In this section we present our new parallel inertial Tseng type algorithm (PITTA) for solving (1.9). For the convergence analysis of the proposed method, we assume the following assumptions for all \(i=1, 2,\dots, K\).

### Assumption 1

\(\mathcal{H}\) is a real Hilbert space, \(F_{i}: \mathcal{H}\to \mathcal{H}\) is an \(\mathcal{L}_{i}\)-Lipschitz continuous and monotone mapping and \(G_{i}: \mathcal{H}\to 2^{\mathcal{H}}\) is a maximal monotone operator.

### Assumption 2

\(\Phi:= \bigcap_{i=1}^{K}(F_{i} + G_{i})^{-1}(0)\) is nonempty.

### Assumption 3

\(\{\xi _{k}\}\subset [0, \xi )\), \(\{b_{n}\}\subset (b^{*}, b^{\prime })\subset (0, 1-a_{n})\) for some \(\xi > 0, b^{*} >0, b^{\prime } >0\), and \(\{a_{n}\}\subset (0, 1)\) satisfies \(\lim_{k\rightarrow \infty }a_{k}=0 \text{ and } \sum_{k=1}^{ \infty }a_{n}=\infty \).

### Assumption 4

\(\varphi: \mathcal{H}\to \mathcal{H}\) is a *ρ*-contractive mapping.

Next the algorithm is presented.

### Lemma 3.1

*Assume that Assumptions *1*–*4*hold*, *then any sequence* \(\{\gamma _{k}^{i}\}\) *in Algorithm PITTA is nonincreasing and converges to* \(\gamma _{i}\) *such that* \(\min \lbrace \gamma _{1}^{i}, \frac{\lambda _{i}}{\mathcal{L}_{i}} \rbrace \leq \gamma _{i}\) *for all* \(i=1,2,\dots,K\).

### Proof

See [8, Lemma 5]. □

### Lemma 3.2

*Let* \(u\in \Phi \). *Then under Assumptions *1*–*4, *we have*, *for all* \(i=1,2,\dots,K\),

$$\begin{aligned} \bigl\Vert t_{k}^{i}-u \bigr\Vert ^{2}\leq \Vert r_{k}-u \Vert ^{2}- \bigl[1- \bigl( \varrho _{k}^{i} \bigr)^{2} \bigr] \bigl\Vert r_{k}-s_{k}^{i} \bigr\Vert ^{2} \end{aligned}$$

(3.1)

*and*

$$\begin{aligned} \bigl\Vert t_{k}^{i}-s_{k}^{i} \bigr\Vert \leq \varrho _{k}^{i} \bigl\Vert r_{k}-s_{k}^{i} \bigr\Vert , \end{aligned}$$

(3.2)

*where* \(\varrho _{k}^{i}=\lambda _{i} \frac{\gamma _{k}^{i}}{\gamma _{k+1}^{i}}\).

### Proof

In the same manner as [8, Lemma 6], we obtain that inequalities (3.1) and (3.2) hold. □

### Lemma 3.3

*Suppose that* \(\lim_{k\rightarrow \infty }\|r_{k}-s_{k}^{i}\|=0\) *for all* \(i=1,2,\dots,K\). *If there exists a weakly convergent subsequence* \(\{r_{k_{j}}\}\) *of* \(\{r_{k}\}\), *then under Assumptions *1*–*4, *we have that the limit of* \(\{r_{k_{j}}\}\) *belongs to* Φ.

### Proof

The proof is similar to the proof of [8, Lemma 7]. □

With the above results we are now ready for the main convergence theorem.

### Theorem 3.4

*Suppose that* \(\lim_{k\rightarrow \infty }\frac{\xi _{k}}{a_{k}}\|u_{k}-u_{k-1} \|=0\), *then under Assumptions *1*–*4, *we have* \(u_{k}\to \mu \) *as* \(k\to \infty \), *where* \(\mu = P_{\Phi }\circ \varphi (\mu )\).

### Proof

First, since \(\lim_{k\to \infty } [1- (\varrho _{k}^{i} )^{2} ]=1-\lambda _{i}^{2}>0\), one can find \(m_{i}\in \mathbb{N}\) such that \(1- (\varrho _{k}^{i} )^{2}>0\) for all \(k\geq k_{0}\), where \(k_{0} = \max_{i=1,2,\dots,K} m_{i}\). Let \(u\in \Phi \), from (3.1), we get

$$\begin{aligned} \bigl\Vert t_{k}^{i}-u \bigr\Vert \leq \Vert r_{k}-u \Vert \end{aligned}$$

(3.3)

for all \(k\geq k_{0}\). Next, we divide the proof into the following claims.

