It is well known that error bounds play important roles in the study of variational inequality problems. They allow one to estimate how far a feasible element is from the solution set without even having computed a single solution of the associated variational inequality. In [28], Aussel et al. provided the following two error bounds.
Theorem DA1
(Theorem 1 of [28])
Let
\(g, A:R^{n}\to R^{n}\)
be Lipschitz continuous on
\(R^{n}\) (with constants
l
and
L, respectively), and let
\(K:R^{n}\to2^{R^{n}}\)
be a set-valued map such that
\(K(x)\)
is a closed convex set in
\(R^{n}\), for each
\(x\in R^{n}\). Suppose the following hold:
-
(a)
\((A, g)\)
is a strongly monotone couple on
\(R^{n}\)
with constant
μ,
-
(b)
there exists
\(0< k<\frac{\mu}{l}\)
such that, for any
\(\theta >\frac{Lk}{\mu-lk}\),
$$ \bigl\Vert P_{K(x)}^{\theta}z-P_{K(y)}^{\theta}z \bigr\Vert \leq k\Vert x-y\Vert ,\quad\forall x,y,z\in R^{n}. $$
If
\(x^{*}\)
is the solution of (2.2), then, for any
\(x\in R^{n}\)
and any
\(\theta>\frac{Lk}{\mu-lk}\), we have
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{(\theta l+L)}{\theta\mu-(\theta l+L)k}\bigl\Vert R^{\theta}(x)\bigr\Vert , $$
where
\(R^{\theta}(x)=Ax-P_{K(x)}[Ax-\theta g(x)]\).
Theorem DA2
(Lemma 1 of [28])
Let
\(A:R^{n}\to R^{n}\)
be Lipschitz continuous on
\(R^{n}\) (with constant
L) on
\(R^{n}\)
and let
\(K:R^{n}\to2^{R^{n}}\)
be a set-valued map such that
\(K(x)\)
is a closed convex set in
\(R^{n}\), for each
\(x\in R^{n}\). Assume the following hold:
-
(a)
A is strongly monotone on
\(R^{n}\)
with constant
μ,
-
(b)
there exists
\(0< k<\mu\)
such that, for any
\(\theta>\frac {L(8k+L)}{4(\mu-k)}\),
$$ \bigl\Vert P_{K(x)}^{\theta}z-P_{K(y)}^{\theta}z \bigr\Vert \leq k\Vert x-y\Vert ,\quad\forall x,y,z\in R^{n}. $$
If
\(x^{*}\)
is the solution of (2.3), then, for any
\(x\in R^{n}\)
and any
\(\theta>\frac{L(8k+L)}{4(\mu-k)}\), we have
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{4\theta}{4\theta(\mu-k)-L(8k+L)}\bigl\Vert R^{\theta}(x)\bigr\Vert , $$
where
\(R^{\theta}(x)=Ax-P_{K(x)}(Ax-\theta x)\).
In this section, we will develop some error bounds measuring the distance between any point and the exact solution of IMQVI (2.1) in Hilbert spaces. Let
$$ e(x,\rho)=Ax-P_{K(x)}^{f,\rho}\bigl[Ax-\rho g(x)\bigr] $$
denote the residue of the generalized f-projection equation (4.1). Then solving an IMQVI (2.1) problem is equivalent to finding a zero point of \(e(x,\rho)\). For any given \(x\in H\), the magnitude of \(\Vert e(x,\rho)\Vert \) depends on the value of ρ. Now we give the error bounds in terms of the residual function \(\Vert e(x,\rho)\Vert \).
