb {N}$ be of class $C^1$. We say that the Constant Rank Condition Plus (CRC+ in short) holds for $(f_{i})_{i\in J_1}$ at $x_0\in E$ if there exist a neighbourhood $V(x_0)$ and a subset $J_2\subset J_1$ such that

1.:

$(Df_i(x))_{i\in J_2}$ forms a Schauder basis for $\overline{{\text {span}}}(D f_{i}(x), i \in J_1)$ for $x\in V(x_0)$,

2.:

for all $x\in V(x_0)$, there exists a topological isomorphism

$$\begin{aligned} z_x: \overline{{\text {span}}}( D f_{i}(x), i \in J_2) \rightarrow \overline{{\text {span}}}(D f_{i}(x_0), i \in J_2), \text{ for } \text{ all } x\in V(x_0) \end{aligned}$$

such that $z_x(D f_{i}(x))= D f_{i}(x_0)$, $i \in J_2$

and additionally

3.:

$(Df_i(x_0))_{i\in J_2}(E)$ is closed in $Y(J_2)$ defined in (14) with $v_i=b_i$, $i\in J_2$,

4.:

$(Df_i(x_0))_{i\in J_2}$ forms a basis which is shrinking and boundedly-complete for $\overline{{\text {span}}}(Df_i(x_0),\ i\in J_1)$, equivalently $\overline{{\text {span}}}(Df_i(x_0),\ i\in J_1)$ is reflexive and $(Df_i(x_0))_{i\in J_2}$ forms a basis of this space,

5.:

$(Df_i(x_0))_{i\in J_2}$ is Besselian for $\overline{{\text {span}}}(Df_i(x_0),\ i\in J_1)$,

6.:

$Df_i(x_0)^*\in E$, $i\in J_2$.

In view of Remark 7 if $(b_{i})_{i\in J_{2}}$ is orthonormal, then, in Definition 9, 3. we have $Y(J_{2})=\ell _{2}(J_{2})$.

Remark 9

If E is a reflexive space, then CRC+ for $(f_{i})_{i\in J_1}:\ E \rightarrow Y(J_1)$, $J_1\subset \mathbb {N}$ takes the following form: there exist a neighbourhood $V(x_0)$ and a subset $J_2\subset J_1$ such that

1.:

$(Df_i(x))_{i\in J_2}$ forms a Schauder basis for $\overline{{\text {span}}}(D f_{i}(x), i \in J_1)$ for all $x\in V(x_0)$,

2.:

for any $x\in V(x_0)$, there exists a topological isomorphism

$$\begin{aligned} z_x: \overline{{\text {span}}}(D f_{i}(x), i \in J_2) \rightarrow \overline{{\text {span}}}(D f_{i}(x_0), i \in J_2), \end{aligned}$$

such that $z_x(D f_{i}(x_0))= D f_{i}(x)$, $i \in J_2$,

and

3.:

$(Df_i(x_0))_{i\in J_2}(E)$ is closed in $Y(J_2)$ defined in (14), with $v_i=b_i$, $i\in J_2$,

4.:

$(Df_i(x_0))_{i\in J_2}$ is Besselian for $\overline{{\text {span}}}(Df_i(x_0),\ i\in J_1)$.

Remark 10

Let us remark that, if $J_1$ is finite and E is reflexive, then 3., 4., 5., 6. of Definition 9 are automatically satisfied.

In the definition of CRC+ we assume the closedness of $((Df_i(x_0))_{i\in J_2})(E)$. Let us recall that the following sets do not have to be closed, as shown in the example below.

Example 3

([10]) Let $G:\ \ell _2 \rightarrow \ell _2$ be defined as $G(x)=(\frac{1}{2i}x_i^2)_{i\in \mathbb {N}}$. Then $DG(\cdot ):=\sum _{n=1}^{\infty } \frac{1}{n} e_n \langle e_n \mid \cdot \rangle$ and

• 1. $DG(x)\in L(\ell _2,\ell _2)$ (with $\Vert DG(x)\Vert =1$),

• 2. $v\in DG(x)(\ell _2)$ if and only if $\sum _{n\in \mathbb {N}} n^2 |\langle e_n \mid v \rangle |^2<+\infty$,

• 3. with $v_0:=\sum _{n\in \mathbb {N}} \frac{1}{n^{\frac{3}{2}}}e_n$ and $\{v_j\}:=\sum _{n\in \mathbb {N}} \frac{1}{n^{\frac{3}{2}+\frac{1}{j}}}e_n$ we have $v_0\notin DG(x)(\ell _2)$ yet $v_j\in DG(x)(\ell _2)$ and $v_j\rightarrow v_0$ as $j\rightarrow +\infty$.

4 Complemented Kernels and Isomorphisms

In this section we prove Propositions 7 and 8 which, together with CRC+ will be used in the next sections. We believe that the mentioned propositions are of independent interest.

We start with Proposition 7 providing conditions for a $C^{1}$ mapping f ensuring that the kernel of its derivative $Df(x_{0})$ is complemented. Next, it is used in the proof of Lemma 2 which is, in turn, used in the proof of Proposition 8.

Proposition 7

Let E be a Banach space, F be Banach space with a Besselian Schauder basis $(b_i)_{i\in \mathbb {N}}$. Let $(f_{i})_{i\in J}:\ E \rightarrow Y(J)$, $J\subset \mathbb {N}$ be of class $C^1$, $f:\ E \rightarrow F$, $f(x):=\sum _{i\in J} f_i(x)b_i$, $x\in E$. Let $x_0\in E$ and $E_2=\ker Df(x_0)$, $X_1:=\overline{{\text {span}}}(Df_i(x_0),\ {i\in J})$, and assume that:

• (A) $(Df_i(x_0))_{i\in J}$ forms a shrinking and boundedly-complete basis for $X_1$,

• (B) $(Df_i(x_0))_{i\in J}$ is Besselian for $X_1$,

• (C) $Df_i(x_0)^*\in E$, $i\in J$.

Then $E=E_1 \oplus E_2$, where $E_1= \overline{{\text {span}}}(Df_i(x_0)^*,\ {i\in J}).$ Moreover, $X_1$ is a reflexive space.

Proof

By (A), the fact that $X_1$ is a reflexive space follows immediately from James Theorem (Theorem 2). Let $v_{i}:=D f_i(x_0)\in E^*$, $i\in J$. Since $(Df_i(x_0))_{i\in J}$ forms a boundedly-complete basis for $X_1:=\overline{{\text {span}}}(Df_i(x_0),\ {i\in J})$, by Proposition 5, there exists canonical isomorphism ${\text {eval}}_{X_1}:\ X_1\rightarrow \overline{{\text {span}}}( (Df_i(x_0)^*,\ {i\in J})^*$ defined as

$$\begin{aligned} {\text {eval}}_{X_1}(v)(u^*)= u^* (v) \quad \text {for every} \ v\in X_1,\ u^*\in \overline{{\text {span}}}(v_n^*). \end{aligned}$$

We have $v_{i}^{*}(v_{j})=v_{j}(v_{i}^{*})=Df_i(x_0)Df_{j}(x_{0})^{*}$ for all $i\in J$. By (1), (C), Fact 1 and Proposition 5

$$\begin{aligned} \begin{aligned}&Df(x_0)(Df_i(x_0)^*)= \sum _{j\in J} Df_j(x_0)(Df_i(x_0)^*)b_j=b_i ,\ i\in J. \end{aligned} \end{aligned}$$

(17)

Now we show that $E=X_{1}^{*}\oplus \text {ker }Df(x_{0})$. For any $x\in E$, $Df(x_{0})(x)= \sum _{j\in J} Df_j(x_0)(x)b_j=\sum _{i\in J} \alpha _i(x) b_i$, $\alpha _i(x)=Df_i(x_0)(x)\in \mathbb {R}$. Since $(b_i)_{i\in \mathbb {N}}$ is Besselian for F, $\sum _{i\in J} (\alpha _i(x))^2<+\infty$. Let $m:=\sum _{i\in J}\alpha _i(x) Df_{i}(x_{0})^{*}$. By (B) and Lemma 1, $(Df_i(x_0)^*)_{i\in J}$ is Hilbertian for $X_1^*$, m is well defined, i.e., $m\in X_{1}^{*}$.

By (17), we have

$$\begin{aligned} Df(x_{0})(x-m)&=Df(x_{0})(x)-Df(x_{0})(m)\\&= \sum _{i\in J}\alpha _{i}(x)b_{i}- Df(x_0)\left( \alpha _i(x)\sum _{i\in J} Df_i(x_0)^*\right) \\&=\sum _{i\in J}\alpha _{i}(x)b_{i}-\sum _{i\in J}\alpha _i(x)Df(x_{0})(Df_{i}(x_{0})^{*})\\&=\sum _{i\in J}\alpha _{i}(x)b_{i}-\sum _{i\in J}\alpha _i(x)b_i=0. \end{aligned}$$

This shows that $x-m\in \text {ker}\, Df(x_{0})$ which proves the assertion with $E_{1}:=X_{1}^{*}$ and $E_{2}:=\text {ker}\, Df(x_{0}).$ $\square$

Remark 11

Let us note that space $\ell _1$ contains no infinite-dimensional reflexive subspaces.

Corollary 1

Let E be a reflexive Banach space, F be Hilbert space with a Besselian Schauder basis $(b_i)_{i\in \mathbb {N}}$. Let $(f_{i})_{i\in J}:\ E \rightarrow Y(J)$, $J\subset \mathbb {N}$ be of class $C^1$, $f:\ E \rightarrow F$, $f(x):=\sum _{i\in J} f_i(x)b_i$, $x\in E$. Let $x_0\in E$ and $E_2=\ker Df(x_0)$, $X_1:=\overline{{\text {span}}}(Df_i(x_0),\ {i\in J})$ and assume that:

• (A) $(Df_i(x_0))_{i\in J}$ is a basis for $X_1$,

• (B) $(Df_i(x_0))_{i\in J}$ is Besselian for $X_1$.