### Claim 1

\(\{u_{k}\}\) *is a bounded sequence*.

By the sequence \(\lbrace \frac{\xi _{k}}{a_{k}}\|u_{k}-u_{k-1}\| \rbrace \) converges to 0, we have that there exists a constant \(M_{*}\geq 0\) such that, for all \(k\in \mathbb{N}\),

$$\begin{aligned} \frac{\xi _{k}}{a_{k}} \Vert u_{k}-u_{k-1} \Vert \leq M_{*}. \end{aligned}$$

(3.4)

From the definition of \(r_{k}\) and combining (3.3) and (3.4), we obtain, for all \(k\geq k_{0}\),

$$\begin{aligned} \bigl\Vert t_{k}^{i}-u \bigr\Vert \leq \Vert r_{k}-u \Vert &= \bigl\Vert u_{k}+\xi _{k}(u_{k}-u_{k-1})-u \bigr\Vert \\ &\leq \Vert u_{k}-u \Vert +\frac{\xi _{k}}{a_{k}} \Vert u_{k}-u_{k-1} \Vert a_{k} \\ &\leq \Vert u_{k}-u \Vert +a_{k}M_{*}. \end{aligned}$$

From the definition of *i*, we get, for all \(k\geq k_{0}\),

$$\begin{aligned} \Vert \bar{t}_{k}-u \Vert \leq \Vert r_{k}-u \Vert \end{aligned}$$

(3.5)

and

$$\begin{aligned} \Vert \bar{t}_{k}-u \Vert \leq \Vert u_{k}-u \Vert +a_{k}M_{*}. \end{aligned}$$

(3.6)

By Assumption 4 and using (3.6), the following relation is obtained for all \(k\geq k_{0}\):

$$\begin{aligned} \Vert u_{k+1}-u \Vert &= \bigl\Vert a_{k} \bigl( \varphi (u_{k})-u \bigr)+(1-a_{k}-b_{k}) (u_{k}-u)+b_{k}( \bar{t}_{k}-u) \bigr\Vert \\ &\leq a_{k} \bigl\Vert \varphi (u_{k})-u \bigr\Vert +(1-a_{k}-b_{k}) \Vert u_{k}-u \Vert +b_{k} \Vert \bar{t}_{k}-u \Vert \\ &\leq a_{k} \bigl\Vert \varphi (u_{k})-\varphi (u) \bigr\Vert +a_{k} \bigl\Vert \varphi (u)-u \bigr\Vert +(1-a_{k}) \Vert u_{k}-u \Vert +a_{k}b_{k}M_{*} \\ &\leq \bigl[1-a_{k}(1-\rho ) \bigr] \Vert u_{k}-u \Vert +a_{k} \bigl( \bigl\Vert \varphi (u)-u \bigr\Vert +M_{*} \bigr) \\ &= \bigl[1-a_{k}(1-\rho ) \bigr] \Vert u_{k}-u \Vert +a_{k}(1-\rho ) \frac{ \Vert \varphi (u)-u \Vert +M_{*}}{1-\rho } \\ &\leq \max \biggl\lbrace \Vert u_{k}-u \Vert , \frac{ \Vert \varphi (u)-u \Vert +M_{*}}{1-\rho } \biggr\rbrace . \end{aligned}$$

This leads to a conclusion that \(\|u_{k+1}-u\|\leq \max \lbrace \|u_{k_{0}}-u\|, \frac{\|\varphi (u)-u\|+M_{*}}{1-\rho } \rbrace \) for any \(k\geq k_{0}\). Consequently, the sequence \(\{u_{k}\}\) is bounded. In addition, \(\{\varphi (u_{k})\}\) is also bounded. Since Φ is a closed and convex set, \(P_{\Phi }\circ \varphi \) is a *ρ*-contractive mapping. Now, we can uniquely find \(\mu \in \Phi \) with \(\mu = P_{\Phi }\circ \varphi (\mu )\) due to the Banach fixed point theorem. We also get that, for any \(u\in \Phi \),

$$\begin{aligned} \bigl\langle \varphi (\mu )-\mu, u-\mu \bigr\rangle \leq 0. \end{aligned}$$

(3.7)

Now, for each \(k\in \mathbb{N}\), set \(\Xi _{k}:=\|u_{k}-\mu \|^{2}\).

