Characterizations of the existence of solutions for variational inequality problems in Hilbert spaces

Abstract

In this work, we give necessary and sufficient conditions for the existence of solutions to the variational inequality problem: find $x_0 \in K$ such that $\langle F(x_0),y-x_0 \rangle \ge 0$, for every $y \in K$, where K is a nonempty closed convex subset of a real Hilbert space H and $F:K \rightarrow H$ is a monotone and continuous operator. These characterizations are given in terms of approximate fixed points sequences, as well as by Leray–Schauder condition. We apply our obtained results in a constrained convex minimization problem.

1 Introduction

Let $(H, \langle \cdot , \cdot \rangle )$ be a real Hilbert space and let K be a nonempty closed convex subset of H. Given a mapping $F:K \rightarrow H$, the variational inequality problem VIP(F, K) defined by F and K is formulated as:

$$\begin{aligned} \left\{ \begin{array}{ll} \text {Find} \ x_0 \in K &{}\text {such that} \\ \langle F(x_0),y-x_0 \rangle \ge 0, &{}\text {for all} \ y \in K. \end{array}\right. \end{aligned}$$

Variational inequalities problems were initially studied by Browder in [1], and by Hartman and Stampacchia in [2], and have many applications in physics, economics and engineering. They were originally developed as a tool for the study of certain classes of partial differential equations, particularly those related to the calculus of variations associated with minimization of functionals [3]. Shortly after the works of Browder, Hartman and Stampacchia, Minty studied in [4] the variational inequality problem and applied his results to the solution of Hammerstein’s nonlinear integral equations. Variational inequalities problems have been widely studied, since they are specially useful in the study of several nonlinear problems, such as optimization and equilibrium. They also have a close relationship with Economical Systems, as well as with Projected Dynamical Systems on infinite-dimensional Hilbert spaces, since the equilibrium points of a Economical System, and of a Projected Dynamical System, are exactly the solutions of a variational inequality, and of a problem of evolutionary variational inequalities, respectively (see, for instance, [5,6,7,8,9].) In this work, we give necessary and sufficient conditions for the existence of solutions to the $\textrm{VIP}(F,K)$, where K is a nonempty closed convex subset of a real Hilbert space H and F is a monotone and continuous operator. The main motivation for giving these characterizations is that they could be used to find other conditions, more general or independent of the classical ones (as coercivity, strong monotonicity or compactness), that guarantee the existence of solutions for variational inequalities determined by continuous and monotone mappings. These characterizations are given in terms of approximate fixed points sequences, as well as by Leray–Schauder condition. There are some works that use the Leray–Schauder condition in the study of the variational inequality problem, but determined by completely continuous fields (see [10]) or by completely upper semi-continuous field (see [11]), providing sufficient and sometimes necessary conditions for the problem to have a solution. On the other hand, as far as we know, there are no works that provide similar results using approximate fixed points sequences in the study of the variational inequality problem. The closest thing we have found has been the use of the concept of exceptional family of elements (EFE) for a certain class of operators, such as completely continuous fields (see, for instance, [12, Chapter 8] and [13]), completely upper semi-continuous field (see [11]), and k-set fields (see [14]). For this reason, we believe that this is one of the novelties of this work. Finally, we apply our obtained results in a constrained convex minimization problem, involving convex and continuously differentiable functions. We think that another novelty of this work is that we found a close relationship between minimizing sequences and approximate fixed points sequences. Specifically, given a continuously differentiable convex function, we construct a sequence that turns out to be a minimizing sequence of this function and, in turn, an approximate fixed points sequence of an auxiliary operator, whenever this sequence is bounded. Thus we relate an essential concept of the minimization theory with an essential concept of the fixed point theory. We have not found a similar result in the literature.

2 Preliminaries

Let $(H, \langle \cdot , \cdot \rangle )$ be a real Hilbert space and let K be a nonempty closed convex subset of H. Given a mapping $F:K \rightarrow H$, the variational inequality problem $\textrm{VIP}(F,K)$ defined by F and K is formulated as:

$$\begin{aligned} \left\{ \begin{array}{l} \text {Find} \ x_0 \in K \ \text {such that} \\ \langle F(x_0),y-x_0 \rangle \ge 0, \ \text {for all} \ y \in K. \end{array}\right. \end{aligned}$$

Monotone type mappings are especially useful for studying this problem. A mapping $F:D(F) \subset H \rightarrow H$ is said to be monotone if

$$\begin{aligned} \langle F(x)-F(y), x-y \rangle \ge 0, \text {for every} \ x,y \in D(F). \end{aligned}$$

F is said to be strongly monotone if there exists a constant $\mu >0$ such that

$$\begin{aligned} \langle F(x)-F(y), x-y \rangle \ge \mu \Vert x-y\Vert ^2, \text {for every} \ x,y \in D(F). \end{aligned}$$

Browder in [1], and Hartman and Stampacchia in [2], proved independently that if F is continuous and strongly monotone, then the $\textrm{VIP}(F,K)$ has a unique solution. We have the following theorem for monotone mappings, which will be useful in our main theorem.

Lemma 2.1

Let K be a nonempty closed convex subset of a real Hilbert space H and $F:K \rightarrow H$ be a continuous and monotone mapping. Then $x \in K$ is solution of VIP(F, K) if and only if

$$\begin{aligned} \langle F(y),y-x \rangle \ge 0, \text {for all} \ y \in K. \end{aligned}$$

Let K be a nonempty closed convex subset of a Hilbert space H. The metric projection of H onto K is the mapping $P_K:H \rightarrow K$ which assigns to each $x \in H$ the unique element $P_K(x)$ in K with the property

$$\begin{aligned} \Vert x-P_K(x)\Vert =\min \{ \Vert x-z\Vert : z \in K \}. \end{aligned}$$

The solutions of the $\textrm{VIP}(F,K)$ are characterized in terms of the metric projection as follows.

Lemma 2.2

Let K be a nonempty closed convex subset of a real Hilbert space H and $F:K \rightarrow H$ be a mapping. Then $x \in K$ is solution of VIP(F, K) if and only if x is a fixed point of the mapping $P_K \circ (I-\lambda F):K \rightarrow K$, for every $\lambda >0$.

We have the following characterization for the metric projection.

