We next consider an iterative scheme of resolvents of a monotone operator which is different type of Section 3, in a Banach space. Let C be a nonempty closed convex subset of a reflexive, smooth, and strictly convex Banach space E, let \(r>0\) and let \(A\subset E\times E^{*}\) be a monotone operator satisfying
$$ D(A) \subset C \subset J^{1}R(J+rA) $$
(4.1)
for \(r>0\). It is well known that if A is maximal monotone operator, then \(J^{1}R(J+rA)=E\); see [25–27]. Hence, if A is maximal monotone, then (4.1) holds for \(C=\overline{D(A)}\). We also know that \(\overline{D(A)}\) is convex; see [28]. If A satisfies (4.1) for \(r>0\), then we can define the resolvent (of type (Q)) \(Q_{r}:C\to D(A)\) of A by
$$ Q_{r} x =\{z\in E : Jx\in Jz+rAz \} $$
(4.2)
for all \(x\in C\). In other words, \(Q_{r}x=(J+rA)^{1}Jx\) for all \(x\in C\). We know the following; see, for instance, [17, 18]:

(1)
\(Q_{r}\) is mapping of type (Q) from C into \(D(A)\);

(2)
\((JxJQ_{r}x)/r \in AQ_{r} x\) for all \(x \in C\);

(3)
\(F(Q_{r})=A^{1}0\).
Before our result, we need the following lemma.
Lemma 4.1
Let
E
be a reflexive, smooth, and strictly convex Banach space, and let
\(A\subset E\times E^{*}\)
be a monotone operator. Let
\(r>0\)
and
C
be a closed convex subset of
E
satisfying (4.1) for
\(r>0\). Then the following holds:
$$V(x,Q_{r}x)+V(Q_{r}x,x)\leq2r\bigl\langle xQ_{r}x, x^{*}\bigr\rangle $$
for all
\((x,x^{*})\in A\).
Proof
Let \((x,x^{*})\in A\). Since \((JxJQ_{r} x)/r \in AQ_{r} x\), we see that
$$\begin{aligned}& 0 \leq \biggl\langle xQ_{r} x, x^{*}\frac{JxJQ_{r} x}{r} \biggr\rangle , \\& \biggl\langle xQ_{r} x,\frac{JxJQ_{r} x}{r} \biggr\rangle \leq \bigl\langle xQ_{r} x, x^{*}\bigr\rangle , \\& \langle xQ_{r} x,JxJQ_{r} x\rangle \leq r\bigl\langle xQ_{r} x, x^{*}\bigr\rangle . \end{aligned}$$
From the property of V, we see that
$$V(x,Q_{r}x)+V(Q_{r}x,x) = 2\langle xQ_{r}x, JxJQ_{r}x\rangle \leq2r\bigl\langle xQ_{r} x, x^{*}\bigr\rangle $$
for all \((x,x^{*})\in A\). □
We obtain an approximation theorem for a zero point of a monotone operator in a smooth and uniformly convex Banach space by using the resolvent of type (Q).
Theorem 4.2
Let
E
be a uniformly smooth and uniformly convex Banach space and let
\(A\subset E\times E^{*}\)
be a monotone operator with
\(A^{1}0\ne \emptyset\). Let
\(\{r_{n}\}\)
be a positive real numbers such that
\(\liminf_{n} r_{n} >0\), let
C
be a nonempty bounded closed convex subset of
E
satisfying
$$D(A) \subset C \subset J^{1}R(J+r_{n}A) $$
for all
\(n\in \mathbb {N}\)
and let
\(r\in\mathopen]0,\infty\mathclose[\)
such that
\(C \subset B_{r}\). Let
\(\{\delta_{n}\}\)
be a nonnegative real sequence and let
\(\delta_{0}=\limsup_{n} \delta_{n}\). For a given point
\(u\in E\), generate a sequence
\(\{x_{n}\}\)
by
\(x_{1} = x \in C\), \(C_{1} =C\), and
$$\begin{aligned} &y_{n}=Q_{r_{n}}x_{n}, \\ &C_{n+1} = \bigl\{ z \in C : \langle y_{n}  z, Jx_{n}  Jy_{n} \rangle\geq0\bigr\} \cap C_{n}, \\ &x_{n+1} \in\bigl\{ z \in C : \Vert uz\Vert ^{2} \leq d(u,C_{n+1})^{2}+\delta _{n+1}\bigr\} \cap C_{n+1}, \end{aligned}$$
for all
\(n \in \mathbb {N}\). Then
$$\limsup_{n\to\infty} \Vert x_{n}y_{n}\Vert \leq\underline{g}_{r}^{1}\bigl(\overline{g}_{r} \bigl(\underline{g}_{r}^{1}(\delta_{0})\bigr) \bigr). $$
Moreover, if
\(\delta_{0}=0\), then
\(\{x_{n}\}\)
converges strongly to
\(P_{A^{1}0}u\).