Then $E=E_1 \oplus E_2$, where $E_1=\overline{{\text {span}}}(Df_i(x_0)^*,\ {i\in J}).$

Proof

The proof follows directly from James Theorem applied to $X_1$ and Proposition 7.$\square$

Remark 12

Assume that CRC holds for $(f_i)_{i\in J}$ at $x_0$ with $V(x_{0})$ as in Definition 7. Then, by 2. of CRC

$$\begin{aligned} \forall _{k\in J_2}\ \forall _{e\in E}\ z_{x}\circ (b^{*}_{k}(Df(x_{0}))(e)=b^{*}_{k}(Df(x)(e)) \end{aligned}$$

and

$$\begin{aligned} b_l^*(Df(x)(e))=\sum _{i\in J_2} \beta _{i} Df_i(x)(e), \quad l\in J\setminus J_2,\ x\in V(x_0). \end{aligned}$$

Hence

$$\begin{aligned} e\in \ker Df(x)\Leftrightarrow & {} \forall _{k\in J}\ b_{k}^{*}(Df(x)(e))=0\nonumber \\\Leftrightarrow & {} \forall _{k\in J_2}\ z_{x}\left( b_{k}^{*}(Df(x_{0})(e))\right) =0\nonumber \\\Leftrightarrow & {} \forall _{k\in J}\ b_{k}^{*}(Df(x_{0})(e))=0 \ \Leftrightarrow e\in \ker Df(x_{0}). \end{aligned}$$

(18)

For any $f:E\rightarrow F$, and $J_1\subset \mathbb {N}$, we have $f(x)=\sum _{i\in J_1} b_i^*(f(x))b_i$, $x\in E$, where $f_{i}(x)=b_i^*(f(x))=0$ for $i\in \mathbb {N}{\setminus } J_1$. Consequently, for any $e\in E$,

$$\begin{aligned} Df(x)(e)=\sum _{j\in J_1}b_j^*(Df(x)e)b_i=\sum _{j\in J_1}(Df_{j}(x)e)b_j, \end{aligned}$$

(19)

where, by Fact 1, $Df_{j}(x)e=b_j^*(Df(x)e)$, $j\in J_1$.

In the lemma below we investigate the coefficients $Df_{j}(x)e=b_j^*(Df(x)e)$, $j\in J_1$, $e\in E$, in a neighbourhood of $x_{0}$ at which CRC+ holds.

Lemma 2

Let E be a Banach space, F be a Hilbert space and $(b_i)_{\in \mathbb {N}}$ be a Besselian basis of F. Let $(f_{i})_{i\in J}:\ E \rightarrow Y(J)$, $J\subset \mathbb {N}$ be of class $C^1$, $f:\ E \rightarrow F$, $f(x):=\sum _{i\in J} f_i(x)b_i$, $x\in E$. Assume that CRC+ holds for $(f_i)_{i\in J_1}$ at $x_0$ with $J_2\subset J_1$ and neighbourhood $V(x_0)\subset U$.

Let $x\in V(x_0)$, $e\in E$. Then there exist scalars $\beta _{j}=\beta _{j}(e)$, $j \in J_{2}$ depending on e but not on x such that

$$\begin{aligned} Df_i(x)(e)=\sum _{j\in J_2}\beta _j(e)w_j^i(x), \quad i\in J_1, \end{aligned}$$

(20)

where

$$\begin{aligned} w_j^i(x):=b_i^*(Df(x) (D f_{j}(x_0)^*)),\quad i,j\in J_2 \end{aligned}$$

and $D f_{i}(x_0)^*\in E$, $i\in J_2,$ (see Proposition 1) are such that

$$\begin{aligned} w_{j}^{i}(x_{0})=D f_{i}(x_0)^*(D f_{j}(x_0))=\left\{ \begin{array}{ll} 1 &{}\quad \text {if } i=j,\\ 0 &{}\quad \text {if } i\ne j \end{array} \quad i,j \in J_2. \right. \end{aligned}$$

(21)

Proof

The existence of $Df_i(x_0)^*\in E^{**}$, $i\in J_2$ is ensured by Proposition 1. By 6. of CRC+, $Df_i(x_0)^*\in E$, $i\in J_2$. Since, by 5. of CRC+, $D f_{i}(x_0)$, $i\in J_2$ forms a basis of $\overline{{\text {span}}}( Df_i(x_0),\ i \in J_1)$ by Proposition 1, $D f_{i}(x_0)^*$, $i\in J_2$ forms a basis of $(\overline{{\text {span}}}( Df_i(x_0),\ i \in J_1))^*$.

By Proposition 7,

$$\begin{aligned} \begin{aligned}&e=e_{1}+e_{2},\ e_{2}\in \text {ker } Df(x_{0}),\\&e_{1}=\sum _{k\in J_{2}}\beta _{k}(e)Df_{k}^{*}(x_{0}),\ \beta _{k}(e)\in \mathbb {R},\ k\in J_2. \end{aligned} \end{aligned}$$

(22)

By 2. of CRC+ and (18), $e_{2}\in \text {ker } Df(x)$ for $x\in V(x_{0})$ and for $i\in J_1$ we have

$$\begin{aligned} \begin{aligned} Df_i(x)(e)&=b_i^*(Df(x)(e_1+e_2))=b_i^*\left( Df(x)\sum _{k\in J_{2}}\beta _{k}(e)Df_{k}^{*}(x_{0})\right) \\&=\sum _{k\in J_{2}}\beta _{k}(e)b_{i}^{*}(Df(x)Df_{k}^{*}(x_{0})). \end{aligned} \end{aligned}$$

(23)

$\square$

Remark 13

In other words, (20) means that for $x\in V(x_{0})$ and $i\in J_{2}$

$$\begin{aligned} Df_i(x)(e)=\sum _{j\in J_2}\beta _j(e)w_j^{i}(x)=\sum _{j\in J_2}\beta _j(e)b_i^*(Df(x) (D f_{j}(x_0)^*)). \end{aligned}$$

(24)

In particular, for $x=x_{0}$ and $i\in J_{2}$,

$$\begin{aligned} Df_i(x_{0})(e)=\sum _{j\in J_2}\beta _j(e)w_j^{i}(x_{0})=\sum _{j\in J_2}\beta _j(e)b_i^*(Df(x_{0}) (D f_{j}(x_0)^*))=\beta _{i}(e).\nonumber \\ \end{aligned}$$

(25)

Remark 14

By Proposition 1, vectors $w_j(x_0)$, $j\in J_2$ are such that

$$\begin{aligned} (w_{j}^{i}(x_0))_{i\in J_2}= \left\{ \begin{array}{rc} 1, &{}\quad i=j, \\ 0, &{}\quad i\ne j. \end{array} \right. \end{aligned}$$

Observe that by (23), for $x=x_{0}$ and $j\in J_{1}{\setminus } J_{2}$

$$\begin{aligned} \sum _{k\in J_{2}}\beta _{k}(e)b_{j}^{*}(Df(x_{0})Df_{k}^{*}(x_{0}))=0 \end{aligned}$$

because $Df(x_{0})Df_{k}^{*}(x_{0})=\sum _{i\in J_1} Df_i(x_0)(Df_k^*(x_0))b_i=b_{k}$ for $k\in J_{2}$ and, by the definition of $b^{*}_{j}$,

$$\begin{aligned} b^{*}_{j}(b_{k})=0\ \ \text {for}\ \ j\in J_{1}\setminus J_{2}. \end{aligned}$$

Hence,

$$\begin{aligned} Df(x_{0})(e)=\sum _{j\in J_1}Df_{j}(x_{0})(e)b_{j} =\sum _{j\in J_2}\beta _j(e)b_{j}. \end{aligned}$$

(26)

In the following proposition we prove that CRC+ ensures that the mapping defined by (27) is an isomorphism. This proposition together with the Rank Theorem (Theorem 3) will allow us to prove Proposition 9 which is crucial in the proof of the main result in Sect. 7.

Proposition 8

Let E be a Banach space, F be a Hilbert space and assume that $(b_i)_{i\in \mathbb {N}}$ is a Besselian basis of F and $J_1\subset \mathbb {N}$. Let $(f_i)_{i\in J_1}:\ U \rightarrow Y(J_1)$, $U\subset E$ open, be a $C^{1}$ mapping. Assume that CRC+ holds for $(f_i)_{i\in J_1}$ at $x_0$ with a neighbourhood $V(x_0)$ with index set $J_2\subset J_1$.

Then

$$\begin{aligned} t_x:=((Df_i(x))_{i\in J_1})\big |_{E_1}: E_1 \rightarrow (Df_i(x))_{i\in J_1}(E),\quad \forall x\in V(x_0) \end{aligned}$$

(27)

is an isomorphism, where $E_1:=X_1^*=\overline{{\text {span}}}(Df_i(x_0)^*,\ i\in J_2)$.

Proof

By 1. of CRC+, for any $x\in V(x_0)$ for any $l\in J_1{\setminus } J_2$, $Df_l(x)(e_1)$ can be expressed by $Df_i(x)(e_1)$, $i\in J_2$, i.e., there exists scalars $\alpha _i^l (x)$, $i\in J_2$, $l\in J_1{\setminus } J_2$ such that

$$\begin{aligned} Df_l(x)(e_1)=\sum _{i\in J_2}\alpha _i^l (x)Df_i(x)(e_1). \end{aligned}$$

Since $e_1\in E_1$, by Lemma 2, we have

$$\begin{aligned} Df_i(x)(e_1) = \sum _{j\in J_2} \beta _j(e_1) Df_i(x) Df_j(x_0)^*, \quad i\in J_2. \end{aligned}$$

Since $(f_i)_{i\in J_1}$ is a $C^{1}$ mapping, $t_{x}$ is a continuous (linear) mapping. Now, we show that $t_{x}$, $x\in V(x_{0})$, is a bijection.