Theorem 5.1
Let
H
be a real Hilbert space, and
\(g,A:H\to H\)
be Lipschitz continuous on
H (with constants
α
and
β, respectively). Let
\(K:H\to 2^{H}\)
be a set-valued mapping such that, for each
\(x\in H\), \(K(x)\subset H\)
is a closed convex set and
\(f:H\to R\cup\{+\infty\}\)
be proper, convex, and lower semicontinuous on
\(K(x)\). Assume that
-
(i)
\((A,g)\)
is a
μ-strongly monotone couple on
H;
-
(ii)
there exists
\(0< k<\frac{\mu}{\alpha}\)
such that, for any
\(\rho>\frac{\beta k}{\mu-\alpha k}\),
$$ \bigl\Vert P_{K(x)}^{f,\rho}z-P_{K(y)}^{f,\rho}z \bigr\Vert \leq k\Vert x-y\Vert ,\quad\forall x,y\in H,z\in\bigl\{ v|v=Ax- \rho g(x),x\in H\bigr\} . $$
If
\(x^{*}\)
is the solution of IMQVI (2.1), then, for any
\(x\in H\)
and any
\(\rho>\frac{\beta k}{\mu-\alpha k}\), we have
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{(\alpha\rho+\beta)}{\mu\rho-(\alpha\rho+\beta)k} \bigl\Vert e(x,\rho)\bigr\Vert . $$
Proof
Denote \(u=P_{K(x^{*})}^{f,\rho}[Ax-\rho g(x)]\). Since \(x^{*}\) is the solution of IMQVI (2.1), it follows that
$$\begin{aligned} \bigl\langle \rho g\bigl(x^{*}\bigr),u-Ax^{*}\bigr\rangle +\rho f(u)-\rho f\bigl(Ax^{*}\bigr)\geq0, \end{aligned}$$
(5.1)
for all \(\rho>0\). From the definition of u and \(Ax^{*}\in K(x^{*})\), we know that
$$\begin{aligned} \bigl\langle u-\bigl[Ax-\rho g(x)\bigr],Ax^{*}-u\bigr\rangle +\rho f\bigl(Ax^{*} \bigr)-\rho f(u)\geq0. \end{aligned}$$
(5.2)
By (5.1) and (5.2), we have
$$\begin{aligned} 0 \leq&\bigl\langle \rho\bigl[g\bigl(x^{*}\bigr)-g(x)\bigr]+Ax-u,u-Ax^{*}\bigr\rangle \\ =&\rho\bigl\langle g\bigl(x^{*}\bigr)-g(x),u-Ax\bigr\rangle +\rho\bigl\langle g \bigl(x^{*}\bigr)-g(x),Ax-Ax^{*}\bigr\rangle \\ &{}+\langle Ax-u,u-Ax\rangle+\bigl\langle Ax-u,Ax-Ax^{*}\bigr\rangle . \end{aligned}$$
(5.3)
Since \((A,g)\) is a μ-strongly monotone couple, it follows from (5.3) that
$$\begin{aligned} \rho\bigl\langle g\bigl(x^{*}\bigr)-g(x),u-Ax\bigr\rangle +\bigl\langle Ax-u,Ax-Ax^{*}\bigr\rangle \geq \rho \mu\bigl\Vert x^{*}-x\bigr\Vert ^{2}+\Vert Ax-u\Vert ^{2}. \end{aligned}$$
(5.4)
In light of the facts that g is α-Lipschitz continuous and A is β-Lipschitz continuous, (5.4) implies that
