Proof. It follows from the inequality

$$ \begin{align*} \begin{aligned} \|x_{k+1} - w \| &= \|c_k T_{t_{k+1}}(x_{k+1})+(1-c_k)x_k - w \| \\ &\leq c_k \|T_{t_{k+1}}(x_{k+1}) - T_{t_{k+1}}(w)\| + (1-c_k) \|x_k - w \| \\ &\leq c_k (1+b_{t_{k}}) \|x_{k+1} - w \| + (1-c_k) \|x_k - w \| \end{aligned} \end{align*} $$

that

$$ \begin{align*} \|x_{k+1} - w \| \leq \frac{c_k}{1 - c_k} b_{t_{k}} \|x_{k+1} - w \| + \|x_k - w \| \leq \frac{\beta}{1 - \beta} b_{t_{k}} \textrm{diam}(C) + \|x_k - w \|. \end{align*} $$

Set $d_{k,n} = ({\beta }/{(1 - \beta )}) \textrm {diam}(C) \sum ^{k+n}_{i=k+1}b_{t_{i}}$ and note that $\|x_{k+n}-w\| \leq \|x_{k}-w\|+ d_{k,n}$ for every $n\in \mathbb {N}$ . Since $\sum ^{+\infty }_{i=1}b_{t_{i}} < \infty $ , it follows that $\limsup _{k\rightarrow \infty } \limsup _{n\rightarrow \infty }d_{k,n} = 0$ . Consequently, by using Lemma 2.1 with $r_k=\|x_k-w\|$ , we infer that there exists an $r \in \mathbb {R}$ such that $\lim _{k\rightarrow \infty }\|x_k-w\|=r$ . The proof is complete.

Proof. Fix any t with $0 < t < \limsup _{n \rightarrow \infty } t_n$ . According to Lemma 2.3, there exists a subsequence $\{t_{k_{n}}\}$ of $\{t_k\}$ such that

(3.1) $$ \begin{align} \lim_{n \rightarrow \infty}t_{k_{n}} = t. \end{align} $$

We will prove that

(3.2) $$ \begin{align} \lim_{n \rightarrow \infty} \| T_{t_{k_{n}} } (x_{k_{n}}) - x_{k_{n}} \|= 0. \end{align} $$

Fix temporarily a $w \in F(\mathcal {T})$ , which exists according to Theorem 1.1. By Lemma 2.4, there exists $r \in \mathbb {R}$ such that $\lim _{k \rightarrow \infty } \|x_k-w\| = r$ . Since $a_{t_{k}} \rightarrow 1$ , it follows that

$$ \begin{align*} \limsup_{k \rightarrow \infty} \|T_{t_{k}}(x_k)-w\| = \limsup_{k \rightarrow \infty} \|T_{t_{k}}(x_k)-T_{t_{k}}(w)\| \leq \limsup_{k \rightarrow \infty} a_{t_{k}}\|x_k-w\| = r. \end{align*} $$

Denote $v_k=x_{k-1}-w$ , $u_k=T_{t_{k}}(x_k)-w$ , and observe that $\lim _{k \rightarrow \infty }\|v_k\|=r$ , $\limsup _{k \rightarrow \infty } \|u_k\| \leq r$ and $\lim _{k \rightarrow \infty } \|c_k u_k+(1-c_k) v_k\| = \lim _{k \rightarrow \infty }\|x_k-w\| = r$ . We can, therefore, apply Lemma 2.2 to obtain $\lim _{k \rightarrow \infty }\|T_{t_{k}}(x_k)-x_{k-1} \| = \lim _{k \rightarrow \infty } \|u_k - v_k \| = 0$ , which implies that $\lim _{k \rightarrow \infty } \|x_{k+1}-x_k\| = 0$ because of the equality

$$ \begin{align*} \|x_{k+1}-x_k\| = \|c_k T_{t_{k+1}}(x_{k+1})+(1-c_k)x_k-x_k \| = c_k \|T_{t_{k+1}}(x_{k+1}) - x_k \|. \end{align*} $$