1. Step: injectivity. Let $x\in V(x_0)$. Let us take $x_1, x_2 \in E_1$. Suppose that $(Df_i(x))_{i\in J_1}(x_1)=(Df_i(x))_{i\in J_1}(x_2)$. Then $(Df_i(x))_{i\in J_1}(x_1-x_2)=0$ and, by Remark 12, $x_1-x_2\in \ker (Df_i(x)_{i\in J_2})=\ker (Df_i(x_0)_{i\in J_1})=\bigcap _{i\in J_1} Df_i(x_0)$. On the other hand $x_1-x_2\in E_1$, hence $x_1=x_2$.

2. Step: surjectivity. Let $e\in E$ and $x\in V(x_0)$. By 1. of CRC+,

$$\begin{aligned} Df_l(x)(e)=\sum _{i\in J_2} \alpha _i^l(x)Df_i(x)(e),\quad l\in J_1\setminus J_2. \end{aligned}$$

For $i\in J_2$

$$\begin{aligned} \begin{aligned} Df_i(x)(e)&=\sum _{j\in J_2} \beta _j(e) (Df_i(x)Df_j(x_0)^*)\\&=Df_i(x)\left( \sum _{j\in J_2} \beta _j(e) Df_j(x_0)^*\right) \end{aligned} \end{aligned}$$

(28)

and for $l\in J_1\setminus J_2$

$$\begin{aligned} \begin{aligned} Df_l(x)(e)&= \sum _{i\in J_2} \alpha _i^l(x)Df_i(x)(e)\\&= \sum _{i\in J_2} \alpha _i^l(x)Df_i(x) \left( \sum _{j\in J_2} \beta _j(e) Df_j(x_0)^*\right) \\&= \sum _{i\in J_2} \alpha _i^l(x)Df_i(x) \left( e_1 \right) \\&=Df_l(x)\left( e_1 \right) , \end{aligned} \end{aligned}$$

(29)

where $e_1=\sum _{j\in J_2} \beta _j(e)Df_j(x_0)^*\in E_1$ (hence $Df_i(x)(e)=Df_i(x)(e_1)$, $i\in J_2$). Since $Df_i(x)(e)=Df_i(x)(e_1)$, $i\in J_2$ and $Df_l(x)(e)=Df_l(x)(e_1)$, $l\in J_1\setminus J_2$, we obtain $(Df_i(x))_{i\in J_1}(e)=(Df_i(x))_{i\in J_1}(e_1)$. $\square$

5 CRC+ and Functional Dependence

In this section we use CRC+ to prove Proposition 9 which provides conditions for the functional dependence in the form of formula (30) and is based on Propositions 7, 8, and Rank Theorem (Theorem 3). Proposition 9 will be used in the proof of the main result, Theorem 6 of Sect. 7.

For convenience of the reader, we start by recalling the rank and the local representation theorems.

Theorem 3

(Rank Theorem, see [1, Theorem 2.5.15]) Let $E,\ Y$ be Banach spaces. Let $x_0\in U$, where U is an open subset of E and $f:\ U\rightarrow Y$ be of class $C^1.$

Assume that $Df(x_0)$ has a closed split image $Y_1$ with closed component $Y_2$ and a split kernel $E_2$ with closed component $E_1$ and that for all x in some neighbourhood of $x_0$, $Df(x)|E_1:\ E_1 \rightarrow Df(x)(E)$ is an isomorphism.

Then there exist open sets $U_1\subset Y_1\oplus E_2$, $U_2\subset E$, $V_1\subset Y$, $V_2\subset Y_1\oplus E_2$ and diffeomorphisms of class $C^1$, $\varphi :\ V_1\rightarrow V_2$ and $\psi :\ U_1\rightarrow U_2$, $x_0=(x_{01},x_{02})\in U_2\subset U\subset E_1\oplus E_2$, i.e. $x_{01}\in E_1$, $x_{02}\in E_2$, $f(x_0)\in V_1$ satisfying

$$\begin{aligned} (\varphi \circ f \circ \psi )(w,e)=(w,0),\quad \text {where}\ w\in Y_1,\ e\in E_2 \end{aligned}$$

for all $(w,e)\in U_1$.

Theorem 4

([1, Theorem 2.5.14] Local Representation Theorem) Let $E,\ Y$ be Banach spaces. Let $f:\ U \rightarrow Y$ be of class $C^r$, $r\ge 1$ in a neighbourhood of $x_0\in U$, $U\subset E$ open set. Let $Y_1$ be a closed split image of $Df(x_0)$ with closed complement $Y_2$. Suppose that $Df(x_0)$ has a split kernel $E_2=\ker Df(x_0)$ with closed complement $E_1$. Then there are open sets $U_1\subset U \subset E_1\oplus E_2$ and $U_2 \subset Y_1 \oplus E_2$, $x_0\in U_2$ and a $C^r$ diffeomorphism $\psi :\ U_2\rightarrow U_1$ such that $(f\circ \psi )(u,v)=(u,\eta (u,v))$ for any $(u,v)\in U_1$, where $u\in E_1$, $v\in E_2$ and $\eta :\ U_2\rightarrow E_2$ is a $C^r$ map satisfying $D\eta (\psi ^{-1}(x_0))=0$.

Theorems 3 and 4 allow us to prove the following functional dependence result.

Proposition 9

Let E be a Banach space, F be a Hilbert space with Besselian and Hilbertian basis $(b_i)_{i\in \mathbb {N}}$. Let $(f_i)_{i\in J_1}:\ U \rightarrow Y(J_1)$, $U\subset E$ open, be a $C^{1}$ mapping. Let $E_2=\text {ker}\, (Df_i(x_0)_{i\in J_1})$. Assume that CRC+ holds for $(f_i)_{i\in J_1}$, at $x_0$ with the index set $J_2\subset J_1$ and a neighbourhood $V(x_0)$.

Then there exist functions $h_l:\ Y(J_2) \rightarrow \mathbb {R}$, $l\in J_1\setminus J_2$ of class $C^1$ such that

$$\begin{aligned} f_l(x)=h_l((f_i(x))_{i\in J_2})\quad \text {for all } x \text { in some neighbourhood of } x_0. \end{aligned}$$

(30)

Proof

By 3. of CRC+, $Y_1:=(Df_i(x_0)(E))_{i\in J_2}$ is a closed subset of $Y(J_2)$. By Proposition 7, applied to $(f_{i})_{i\in J_2}:\ E \rightarrow Y(J_2)$, $(Df_i)_{i\in J_2}(x_0)$ has a split kernel $E_2=\ker ((Df_i(x_0))_{i\in J_2})=\ker ((Df_i(x_0))_{i\in J_1})$ with closed complement $E_1=\overline{{\text {span}}}(Df_i(x_0)^*,i\in J_2)=\overline{{\text {span}}}(Df_i(x_0)^*,i\in J_1)$, $E=E_1 \oplus E_2$. By Proposition 8, $(Df_i(x)_{i\in J_1})_{|E_1}:\ E_1\rightarrow (Df_i(x)_{i\in J_1})(E)$, $x\in V(x_0)$ is an isomorphism. By 1. of CRC+,

$$\begin{aligned} Df_i(x_0)(e)= \sum _{j\in J_2}( \beta _{j}^i Df_j(x_0)(e) ),\quad i\in J_1\setminus J_2, \end{aligned}$$

where $\beta _{j}^i\in \mathbb {R}$, $i\in J_1 {\setminus } J_2$, $j\in J_2$.

Since $Y_1$ is closed, by Proposition 4, $\overline{{\text {span}}}(b_i, i\in J_1)$ splits, i.e., there exists $Y_2$ such that $\overline{{\text {span}}}(b_i, i\in J_1)=Y_1\oplus Y_2$.

By Rank Theorem 3, there are open sets $U_2\subset U \subset E_1\oplus E_2$ and $U_1 \subset Y_1 \oplus E_2$, $x_0\in U_2$, $V_1\subset \overline{{\text {span}}}( b_i,\ i\in J_1)$, $V_2\subset Y_1\oplus E_2$ and $C^1$ diffeomorphisms $\psi :\ U_1\rightarrow U_2$, $\varphi :\ V_1\rightarrow V_2$ such that for all $x\in U_2$ there exists $(q,e)\in U_1$ such that $x=\psi (q,e)$ and

$$\begin{aligned} (f_i)_{\in J_1}\circ \psi (q,e)=\varphi ^{-1}(q,0). \end{aligned}$$

(31)

By Local Representation Theorem 4, there exists a function $\eta :\ U_1\rightarrow Y_2$ of class $C^1$ such that for all $(q,e)\in U_1$

$$\begin{aligned} (f_i)_{i\in J_1} \circ \psi (q,e)=(q,\eta (q,e)) \in Y(J_1). \end{aligned}$$

(32)

Let us put $(\bar{f}_i)_{i\in J_1}(q):=(q,\eta (q,e))=\varphi ^{-1}(q,0)$ for $(q,e)\in U_1$, where $\varphi ^{-1}(q,0)=(\varphi _i^{-1}(q,0))_{i\in J_1}$.

Then, by (31) and (32) we have that

$$\begin{aligned}&q_i:=\bar{f}_i(q)=\varphi _i^{-1}(q,0)=f_i\circ \psi (q,e)=f_i(x),\quad i\in J_2 \end{aligned}$$

and for any $l\in J_1\setminus J_2$,

$$\begin{aligned} f_l(x)&=f_l\circ \psi (q,e)=\bar{f}_l(q)=f_l\circ \psi (q,e)\\&= b_l^*(f\circ \psi (q,e)) ) = b_l^*(q,\eta (q,e))\\&= \eta _\ell (q,e)= \eta _\ell (q,\bar{e}). \end{aligned}$$

Therefore,

$$\begin{aligned} \eta _\ell (q,\bar{e}) =:h_l((f_i(x))_{i\in J_2}),\quad l\in J_1\setminus J_2, \end{aligned}$$

where $h_l:\ Y(J_2)\rightarrow \mathbb {R}$. $\square$

6 Relaxed Constant Rank Constraint Qualification Plus (RCRCQ+)

In this section we introduce the concept of Relaxed Constant Rank Constraint Qualification Plus, which is the crucial assumption of our main result, Theorem 6 of Sect. 7.