Lemma 2.3

Let H be a real Hilbert space, K be a closed convex subset of H and $x \in H$. We have that $y=P_K(x)$ if and only if $y \in K$ and

$$\begin{aligned} \langle y-x, z-y \rangle \ge 0, \quad \forall \, z \in K. \end{aligned}$$

As a consequence of Lemma 2.3, we have that the metric projection is firmly nonexpansive. We say that a mapping $T:D(T)\subset H \rightarrow H$ is firmly nonexpansive if

$$\begin{aligned} \Vert T(x)-T(y)\Vert ^2 \le \langle T(x)-T(y),x-y \rangle , \quad \text {for every} \ x,y \in D(T). \end{aligned}$$

All the above concepts and results, and other important facts regarding variational inequalities, can be seen, for instance, in [3, 12, 15].

A mapping $T:D(T) \subset H \rightarrow H$ is said to be pseudocontractive if

$$\begin{aligned} \langle (I-T)(x)-(I-T)(y),x-y \rangle \ge 0, \end{aligned}$$

for every $x,y \in D(T)$, that is, if $I-T$ is monotone. Pseudocontractive mappings are easily seen to be more general than nonexpansive ones. Recall that a mapping $T:D(T) \subset H \rightarrow H$ is said to be nonexpansive if the inequality $\Vert T(x)-T(y) \Vert \le \Vert x-y\Vert$ holds for every $x,y \in D(T)$. We say that a mapping $T:D(T) \subset H \rightarrow H$ is weakly inward on D(T) if

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+} \frac{1}{\lambda } \ d( (1-\lambda ) x + \lambda T(x), D(T) )=0, \end{aligned}$$

for all $x \in D(T)$. Such condition is weaker than the assumption of T maps the boundary of D(T) into D(T). All these last concepts can be seen in [16], in the context of Banach spaces.

3 Existence theorems for the variational inequality problem

Let H be a real Hilbert space, K be a closed convex subset of H and $F:K \rightarrow H$ be a continuous and monotone mapping. Since for each $\alpha >0$ the mapping $\alpha I+ F$ is strongly monotone, then $\textrm{VIP}(\alpha I+ F, K)$ has a unique solution, that is, for each $\alpha >0$ there exists a unique $x_\alpha \in K$ such that

$$\begin{aligned} \biggl \langle \alpha x_\alpha + F(x_\alpha ), y-x_\alpha \biggr \rangle \ge 0, \quad \text {for every} \ y \in K. \end{aligned}$$

(3.1)

Equivalently, $x_\alpha$ is the unique fixed point of the function $T_\alpha :=P_K \circ \bigl ( (1-\alpha ) I- F \bigr ): K \rightarrow K$. Notice that if the subset $\{ x_\alpha : \alpha >0 \}$ is bounded, then $\inf _{x \in K} \Vert x-Tx\Vert =0$, where $T:=P_K \circ (I-F): K \rightarrow K$. Indeed, since $P_K$ is nonexpansive, then

$$\begin{aligned} \Vert x_\alpha - Tx_\alpha \Vert&= \Vert T_\alpha x_\alpha - Tx_\alpha \Vert \\ {}&= \bigl \Vert P_K \bigl ( (1-\alpha ) \, x_\alpha - F(x_\alpha ) \bigr ) - P_K \bigl ( x_\alpha - F(x_\alpha ) \bigr ) \bigr \Vert \\&\le \alpha \Vert x_\alpha \Vert \rightarrow 0, \text {as} \ \alpha \rightarrow 0. \end{aligned}$$

We can consider the sequence $(y_n)_n$ in K given by $y_n=x_{1/n}$, for each $n \in \mathbb {N}$. By the above, we have that $y_n$ is the unique solution of the $\textrm{VIP}(\frac{1}{n}I+F, K)$:

$$\begin{aligned} \biggl \langle \frac{1}{n} y_n + F(y_n), y-y_n \biggr \rangle \ge 0, \quad \text {for every} \ y \in K. \end{aligned}$$

(3.2)

Moreover, if the sequence $(y_n)$ is bounded, then it is a bounded approximate fixed point sequence of the function $T:=P_K \circ (I-F): K \rightarrow K$, that is,

$$\begin{aligned} \Vert y_n-Ty_n\Vert \rightarrow 0, \quad \text {as} \ n \rightarrow \infty . \end{aligned}$$

(3.3)

In the following proposition, we will show that the existence of solution for the $\textrm{VIP}(F,K)$ is characterized by the boundedness of the sequence $(y_n)$ given in (3.2).

Proposition 3.1

Let H be a real Hilbert space, K be a closed and convex subset of H and $F:K \rightarrow H$ be a continuous and monotone mapping. The following conditions are equivalent:

• (a) The variational inequality problem VIP(F, K) has a solution.

• (b) The subset $\{ x_\alpha : \alpha >0 \}$ obtained in (3.1) is bounded.

• (c) The sequence $(y_n)$ obtained in (3.2) is bounded.

Furthermore, if one of the above conditions holds, passing to a subsequence if necessary, we have that $(y_n)$ converges weakly to a solution of the VIP(F, K).

Proof

$(a) \Rightarrow (b)$. Let $\omega$ be a solution of the VIP(F, K) and let $\alpha >0$. Since $\omega$ is the unique fixed point of the operator $T:=P_K \circ (I-F): K \rightarrow K$, and $x_\alpha$ is the unique fixed point of the operator $T_\alpha :=P_K \circ \bigl ((1-\alpha ) I-F \bigr ): K \rightarrow K$, by firmly nonexpansiveness of $P_K$ we have that

$$\begin{aligned} \Vert \omega - x_\alpha \Vert ^2&= \Vert P_K \bigl ( \omega - F(\omega ) \bigr ) - P_K \bigl ( (1-\alpha ) x_\alpha - F(x_\alpha ) \bigr ) \Vert ^2 \\ {}&\le \bigl \langle \omega - F(\omega ) - (1-\alpha ) x_\alpha + F(x_\alpha ) , \omega - x_\alpha \bigr \rangle \\ {}&= \bigl \langle \alpha \omega + (1-\alpha ) \omega - (1-\alpha ) x_\alpha - \bigl ( F(\omega ) - F(x_\alpha ) \bigr ) , \omega - x_\alpha \bigr \rangle \\ {}&= \alpha \langle \omega , \omega - x_\alpha \rangle + (1-\alpha ) \Vert \omega - x_\alpha \Vert ^2 - \bigl \langle F(\omega ) - F(x_\alpha ) , \omega - x_\alpha \bigr \rangle . \end{aligned}$$

Since F is monotone, by the above inequality we obtain that

$$\begin{aligned} \Vert \omega - x_\alpha \Vert ^2&\le \langle \omega , \omega - x_\alpha \rangle - \frac{1}{\alpha } \bigl \langle F(\omega ) - F(x_\alpha ) , \omega - x_\alpha \bigr \rangle \\&\le \langle \omega , \omega - x_\alpha \rangle \le \Vert \omega \Vert \, \Vert \omega - x_\alpha \Vert , \end{aligned}$$

and consequently, $\Vert \omega - x_\alpha \Vert \le \Vert \omega \Vert$. Hence, $\{ x_{\alpha }: \alpha >0 \}$ is bounded.