Proof
Since \(C_{n}\) includes \(A^{1}0\ne\emptyset\) for all \(n\in \mathbb {N}\), \(\{C_{n}\}\) is a sequence of nonempty closed convex subsets and, by definition, it is decreasing with respect to inclusion. Let \(p_{n}=P_{C_{n}}u\) for all \(n\in \mathbb {N}\). Then, by Theorem 2.4, we see that \(\{p_{n}\}\) converges strongly to \(p_{0}=P_{C_{0}}u\), where \(C_{0}=\bigcap_{n=1}^{\infty}C_{n}\). Since \(x_{n}\in C_{n}\) and \(d(u,C_{n})=\Vert up_{n} \Vert \), we see that
$$\Vert ux_{n} \Vert ^{2} \leq \Vert up_{n} \Vert ^{2} +\delta_{n} $$
for every \(n\in \mathbb {N}\setminus\{1\}\). From Theorem 2.6(i), we see that for \(\alpha\in\mathopen]0,1\mathclose[\),
$$\begin{aligned} \Vert p_{n} u \Vert ^{2} &\leq\bigl\Vert \alpha p_{n} +(1\alpha)x_{n}u \bigr\Vert ^{2} \\ &\leq\alpha \Vert p_{n} u \Vert ^{2} +(1\alpha)\Vert x_{n}u \Vert ^{2}  \alpha(1\alpha)\underline{g}_{r} \bigl(\Vert p_{n} x_{n} \Vert \bigr) \end{aligned}$$
and thus
$$\alpha\underline{g}_{r} \bigl(\Vert p_{n} x_{n} \Vert \bigr) \leq \Vert x_{n}u \Vert ^{2}  \Vert p_{n} u \Vert ^{2} \leq \delta_{n}. $$
As \(\alpha\to1\), we see that \(\underline{g}_{r} (\Vert p_{n} x_{n} \Vert )\leq\delta_{n}\) and thus \(\Vert p_{n} x_{n} \Vert \leq\underline{g}_{r}^{1}(\delta_{n})\). Using the definition of \(p_{n}\), we see that \(p_{n+1}\in C_{n+1}\) and thus
$$\langle y_{n}  p_{n+1}, Jx_{n}  Jy_{n} \rangle\geq0. $$
From the property of the function V, we see that
$$\begin{aligned} 0 &\leq2\langle y_{n}  p_{n+1}, Jx_{n}  Jy_{n} \rangle \\ &= 2\langle p_{n+1}  y_{n}, Jy_{n}  Jx_{n} \rangle \\ &= V(p_{n+1}, x_{n})V(p_{n+1}, y_{n})V(y_{n}, x_{n}) \\ &\leq V(p_{n+1}, x_{n})V(y_{n}, x_{n}). \end{aligned}$$
By Theorem 2.7, we obtain
$$\begin{aligned} V(y_{n}, x_{n}) &\leq V(p_{n+1}, x_{n}) \\ &= V(p_{n+1}, p_{n})+V(p_{n}, x_{n})+2 \langle p_{n+1}p_{n}, Jp_{n}Jx_{n}\rangle \\ &\leq V(p_{n+1}, p_{n})+\overline{g}_{r}\bigl( \Vert p_{n}x_{n}\Vert \bigr) +2\langle p_{n+1}p_{n}, Jp_{n}Jx_{n}\rangle \\ &\leq V(p_{n+1}, p_{n})+\overline{g}_{r}\bigl( \underline{g}_{r}^{1}(\delta_{n})\bigr) +2\langle p_{n+1}p_{n}, Jp_{n}Jx_{n}\rangle. \end{aligned}$$
Since \(\limsup_{n}\delta_{n}=\delta_{0}\) and \(p_{n} \to p_{0}\), we see that
$$\limsup_{n\to\infty} V(y_{n}, x_{n}) \leq \overline{g}_{r}\bigl(\underline {g}_{r}^{1}( \delta_{0})\bigr). $$