Definition 10

Let $(g_{i})_{i\in I_0\cup I_1}:\ E \rightarrow Y(I_0\cup I_1)$, $I_0\cup I_1\subset \mathbb {N}$, $I_0\cap I_1=\emptyset$ be of class $C^1$. We say that Relaxed Constant Rank Constraint Qualification Plus (RCRCQ+ in short) holds for set ${{\mathcal {F}}}$, given by (5), at $x_0$ if there exists a neighbourhood $V(x_0)$ such that for all J, $I_{0}\subset J\subset I_0\cup I(x_0)$, CRC+ holds for $(g_i)_{i\in J}$ at $x_0$ with neighbourhood $V(x_0)$, i.e., exists $J_2\subset J$ such that

1.:

$(Dg_i(x))_{i\in J_2}$ forms a Schauder basis for $\overline{{\text {span}}}(D g_{i}(x), i \in J)$ for all $x\in V(x_0)$,

2.:

for any $x\in V(x_0)$, there exists a topological isomorphism

$$\begin{aligned} z_x: \overline{{\text {span}}}(D g_{i}(x), i \in J) \rightarrow \overline{{\text {span}}}(D g_{i}(x_0), i \in J), \end{aligned}$$

such that $z_x(D g_{i}(x_0))= D g_{i}(x)$, $i \in J_2$,

and additionally

3.:

$(Dg_i(x_0))_{i\in J_2}(E)$ is closed in $Y(J_2)$ defined in (14), with $v_i=b_i$, $i\in J_2$,

4.:

$(Dg_i(x_0))_{i\in J_2}$ forms a shrinking and boundedly-complete basis for $X_1:=\overline{{\text {span}}}(Dg_i(x_0),\ i\in J)$,

5.:

$(Dg_i(x_0))_{i\in J_2}$ is Besselian for $\overline{{\text {span}}}(Dg_i(x_0),\ i\in J)$,

6.:

$Dg_i(x_0)^*\in E$, $i\in J_2$.

Remark 15

Let us note that if there is no inequality constraints, i.e. $I_1=\emptyset$, then the Relaxed Constant Rank Constraint Qualification Plus for set ${{\mathcal {F}} }$ at $x_0$ is equivalent to Constant Rank Condition Plus for $(g_i)_{i\in I_0}$ at $x_0$.

Remark 16

If E is a reflexive space, then RCRCQ+ for the set ${{\mathcal {F}}}$ given by (5) takes the following form: there exists a neighbourhood $V(x_0)$ such that for all J, $I_{0}\subset J\subset I_0\cup I(x_0)$, exists $J_2\subset J$ such that

1.:

$(Dg_i(x))_{i\in J_2}$ forms a Schauder basis for $\overline{{\text {span}}}(D g_{i}(x), i \in J)$ for all $x\in V(x_0)$,

2.:

for any $x\in V(x_0)$, there exists a topological isomorphism

$$\begin{aligned} z_x: \overline{{\text {span}}}(D g_{i}(x), i \in J) \rightarrow \overline{{\text {span}}}(D g_{i}(x_0), i \in J), \end{aligned}$$

such that $z_x(D g_{i}(x_0))= D g_{i}(x)$, $i \in J_2$,

and

3.:

$(Dg_i(x_0))_{i\in J_2}(E)$ is closed in $Y(J_2)$ defined in (14), with $v_i=b_i$, $i\in J_2$,

4.:

$(Dg_i(x_0))_{i\in J_2}$ is Besselian for $\overline{{\text {span}}}(Dg_i(x_0),\ i\in J)$.

Let us note that if $J\subset \mathbb {N}$ is finite, $|J|=n$ and $g_i:\ E \rightarrow \mathbb {R}$, $i\in J$ are of class $C^1$ in some neighbourhood of $x_0$ then $((Dg_i(x_0))_{i\in J})(E)$ is closed in $\mathbb {R}^n$.

7 Main Result

This section contains our main result of the paper, which is Theorem 6. The following condition will be used in our main theorem.

• (H1) For all $d\in \varGamma _{{\mathcal {F}}}(x_0)$ and for any vector function $r:\ (0,1)\rightarrow E$ such that $\Vert r(t)\Vert t^{-1}\rightarrow 0$, as $t\downarrow 0$, there exists a number $\varepsilon _0>0$ such that

  $$\begin{aligned} g_i(x_0+td+r(t))\le 0 \text { for all } i\in I_1 \setminus I(x_0,d) \text { and for all } t\in (0,\varepsilon _0), \end{aligned}$$

  where $I(x_0,d):=\{ i\in I(x_0)\mid \langle D g_i(x_0) \,, \, d \rangle =0 \}$.

Note that if for some $d\in \varGamma _{{\mathcal {F}}}(x_0)$, set $I_1\setminus I(x_0,d)$ is finite, then the Condition (H1) is satisfied, cf. [7, Lemma 6.4].

The following examples illustrates condition (H1).

Example 4

Let $h:\ \mathbb {R}\rightarrow {{\mathcal {X}}}$, ${{\mathcal {X}}}$ is sequence space $\ell _p$, $p\ge 1$, $h_i(x)=a_i x^2 + b_i x + c_i$, $i=1,\dots$ where $a_i\le 0$, $i=1,\dots$ and

$$\begin{aligned} c_i=\left\{ \begin{array}{ll} 0, &{}\quad i - \text {even}, \\ <0, &{}\quad i - \text {odd} \end{array} \right. , \quad i=1,2, \dots . \end{aligned}$$

Let ${{\mathcal {F}}}:= \{ x \in \mathbb {R} \mid h(x)\in {{\mathcal {X}}}_{-} \}$, where $h(x)=(h_i(x))_{i\in I_0\cup I_1}$ and $x_0=0$. Then $I(x_0)=\{2k,\ k\in \mathbb {N}\}$. Assume that $b_i> 0$, $i=1,\ldots$. Then

$$\begin{aligned}&\varGamma _{{\mathcal {F}}}(0)=\{ d\in \mathbb {R} \mid b_i \cdot d \le 0,\ i\in I(x_0)\} = \mathbb {R}_{-},\\&{{\mathcal {T}}}_{{\mathcal {F}}}(0)=\{ d \in \mathbb {R} \mid \exists r(t),\ r(t)/t\rightarrow 0^+ \ \exists \varepsilon _0>0\ \forall t\in [0,\varepsilon _0) \quad h(x_0+td+r(t))\in {{\mathcal {X}}}_{-}\}. \end{aligned}$$

Let us take $d=-1$. We have

$$\begin{aligned}&a_i(-t+r(t))^2+b_i(-t+r(t))+c_i\le 0, \quad i=1,\ldots \\&\iff t^2 a_i\left( -1+\frac{r(t)}{t}\right) ^2+tb_i\left( -1+\frac{r(t)}{t}\right) +c_i \le 0, \quad i=1,\ldots . \end{aligned}$$

Since $a_i,c_i\le 0$ and $b_i\ge 0$, $i=1,\dots$ we obtain that $-1 \in {{\mathcal {T}}}_{{\mathcal {F}}}(0)$. Since ${{\mathcal {T}}}_{{\mathcal {F}}}(0)\subset \varGamma _{{\mathcal {F}}}(0)$, Abadie constraint qualification holds, i.e., ${{\mathcal {T}}}_{{\mathcal {F}}}(0)= \varGamma _{{\mathcal {F}}}(0)$. Let us also prove that $h_i$ also satisfy condition (H1). Indeed, if $i\in I(0) {\setminus } I(0;d)$, we have $b_i \cdot d <0.$ So, inequality

$$\begin{aligned} h_i(td+r(t)) = a_i(td+r(t))^2 + t b_i \left( d + \frac{r(t)}{t}\right) <0 \end{aligned}$$

is satisfied for any function r(t) s.t. $r(t)/t\rightarrow 0^+$ for $t\in (0, \varepsilon _0)$ for some $\varepsilon _0$.

Note that condition PMFCQ from [25] is not satisfied for functions in Example 4.

In the following example RCRCQ+ and Abadie constraint qualification are not satisfied.

Example 5

Let $g_j:\ \mathbb {R}^2 \rightarrow \mathbb {R}$, $j\in \mathbb {N}$ be defined by

$$\begin{aligned}&g_1(x_1,x_2)=-x_1,\\&g_2(x_1,x_2)=-x_2,\\&g_j(x_1,x_2):=jx_1x_2, \quad j=3,4,\dots . \end{aligned}$$

Functions $g_j:\ \mathbb {R}^2 \rightarrow \mathbb {R}$, $j\in \mathbb {N}$ are differentiable, and

$$\begin{aligned} {{\mathcal {F}}}:&= \left\{ (x_1,x_2)\in \mathbb {R}^2\mid g_j(x_1,x_2)\le 0,\ j\in \mathbb {N}\right\} \\&=\left\{ (x_1,x_2)\in \mathbb {R}^2\mid x_1=0, x_2\ge 0\} \cup \{ (x_1,x_2)\in \mathbb {R}^2\mid x_1\ge 0, x_2= 0\right\} . \end{aligned}$$

Then

$$\begin{aligned} {{\mathcal {T}}}_{{\mathcal {F}}}(0,0)=\left\{ (u_1,u_2)\in \mathbb {R}^2 \mid u_1\ge 0,\ u_2\ge 0,\ u_1\cdot u_2 =0 \right\} \end{aligned}$$

and

$$\begin{aligned} \varGamma _{{\mathcal {F}}}(0,0)=\mathbb {R}_+^2. \end{aligned}$$

Hence, Abadie constraint qualification does not hold. Moreover, RCRCQ+ does not hold, since

$$\begin{aligned} (Dg_i(0,0))_{i\in \{1,3\}}\quad \text {is not isomorphic to} \quad (Dg_i(x_1,x_2))_{i\in \{1,3\}} \end{aligned}$$

for any $(x_1,x_2)$ s.t. $x_2\ne 0$. Note that for any $d\in \varGamma _{{\mathcal {F}}}(0,0)$, $I_1 {\setminus } I((0,0),d)$ is a finite set, thus it is easy to check that condition (H1) is satisfied.