$(b) \Rightarrow (c)$. Obvious.

$(c) \Rightarrow (a)$. Suppose that the sequence $(y_n)$ obtained in (3.2) is bounded. By Inequality (3.2) and due to the monotonicity of F, we have that

$$\begin{aligned} \frac{1}{n} \langle y_n,y_n-y \rangle \le \langle F(y_n), y-y_n \rangle \le \langle F(y), y-y_n \rangle , \quad \forall \, n \in \mathbb {N}, \ \forall \, y \in K. \end{aligned}$$

Since $(y_n)$ is bounded, then for each $y \in K$ we have that $\frac{1}{n} \langle y_n,y_n-y \rangle \rightarrow 0$ as $n \rightarrow \infty$. Furthermore, there exists a subsequence $(y_{n_k})_k$ of $(y_n)_n$ such that $y_{n_k} \rightharpoonup \omega$ for some $\omega \in K$. Hence,

$$\begin{aligned} 0 \le \underset{k \rightarrow \infty }{\lim }\ \langle F(y), y-y_{n_k} \rangle = \langle F(y), y-\omega \rangle , \quad \text {for every} \ y \in K. \end{aligned}$$

Finally, since F is continuous and monotone, Lemma 2.1 and the last inequality imply that $\omega$ is a solution of the $\textrm{VIP}(F,K)$. $\square$

In the following theorem, we characterize the existence of solution for the $\textrm{VIP}(F,K)$ in terms of bounded approximate fixed points sequences.

Theorem 3.2

Let H be a real Hilbert space, K be a closed convex subset of H and $F:K \rightarrow H$ be a continuous and monotone mapping. Consider the following conditions:

• (i) The variational inequality problem VIP(F, K) has a solution.

• (ii) The mapping $T=P_K \circ \bigl ( I- F \bigr ): K \rightarrow K$ has a bounded approximate fixed point sequence.

Then Condition (i) implies Condition (ii). If F is uniformly continuous, then Conditions (i) and (ii) are equivalent. Furthermore, if Condition (i) holds, or if Condition (ii) and the uniform continuity of F hold, then the sequence $(y_n)$ obtained in (3.2) is a bounded approximate fixed point sequence of T which, passing to a subsequence if necessary, converges weakly to a solution of the VIP(F, K).

Proof

$(i) \Rightarrow (ii)$ If the $\textrm{VIP}(F,K)$ has a solution, by Proposition 3.1 we have that the sequence $(y_n)$ given in (3.2) is bounded and, passing to a subsequence if necessary, it converges weakly to a solution of the $\textrm{VIP}(F,K)$. By (3.3) we have that $(y_n)$ is an approximate fixed point sequence of T.

$(ii) \Rightarrow (i)$ Let $(\xi _n)_n$ be a bounded approximate fixed point sequence of $T=P_K \circ (I-F): K \rightarrow K$, that is,

$$\begin{aligned} \Vert \xi _n - T \xi _n \Vert \rightarrow 0, \text {as} \ n \rightarrow \infty . \end{aligned}$$

For each $n \in \mathbb {N}$, define $\eta _n=T \xi _n=P_K(\xi _n-F \xi _n)$. Notice that the sequence $(\eta _n)$ is also bounded. By Lemma 2.3 we have that

$$\begin{aligned} \langle \eta _n-(\xi _n-F(\xi _n)), z-\eta _n \rangle \ge 0, \quad \forall \, z \in K, \ \forall \, n \in \mathbb {N}, \end{aligned}$$

that is,

$$\begin{aligned} \langle F(\xi _n), z-\eta _n \rangle \ge \langle \xi _n-\eta _n, z-\eta _n \rangle , \quad \forall \, z \in K, \ \forall \, n \in \mathbb {N}. \end{aligned}$$

(3.4)

Let $z \in K$. We shall prove that $\liminf \nolimits _{n \rightarrow \infty }\langle F(\eta _n), z-\eta _n \rangle \ge 0$. Since $\Vert \xi _n-\eta _n\Vert \rightarrow 0$ as $n \rightarrow \infty$, and $(z-\eta _n)_n$ is bounded, we have that $\lim \nolimits _{n \rightarrow \infty }| \langle \xi _n-\eta _n, z-\eta _n \rangle | = \lim \nolimits _{n \rightarrow \infty } \Vert \xi _n - \eta _n\Vert \Vert z-\eta _n\Vert = 0$. By this last equality and by Inequality (3.4), we obtain that

$$\begin{aligned} \liminf _{n \rightarrow \infty } \langle F(\xi _n), z-\eta _n \rangle \ge 0. \end{aligned}$$

(3.5)

Now let $\varepsilon >0$ and $M>0$ be such that $\Vert z-\eta _n\Vert \le M$ for every $n \in \mathbb {N}$. Since F is uniformly continuous, there exists $\delta >0$ such that $\Vert F(x)-F(y)\Vert \le \frac{\varepsilon }{M}$, whenever $x,y \in K$ and $\Vert x-y\Vert \le \delta$. On the other hand, since $\Vert \xi _n-\eta _n\Vert \rightarrow 0$ as $n \rightarrow \infty$, there exists $N \in \mathbb {N}$ such that $\Vert \xi _n-\eta _n\Vert \le \delta$ for every $n \ge N$, and hence, $\Vert F(\xi _n)-F(\eta _n)\Vert \le \frac{\varepsilon }{M}$ for every $n \ge N$. Therefore,

$$\begin{aligned} | \langle F(\eta _n)-F(\xi _n), z-\eta _n \rangle | \le \Vert F(\eta _n)-F(\xi _n)\Vert \Vert z-\eta _n \Vert \le \varepsilon , \quad \forall \, n \ge N, \end{aligned}$$

that is, $\lim \nolimits _{n \rightarrow \infty }\langle F(\eta _n)-F(\xi _n), z-\eta _n \rangle =0$. Hence, by this equality and Inequality (3.5) we obtain that

$$\begin{aligned} \begin{aligned} \underset{{n \rightarrow \infty }}{\liminf } \langle F(\eta _n), z-\eta _n \rangle =&\underset{{n \rightarrow \infty }}{\lim } \langle F(\eta _n) - F(\xi _n), z-\eta _n \rangle + \\&\underset{{n \rightarrow \infty }}{\liminf } \langle F(\xi _n), z-\eta _n \rangle \ge 0, \end{aligned} \end{aligned}$$