Finally,

$$ \begin{align*} \lim_{k \rightarrow \infty} \|T_{t_{k}}(x_k)-x_k\| \leq \lim_{k \rightarrow \infty}\|T_{t_{k}}(x_k)-x_{k-1} \| + \lim_{k \rightarrow \infty} \|x_{k-1}-x_k\| =0, \end{align*} $$

proving (3.2). Since C is compact, there exists a subsequence $\{x_{k_{n_{i}}}\}$ of $\{x_{k_{n}}\}$ and an element of $x \in C$ such that

(3.3) $$ \begin{align} \lim_{i \rightarrow \infty} \| T_{t_{k_{n_{i}}} } (x_{k_{n_{i}}}) - x \|= 0. \end{align} $$

Denote $s_i = t_{k_{n_{i}}}$ , $w_i = x_{k_{n_{i}}}$ . From (3.2), (3.3) and the inequality

$$ \begin{align*} \|w_i - x\| \leq \|w_i - T_{s_{i}} (w_i)\| + \| T_{s_{i}} (w_i) - x \|, \end{align*} $$

we derive $\|w_i - x\| \rightarrow 0$ . Therefore,

(3.4) $$ \begin{align} \| T_{s_{i}}(x) - x \| &\leq \| T_{s_{i}}(x) - T_{s_{i}} (w_i)\| + \| T_{s_{i}} (w_i) - w_i \| + \|w_i - x \| \notag\\ &\leq a_{s_{i}}(x) \|x - w_i \| + \| T_{s_{i}} (w_i) - w_i \| + \|w_i - x \| \notag \\ &\leq \gamma \|x - w_i \| + \| T_{s_{i}} (w_i) - w_i \| + \|w_i - x \| \rightarrow 0, \end{align} $$

as $i \rightarrow \infty $ . From (3.4), (3.1) and the continuity of the semigroup $\mathcal {T}$ , we conclude that

$$ \begin{align*} \|T_t(x) - x \| \leq \|T_t(x) - T_{s_{i}}(x) \| + \| T_{s_{i}}(x) - x \| \rightarrow 0. \end{align*} $$

Thus, $T_t(x) = x$ . We now need to prove this equality for any $s>0$ . Observe that there exist $t, u$ with $0 < t < \limsup _{n \rightarrow \infty } t_n$ , $0 < u < \limsup _{n \rightarrow \infty } t_n$ and $k \in \mathbb {N}_0$ such that $s = t + ku$ . Therefore, $T_s(x) = x$ , because

$$ \begin{align*} T_s(x) = T_{ku} (T_t(x)) = T_{ku}(x) = T_{u + ... + u} (x) = x. \end{align*} $$

We conclude that $x \in F(\mathcal {T})$ . It only remains to show that $\lim _{k \rightarrow \infty } \|x_k - x\| = 0$ . To achieve this, note that, according to Lemma 2.4, there exists a nonnegative number r such that $\lim _{k \rightarrow \infty } \|x_k - x\| = r$ . We previously established that $\lim _{i \rightarrow \infty } \|x_{k_{n_{i}}} - x\| = \lim _{i \rightarrow \infty } \|w_i - x\| = 0$ , which requires $r=0$ . Thus, the proof is complete.

Proof. Let us fix an arbitrary $t>0$ . Without any loss of generality, we can assume that $s_i < t$ for every $i \in \mathbb {N}$ . Denoting for simplicity $ p_i = [ {t}/{s_i} ]$ ,

$$ \begin{align*} \|x_i-T_t(x_i)\| &\leq \sum^{p_{i}-1}_{k=0} \|T_{(k+1)s_i}(x_i) - T_{k s_i}(x_i) \| + \| T_{p_i s_i}(x_i) - T_{t}(x_i) \| \\ &\leq \|T_{s_{i}}(x_i)-x_i \| \sum^{p_{i}-1}_{k=0} a_{ks_{i}} + \| T_{p_i s_i}(x_i) - T_{t}(x_i) \| \\ &\leq \frac{t}{s_i} \gamma \|T_{s_{i}}(x_i)-x_i \| + \|T_{t-p_{i} s_{i}} (x_i) - x_i \|. \end{align*} $$

Since $({t}/{s_i}) \|T_{s_{i}}(x_i)-x_i \| \rightarrow 0$ by (4.3) and $t-p_{i} s_{i} \rightarrow 0$ , it follows from the equicontinuity of $\mathcal {T}$ that $\|x_i-T_t(x_i)\| \rightarrow 0$ . Therefore, the sequence $\{x_i\}$ is an approximate fixed point sequence for every $t \in J$ . By assumption, $x_i \rightharpoonup w$ . Consequently, it follows from the demiclosedness principle (Theorem 2.1) that $w \in F(\mathcal {T})$ , as claimed.