### Claim 2

*There is* \(M_{0}>0\) *such that*

$$\begin{aligned} b_{k}(1-a_{k}-b_{k}) \Vert u_{k}- \bar{t}_{k} \Vert ^{2} &\leq \Xi _{k}-\Xi _{k+1}+a_{k} \bigl( \bigl\Vert \varphi (u_{k})- \mu \bigr\Vert ^{2}+M_{0} \bigr) \end{aligned}$$

*for all* \(k\geq k_{0}\).

Applying (3.6), we have, for all \(k\geq k_{0}\),

$$\begin{aligned} \Vert \bar{t}_{k}-\mu \Vert ^{2}&\leq \bigl( \Vert u_{k}-\mu \Vert +a_{k}M_{*} \bigr) ^{2} \\ &=\Xi _{k}+a_{k} \bigl( 2M_{*} \Vert u_{k}-\mu \Vert +a_{k}M_{*}^{2} \bigr) \\ &\leq \Xi _{k}+a_{k}M_{0} \end{aligned}$$

(3.8)

for some \(M_{0}>0\). For any \(k\geq k_{0}\), it follows from the assumption on *φ* and (3.8) that

$$\begin{aligned} \Xi _{k+1}&= \bigl\Vert a_{k} \bigl(\varphi (u_{k})-\mu \bigr)+(1-a_{k}-b_{k}) (u_{k}-\mu )+b_{k}( \bar{t}_{k}-\mu ) \bigr\Vert ^{2} \\ &\leq a_{k} \bigl\Vert \varphi (u_{k})-\mu \bigr\Vert ^{2}+(1-a_{k}-b_{k})\Xi _{k}+b_{k} \Vert \bar{t}_{k}-\mu \Vert ^{2}-b_{k}(1-a_{k}-b_{k}) \Vert u_{k}-\bar{t}_{k} \Vert ^{2} \\ &\leq a_{k} \bigl\Vert \varphi (u_{k})-\mu \bigr\Vert ^{2}+(1-a_{k})\Xi _{k}+a_{k}b_{k}M_{0}-b_{k}(1-a_{k}-b_{k}) \Vert u_{k}-\bar{t}_{k} \Vert ^{2} \\ &\leq \Xi _{k}+a_{k} \bigl( \bigl\Vert \varphi (u_{k})-\mu \bigr\Vert ^{2}+M_{0} \bigr) -b_{k}(1-a_{k}-b_{k}) \Vert u_{k}- \bar{t}_{k} \Vert ^{2} . \end{aligned}$$

Therefore, Claim 2 is obtained.

### Claim 3

*There is* \(\bar{M}>0\) *such that*

$$\begin{aligned} \Xi _{k+1}\leq {}&\bigl[1-a_{k}(1-\rho ) \bigr]\Xi _{k}+a_{k}(1-\rho ) \biggl[ \frac{3\bar{M}}{1-\rho }\cdot \frac{\xi _{k}}{a_{k}} \Vert u_{k}-u_{k-1} \Vert \biggr] \\ &{} +a_{k}(1-\rho ) \biggl[\frac{2\bar{M}}{1-\rho } \Vert u_{k}- \bar{t}_{k} \Vert +\frac{2}{1-\rho } \bigl\langle \varphi (\mu )- \mu, u_{k+1}-\mu \bigr\rangle \biggr] \end{aligned}$$

*for all* \(k\geq k_{0}\).

Indeed, setting \(c_{k}=(1-b_{k})u_{k} + b_{k}\bar{t}_{k}\). From inequality (3.5) and the definition of \(r_{k}\), we have

$$\begin{aligned} \Vert c_{k}-\mu \Vert &\leq (1-b_{k}) \Vert u_{k}-\mu \Vert +b_{k} \Vert \bar{t}_{k}-\mu \Vert \\ &\leq (1-b_{k}) \Vert u_{k}-\mu \Vert +b_{k} \Vert r_{k}-\mu \Vert \\ &\leq \Vert u_{k}-\mu \Vert +b_{k}\xi _{k} \Vert u_{k}-u_{k-1} \Vert \end{aligned}$$

(3.9)

and

$$\begin{aligned} \Vert u_{k}-c_{k} \Vert =b_{k} \Vert u_{k}-\bar{t}_{k} \Vert \end{aligned}$$