(3.6)

as we claim. Now, due to the monotonicity of F, we have that

$$\begin{aligned} \langle F(\eta _n), z-\eta _n \rangle \le \langle F(z), z-\eta _n \rangle , \quad \forall \, n \in \mathbb {N}, \ \forall \, z \in K. \end{aligned}$$

(3.7)

Since $(\eta _n)$ is bounded, there exists a subsequence $(\eta _{n_k})_k$ of $(\eta _n)_n$ such that $\eta _{n_k} \rightharpoonup \omega$ for some $\omega \in K$. By this last fact and by Inequalities (3.6) and (3.7) we conclude that

$$\begin{aligned} 0 \le \liminf _{k \rightarrow \infty } \langle F(\eta _{n_k}), z-\eta _{n_k} \rangle \le \underset{k \rightarrow \infty }{\lim }\ \langle F(z), z-\eta _{n_k} \rangle = \langle F(z), z-\omega \rangle , \end{aligned}$$

for every $z \in K$. Finally, since F is continuous and monotone, Lemma 2.1 and the last inequality imply that $\omega$ is a solution of the $\textrm{VIP}(F,K)$. $\square$

For the next characterization of the existence of solution for $\textrm{VIP}(F,K)$, we need to introduce the Leray–Schauder condition. We say that a mapping $T:K \rightarrow K$ satisfies the Leray–Schauder condition if there exists $R>0$ such that $Tx \ne \lambda x$, for every $x \in K \cap S_R(0)$ and every $\lambda >1$, where $S_R(0):=\{ x \in H: \Vert x\Vert =R \}$. To know more about Leray–Schauder type alternatives related to variational inequality problems, see [12, 17].

The following characterization is given in closed convex cones. Recall that a nonempty subset C of a real linear space X is called a cone if for every $x \in C$ and every $\lambda \ge 0$ we have that $\lambda x \in C$.

Theorem 3.3

Let H be a real Hilbert space, K be a closed convex cone of H and $F:K \rightarrow H$ be a continuous and monotone mapping. The following conditions are equivalent:

• (i) The variational inequality problem VIP(F, K) has a solution.

• (ii) The mapping $T=P_K \circ \bigl ( I- F \bigr ): K \rightarrow K$ satisfies the Leray–Schauder condition.

Furthermore, if one of the above conditions holds, then the sequence $(y_n)$ obtained in (3.2) is a bounded approximate fixed point sequence of T which, passing to a subsequence if necessary, converges weakly to a solution of the VIP(F, K).

Proof

$(i) \Rightarrow (ii)$. Let $\omega$ be a solution of the $\textrm{VIP}(F,K)$. We proceed by contradiction supposing that T does not satisfy Leray–Schauder condition. Let $R >\Vert \omega \Vert$ and choose $x \in K \cap S_R(0)$ and $\lambda >1$ such that $Tx=\lambda x$. By Lemma 2.3 we have that

$$\begin{aligned} \biggl \langle P_K \bigl ( x- F(x) \bigr )-\bigl ( x- F(x) \bigr ), z-P_K \bigl ( x- F(x) \bigr ) \biggr \rangle \ge 0, \quad \forall \, z \in K. \end{aligned}$$

(3.8)

Replacing $P_K\bigl ( x- F(x) \bigr )$ by $\lambda x$ in Inequality (3.8) we obtain that

$$\begin{aligned} \bigl \langle (\lambda -1) x + F(x), z-\lambda x \bigr \rangle \ge 0, \quad \forall \, z \in K. \end{aligned}$$

(3.9)

Since K is a cone, for every $z \in K$, we have that $\lambda z \in K$. Therefore, Inequality (3.9) implies that

$$\begin{aligned} \bigl \langle (\lambda -1) x + F(x), z-x \bigr \rangle \ge 0, \quad \forall \, z \in K, \end{aligned}$$

and hence,

$$\begin{aligned} \bigl \langle F(x), z-x \bigr \rangle \ge (\lambda -1) \bigl \langle x, x-z \bigr \rangle , \quad \forall \, z \in K. \end{aligned}$$

(3.10)

Since F is monotone, taking $z= \omega$ in Inequality (3.10) we conclude that

$$\begin{aligned} \langle F(\omega ), \omega -x \rangle&\ge \langle F(x), \omega -x \rangle \\&\ge (\lambda -1) \bigl \langle x, x-\omega \bigr \rangle \\&= (\lambda -1) \bigl ( \Vert x\Vert ^2 - \langle x,\omega \rangle \bigr )\\&\ge (\lambda -1) \bigl ( \Vert x\Vert ^2 - \Vert x\Vert \Vert \omega \Vert \bigr ) \\&= (\lambda -1) \bigl ( R^2 - R \Vert \omega \Vert \bigr ) > 0, \end{aligned}$$

that is, $\langle F(\omega ), x-\omega \rangle <0$, contradicting the fact that $\omega$ is solution of the $\textrm{VIP}(F,K)$.