Proof. Since $P(C,\mathcal {T},x_0,\{c_k\},\{t_k\})$ is normalised, it follows that $t_{k} \rightarrow 0$ and that

(4.4) $$ \begin{align} \lim_{k \rightarrow \infty} \frac{1}{t_{k}} \|T_{t_{k}}(x_{k}) - x_{k} \| = 0. \end{align} $$

This implies that

$$ \begin{align*} \lim_{k \rightarrow \infty} \|T_{t_{k}}(x_{k}) - x_{k} \| = 0. \end{align*} $$

Consider $y,z \in C$ , which are two weak cluster points of the sequence $\{x_k\}$ . Then, there exist two subsequences $\{y_k\}$ and $\{z_k\}$ of $\{x_k\}$ such that $y_k\rightharpoonup y$ and $z_k\rightharpoonup z$ . It follows from Lemma 4.3 that $y,z \in F(\mathcal {T})$ . According to Lemma 2.4, the limits

$$ \begin{align*} r_1=\lim_{k \rightarrow \infty} \|x_k-y\|,\quad r_2=\lim_{k \rightarrow \infty} \|x_k-z\| \end{align*} $$

exist. We assert that $y=z$ . For a contradiction, assume $y\neq z$ . Due to the Opial property,

$$ \begin{align*} r_1 = \liminf_{k\rightarrow \infty}\|y_k-y\| &< \liminf_{k\rightarrow \infty}\|y_k-z\|=r_2 \notag \\ &= \liminf_{k\rightarrow \infty}\|z_k-z\| < \liminf_{k\rightarrow \infty}\|z_k-y\|=r_1. \end{align*} $$

This contradiction indicates that $y=z$ , showing that the sequence $\{x_k\}$ has no more than one weak cluster point. Since C is weakly sequentially compact, it follows that $\{x_k\}$ indeed has exactly one weak cluster point $w \in C$ , meaning that $x_k\rightharpoonup w$ . Since the process is normalised, we also have (4.4). Consequently, by applying Lemma 4.3, we can conclude that $w \in \mathcal {T}$ . This concludes the proof.

Proof. For each $k \in \mathbb {N}$ , define the mapping $Z_k : C \rightarrow C$ by $Z_k(w) = z_{k,w}$ , where $z_{k,w}$ is the unique fixed point of the contractive mapping

$$ \begin{align*} P_{k,w}(u) = c_k T_{t_{k+1}}(u) + (1-c_k) w. \end{align*} $$

By straightforward calculation, we obtain $u_{k+1} = Z_k(u_k)$ for $k \in \mathbb {N}$ . It is easy to prove that $F(\mathcal {T}) \subset \bigcap ^{\infty }_{k=1} F(Z_k)$ . We will prove that the sequence $\{Z_k\} \in \mathcal {A}_{c}(C)$ . Let us take any $v_1, v_2 \in C$ . Observe that, for $i=1,2$ , the equation

(4.5) $$ \begin{align} Z_k(v_i)=u_i \end{align} $$

is equivalent to

(4.6) $$ \begin{align} u_i = c_k T_{t_{k+1}}(u_i) + (1-c_k) v_i. \end{align} $$

Using the notation introduced in (4.5) and (4.6),

(4.7) $$ \begin{align} \|Z_k(v_1)- Z_k(v_2)\| & = \|u_1-u_2\| \notag \\ & \leq c_k \|T_{t_{k+1}}(u_1) - T_{t_{k+1}}(u_2) \| + (1-c_k) \|v_1-v_2\| \notag\\ & \leq c_k a_{t_{k}} \|u_1-u_2\| + (1-c_k) \|v_1-v_2\|. \end{align} $$

It follows from (4.7) that

$$ \begin{align*} (1 - c_k a_{t_{k}}) \|u_1-u_2\| \leq (1-c_k) \|v_1-v_2\|, \end{align*} $$

which implies

$$ \begin{align*} \|Z_k(v_1)- Z_k(v_2)\| = \|u_1-u_2\| \leq \frac{1-c_k}{1 - c_k a_{t_{k}}} \|v_1-v_2\|. \end{align*} $$