(3.10)

for all \(k\geq k_{0}\). Hence, from the assumption on *φ*, and (3.2), (3.9), and (3.10), we obtain, for all \(k\geq k_{0}\),

$$\begin{aligned} \Xi _{k+1}= {}&\bigl\Vert (1-a_{k}) (c_{k}-\mu )+a_{k} \bigl(\varphi (u_{k})-\varphi ( \mu ) \bigr)-a_{k}(u_{k}-c_{k})-a_{k} \bigl( \mu -\varphi (\mu ) \bigr) \bigr\Vert ^{2} \\ \leq {}&\bigl\Vert (1-a_{k}) (c_{k}-\mu )+a_{k} \bigl(\varphi (u_{k})-\varphi (\mu ) \bigr) \bigr\Vert ^{2}-2a_{k} \bigl\langle u_{k}-c_{k}+ \mu -\varphi (\mu ), u_{k+1}-\mu \bigr\rangle \\ \leq{}& (1-a_{k}) \Vert c_{k}-\mu \Vert ^{2}+a_{k} \bigl\Vert \varphi (u_{k})-\varphi ( \mu ) \bigr\Vert ^{2}+2a_{k}\langle c_{k}-u_{k}, u_{k+1}-\mu \rangle \\ &{} +2a_{k} \bigl\langle \varphi (\mu )-\mu, u_{k+1}-\mu \bigr\rangle \\ \leq{}& (1-a_{k}) \bigl( \Vert u_{k}-\mu \Vert +b_{k}\xi _{k} \Vert u_{k}-u_{k-1} \Vert \bigr)^{2} +a_{k}\rho ^{2}\Xi _{k}+2a_{k} \Vert c_{k}-u_{k} \Vert \Vert u_{k+1}- \mu \Vert \\ &{} +2a_{k} \bigl\langle \varphi (\mu )-\mu, u_{k+1}-\mu \bigr\rangle \\ \leq{}& (1-a_{k})\Xi _{k}+2\xi _{k} \Vert u_{k}-\mu \Vert \Vert u_{k}-u_{k-1} \Vert + \xi _{k}^{2}| \Vert u_{k}-u_{k-1} \Vert ^{2} +a_{k}\rho \Xi _{k} \\ &{} +2a_{k}b_{k} \Vert u_{k}- \bar{t}_{k} \Vert \Vert u_{k+1}-\mu \Vert +2a_{k} \bigl\langle \varphi (\mu )-\mu, u_{k+1}-\mu \bigr\rangle \\ \leq{}& \bigl[1-a_{k}(1-\rho ) \bigr]\Xi _{k}+\xi _{k} \Vert u_{k}-u_{k-1} \Vert \bigl( 2 \Vert u_{k}- \mu \Vert +\xi \Vert u_{k}-u_{k-1} \Vert \bigr) \\ &{} +2a_{k}b_{k} \Vert u_{k}- \bar{t}_{k} \Vert \Vert u_{k+1}-\mu \Vert +2a_{k} \bigl\langle \varphi (\mu )-\mu, u_{k+1}-\mu \bigr\rangle \\ \leq{}& \bigl[1-a_{k}(1-\rho ) \bigr]\Xi _{k}+3\bar{M}\xi _{k} \Vert u_{k}-u_{k-1} \Vert +2 \bar{M}a_{k}b_{k} \Vert u_{k}- \bar{t}_{k} \Vert \\ &{}+2a_{k} \bigl\langle \varphi (\mu )- \mu, u_{k+1}-\mu \bigr\rangle \\ \leq{}& \bigl[1-a_{k}(1-\rho ) \bigr]\Xi _{k}+a_{k}(1- \rho ) \biggl[ \frac{3\bar{M}}{1-\rho }\cdot \frac{\xi _{k}}{a_{k}} \Vert u_{k}-u_{k-1} \Vert \biggr] \\ &{} +a_{k}(1-\rho ) \biggl[\frac{2\bar{M}}{1-\rho } \Vert u_{k}- \bar{t}_{k} \Vert +\frac{2}{1-\rho } \bigl\langle \varphi (\mu )- \mu, u_{k+1}-\mu \bigr\rangle \biggr] \end{aligned}$$

for \(\bar{M}:= \sup_{n\in \mathbb{N}} \lbrace \|u_{k}- \mu \|, \xi \|u_{k}-u_{k-1}\| \rbrace > 0\). Recall that our task is to show that \(u_{k}\to \mu \), which is now equivalent to showing that \(\Xi _{k}\to 0\) as \(k\to \infty \).

### Claim 4

\(\Xi _{k}\to 0\) *as* \(k\to \infty \).

The proof is divided into the following two cases.