$$ \begin{aligned}[b] \mu\rho\bigl\Vert x^{*}-x\bigr\Vert ^{2}\leq{}&\rho\bigl\Vert g \bigl(x^{*}\bigr)-g(x)\bigr\Vert \Vert u-Ax\Vert +\Vert Ax-u\Vert \bigl\Vert Ax-Ax^{*}\bigr\Vert \\ \leq{}&(\rho\alpha+\beta)\bigl\Vert x^{*}-x\bigr\Vert \bigl\Vert P_{K(x^{*})}^{f,\rho}\bigl[Ax-\rho g(x)\bigr]-Ax\bigr\Vert \\ \leq{}&(\rho\alpha+\beta)\bigl\Vert x^{*}-x\bigr\Vert \bigl(\bigl\Vert P_{K(x^{*})}^{f,\rho}\bigl[Ax-\rho g(x)\bigr]-P_{K(x)}^{f,\rho} \bigl[Ax-\rho g(x)\bigr]\bigr\Vert \\ &{}+\bigl\Vert P_{K(x)}^{f,\rho}\bigl[Ax-\rho g(x)\bigr]-Ax\bigr\Vert \bigr) \\ \leq{}&(\rho\alpha+\beta)\bigl\Vert x^{*}-x\bigr\Vert \bigl(k\bigl\Vert x^{*}-x \bigr\Vert +\bigl\Vert e(x,\rho)\bigr\Vert \bigr), \end{aligned} $$
(5.5)
for any \(\rho>\frac{\beta k}{\mu-\alpha k}\). Since \(\alpha k<\mu\) and \(\rho>\frac{\beta k}{\mu-\alpha k}\), it follows from (5.5) that
$$ \bigl\Vert x^{*}-x\bigr\Vert \leq\frac{(\alpha\rho+\beta)}{\mu\rho-(\alpha\rho+\beta)k} \bigl\Vert e(x,\rho)\bigr\Vert . $$
This completes the proof. □
If \(H=R^{n}\) and \(f(x)=0\) for all \(x\in R^{n}\), from Theorem 5.1, we obtain the following theorem.
Theorem 5.2
Let
\(g,A:R^{n}\to R^{n}\)
be Lipschitz continuous on
\(R^{n}\) (with constants
l
and
L, respectively). Let
\(K:R^{n}\to2^{R^{n}}\)
be a set-valued mapping such that, for each
\(x\in R^{n}\), \(K(x)\subset R^{n}\)
is a closed convex set. Assume that
-
(i)
\((A,g)\)
is a
μ-strongly monotone couple on
\(R^{n}\);
-
(ii)
there exists
\(0< k<\frac{\mu}{l}\)
such that, for any
\(\theta >\frac{Lk}{\mu-lk}\),
$$ \Vert P_{K(x)}z-P_{K(y)}z\Vert \leq k\Vert x-y\Vert , \quad\forall x,y\in R^{n},z\in \bigl\{ v|v=Ax-\theta g(x),x\in R^{n}\bigr\} . $$
If
\(x^{*}\)
is the solution of (2.2), then, for any
\(x\in R^{n}\)
and any
\(\theta>\frac{Lk}{\mu-lk}\), we have
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{(\theta l+L)}{\theta\mu-(\theta l+L)k}\bigl\Vert R^{\theta}(x)\bigr\Vert , $$
where
\(R^{\theta}(x)=Ax-P_{K(x)}[Ax-\theta g(x)]\).
For any \(x\in H\), based on Theorem 3.3, we know that \(P_{K(x)}^{f,\rho }\) is co-coercive mapping with modulus 1 on H. Applying the co-coercivity of \(P_{K(x)}^{f,\rho}\), we prove another error bound for IMQVI (2.1).