Therefore, by Theorem 2.7, we see that
$$\limsup_{n\to\infty} \Vert x_{n}y_{n}\Vert \leq\limsup_{n\to\infty} \underline{g}_{r}^{1} \bigl(V(y_{n}, x_{n})\bigr) \leq\underline{g}_{r}^{1} \bigl(\overline{g}_{r}\bigl(\underline{g}_{r}^{1}( \delta_{0})\bigr)\bigr). $$
For the latter part of the theorem, suppose that \(\delta_{0}=0\). Then we see that
$$\limsup_{n\to\infty} \Vert x_{n}y_{n}\Vert \leq \underline{g}_{r}^{1}\bigl(\overline{g}_{r} \bigl(\underline{g}_{r}^{1}(0)\bigr)\bigr)=0 $$
and
$$\limsup_{n\to\infty}\underline{g}_{r}\bigl(\Vert x_{n}p_{n}\Vert \bigr) \leq\limsup_{n\to\infty} \delta_{n}=0. $$
Therefore, we obtain
$$\lim_{n\to\infty} \Vert x_{n}y_{n}\Vert =0 \quad \text{and}\quad \lim_{n\to\infty} \Vert x_{n}p_{n} \Vert =0. $$
Hence, we also obtain
$$ \lim_{n\to\infty} x_{n} = p_{0} \quad \text{and}\quad \lim_{n\to\infty} y_{n} = p_{0}. $$
(4.3)
Since E is uniformly smooth, the duality mapping J is uniformly normtonorm continuous on each bounded subset on E. Therefore, we obtain
$$ \lim_{n\to\infty} \Vert Jx_{n}Jy_{n} \Vert =0. $$
(4.4)
From Lemma 4.1 we see that
$$V(y_{n}, Q_{r_{1}}y_{n}) \leq V(y_{n}, Q_{r_{1}}y_{n})+V(Q_{r_{1}}y_{n}, y_{n}) \leq2r_{1}\bigl\langle y_{n}Q_{r_{1}}y_{n}, x^{*}\bigr\rangle $$
for all \(x^{*}\in Ay_{n}\). From \(y_{n}\), \(Q_{r_{1}}y_{n} \in D(A) \subset C \subset B_{r}\) and \((Jx_{n}Jy_{n})/r_{n} \in Ay_{n}\), we see that
$$\begin{aligned} V(y_{n}, Q_{r_{1}}y_{n}) \leq& 2r_{1} \biggl\langle y_{n}Q_{r_{1}}y_{n}, \frac{Jx_{n}Jy_{n}}{r_{n}} \biggr\rangle \\ \leq& 2r_{1}\Vert y_{n}Q_{r_{1}}y_{n} \Vert \biggl\Vert \frac{Jx_{n}Jy_{n}}{r_{n}}\biggr\Vert \\ \leq& 2r_{1} \bigl(\Vert y_{n}\Vert +\Vert Q_{r_{1}}y_{n} \Vert \bigr) \biggl\Vert \frac{Jx_{n}Jy_{n}}{r_{n}} \biggr\Vert \\ = & 4r_{1} r \biggl\Vert \frac{Jx_{n}Jy_{n}}{r_{n}}\biggr\Vert . \end{aligned}$$
Since \(\liminf_{n} r_{n} > 0\) and (4.4), we obtain
$$\limsup_{n\to\infty} V(y_{n},Q_{r_{1}}y_{n}) \leq0. $$
This implies \(\lim_{n} V(y_{n},Q_{r_{1}}y_{n}) = 0\). From Theorem 2.5, we see that
$$\lim_{n\to\infty} \Vert y_{n} Q_{r_{1}}y_{n} \Vert =0. $$
Then, by Lemma 2.3 and (4.3), we see that \(x_{n}\to p_{0} \in\hat{F}(Q_{r_{1}})=F(Q_{r_{1}})=A^{1}0\). Since \(A^{1}0\subset C_{0}\), we get \(p_{0}=P_{C_{0}}u=P_{A^{1}0}u\), which completes the proof. □