$(ii) \Rightarrow (i)$. Suppose that $P_K \circ (I-F)$ satisfies the Leray–Schauder condition and let $R>0$ be such that $P_K \bigl ( x-F(x) \bigr ) \ne \lambda x$ for every $x \in K \cap S_R(0)$ and $\lambda >1$. For each $\alpha >0$, let $x_\alpha$ be the unique solution of the $\textrm{VIP}(\alpha I+F, K)$, that is, $x_\alpha$ is the unique fixed point of the operator $T_\alpha :=P_K \circ \bigl ((1-\alpha ) I-F \bigr ): K \rightarrow K$. We will prove that $\Vert x_\alpha \Vert < R$ for all $\alpha >0$. First, we shall prove the function $\alpha \mapsto x_\alpha$, $\alpha >0$, is continuous. Indeed, let $\alpha$, $\beta \in (0,+\infty )$. Since $P_K$ is firmly nonexpansive, we have that

$$\begin{aligned} \Vert x_\alpha - x_\beta \Vert ^2&= \Vert P_K \bigl ( (1-\alpha ) x_\alpha - F(x_\alpha ) \bigr ) - P_K \bigl ( (1-\beta ) x_\beta - F(x_\beta ) \bigr ) \Vert ^2 \\ {}&\le \bigl \langle (1-\alpha ) x_\alpha - F(x_\alpha ) - (1-\beta ) x_\beta + F(x_\beta ) , x_\alpha - x_\beta \bigr \rangle \\&= \bigl \langle (1-\alpha ) x_\alpha - (1-\beta ) x_\alpha + (1-\beta ) x_\alpha - (1-\beta ) x_\beta \\ {}&\ \ \ - \bigl ( F(x_\alpha ) - F(x_\beta ) \bigr ) , x_\alpha - x_\beta \bigr \rangle \\&= (\beta - \alpha ) \langle x_\alpha , x_\alpha - x_\beta \rangle + (1-\beta ) \Vert x_\alpha - x_\beta \Vert ^2 \\&\ \ \ - \bigl \langle F(x_\alpha ) - F(x_\beta ) , x_\alpha - x_\beta \bigr \rangle . \end{aligned}$$

Since F is monotone, by the above inequality, we obtain that

$$\begin{aligned} \Vert x_\alpha - x_\beta \Vert ^2&\le \biggl (1-\frac{\alpha }{\beta } \biggr ) \langle x_\alpha , x_\alpha - x_\beta \rangle - \frac{1}{\beta } \bigl \langle F(x_\alpha ) - F(x_\beta ) , x_\alpha - x_\beta \bigr \rangle \\ {}&\le \biggl (1-\frac{\alpha }{\beta } \biggr ) \langle x_\alpha , x_\alpha - x_\beta \rangle \\ {}&\le \biggl | 1-\frac{\alpha }{\beta } \biggr | \, \Vert x_\alpha \Vert \, \Vert x_\alpha - x_\beta \Vert , \end{aligned}$$

and consequently, $\Vert x_\alpha - x_\beta \Vert \le \bigl | 1-\frac{\alpha }{\beta } \bigr | \Vert x_\alpha \Vert \rightarrow 0$ as $\beta \rightarrow \alpha$. Hence, the function $\alpha \mapsto x_\alpha$ is continuous in $(0,+\infty )$. As a consequence, we obtain that the function $g:(0,+\infty ) \rightarrow [0,+\infty )$ given by $g(\alpha )=\Vert x_\alpha \Vert$, is also continuous.

Now we will prove that $\Vert x_\alpha \Vert < R$ for every $\alpha >0$. First, assume that $\alpha > \frac{\Vert F(0)\Vert }{R}$. Since F is monotone, we have that $\langle F(0)-F(x_\alpha ), -x_\alpha \rangle \ge 0$, and by definition of $x_\alpha$, it follows that

$$\begin{aligned} \alpha \Vert x_\alpha \Vert ^2 \le \langle F(x_\alpha ), -x_\alpha \rangle \le \langle F(0), -x_\alpha \rangle \le \Vert F(0)\Vert \, \Vert x_\alpha \Vert \end{aligned}$$

and therefore, $\Vert x_\alpha \Vert \le \frac{\Vert F(0)\Vert }{\alpha } < R$. Now we shall prove that $\Vert x_\alpha \Vert < R$ for every $0<\alpha \le \frac{\Vert F(0)\Vert }{R}$. We proceed by contradiction supposing that there exists $0<\beta \le \frac{\Vert F(0)\Vert }{R}$ such that $\Vert x_{\beta }\Vert \ge R$. Since g is continuous, $g(\beta ) \ge R$ and $g(\alpha ) < R$ for every $\alpha > \frac{\Vert F(0)\Vert }{R}$, there exists $0<\gamma \le \frac{\Vert F(0)\Vert }{R}$ such that $\Vert x_{\gamma }\Vert = R$.

Since K is a cone, then for each $z \in K$ we have that $\frac{1}{1+\gamma } \, z \in K$. Therefore,

$$\begin{aligned} \biggl \langle \gamma x_\gamma + F(x_\gamma ), \frac{1}{1+\gamma } \, z-x_\gamma \biggr \rangle \ge 0, \quad \text {for every} \ z \in K. \end{aligned}$$

(3.11)

Multiplying both sides of Inequality (3.11) by $1+\gamma$, we obtain that

$$\begin{aligned} \biggl \langle (1+\gamma ) \, x_\gamma - \bigl ( x_\gamma - F(x_\gamma ) \bigr ), z - \, (1+\gamma ) \, x_\gamma \biggr \rangle= & {} \biggl \langle \gamma x_\gamma + F(x_\gamma ), z - \, (1+\gamma ) \, x_\gamma \biggr \rangle \nonumber \\\ge & {} 0, \end{aligned}$$

(3.12)

for every $z \in K$. Since $(1+\gamma ) \, x_\gamma \in K$, by Inequality (3.12) and by Lemma 2.3, we conclude that $P_K\bigl (x_\gamma -F(x_{\gamma })\bigr )=(1+\gamma ) \, x_\gamma$. This fact and equality $\Vert x_\gamma \Vert =R$ contradict Leray–Schauder condition. Therefore, $\Vert x_\alpha \Vert < R$ for every $\alpha >0$ and hence, the set $\{x_\alpha : \alpha >0 \}$ is bounded. Consequently, by Proposition 3.1 we conclude that the $\textrm{VIP}(F,K)$ has a solution.

Finally, if the $\textrm{VIP}(F,K)$ has a solution, by Proposition 3.1, we have that the sequence $(y_n)$ given in (3.2) is bounded and, passing to a subsequence if necessary, it converges weakly to a solution of the $\textrm{VIP}(F,K)$. By (3.3), we have that $(y_n)$ is an approximate fixed point sequence of T. $\square$

Recall that, given a normed space X, we say that a mapping $T:D(T) \subset X \rightarrow X$ is nonexpansive if $\Vert Tx -Ty\Vert \le \Vert x-y\Vert$, for every $x,y \in D(T)$. The following theorem gives necessary and sufficient conditions for the nonexpansive mapping T, defined in a Hilbert space, to have a fixed point.