Theorem 5.3
Let
H
be a real Hilbert space, and
\(g,A:H\to H\)
be Lipschitz continuous on
H (with constants
α
and
β, respectively). Let
\(K:H\to 2^{H}\)
be a set-valued mapping such that, for each
\(x\in H\), \(K(x)\subset H\)
is a closed convex set and
\(f:H\to R\cup\{+\infty\}\)
be proper, convex, and lower semicontinuous on
\(K(x)\). Assume that
-
(i)
\((A,g)\)
is a
μ-strongly monotone couple on
H;
-
(ii)
there exists
\(0< k<\frac{\mu}{\alpha}\)
such that, for any
\(\rho>\frac{\beta(\beta+8k)}{4(\mu-\alpha k)}\),
$$ \bigl\Vert P_{K(x)}^{f,\rho}z-P_{K(y)}^{f,\rho}z \bigr\Vert \leq k\Vert x-y\Vert ,\quad\forall x,y\in H,z\in\bigl\{ v|v=Ax- \rho g(x),x\in H\bigr\} . $$
If
\(x^{*}\)
is the solution of IMQVI (2.1), then, for any
\(x\in H\)
and any
\(\rho>\frac{\beta(\beta+8k)}{4(\mu-\alpha k)}\), we have
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{4\alpha\rho}{4\rho(\mu-\alpha k)-\beta(\beta+8 k)} \bigl\Vert e(x,\rho)\bigr\Vert . $$
Proof
Denote \(v=Ax-\rho g(x)\) and \(v^{*}=Ax^{*}-\rho g(x^{*})\). From the definition of \(e(x,\rho)\), we know that
$$\begin{aligned} &\bigl\langle e(x,\rho),g(x)-g\bigl(x^{*}\bigr)\bigr\rangle \\ &\quad =\bigl\langle e(x,\rho)-e\bigl(x^{*},\rho\bigr),g(x)-g\bigl(x^{*}\bigr)\bigr\rangle \\ &\quad =\bigl\langle P_{K(x^{*})}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v,g(x)-g \bigl(x^{*}\bigr)\bigr\rangle +\bigl\langle Ax-Ax^{*},g(x)-g\bigl(x^{*}\bigr)\bigr\rangle \\ &\quad =\frac{1}{\rho}\bigl\langle P_{K(x^{*})}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v,Ax-Ax^{*} \bigr\rangle +\frac{1}{\rho}\bigl\langle P_{K(x^{*})}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v,v^{*}-v \bigr\rangle \\ &\qquad {}+\bigl\langle Ax-Ax^{*},g(x)-g\bigl(x^{*}\bigr)\bigr\rangle \\ &\quad =\frac{1}{\rho}\bigl\langle P_{K(x)}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v,Ax-Ax^{*} \bigr\rangle -\frac{1}{\rho }\bigl\langle P_{K(x)}^{f,\rho}v^{*}-P_{K(x^{*})}^{f,\rho}v^{*},Ax-Ax^{*} \bigr\rangle \\ &\qquad {}+\frac{1}{\rho}\bigl\langle P_{K(x)}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v,v^{*}-v \bigr\rangle -\frac{1}{\rho }\bigl\langle P_{K(x)}^{f,\rho}v^{*}-P_{K(x^{*})}^{f,\rho}v^{*},v^{*}-v \bigr\rangle \\ &\qquad {}+\bigl\langle Ax-Ax^{*},g(x)-g\bigl(x^{*}\bigr)\bigr\rangle . \end{aligned}$$
(5.6)
Since \((A,g)\) is a μ-strongly monotone couple, it follows from (5.6) and Theorem 3.3 that