Our main results exploit the Theorem of Lyusternik-Graves.

Theorem 5

(Lyusternik–Graves) Let X and Y be Banach spaces, let U be a neighborhood of a point $x_0\in X$, and let $f:\ U \rightarrow Y$ be a Fréchet differentiable mapping. Assume that f is regular at $x_0$, i.e., that $\text {Im}\, Df(x_0)= Y$, and that its derivative is continuous at this point (in the uniform operator topology of the space $\varGamma (X, Y)$). Then the tangent space ${{\mathcal {T}}}_M(x_0)$ to the set

$$\begin{aligned} M = \{x \in U \mid f(x) = f(x_0)\} \end{aligned}$$

at the point $x_0$ coincides with the kernel of the operator $Df(x_0)$, i.e.,

$$\begin{aligned} {{\mathcal {T}}}_M(x_0) = \ker \, Df(x_0). \end{aligned}$$

(33)

Moreover, if the assumptions of the theorem are satisfied, then there exist a neighbourhood $U'\subset U$ of the point $x_0$, a number $K>0$, and a mapping $\xi \rightarrow x (\xi )$ of the set $U'$ into X such that

$$\begin{aligned} \begin{aligned}&f(\xi + x(\xi ))= f(x_0),\\&\Vert x(\xi )\Vert \le K \Vert f(\xi ) - f(x_0)\Vert \end{aligned} \end{aligned}$$

(34)

for all $\xi \in U'$.

Assertion (33) follows from (34), see [20]. For discussion, see e.g. [13, 19]. The assertion (34) is called the generalized Ljusternik Theorem in [12].

Now we are ready to establish our main theorem.

Theorem 6

Let E be a Banach space, F be a Hilbert space and assume that $(b_i)_{i\in \mathbb {N}}$ is a Besselian and Hilbertian basis of F. Let ${{\mathcal {F}}}\subset E$ be given as in (5).

Assume that

• (i) RCRCQ+ holds for ${{\mathcal {F}}}$ at $x_0\in {{\mathcal {F}}}$,

• (ii) condition (H1) is satisfied at $x_0$.

Then Abadie constraint qualification holds, i.e., $\varGamma _{{\mathcal {F}}}(x_0)={{\mathcal {T}}}_{{\mathcal {F}}}(x_0)$.

Moreover, for each $d\in {{\mathcal {T}}}_{{\mathcal {F}}}(x_{0})$ there is a vector function $r:\ (0,1)\rightarrow E$, $\Vert r(t)\Vert /t\rightarrow 0$ when $t\downarrow 0$, such that for all t sufficiently small

$$\begin{aligned} \begin{array}{l} g_i(x_{0}+td+r(t))=0,\ i\in J(d),\\ g_{\ell }(x_{0}+td+r(t))\le 0,\ \ell \in I_1\setminus J(d), \end{array} \quad J(d):=I_{0}\cup I(x_{0},d). \end{aligned}$$

(35)

Proof

The inclusion ${{\mathcal {T}}}_{{\mathcal {F}}}(x_0)\subset \varGamma _{{\mathcal {F}}}(x_0)$ is immediate. To see the converse, take any $d\in \varGamma _{{\mathcal {F}}}(x_0)$, where by Proposition 3,

$$\begin{aligned} \varGamma _{{{\mathcal {F}}}}(x_{0}) \subset \{d\in E \mid \langle Dg_{i}(x_{0})\mid d\rangle =0,\ i\in I_0,\ \langle Dg_{i}(x_{0})\mid d\rangle \le 0,\ \ i\in I(x_{0})\}. \end{aligned}$$

Recall that $I(x):=\{ i\in I_1 \mid g_i(x)=0 \}$.

We start by considering the case $J:=J(d)\ne \emptyset$, $|J|=+\infty$.

By RCRCQ+ of ${{\mathcal {F}}}$ at $x_0$, there exist $V(x_0)$ and $J_2\subset J$ such that $(Dg_i(x_0))_{i\in J_2}$ forms a boundedly-complete and shrinking basis for $\overline{{\text {span}}}(Dg_i(x_0),i\in J)$ and there exists a topological isomorphism z

$$\begin{aligned} z_{t,r}:\ \overline{{\text {span}}}( Dg_i(x_0+td+r), i\in J )\rightarrow \overline{{\text {span}}}(D g_i(x_0), i\in J ), \end{aligned}$$

for (t, r) in some neighbourhood of $(0,0)\in \mathbb {R}\times E$ such that $z_{t,r}(Dg_i(x_0+td+r))=Dg_i(x_0), i \in J_2$ for all $(t,r)\in \mathbb {R}\times E$ such that $x_0+td+r\in V(x_0)$.

Let $f:E\rightarrow Y(J_2)$ be defined as $f(x):=(g_i(x))_{i\in J_2}$, where $Y(J_2)$ is defined by (14). By (22), the derivative $Df(x_0)$ is onto $Y(J_2)$.

Let us define

$$\begin{aligned} M:= \{ x\in E \mid g_{i}(x)=g_i(x_0)=0,\ i\in J_2 \}. \end{aligned}$$

(36)

By Lyusternik–Graves Theorem 5 applied to the set M,

$$\begin{aligned} \ker Dg_i(x_0)_{i\in J_2} = {{\mathcal {T}}}_M(x_0). \end{aligned}$$

By applying Lyusternik-Graves Theorem 5 with f at $x_0$, we obtain that $d\in {{\mathcal {T}}}_{M}(x_0)$.

• Case 1 If $J_2=J$, then $g_i(x_0+td+r(t))=0$, $i\in J$ for $t\in [0,\varepsilon ]$, where $\varepsilon >0$ and r(t) is given by Lyusternik-Graves Theorem.

• Case 2 If $J_2\subsetneq J$ then, by Proposition 9, applied to $g_i$, $i\in J$, there exist functions $h_l$, $l\in J{\setminus } J_2$ of class $C^1$, such that

  $$\begin{aligned} g_l(x_0+td+r)=h_l((g_i(x_0+td+r))_{i\in J_2}), \end{aligned}$$

  (37)

  for (t, r) in some neighbourhood of $(0,0)\subset \mathbb {R}\times E$.

  Consider the system

  $$\begin{aligned} g_i(x_0+td+r)=0,\quad i\in J \end{aligned}$$

  (38)

  with respect to variables t, r. Let us note that system (38) is satisfied for $(t,r)=(0,0).$

  Obviously, by Proposition 9, in some neighbourhood of (0, 0), system (38) is equivalent to

  $$\begin{aligned} g_i(x_0+td+r)=0, \quad i\in J_2 \end{aligned}$$

  (39)

  with additional condition

  $$\begin{aligned} g_l(x_0+td+r)=h_l((g_i(x_0+td+r))_{i\in J_2})=0,\ l\in J\setminus J_2. \end{aligned}$$

  (40)

  Note that $h_l((g_i(x_0))_{i\in J_2})=0$, $l\in J{\setminus } J_2$ since $g_l(x_0)=0=h_l((g_i(x_0))_{i\in J_2})=h_l(0)$.

In both cases there exist $\varepsilon >0$ and a function $r:\ [0,\varepsilon )\rightarrow E$, $\Vert r(t)\Vert t^{-1}\rightarrow 0$, $t\downarrow 0,$ such that

$$\begin{aligned} g_i(x_0+td+r(t))=0, \quad i\in J, \end{aligned}$$

i.e.,

$$\begin{aligned} d\in {{\mathcal {T}}}_{\tilde{M}}(x_0),\quad \text {where}\ \tilde{M}:= \{ x\in E \mid g_{i}(x)=g_i(x_0)=0,\ i\in J \}. \end{aligned}$$

(41)

By condition (H1), there exists $\varepsilon _0>0$ such that

$$\begin{aligned} g_i(x_0+td+r(t))\le 0 \text { for all } i\in (I_0\cup I_1)\setminus I(x_0,d) \text { and for all } t\in (0,\varepsilon _0), \end{aligned}$$

therefore

$$\begin{aligned} x_0+td+r(t)\in {{\mathcal {F}}}\quad t\in [0,\min \{\varepsilon _0,\varepsilon \}]. \end{aligned}$$

(42)

Thus, $d\in {{\mathcal {T}}}_{{\mathcal {F}}}(x_0)$.

Now, let us consider the case $J=\emptyset$ (i.e. the case when both $I_0=\emptyset$ and $I(x_0,d)=\emptyset$). Then, by condition (H1), for any vector function $r:\ (0,1)\rightarrow E$, $\Vert r(t)\Vert /t\rightarrow 0$ when $t\downarrow 0$ there exists $\varepsilon >0$ such that

$$\begin{aligned} x_0+td+r(t)\in {{\mathcal {F}}}\quad t\in [0,\varepsilon ], \end{aligned}$$

(43)

i.e., $d\in {{\mathcal {T}}}_{{\mathcal {F}}}(x_0)$. $\square$

8 RCRCQ+ and Lagrange Multipliers

In this section, using [11] and RCRCQ+ condition we will prove non-emptiness of the Lagrange multipliers set.

The Lagrange function or Lagrangian corresponding to problem (P) is a function $L:\ E \times F\rightarrow \mathbb {R}$,

$$\begin{aligned} L(x,\lambda ):=f(x)+\langle \lambda \mid G(x) \rangle \quad x\in E,\ \lambda \in F. \end{aligned}$$

Definition 11

Let $Q\subset F$ be a closed, convex set. The normal cone to Q at $\bar{y} \in F$ is the set

$$\begin{aligned} {{\mathcal {N}}}_Q(\bar{y})= \{ y\in F \mid \langle y \mid k-\bar{y} \rangle \le 0 \ \forall k\in Q \}. \end{aligned}$$

Definition 12

A feasible point $\bar{x}\in {{\mathcal {F}}}$ of (P) is called a KKT point if there exists $\bar{\lambda }\in {{\mathcal {N}}}_{K}(G(\bar{x}))$ such that

$$\begin{aligned} D_{\bar{x}}L(\bar{x},\bar{\lambda })=0, \end{aligned}$$

where K is defined by (2). In this case $\bar{\lambda }$ is called Lagrange multiplier of (P) at $\bar{x}\in {{\mathcal {F}}}$.