Since each $a_{t_{k}} \geq 1$ and $a_{t_{k}} \rightarrow 1$ , it follows that each $A_k$ , defined by

$$ \begin{align*} A_k = \frac{1-c_k}{1 - c_k a_{t_{k}}}, \end{align*} $$

possesses the same properties. Using the constants $\beta $ and $\gamma $ from Definition 1.4, we obtain

(4.8) $$ \begin{align} B_k := 1 - A_k = c_k \frac{b_{t_{k}}} {1 - c_k a_{t_{k}}} \leq \frac{\beta} {1 - \beta \gamma} b_{t_{k}}. \end{align} $$

Since $\beta \gamma < 1$ and $\sum _{k=1}^{\infty } b_{t_{k}} < \infty $ , it follows from (4.8) that $\sum _{k=1}^{\infty } B_k < \infty $ . Finally, we conclude that the sequence $\{Z_k\}$ belongs to $\mathcal {A}_{c}(C)$ , as claimed.

Proof. Let $Z_k : C \rightarrow C$ be the mappings defined in Lemma 4.4. Recall that we established there that $x_{k+1} = Z_k(x_k)$ , $F(\mathcal {T}) \subset \bigcap ^{\infty }_{k=1} F(Z_k)$ and that the sequence $\{Z_k\}$ belongs to $ \mathcal {A}_{c}(C)$ . Therefore, by applying Lemma 2.6, we conclude that the equality (4.9) holds for any two weak cluster points $y,\ z$ of the sequence $\{x_k\}$ .

Proof. Since $P(C,\mathcal {T},x_0,\{c_k\},\{t_k\})$ is normalised, it follows that $t_{k} \rightarrow 0$ and that

(4.10) $$ \begin{align} \lim_{k \rightarrow \infty} \frac{1}{t_{k}} \|T_{t_{k}}(x_{k}) - x_{k} \| = 0. \end{align} $$

Consider $y, z \in C$ , two weak cluster points of the sequence $\{x_{k}\}$ . There exist two subsequences $\{x_{\alpha _{n}}\}$ and $\{x_{\beta _{n}}\}$ of the sequence $\{x_k\}$ such that $x_{\alpha _{n}} \rightharpoonup y$ , $\ x_{\beta _{n}} \rightharpoonup z$ . From (4.10), we conclude that

$$ \begin{align*} \lim_{n \rightarrow \infty} \frac{1}{t_{\alpha_{n}}} \|T_{t_{\alpha_{n}}}(x_{\alpha_{n}})-x_{\alpha_{n}}\| = 0,\quad \lim_{n \rightarrow \infty} \frac{1}{t_{\beta_{n}}} \|T_{t_{\beta_{n}}}(x_{\beta_{n}})-x_{\beta_{n}}\| = 0. \end{align*} $$

From Lemma 4.3, it follows that $y \in F(\mathcal {T})$ and $z \in F(\mathcal {T})$ . According to Lemma 4.5, this implies $\|y-z\|^2 = \langle y-z,J(y-z) \rangle = 0$ , leading us to conclude that $y=z$ . Therefore, the sequence $\{x_{k}\}$ has at most one weak cluster point. Since C is weakly sequentially compact, it follows that $\{x_{k}\}$ has exactly one weak cluster point $w \in C$ . This means that $x_{k} \rightharpoonup w$ . It follows from Lemma 4.3 that $w \in F(\mathcal {T})$ . The proof is now complete.

Proof. Recall that $u_{k+1} = Z_k(u_k)$ , where each $u_k$ is generated by the implicit iteration process $\{u_k\}=P(C,\mathcal {T},u_1,\{c_k\},\{t_k\})$ , and $u_1 \in C$ was chosen arbitrarily. It can be concluded from Theorem 4.1 regarding the Opial property case and from Theorem 4.2 for uniformly smooth X that, regardless of the choice of the starting point, the sequence generated by the implicit iteration process converges in the weak topology to a common fixed point of the semigroup $\mathcal {T}$ . By employing [Reference Kozlowski10, Theorem 4.2], we conclude that the iteration process $\{Z_k\}$ is stable under summable errors, as claimed.