Theorem 3.4

Let H be a real Hilbert space, K be a nonempty closed convex subset of H with $0 \in K$, and $T:K \rightarrow K$ be a nonexpansive mapping. The following conditions are equivalent:

(a):

T has a fixed point in K.

(b):

T has a bounded approximate fixed point sequence in K.

(c):

T satisfies the Leray–Schauder condition.

Proof

The proof (a) $\Leftrightarrow$ (b) can be found in [18, Theorem9.1], following the comment made immediately after the proof of [18, Theorem 9.2].

In order to prove (b) $\Rightarrow$ (c), we will use some results from [16]. Notice that if K is bounded, it is well known that conditions (a) and (b) are always satisfied, and condition (c) is also satisfied by an argument of emptiness, taking $R>0$ such that $\Vert x\Vert <R$ for every $x \in K$. Therefore, we can assume that K is unbounded. As a consequence of [16, Proposition 3.8], if C is a closed convex and unbounded subset of H and $S:C \rightarrow H$ is a pseudocontractive mapping with a bounded approximate fixed point sequence in C, then for each $x_0 \in C$ there exist $R>0$ and a mapping $G:H \times H \rightarrow \mathbb {R}$ satisfying that $G(x,x)>0$ for any $x \in H$ with $\Vert x\Vert \ge R$, such that $G(S(x)-x_0,x-x_0) \le G(x-x_0,x-x_0)$ for all $x \in C$. By [16, Lemma 3.1], the mapping G also satisfies that $G(\lambda x,y) = \lambda G(x,y)$ for every $x,y \in H$ and every $\lambda >0$. Since the mapping under consideration $T:K \rightarrow K$ is pseudocontractive with a bounded approximate fixed point sequence in K and $0 \in K$, in particular, we have that there exist $R>0$ and a mapping $G:H \times H \rightarrow \mathbb {R}$ such that $G(T(x),x) \le G(x,x)$ for all $x \in K$. From this, it follows that $Tx \ne \lambda x$ for every $x \in K \cap S_R(0)$ and every $\lambda >1$. Indeed, if $Tx = \lambda x$ for some $x \in K \cap S_R(0)$ and some $\lambda >1$, then we have that $\lambda G(x,x) = G(\lambda x,x) = G(Tx,x) \le G(x,x)$, which is not possible, since $G(x,x)>0$.

The implication (c) $\Rightarrow$ (b) is a direct consequence of [16, Theorem 3.9], since the mapping $T:K \rightarrow K$ is continuous, pseudocontractive and weakly inward on K, and $0 \in K$. $\square$

If we consider a real Hilbert space H, a closed convex cone K of H, and a uniformly continuous and monotone mapping $F: K \rightarrow H$, we notice that the mapping $T=P_K \circ (I-F):K \rightarrow K$ is not necessarily nonexpansive. For instance, if $H=K=\mathbb {R}$ and $F: \mathbb {R} \rightarrow \mathbb {R}$ is given by

$$\begin{aligned} Fx=\left\{ \begin{array}{ll} x, &{} x < 0, \\ x^2, &{}0 \le x \le 3, \\ x+6, &{}x>3, \end{array}\right. \end{aligned}$$

then F is a monotone and uniformly continuous mapping, but

$$\begin{aligned} Tx=P_K (x-Fx)=\left\{ \begin{array}{ll} 0, &{}x < 0, \\ x-x^2, &{}0 \le x \le 3, \\ -6, &{}x>3, \end{array}\right. \end{aligned}$$

is not a nonexpansive mapping, since $|T(x)-T(0)|=x^2-x>x=|x-0|$ for $2<x \le 3$. However, we have seen in Theorems 3.2 and 3.3, that for the function $T=P_K \circ (I-F):K \rightarrow K$, the conditions (a)–(c) of Theorem 3.4 are equivalent, even if T is not nonexpansive.

Regarding the convergence of $(y_n)$ in Theorem 3.3, we will show that if F is strongly monotone, then $(y_n)$ converges strongly to the unique solution of the $\textrm{VIP}(F,K)$.

Theorem 3.5

Let H be a real Hilbert space, K be a closed convex cone of H and $F:K \rightarrow H$ be a continuous and strongly monotone mapping. Then $(y_n)$ converges strongly to the unique solution of the VIP(F, K).

Proof

Due to Theorem 3.3, there exists $R>0$ such that $\Vert x_\alpha \Vert <R$ for every $\alpha >0$. Proceeding as in the proof of Theorem 3.3 and taking into account that F is strongly monotone with parameter $\mu >0$, we have that

$$\begin{aligned} \Vert x_\alpha - x_\beta \Vert ^2&\le (\beta - \alpha ) \langle x_\alpha , x_\alpha - x_\beta \rangle + (1-\beta ) \Vert x_\alpha - x_\beta \Vert ^2 \\ {}&\ \ \ - \bigl \langle F(x_\alpha ) - F(x_\beta ) , x_\alpha - x_\beta \bigr \rangle \\ {}&\le (\beta - \alpha ) \langle x_\alpha , x_\alpha - x_\beta \rangle + (1-\beta -\mu ) \Vert x_\alpha - x_\beta \Vert ^2, \end{aligned}$$

for every $\alpha , \beta >0$, Hence,

$$\begin{aligned} \Vert x_\alpha - x_\beta \Vert ^2 \le \frac{\beta - \alpha }{\beta + \mu } \, \langle x_\alpha , x_\alpha - x_\beta \rangle \le \biggl | \frac{\beta - \alpha }{\beta + \mu } \biggr | \, \Vert x_\alpha \Vert \, \Vert x_\alpha - x_\beta \Vert , \end{aligned}$$

and consequently, $\Vert x_\alpha - x_\beta \Vert \le \bigl | \frac{\beta - \alpha }{\beta + \mu } \bigr | \Vert x_\alpha \Vert \le \bigl | \frac{\beta - \alpha }{\beta + \mu } \bigr | R$. In particular, for every $n,m \in \mathbb {N}$ we obtain that

$$\begin{aligned} \Vert y_n-y_m\Vert = \Vert x_{1/n} - x_{1/m}\Vert \le \biggl | \frac{n-m}{n + \mu mn} \biggr | R \rightarrow 0, \quad \text {as} \ n,m \rightarrow \infty , \end{aligned}$$

that is, $(y_n)$ is a Cauchy sequence and therefore, $(y_n)$ converges strongly to an element of K. Finally, since a subsequence of $(y_n)$ converges weakly to the unique solution $\omega \in K$ of the $\textrm{VIP}(F,K)$, we conclude that $y_n \rightarrow \omega$ as $n \rightarrow \infty$. $\square$

Next we have the following corollaries of Theorem 3.3.