Following the notation introduced in [11] we recall the set corresponding to KKT points,

$$\begin{aligned} {{\mathcal {M}}}(x_0,0):= DG(x_0)^* {{\mathcal {N}}}_{K}(G(x_0)). \end{aligned}$$

In a more general setting, this set has been defined by Hurwicz in [18] and a number of its properties has been shown in [22]. We refer to this set as a Hurwicz set.³

By Proposition 5.6 of [11] we immediately get the following result.

Proposition 10

Let E be a Banach space, F be a Hilbert space and assume that $(b_i)_{i\in \mathbb {N}}$ is a Besselian and Hilbertian basis of F. Let ${{\mathcal {F}}}\subset E$ be given as in (5). Let $x_0\in {{\mathcal {F}}}$ be a local minimizer of (P). Assume RCRCQ+ holds for ${{\mathcal {F}}}$ at $x_0\in {{\mathcal {F}}}$ with a neigbourhood $V(x_0)$. Assume that assumption (H1) is satisfied at $x_0$. Assume that the Hurwicz set ${{\mathcal {M}}}(x_0,0)$ is weakly*-closed. Then the set of Lagrange multipliers at $x_0$ is nonempty.

Proof

By Theorem 6, Abadie constraint qualification holds for ${{\mathcal {F}}}$ at $x_0$. The rest of the proof follows the lines of the proof of Proposition 5.6 of [11]. $\square$

Remark 17

Let us underline the fact that in both papers [4, 11] in the definition of Abadie constraint qualification the Hurwicz set ${{\mathcal {M}}}(x_0,0)$ is weakly*-closed.

The next two lemmas allow us to provide descriptions of elements of the Hurwicz set ${{\mathcal {M}}}(x_0,0)$.

Lemma 3

Let E be a Banach space and F be a Hilbert space with basis $(b_i)\subset F$, $i\in \mathbb {N}$. Let $K=\{ y=\sum _{i\in \mathbb {N} } b_i^*(y)b_i\in F \mid b_i^*(y)\in \mathbb {R},\ i\in I_0, b_i^*(y)\le 0, i\in I_1 \}$, $I_0\cup I_1=\mathbb {N}$, $I_0\cap I_1=\emptyset$. Let $G:\ E\rightarrow F$ be of class $C^1$. Let $x_0\in E$ and $y_0:=G(x_0)\in K$. Then the following hold

$$\begin{aligned}&{{\mathcal {N}}}_K(y_0)\subset \left\{ y \in F \mid y=\sum _{i\in I_0\cup I_1} b_i^{**}(y)b_i^*,\ b_i^{**}(y)=0, i\in I_1\setminus I(x_0),\right. \\&\quad \left. b_i^{**}(y)\ge 0,\ i\in I(y_0),\ b_i^{**}(y)\in \mathbb {R},\ i\in I_0 \right\} , \end{aligned}$$

where $I(x_0)=\{ i\in I_1 \mid g_i(x_0)=0 \}$.

Proof

Let $x_0\in E$ be such that $y_0=G(x_0)\in K$, i.e. $y_0=\sum _{i\in \mathbb {N}} b_i^*(y_0)b_i=\sum _{i\in I_0\cup I_1} b_i^*(G(x_0))b_i$, $b_i^*(G(x_0))\le 0$, $i\in I_1$. Let $y\in {{\mathcal {N}}}_K(y_0)$ be arbitrary.

From the definition of normal cone we have

$$\begin{aligned} \forall k \in K \quad \langle y \mid k - G(x_0) \rangle \le 0 \end{aligned}$$

and in consequence

$$\begin{aligned} \forall k \in K \quad \left\langle \sum _{i\in \mathbb {N}} b_i^{**}(y)b_i^* \mid k - G(x_0) \right\rangle \le 0. \end{aligned}$$

From this, by taking $k=\sum _{j\in I_0\cup I_1} \alpha _j b_j \in K$, $\alpha _j=0$, $j\in I_0$, $\alpha _j\le 0$, $j\in I_1$ we get

$$\begin{aligned}&\left\langle \sum _{i\in \mathbb {N}} b_i^{**}(y)b_i^* \mid \sum _{j\in I_0\cup I_1} (\alpha _j - b_j^*(G(x_0))) b_j \right\rangle \le 0, \end{aligned}$$

and equivalently

$$\begin{aligned}&\sum _{i\in I_0\cup I_1} b_i^{**}(y) (\alpha _i - b_i^*(G(x_0))) \le 0. \end{aligned}$$

Take any fixed $l\in I_0\cup I_1$.

• Case 1 $b_l^{*}(y_0)<0$, i.e. $l\in I_1 {\setminus } I(x_0)$. By taking $\alpha _i=b_i^*(y_0)=0$, $i\in I_0$, $\alpha _i=b_i^*(y_0)\le 0$, $i\in I_1{\setminus } \{ l \}$ and $\alpha _l=0$ we deduce

  $$\begin{aligned} b_l^{**}(y) (-b_l^*(y_0)) \le 0. \end{aligned}$$

  Therefore, $b_l^{*}(y)\le 0$. On the other hand, by taking $\alpha _i=b_i^*(y_0)=0$, $i\in I_0$, $\alpha _i=b_i^*(y_0)\le 0$, $i\in I_1{\setminus } \{ l \}$ and $\alpha _l=2\cdot b_l^{*}(y_0)\le 0$ we obtain

  $$\begin{aligned} b_l^{**}(y) b_l^*(y_0) \le 0, \end{aligned}$$

  i.e., $b_l^{**}(y)\ge 0$. In conclusion, $b_l^{**}(y)= 0$.

• Case 2 $b_l^{*}(y_0)=0$ and $l\in I(x_0)$. By taking $\alpha _i=b_i^*(y_0)=0$, $i\in I_0$, $\alpha _i=b_i^*(y_0)\le 0$, $i\in I_1{\setminus } \{ l \}$ and $\alpha _l=-1$ we obtain

  $$\begin{aligned} b_l^{**}(y) \cdot (-1) \le 0, \end{aligned}$$

  i.e. $b_l^{**}(y)\ge 0$.

• Case 3 $b_l^{*}(y_0)=0$ and $l\in I_0$. Then $b_l^{**}(y_0)\in \mathbb {R}$.

Therefore,

$$\begin{aligned}&{{\mathcal {N}}}_K(G(x_0))\subset \left\{ y \in F \mid y=\sum _{i\in I_0\cup I_1} b_i^{**}(y)b_i^*,\ b_i^{**}(y)=0, i\in I_1\setminus I(x_0),\right. \\&\quad \left. b_i^{**}(y)\ge 0,\ i\in I(x_0),\ b_i^{**}(y)\in \mathbb {R},\ i\in I_0 \right\} . \end{aligned}$$

$\square$

Lemma 4

Let E be a Banach space and F be a Hilbert space with basis $(b_i)\subset F$, $i\in \mathbb {N}$. Let $K=\{ y=\sum _{i\in \mathbb {N} } b_i^*(y)b_i\in F \mid b_i^*(y)\in \mathbb {R},\ i\in I_0, b_i^*(y)\le 0, i\in I_1 \}$, $I_0\cup I_1=\mathbb {N}$, $I_0\cap I_1=\emptyset$. Let $G:\ E\rightarrow F$ be of class $C^1$.

Let $x_0\in E$ and $y_0:=G(x_0)\in K$. Then

$$\begin{aligned} z\in {{\mathcal {M}}}(x_0,0) \iff z= & {} DG^*(x_0)(y),\ y\in {{\mathcal {N}}}_K(G(x_0)) \implies z\\= & {} \sum _{i\in I_0 \cup I(x_0)} a_i Dg_i(x_0), \end{aligned}$$

where $a_i:=b_i^{**}(y)$, $i\in I_0\cup I(x_0)$, $a_i\ge 0$, $i\in I(x_0)$, $a_i\in \mathbb {R}$, $i\in I_0$.

Proof

Let $\langle \cdot , \cdot \rangle _E:\ E\times E^* \rightarrow \mathbb {R}$ be a duality mapping for the pair $E,E^*$. Observe, that for any $x\in E$ and for any $y\in F$

$$\begin{aligned} \langle x, DG^*(x_0)y \rangle _E = \langle DG(x_0)x, y \rangle _F, \end{aligned}$$

where $DG^*(x_0):\ F \rightarrow E^*$ denotes the adjoint operator for $DG(x_0):\ E\rightarrow F$. Let $z\in {{\mathcal {M}}}(x_0,0)$, $z=DG^*(x)(y),\ y\in {{\mathcal {N}}}_K(G(x_0))$, then

$$\begin{aligned} \begin{aligned} \forall x \in E \langle x , z \rangle _E&= \left\langle x , DG^*(x_0)\left( \sum _{i\in I_0\cup I(x_0)} a_i b_i^*\right) \right\rangle _E \\&= \sum _{i\in I_0\cup I(x_0)} a_i\langle x , DG^*(x_0)b_i^* \rangle _E\\&= \sum _{i\in I_0\cup I(x_0)} a_i\langle DG(x_0)x , b_i^* \rangle _F\\&= \sum _{i\in I_0\cup I(x_0)} a_i\left\langle \sum _{j\in I_0\cup I(x_0)} b_j^*(DG(x_0)x)b_j , b_i^*\right\rangle _F\\&= \sum _{i\in I_0 \cup I(x_0)} a_i Dg_i(x_0) ( x)\\&= \left\langle x , \sum _{i\in I_0 \cup I(x_0)} a_i Dg_i(x_0) \right\rangle _E . \end{aligned} \end{aligned}$$

(44)

Since (44) hold for any $x\in E$, we obtain that $z=\sum _{i\in I_0 \cup I(x_0)} a_i Dg_i(x_0)$.$\square$

Let us consider now the case when no inequality are present (i.e. $I_1=\emptyset$), i.e.,

$$\begin{aligned} {{\mathcal {F}}}=\left\{ \begin{array}{ll} x\in E \mid g_i(x) = 0,&i\in I_0 \end{array}\right\} \end{aligned}$$

Such problems has been considered in e.g. in [9, Theorem 4.1]. In this case we are also getting the existence of Lagrange multipliers (see Proposition 11). By Proposition 7 we are getting split of E, which is included in assumption (B) of [9, Theorem 4.1] and by Proposition 8 we are obtaining isomorphism of $((Dg_i(x))_{i\in I_0})|_{E_1}:\ E_1\rightarrow ((Dg_i(x))_{i\in I_0})(E)$, $x\in U(x_0)$, which is included in assumption (C) of [9, Theorem 4.1] (see Remark 2).