*Case a.* We can find \(N\in \mathbb{N}\) satisfying that, for all \(k\geq N\), the inequality \(\Xi _{k+1}\leq \Xi _{k}\) holds. Since each term \(\Xi _{k}\) is nonnegative, it is convergent. Due to the fact that \(\lim_{k\rightarrow \infty } a_{k}=0\) and \(\lim_{k\rightarrow \infty }b_{k} \in (0,1)\), and by Claim 2,

$$\begin{aligned} \lim_{k\rightarrow \infty } \Vert u_{k}- \bar{t}_{k} \Vert =0. \end{aligned}$$

(3.11)

Indeed, we immediately get

$$\begin{aligned} \lim_{k\rightarrow \infty } \Vert u_{k}-r_{k} \Vert = \lim_{k\rightarrow \infty }\frac{\xi _{k}}{a_{k}} \Vert u_{k}-u_{k-1} \Vert a_{k}=0. \end{aligned}$$

(3.12)

In addition, from the definition of \(\bar{t}_{k}\) and by using the triangle inequality, the following inequalities are obtained:

$$\begin{aligned} \bigl\Vert t_{k}^{i}-r_{k} \bigr\Vert &\leq \Vert \bar{t}_{k}-r_{k} \Vert \leq \Vert \bar{t}_{k}-u_{k} \Vert + \Vert u_{k}-r_{k} \Vert \end{aligned}$$

and

$$\begin{aligned} \bigl\Vert r_{k}-s_{k}^{i} \bigr\Vert &\leq \bigl\Vert r_{k}-t_{k}^{i} \bigr\Vert + \bigl\Vert t_{k}^{i}-s_{k}^{i} \bigr\Vert \end{aligned}$$

for all \(i=1,2,\dots,K\). It follows from inequality (3.2) that

$$\begin{aligned} \bigl(1-\varrho _{k}^{i} \bigr) \bigl\Vert r_{k}-s_{k}^{i} \bigr\Vert \leq \Vert \bar{t}_{k}-u_{k} \Vert + \Vert u_{k}-r_{k} \Vert \end{aligned}$$

for all \(i=1,2,\dots,K\). Since \(\lim_{k\to \infty } [1- (\varrho _{k}^{i} )^{2} ]=1-\lambda _{i}^{2}>0\), (3.11) and (3.12),

$$\begin{aligned} \lim_{k\rightarrow \infty } \bigl\Vert r_{k}-s_{k}^{i} \bigr\Vert =0 \end{aligned}$$

(3.13)

for all \(i=1,2,\dots,K\). Note that, for each \(k\in \mathbb{N}\),

$$\begin{aligned} \Vert u_{k+1}-u_{k} \Vert &\leq \Vert u_{k+1}-\bar{t}_{k} \Vert + \Vert \bar{t}_{k}-u_{k} \Vert \\ &\leq a_{k} \bigl\Vert \varphi (u_{k})-u_{k} \bigr\Vert +(2-b_{k}) \Vert u_{k}-\bar{t}_{k} \Vert . \end{aligned}$$

(3.14)

Consequently, since \(\lim_{k\rightarrow \infty }a_{k}=0\) and by (3.14), \(\lim_{k\rightarrow \infty }\|u_{k+1}-u_{k}\|=0\). Next observe that, for the reason that \(\{u_{k}\}\) is bounded, there is \(w\in \mathcal{H}\) such that \(u_{k_{j}}\rightharpoonup w\) as \(j\to \infty \) for some subsequence \(\{u_{k_{j}}\}\) of \(\{u_{k}\}\). By (3.12), we get \(r_{k_{j}}\rightharpoonup w\) as \(j\to \infty \). Then Lemma 3.3 together with (3.13) implies that \(w\in \Phi \). From (3.7), it is straightforward to show that

$$\begin{aligned} \limsup_{k\rightarrow \infty } \bigl\langle \varphi (\mu )-\mu, u_{k}-\mu \bigr\rangle =\lim_{k\rightarrow \infty } \bigl\langle \varphi (\mu )-\mu, u_{k_{j}}- \mu \bigr\rangle = \bigl\langle \varphi (\mu )-\mu, w-\mu \bigr\rangle \leq 0. \end{aligned}$$

Since \(\lim_{k\rightarrow \infty }\|u_{k+1}-u_{k}\|=0\), the following result is obtained:

$$\begin{aligned} \limsup_{k\rightarrow \infty } \bigl\langle \varphi (\mu )-\mu, u_{k+1}- \mu \bigr\rangle &\leq \limsup_{n\rightarrow \infty } \bigl\langle \varphi (\mu )- \mu, u_{k+1}-u_{k} \bigr\rangle + \limsup_{n\rightarrow \infty } \bigl\langle \varphi (\mu )-\mu, u_{k}- \mu \bigr\rangle \\ & \leq 0. \end{aligned}$$

Applying Lemma 2.2 to the inequality from Claim 3, we can conclude that \(\lim_{k\rightarrow \infty }\Xi _{k}= 0\).