Corollary 3.6

Let H be a real Hilbert space, K be a closed convex cone of H and $F:K \rightarrow H$ be a continuous and monotone mapping. If there exists $R>0$ such that

$$\begin{aligned} \langle F(x),x \rangle \ge 0, \quad \forall \, x \in K \cap S_R(0), \end{aligned}$$

then the VIP(F, K) has a solution.

Proof

If the $\textrm{VIP}(F,K)$ has no solution, then T does not satisfy Leray–Schauder condition. Let $R >0$ be arbitrary and choose $x \in K \cap S_R(0)$ and $\lambda >1$ such that $Tx=\lambda x$. By Lemma 2.3, we have that

$$\begin{aligned} \biggl \langle P_K \bigl ( x- F(x) \bigr )-\bigl ( x- F(x) \bigr ), z-P_K \bigl ( x- F(x) \bigr ) \biggr \rangle \ge 0, \quad \forall \, z \in K. \end{aligned}$$

Replacing $P_K\bigl ( x- F(x) \bigr )$ by $\lambda x$ in last inequality, we obtain that

$$\begin{aligned} \bigl \langle (\lambda -1) x + F(x), z-\lambda x \bigr \rangle \ge 0, \quad \forall \, z \in K. \end{aligned}$$

In particular, when $z=x$, we have that

$$\begin{aligned} \bigl \langle (\lambda -1) x + F(x), (1-\lambda ) x \bigr \rangle \ge 0, \end{aligned}$$

since $\lambda >1$ we get that

$$\begin{aligned} \bigl \langle (\lambda -1) x + F(x), x \bigr \rangle \le 0, \end{aligned}$$

that is, $\langle F(x),x \rangle \le (1-\lambda ) \Vert x\Vert ^2<0$. $\square$

Corollary 3.7

Let H be a real Hilbert space, K be a closed convex cone of H and $F:K \rightarrow H$ be a continuous and monotone mapping. If the VIP(F, K) has a solution, then there exists $R>0$ such that

$$\begin{aligned} F(x) \ne \alpha x, \quad \forall \, x \in K \cap S_R(0), \quad \forall \, \alpha <0. \end{aligned}$$

Proof

Suppose that for every $R>0$, there exist $x \in K \cap S_R(0)$ and $\alpha <0$ such that $F(x) = \alpha x$, that is, $x- F(x)=(1-\alpha ) x$. Since K is a cone and $1-\alpha >1$, then $(1-\alpha ) x \in K$ and therefore, $P_K \bigl ( x- F(x) \bigr ) = P_K \bigl ( (1-\alpha ) x \bigr ) = (1-\alpha ) x$, that is, Leray–Schauder condition does not hold for T. Consequently, the $\textrm{VIP}(F,K)$ has no solution. $\square$

4 Application to the convex minimization problem

Let H be a Hilbert space, K be a convex subset of H and $f:K \rightarrow \mathbb {R}$ be a convex and Fréchet differentiable function. In this section, we consider the following constrained convex minimization problem of minimizing f over the constraint set K:

$$\begin{aligned} \text {minimize} \, \{ f(x): x \in K \}. \end{aligned}$$

(4.1)

An element $x \in K$ is a minimizer (or minimal point) of f over K if x is solution of Problem (4.1), that is, if $f(x)=\inf \{ f(y): y \in K \}$. Let us first see the following remarks in order to apply the results obtained previously to Problem (4.1).

Definition 4.1

Let X be a normed space, K be a nonempty subset of X and $f: K \rightarrow \mathbb {R}$ be a bounded below function. We say that a sequence $(x_n)_n$ in K is a minimizing sequence of f if the equality $\lim \nolimits _{n \rightarrow \infty }f(x_n) = \inf \{ f(y): y \in K \}$ holds.

Remark 4.2

Let X be a reflexive space, K be a closed and convex subset of X and $f: K \subseteq X \rightarrow \mathbb {R}$ be a convex and lower semi-continuous function.

• (i) On the one hand, by definition, there is always a minimizing sequence of f in K. However, this sequence is not necessarily bounded. For instance, the sequence $(x_n)_n$ given by $x_n=n$ for each $n \in \mathbb {N}$, is an unbounded minimizing sequence of the function $f:\mathbb {R} \rightarrow \mathbb {R}$ given by $f(x)=e^{-x}$.

• (ii) On the other hand, if there exists a bounded minimizing sequence $(x_n)_n$ of f in K, then f is bounded below and there exists a minimizer of f in K. Indeed, since $(x_n)_n$ is bounded, there exists a subsequence $(x_{n_k})_k$ of $(x_n)_n$ converging weakly to some x, which belongs to K, since K is weakly closed. Since f is convex and lower semicontinuous, then it is weakly lower semicontinuous and hence, $f(x) \le \liminf \nolimits _{k \rightarrow \infty }f(x_{n_k})=\inf \{f(y): y \in K \},$ that is, f is bounded below and x is a minimizer of f in K. Therefore, the existence of a minimizers of f is equivalent to the existence of a bounded minimizing sequence of f.

Remark 4.3

Examples of functions which all its minimizing sequences are bounded, are the coercive functions. Indeed, let K be an unbounded subset of a normed space X, let $f:K \rightarrow \mathbb {R}$ be coercive, that is, $f(x) \rightarrow +\infty$, as $\Vert x\Vert \rightarrow +\infty$, and suppose that $(x_n)$ is an unbounded minimizing sequence of f. Then there exists a subsequence $(x_{n_k})_k$ of $(x_n)_n$ such that $\Vert x_{n_k}\Vert \rightarrow \infty$ as $k \rightarrow \infty$. Since f is coercive, we have that $f(x_{n_k}) \rightarrow \infty$ as $k \rightarrow \infty$, which is not possible, since $f(x_{n_k}) \rightarrow \inf \{ f(y): y \in K \}$ as $k \rightarrow \infty$. As a consequence of this and Remark 4.2, we have that if X is a reflexive space, K is a closed convex unbounded subset of X and $f: K \subseteq X \rightarrow \mathbb {R}$ is a convex, lower semicontinuous and coercive function, then f has a minimizer in K.