Proposition 11

Let E be a Banach space, F be a Hilbert space and assume that $(b_i)_{i\in \mathbb {N}}$ is a Besselian and Hilbertian basis of F. Let ${{\mathcal {F}}}\subset E$ be given as in (5), where $I_1=\emptyset$. Let $x_0\in {{\mathcal {F}}}$ be a local minimizer of (P). Assume CRC+ holds for $(g_i)_{i\in I_0}$ at $x_0\in {{\mathcal {F}}}$ with a neigbourhood $V(x_0)$. Then the set of Lagrange multipliers at $x_0$ is nonempty.

Proof

Let $x_0$ be a local minimizer of problem ($P_0$) with $I_1=\emptyset$. The first-order necessary optimality condition is $Df_0(x_0)h=0$ for all $h\in {{\mathcal {T}}}_{{\mathcal {F}}}(x_0)$. By the proof of Theorem 6 (see (41) with $J=I_0$),

$$\begin{aligned} \ker Dg(x_0) = \ker (Dg_i(x_0))_{i\in I_0} = {{\mathcal {T}}}_{{\mathcal {F}}}(x_0). \end{aligned}$$

Therefore $Df_0(x)h=0$ for any $h\in \ker DG(x_0)$, i.e., $Df_0(x)\in (\ker DG(x_0))^\perp$. By Proposition 7, $(\ker DG(x_0))^\perp =\overline{{\text {span}}}( Dg_i(x_0)^*, i \in I_0 )$. Therefore $Df_0(x_0)\in \overline{{\text {span}}}( Dg_i(x_0)^*, i \in I_0 )$, i.e., there exists $\lambda \in F$ such that $Df_0(x_0)=\langle DG(x_0)^* \mid \lambda \rangle = \lambda (DG(x_0))$. $\square$

9 The Case of ${{\mathcal {M}}}(x_0,0)$, when $I_0=\emptyset$

In this section, we consider the case when $I_0=\emptyset$, i.e., no equalities appear in the description of the constraint set, and provide discussion with NFMCQ condition introduced in [25]. As a corollary of Lemma 3, we obtain the following crucial characterization.

Corollary 2

Let E be a Banach space and F be a Hilbert space with basis $(b_i)\subset F$, $i\in \mathbb {N}$. Let $K=\{ y=\sum _{i\in \mathbb {N} } b_i^*(y)b_i\in F \mid b_i^*(y)\le 0, i\in I_1 \}$, $I_1=\mathbb {N}$. i.e. $I_0=\emptyset$. Let $G:\ E\rightarrow F$ be of class $C^1$. Let $x_0\in E$ and $y_0:=G(x_0)\in K$. Then the following hold

$$\begin{aligned} {{\mathcal {N}}}_K(y_0)&= \left\{ y \in F \mid y=\sum _{i\in I_1} b_i^{**}(y)b_i^*,\ b_i^{**}(y)=0, i\in I_1\setminus I(x_0),\right. \\&\quad \left. b_i^{**}(y)\ge 0,\ i\in I(y_0)\right\} , \end{aligned}$$

where $I(x_0)=\{ i\in I_1 \mid g_i(x_0)=0 \}$.

Proof

In view of Lemma 3 we need only to show that

$$\begin{aligned}&{{\mathcal {N}}}_K(y_0)\supset \left\{ y \in F \mid y=\sum _{i\in I_1} b_i^{**}(y)b_i^*,\ b_i^{**}(y)=0, i\in I_1\setminus I(x_0),\right. \\&\quad \left. b_i^{**}(y)\ge 0,\ i\in I(y_0)\right\} . \end{aligned}$$

Take any $\bar{y}\in \{ y \in F \mid y=\sum _{i\in I_1} b_i^{**}(y)b_i^*,\ b_i^{**}(y)=0, i\in I_1{\setminus } I(x_0),\ b_i^{**}(y)\ge 0,\ i\in I(y_0)\}$. Take any $k\in K$. We want to show that

$$\begin{aligned} \langle \bar{y} \mid k - G(x_0) \rangle \le 0. \end{aligned}$$

We have

$$\begin{aligned} \langle \bar{y} \mid k - G(x_0) \rangle&=\left\langle \sum _{i\in I_1} b_i^{**}(\bar{y})b_i^* \mid \sum _{j\in I_1} (\alpha _j b_j - g_j(x_0)b_j)\right\rangle \\&= \left\langle \sum _{i\in I(x_0)} b_i^{**}(\bar{y})b_i^* \mid \sum _{j\in I(x_0)} (\alpha _j - g_j(x_0))b_j + \sum _{j\in I_1\setminus I(x_0)} (\alpha _j - g_j(x_0))b_j\right\rangle \\&= \sum _{i\in I(x_0)} b_i^{**}(\bar{y}) (\alpha _i - g_i(x_0))= \sum _{i\in I(x_0)} b_i^{**}(\bar{y}) \alpha _i\le 0. \end{aligned}$$

$\square$

Corollary 3

Let E be a Banach space and F be a Hilbert space with basis $(b_i)\subset F$, $i\in \mathbb {N}$. Let $K=\{ y=\sum _{i\in \mathbb {N} } b_i^*(y)b_i\in F \mid b_i^*(y)\le 0, i\in I_1 \}$, $I_1=\mathbb {N}$. Let $G:\ E\rightarrow F$ be of class $C^1$. Let $x_0\in E$ and $y_0:=G(x_0)\in K$. Then

$$\begin{aligned} z\in {{\mathcal {M}}}(x_0,0)&\iff z=DG^*(x_0)(y),\text {for some}\ y\in {{\mathcal {N}}}_K(G(x_0)) \\&\iff z= \sum _{i\in I(x_0)} a_i Dg_i(x_0), \end{aligned}$$

where $a_i:=b_i^{**}(y)$, $i\in I_0\cup I(x_0)$, $a_i\ge 0$, $i\in I(x_0)$, $a_i\in \mathbb {R}$, $i\in I_0$. Moreover, if $I(x_0)$ is finite, then the Hurwicz set ${{\mathcal {M}}}(x_0,0)$ is closed.

Proof

The implication “$\implies$” follows from Lemma 4. Now we proceed to prove the implication “”. Let $z= \sum _{i\in I(x_0)} a_i Dg_i(x_0)$. Then, by (44), $z=DG^*(x_0)(\sum _{i\in I_0\cup I(x_0)} a_i b_i^*)$. By Corollary 2, $y:=\sum _{i\in I_0\cup I(x_0)} a_i b_i^*\in {{\mathcal {N}}}_K(G(x_0))$, hence $z=DG^*(x_0)(y)$, $y\in {{\mathcal {N}}}_K(G(x_0))$.

The fact that the Hurwicz set ${{\mathcal {M}}}(x_0,0)$ is closed, in the case if $I(x_0)$ is finite, follows from closedness of finitely generated cones (see e.g. [3, Corollary 5.25]). $\square$

Definition 13

Let X be a Banach space. We say that $(h_i)_{i\in J}\in X$ is relatively positive with respect to $J_2\subset J$ if for any $z\in X$ the following implication holds

$$\begin{aligned}&z=\sum _{i\in J} \alpha _i h_i,\quad \alpha _i\ge 0, i\in J \implies \exists \tilde{\alpha }_i\ge 0,\ i\in J_2,\quad z=\sum _{i\in J_2} \tilde{\alpha }_i h_i. \end{aligned}$$

Equivalently,

$$\begin{aligned} \left\{ x\in X \mid x=\sum _{i\in J} \alpha _i h_i,\ \alpha _i\ge 0, i \in J\right\} = \left\{ x\in X \mid x=\sum _{i\in J_2} \tilde{\alpha }_i h_i,\ \tilde{\alpha }_i\ge 0, i \in J_2 \right\} . \end{aligned}$$

Remark 18

Suppose that $h_i,i\in J_2$ is a basis for X with biorthogonal system $(h_i,h_i^*)$. By [26, Proposition 5.1], $h_i^*$ is total. By [24], the cone

$$\begin{aligned} K_{\{h_i\}_{i\in J_2}}:=\left\{ x\in X \mid x=\sum _{i\in J_2} \tilde{\alpha }_i h_i,\ \tilde{\alpha }_i\ge 0, i \in J_2 \right\} \end{aligned}$$

is pointed. Moreover, if $(h_i)_{i\in J}\in X$ is relatively positive with respect to $J_2\subset J$, then the set

$$\begin{aligned} \tilde{K}:=\left\{ x\in X \mid x=\sum _{i\in J} \alpha _i h_i,\ \alpha _i\ge 0, i \in J\right\} \end{aligned}$$

is a pointed cone.

Proposition 12

Let $x_0\in {\mathcal {F}}$. Assume that there exists $J_2\subset J:= I(x_0)$ such that

1.:

$X_1:=\overline{{\text {span}}}(Dg_i(x_0),i\in J)=\overline{{\text {span}}}(Dg_i(x_0),i\in J_2)$,

2.:

$(Dg_i(x_0))_{i\in J_2}$ is a basis for $X_1$,

3.:

$(Dg_i(x_0))_{i\in J}\in X_1$ is relatively positive with respect to $J_2\subset J$,

4.:

$(Dg_i(x_0))_{i\in J_2}$ is boundedly complete and shrinking sequence (i.e. $X_1$ is reflexive),

5.:

$Dg_i^*(x_0)\in E$, $i\in J_2$.