$$\begin{aligned} &\bigl\langle e(x,\rho),g(x)-g\bigl(x^{*}\bigr)\bigr\rangle \\ &\quad \geq\frac{1}{\rho}\bigl\langle P_{K(x)}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v,Ax-Ax^{*} \bigr\rangle + \frac{1}{\rho}\bigl\Vert P_{K(x)}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v \bigr\Vert ^{2}+\mu\bigl\Vert x-x^{*}\bigr\Vert ^{2} \\ &\qquad {}-\frac{1}{\rho}\bigl\Vert P_{K(x)}^{f,\rho}v^{*}-P_{K(x^{*})}^{f,\rho}v^{*} \bigr\Vert \bigl\Vert Ax-Ax^{*}\bigr\Vert -\frac{1}{\rho}\bigl\Vert P_{K(x)}^{f,\rho}v^{*}-P_{K(x^{*})}^{f,\rho}v^{*}\bigr\Vert \bigl\Vert v^{*}-v\bigr\Vert \\ &\quad =\frac{1}{\rho}\biggl\Vert P_{K(x)}^{f,\rho}v^{*}-P_{K(x)}^{f,\rho}v+ \frac{1}{2}\bigl(Ax-Ax^{*}\bigr)\biggr\Vert ^{2}- \frac{1}{4\rho}\bigl\Vert Ax-Ax^{*}\bigr\Vert ^{2}+\mu\bigl\Vert x-x^{*}\bigr\Vert ^{2} \\ &\qquad {}-\frac{1}{\rho}\bigl\Vert P_{K(x)}^{f,\rho}v^{*}-P_{K(x^{*})}^{f,\rho}v^{*} \bigr\Vert \bigl(\bigl\Vert Ax-Ax^{*}\bigr\Vert +\bigl\Vert v^{*}-v \bigr\Vert \bigr) \\ &\quad \geq\mu\bigl\Vert x-x^{*}\bigr\Vert ^{2}-\frac{1}{4\rho} \beta^{2}\bigl\Vert x-x^{*}\bigr\Vert ^{2}- \frac{1}{\rho}k(2\beta+\alpha\rho)\bigl\Vert x-x^{*}\bigr\Vert ^{2}, \end{aligned}$$
(5.7)
for any \(\rho>\frac{\beta(\beta+8k)}{4(\mu-\alpha k)}\). On the other hand, we have
$$\begin{aligned} \bigl\langle e(x,\rho),g(x)-g\bigl(x^{*}\bigr)\bigr\rangle \leq&\bigl\Vert e(x, \rho)\bigr\Vert \bigl\Vert g(x)-g\bigl(x^{*}\bigr)\bigr\Vert \\ \leq&\alpha\bigl\Vert e(x,\rho)\bigr\Vert \bigl\Vert x-x^{*}\bigr\Vert . \end{aligned}$$
(5.8)
Since \(\alpha k<\mu\) and \(\rho>\frac{\beta(\beta+8k)}{4(\mu-\alpha k)}\), it follows from (5.7) and (5.8) that
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{4\alpha\rho}{4\rho(\mu-\alpha k)-\beta(\beta+8 k)} \bigl\Vert e(x,\rho)\bigr\Vert . $$
This completes the proof. □
If \(H=R^{n}\), g is identity mapping in \(R^{n}\), and \(f(x)=0\) for all \(x\in R^{n}\), by using Theorem 5.3, we have the following theorem.
Theorem 5.4
Let
\(A:R^{n}\to R^{n}\)
be Lipschitz continuous on
\(R^{n}\) (with constant
L) on
\(R^{n}\)
and let
\(K:R^{n}\to2^{R^{n}}\)
be a set-valued map such that
\(K(x)\)
is a closed convex set in
\(R^{n}\), for each
\(x\in R^{n}\). Assume the following hold:
-
(i)
A is strongly monotone on
\(R^{n}\)
with constant
μ,
-
(ii)
there exists
\(0< k<\mu\)
such that, for any
\(\theta>\frac {L(8k+L)}{4(\mu-k)}\),
$$ \Vert P_{K(x)}z-P_{K(y)}z\Vert \leq k\Vert x-y\Vert , \quad\forall x,y\in H,z\in \bigl\{ v|v=Ax-\theta x,x\in R^{n}\bigr\} . $$
If
\(x^{*}\)
is the solution of (2.3), then, for any
\(x\in R^{n}\)
and any
\(\theta>\frac{L(8k+L)}{4(\mu-k)}\), we have
$$ \bigl\Vert x-x^{*}\bigr\Vert \leq\frac{4\theta}{4\theta(\mu-k)-L(8k+L)}\bigl\Vert R^{\theta}(x)\bigr\Vert , $$
where
\(R^{\theta}(x)=Ax-P_{K(x)}(Ax-\theta x)\).
Remark 5.1
It is easy to see that the condition (ii) in Theorem 5.4 is weaker than the condition (b) in Theorem DA2 and the condition (ii) in Theorem 5.2 is also weaker than the condition (b) in Theorem DA1.