Let H be a Hilbert space, K be a convex subset of H and $f:K \rightarrow \mathbb {R}$ be a convex and continuously differentiable function. It is well known that $x \in K$ is a minimal point of f if and only if

$$\begin{aligned} \langle \nabla f(x), y-x \rangle \ge 0, \quad \text {for every} \ y \in K. \end{aligned}$$

In other words, $x \in K$ is a solution of the Convex Minimization Problem (4.1) if and only if x is solution of the $\textrm{VIP}(\nabla f, K)$. Since $\nabla f:K \rightarrow H$ is monotone and continuous, we can consider the subset $\{x_\alpha : \alpha >0 \}$ and the sequence $(y_n)$ obtained in (3.1) and (3.2), respectively, in the case $F=\nabla f$, that is, $x_\alpha \in K$ is the unique solution of the $\textrm{VIP}(\alpha I+\nabla f, K)$

$$\begin{aligned} \biggl \langle \alpha x_\alpha + \nabla f(x_\alpha ), y-x_\alpha \biggr \rangle \ge 0, \quad \text {for every} \ y \in K, \end{aligned}$$

(4.2)

and $y_n$ is the unique solution of the $\textrm{VIP}(\frac{1}{n}I+\nabla f, K)$:

$$\begin{aligned} \biggl \langle \frac{1}{n} y_n + \nabla f(y_n), y-y_n \biggr \rangle \ge 0, \quad \text {for every} \ y \in K. \end{aligned}$$

(4.3)

Lemma 4.4

Let H be a Hilbert space, K be a convex subset of H and $f:K \rightarrow \mathbb {R}$ be convex and continuously differentiable. If the sequence $(y_n)$ obtained in (4.3) is bounded, then $(y_n)_n$ is a minimizing sequence of f in K.

Proof

By Inequality (4.3) and due to the convexity of f, we have that

$$\begin{aligned} \frac{1}{n} \langle y_n,y_n-y \rangle \le \langle \nabla f(y_n), y-y_n \rangle \le f(y)-f(y_n), \quad \forall \, n \in \mathbb {N}, \ \forall \, y \in K. \end{aligned}$$

Since $(y_n)$ is bounded, then for each $y \in K$ we have that $\frac{1}{n} \langle y_n,y_n-y \rangle \rightarrow 0$ as $n \rightarrow \infty$. Hence,

$$\begin{aligned} \limsup _{n \rightarrow \infty } f(y_n) \le f(y), \quad \forall \, y \in K, \end{aligned}$$

and then,

$$\begin{aligned} \limsup _{n \rightarrow \infty } f(y_n) \le \inf \{ f(y): y \in K \}. \end{aligned}$$

Obviously $\inf \{ f(y): y \in K \} \le \liminf _{n \rightarrow \infty } f(y_n)$ and hence we conclude that $\lim \nolimits _{n \rightarrow \infty } f(y_n) = \inf \{ f(y): y \in K \}$. $\square$

Proposition 3.1 can be applied in order to obtain the following characterization of the existence of solution for the convex minimization problem (4.1).

Corollary 4.5

Let H be a Hilbert space, K be a closed and convex subset of H and $f: K \rightarrow \mathbb {R}$ be a convex and continuously differentiable mapping. The following conditions are equivalent:

• (a) The convex minimization problem (4.1) has a solution.

• (b) The sequence $(y_n)$ obtained in (4.3) is bounded.

• (c) There exists a bounded minimizing sequence of f in K.

Furthermore, if one of the above conditions holds, then $(y_n)$ is a minimizing sequence of f which, passing to a subsequence if necessary, converges weakly to a solution of (4.1).

Proof

The equivalence of conditions (a) and (b) follows from Proposition 3.1, in the particular case of $F=\nabla f$. If the sequence $(y_n)$ obtained in (4.3) is bounded, by Lemma 4.4, we obtain that $(y_n)_n$ is a minimizing sequence of f in K, which proves the implication (b) $\Rightarrow$ (c). Finally, if there exists a bounded minimizing sequence $(x_n)$ of f in K, due to Remark 4.2 (ii), there exists a minimizer of f in K, which is the weak limit of a subsequence of $(x_n)$, proving the implication (c) $\Rightarrow$ (a). $\square$

As a direct consequence of Theorem 3.3, we have the following existence result for the Convex Minimization Problem (4.1) in convex closed cones.

Corollary 4.6

Let H be a Hilbert space, K be a closed convex cone of H and $f: K \rightarrow \mathbb {R}$ be a convex and continuously differentiable mapping. The following conditions are equivalent:

• (i) The convex minimization problem (4.1) has a solution.

• (ii) The mapping $T=P_K \circ \bigl ( I- \nabla f \bigr ): K \rightarrow K$ satisfies the Leray–Schauder condition.

In order to establish the strong convergence of the sequence $(y_n)$, we consider the following concept. We say that $f: K \rightarrow \mathbb {R}$ is strongly convex if there exists $\mu >0$ such that

$$\begin{aligned} f \bigl ( tx+(1-t)y \bigr ) \le tf(x)+(1-t)f(y)- \frac{\mu }{2} t(1-t) \Vert x-y\Vert ^2 \end{aligned}$$

for all $x,y \in K$ and $t \in [0,1]$. It is well known that f is strongly convex if and only if $\nabla f$ is strongly monotone. As a consequence of Theorem 3.5, we have the following convergence result.

Corollary 4.7

Let H be a Hilbert space, K be a closed convex cone of H and $f: K \rightarrow \mathbb {R}$ be a strongly convex and continuously differentiable mapping. Then $(y_n)$ converges strongly to the unique minimizer of f in K.

Finally, applying Corollaries 3.6 and 3.7, we obtain the following existence results for (4.1).

Corollary 4.8

If there exists $R>0$ such that

$$\begin{aligned} \langle \nabla f (x),x \rangle \ge 0, \quad \forall \, x \in K \cap S_R(0), \end{aligned}$$

then the convex minimization problem (4.1) has a solution.

Corollary 4.9

If the convex minimization problem (4.1) has a solution, then there exists $R>0$ such that

$$\begin{aligned} \nabla f(x) \ne \alpha x, \quad \forall \, x \in K \cap S_R(0), \ \forall \, \alpha <0. \\ \end{aligned}$$