Then the Hurwicz set ${{\mathcal {M}}}(x_0,0)$ is weakly*-closed.

Proof

Let $\langle \cdot , \cdot \rangle _E:\ E\times E^* \rightarrow \mathbb {R}$ be a duality mapping for the pair $E,E^*$.

Let $(z_n)$ be a sequence in ${{\mathcal {M}}}(x_0,0)\subset E^*$ weakly*-converging to some $z_0\in E^*$. We want to show that $z_0\in {{\mathcal {M}}}(x_0,0)$, i.e., there exists $y_0\in {{\mathcal {N}}}_K(G(x_0))$ such that $z_0=DG^*(x_0)(y_0)$.

There exist $y_n \in {{\mathcal {N}}}_K(G(x_0))\subset F$, $y_n=\sum _{i\in \mathbb {N}} b_i^{**}(y_n)b_i^*$, $n\in \mathbb {N}$ such that $z_n=DG^*(x_0)(y_n)$, $n\in \mathbb {N}$.

By Corollary 2, $y_n=\sum _{i\in I(x_0)}b_i^{**}(y_n)b_i^*$, where $b_i^{**}(y_n)=0$, $i\in I_1{\setminus } I(x_0)$, $b_i^{**}(y_n)\ge 0$, $i\in I(x_0)$, for any $n\in \mathbb {N}$. Moreover, $z_n=DG^*(x_0)(y_n)=DG^*(x_0)(\sum _{i\in I(x_0)}b_i^{**}(y_n)b_i^*)$, $n\in \mathbb {N}$.

Let $a_i^n:=b_i^{**}(y_n)$, $i\in I_1$, $n\in \mathbb {N}$. Since $(z_n)$ is a sequence in ${{\mathcal {M}}}(x_0,0)$, by Corollary 3, $z_n=\sum _{i\in I(x_0)} a_i^n Dg_i(x_0)\in X_1=\overline{{\text {span}}}( Dg_i(x_0), i\in I(x_0))$ and since $(z_n)$ is weakly*-converging to some $z_0\in E^*$,

$$\begin{aligned} \forall x\in E\quad \langle x, z_0 \rangle _E=\lim _{n\rightarrow +\infty } \langle x, z_n \rangle _E = \lim _{n\rightarrow +\infty } \left\langle x, \sum _{i\in I(x_0)} a_i^n Dg_i(x_0) \right\rangle _E. \end{aligned}$$

(45)

Since $(z_n)$ converges weakly* in reflexive $X_1$, it converges strongly to $z_0\in X_1$. Since $\overline{{\text {span}}}( Dg_i(x_0),\ i\in J_2 )=X_1$ is closed, $z_0=\sum _{i\in J_2} \tilde{\beta }_i Dg_i(x_0)$ for some $\tilde{\beta }_i\in \mathbb {R}$, $i\in J_2$.

Moreover, since $Dg_i(x_0)$, $i\in J=I(x_0)$ is relatively positive with respect to $J_2$ we have that

$$\begin{aligned} z_n=\sum _{i\in I(x_0)} a_i^n Dg_i(x_0)=\sum _{i\in J_2} \tilde{a}_i^n Dg_i(x_0) \end{aligned}$$

(46)

for some $\tilde{a}_i^n$, such that $\tilde{a}_i^n\ge 0$, $i\in J_2$, $n\in \mathbb {N}$.

By assumption, $(Dg_k(x_0))_{k\in J_2}$ is a basis of $X_1=\overline{{\text {span}}}( Dg_i(x_0),\ i\in I(x_0) )=\overline{{\text {span}}}( Dg_i(x_0),\ i\in J_2 )$, hence, by Proposition 1, there exists $Dg_k^*(x_0)\in X_1^*$, $k\in J_2$ such that

$$\begin{aligned} Dg_k^*(x_0)(Dg_i(x_0))=\delta _{ik}\quad i,k \in J_2. \end{aligned}$$

By assumption $(Dg_k^*(x_0))_{k\in J_2}$ is a basis for $\overline{{\text {span}}}( Dg_i^*(x_0),\ i\in J_2 )$.

Let $\tilde{x}=Dg_j^*(x_0)\in X_1^*\subset E^{**}$, $j\in J_2$. By assumption, $\tilde{x}\in E$. By Corollary 3, (45) and (46) with $x=\tilde{x}$ we have

$$\begin{aligned}&\langle Dg_j^*(x_0), \sum _{i\in I(x_0)} a_i^n Dg_i(x_0) \rangle _E\\&\quad =\langle Dg_j^*(x_0), \sum _{i\in J_2} \tilde{a}_i^n Dg_i(x_0) \rangle _E\rightarrow \left\langle Dg_j^*(x_0), \sum _{i\in J_2} \tilde{\beta }_i Dg_i(x_0) \right\rangle _E, \end{aligned}$$

which means that

$$\begin{aligned} \tilde{a}_j^n\rightarrow \tilde{\beta }_j\quad j\in J_2 \quad \text {as}\ n\rightarrow +\infty . \end{aligned}$$

Since $\tilde{a}_j^n\ge 0$, $j\in J_2$, $n\in \mathbb {N}$ we obtain that $\tilde{\beta }_j\ge 0$, $j\in J_2$. Therefore, $z_0=\sum _{i\in J_2} \tilde{\beta }_i Dg_i(x_0)$, $\tilde{\beta }_j\ge 0$, $j\in J_2$. By Corollary 3, $z_0\in {{\mathcal {M}}}(x_0,0)$. $\square$

Remark 19

Let us note that if E is reflexive, assumptions 4. and 5. of Proposition 12 are automatically satisfied.

Definition 14

([25, Definition 3]) We say that the system ${{\mathcal {F}}}$ given in ($P_0$) with $I_0=\emptyset$ satisfies the Nonlinear Farkas-Minkowski Constraint Qualification (NFMCQ) at $x_0$ if the set

$$\begin{aligned} {\text {cone}} \{ (Dg_i(x_0),\langle Dg_i(x_0), x_0 \rangle - g_i(x_0)), i \in I_1 \} \end{aligned}$$

is weak*-closed in the product space $E^* \times \mathbb {R}.$

Corollary 4

Let E be reflexive and assume that NFMCQ holds. Then

$$\begin{aligned} z\in {{\mathcal {M}}}(x_0,0)&\implies z \in {\text {cone}} (Dg_i(x_0),i\in I_1). \end{aligned}$$

Proof

Let $z \in {{\mathcal {M}}}(x_0,0)$ and $I(x_0)=\{i_1,i_2,\dots \}$, $I_n(x_0)=\{i_1,i_2,\dots ,i_n \}$, $n\in \mathbb {N}$. Then

$$\begin{aligned}&z\in {{\mathcal {M}}}(x_0,0) \implies z\in {\text {cl}}\left( \bigcup _{n\in \mathbb {N}} \{ \tilde{z}\in E^*, \tilde{z} = \sum _{I_n(x_0)} \lambda _i Dg_i(x_0) \} \right) \\&\implies z\in {\text {cl}} \left( \bigcup _{n\in \mathbb {N}} {\text {cone}} (Dg_i(x_0),i\in I_n(x_0)\right) \\&\implies z \in {\text {cl}} {\text {cone}} (Dg_i(x_0),i\in I(x_0))\\&\implies z \in {\text {cl}} {\text {cone}} (Dg_i(x_0),i\in I_1). \end{aligned}$$

Assuming that NFMCQ hold and $E^*$ is reflexive, we obtain

$$\begin{aligned}&z \in {\text {cl}} {\text {cone}} (Dg_i(x_0),i\in I_1)\\&\implies z \in {\text {cone }} (Dg_i(x_0), i\in I(x_0). \end{aligned}$$

$\square$

Remark 20

Let E be a Banach space and F be a Hilbert space with basis $(b_i)\subset F$, $i\in \mathbb {N}$. Let $K=\{ y=\sum _{i\in \mathbb {N} } b_i^*(y)b_i\in F \mid b_i^*(y)\le 0, i\in I_1 \}$, $I_1=\mathbb {N}$. Let $G:\ E\rightarrow F$ be of class $C^1$. Let $x_0\in E$ and $y_0:=G(x_0)\in K$. By Corollary 2, for any $z\in E^*$,

$$\begin{aligned} z \in {\text {cone}} (Dg_i(x_0),i\in I(x_0)) \implies z\in {{\mathcal {M}}}(x_0,0). \end{aligned}$$

The following example shows that NFMCQ may not hold, when ${{\mathcal {M}}}(x_0,0)$ is weak*-closed.

Example 6

Let $E=\ell _2$, $F=\ell _2$ with basis $b_i=e_i$, $i\in \mathbb {N}$. Let $x_0=0\in E$ and consider

$$\begin{aligned} {{\mathcal {F}}}:= \{ x \in E \mid g_i(x)\le 0, i\in \mathbb {N} \}, \end{aligned}$$

where $g_i(x)=\frac{1}{i}x_i$, $i\in \mathbb {N}$.

In this case $I(x_0)=\mathbb {N}$, $Dg_i(x_0)=(0,\dots ,0,\frac{1}{i},0,\dots )$. By Corollary 2, $y=(\frac{1}{2},\frac{1}{2^2},\dots )\in {{\mathcal {N}}}_K(G(x_0))$. Moreover, by Corollary 3, $z:=DG^*(x_0)(y)=(\frac{1}{2},\frac{1}{2\cdot 2^i},\frac{1}{3\cdot 2^3},\dots ) \in {{\mathcal {M}}}(x_0,0)$. We have $z\notin {\text {cone }} (Dg_i(x_0), i\in I_1)$, and since E is reflexive, we conclude by Corollary 4 that NFMCQ does not hold in this case. However, the assumptions of Proposition 12 hold for $J_2=I(x_0)=\mathbb {N}$, therefore ${{\mathcal {M}}}(x_0,0)$ is weak*-closed.

We close the paper by illustrating our results with the following graph (Fig. 1).

Fig. 1

Relationships between the relevant concepts of constraint qualifications