*Case b.* We can find \(k_{n}\in \mathbb{N}\) satisfying that \(k_{n}\geq n\) and \(\Xi _{k_{n}}<\Xi _{k_{n}+1}\) for all \(n\in \mathbb{N}\). According to Lemma 2.3, the inequality \(\Xi _{\psi (k)}\leq \Xi _{\psi (k)+1}\) is obtained, where \(\psi: \mathbb{N}\rightarrow \mathbb{N}\) is defined by (2.1), and \(k\geq k^{*}\) for some \(k^{*}\in \mathbb{N}\). This implies, by Claim 2, for all \(k\geq \max \{k_{0}, k^{*}\}\), that

$$\begin{aligned} &b_{\psi (k)}(1-a_{\psi (k)}-b_{\psi (k)}) \Vert u_{\psi (k)}- \bar{t}_{ \psi (k)} \Vert ^{2} \\ &\quad\leq \Xi _{\psi (k)}-\Xi _{\psi (k)+1}+a_{\psi (k)} \bigl( \bigl\Vert \varphi (u_{\psi (k)})- \mu \bigr\Vert ^{2}+M_{0} \bigr). \end{aligned}$$

Similar to Case a, since \(a_{k}\to 0\) as \(k\to \infty \), we obtain

$$\begin{aligned} \lim_{k\rightarrow \infty } \Vert u_{\psi (k)}-\bar{t}_{\psi (k)} \Vert =0. \end{aligned}$$

Furthermore, an argument similar to the one used in Case a shows that

$$\begin{aligned} \limsup_{k\rightarrow \infty } \bigl\langle \varphi (\mu )-\mu, u_{ \psi (k)+1}-\mu \bigr\rangle \leq 0. \end{aligned}$$

(3.15)

Finally, from the inequality \(\Xi _{\psi (k)}\leq \Xi _{\psi (k)+1}\) and by Claim 3, for all \(k\geq \max \{k_{0}, k^{*}\}\), we obtain

$$\begin{aligned} \Xi _{\psi (k)+1} \leq{}& \bigl[1-a_{\psi (k)}(1-\rho ) \bigr]\Xi _{\psi (k)+1}+a_{ \psi (k)}(1-\rho ) \biggl[ \frac{3\bar{M}}{1-\rho }\cdot \frac{\xi _{\psi (k)}}{a_{\psi (k)}} \Vert u_{\psi (k)}-u_{\psi (k)-1} \Vert \biggr] \\ &{} +a_{\psi (k)}(1-\rho ) \biggl[\frac{2\bar{M}}{1-\rho } \Vert u_{ \psi (k)}- \bar{t}_{\psi (k)} \Vert +\frac{2}{1-\rho } \bigl\langle \varphi (\mu )- \mu, u_{\psi (k)+1}-\mu \bigr\rangle \biggr]. \end{aligned}$$

Some simple calculations yield

$$\begin{aligned} \Xi _{\psi (k)+1} \leq{}& \frac{3\bar{M}}{1-\rho }\cdot \frac{\xi _{\psi (k)}}{a_{\psi (k)}} \Vert u_{\psi (k)}-u_{\psi (k)-1} \Vert + \frac{2\bar{M}}{1-\rho } \Vert u_{\psi (k)}-\bar{t}_{\psi (k)} \Vert \\ &{} +\frac{2}{1-\rho } \bigl\langle \varphi (\mu )-\mu, u_{\psi (k)+1}- \mu \bigr\rangle . \end{aligned}$$

(3.16)

From this it follows that \(\limsup_{k\rightarrow \infty }\Xi _{\psi (k)+1}\leq 0\). Thus, \(\lim_{k\rightarrow \infty }\Xi _{\psi (k)+1}=0\). In addition, by Lemma 2.3,

$$\begin{aligned} \lim_{k\to \infty }\Xi _{k}\leq \lim_{k\to \infty } \Xi _{\psi (k)+1}=0. \end{aligned}$$

Hence, we can conclude that \(u_{k}\) converges strongly to *μ*. □