For x₀, x₁ in (S₂), if x₀ = x₁ or x₁ ∊ T x₁, then x₁ is a fixed point of T. We have nothing to prove.

So, we assume that x₀ ≠ x₁ and x₁ ∉ Tx₁. Since x₁ ∊ Tx₀ and α(x₀,x₁) ≥ 1, by the strictly weakly (α, ψ, ξ)-contractive condition of T, we get $$\begin{array}{l}0<\xi \left(d\left(x_1,T{x}_1\right)\right)\\ \le \psi \left(\xi \left(\text{max}\left\{d\left(x_0,x_1\right),d\left(x_0,T{x}_0\right)d\left(x_1,T{x}_1\right),\frac{d\left(x_0,T{x}_1\right)d\left(x_1,T{x}_0\right)}{2}\right\}\right)\right)\\ =\psi \left(\xi \left(\text{max}\left\{d\left(x_0,x_0\right),d\left(x_1,T{x}_1\right),\frac{d\left(x_0,T{x}_1\right)}{2}\right\}\right)\right)\\ \le \psi \left(\xi \left(\text{max}\left\{d\left(x_0,x_1\right),d\left(x_1,T{x}_1\right),\frac{d\left(x_0,T{x}_1\right)d\left(x_1,T{x}_1\right)}{2}\right\}\right)\right)\\ =\psi \left(\xi \left(\text{max}\left\{d\left(x_0,x_1\right),d\left(x_1,T{x}_1\right)\right\}\right)\right).\end{array}$$(3)

By the property of ψ, the above relation is impossible if max{d(x₀,x₁),d(x₁,Tx₁)} = d(x₁,Tx₁). Hence, from (3), it follows that $$0<\xi \left(d\left(x_1,T{x}_1\right)\right)\le \psi \left(\xi \left(d\left(x_0,x_1\right)\right)\right).$$(4)

Using Lemma 1.6, there exists x₂ ∈ Tx₁ such that $$\xi \left(d\left(x_1,x_2\right)\right)<q\xi \left(d\left(x_1,T{x}_1\right)\right),$$(5)

where q is a fixed real number such that q > 1. If x₁ = x₂ or x₂ ∈ Tx₂, then x₂ is a fixed point of T and hence we have noting to prove. Now, we may assume that x₁ ≠ x₂ and x₂ ∉ Tx₂. From (4) and (5), we obtain that $$0<\xi \left(d\left(x_1,x_2\right)\right)<q\psi \left(\xi \left(d\left(x_0,x_1\right)\right)\right).$$(6)

Applying ψ in the above inequality, we get $$0<\psi (\xi (d(x_1,x_2)))<\psi (q\psi (\xi (d(x_0,x_1))))$$(7)

and so $$q_1:=\frac{\psi (q\psi (\xi (d(x_0,x_1))))}{\psi (\xi (d(x_1,x_2)))}>1.$$(8)

Since T is an α-admissible multi-valued mapping, we get α(x₁, x₂) ≥ 1. From the weakly (α, ψ, ξ)-contractive condition of T, we have $$\begin{array}{l}0<\xi (d(x_2,T{x}_2))\\ \le \psi \left(\xi \left(\text{max}\left\{d\left(x_1,x_2\right),d\left(x_1,T{x}_1\right)d\left(x_2,T{x}_2\right),\frac{d\left(x_1,T{x}_2\right)d\left(x_2,T{x}_1\right)}{2}\right\}\right)\right)\\ =\psi \left(\xi \left(\text{max}\left\{d\left(x_1,x_2\right),d\left(x_2,T{x}_2\right),\frac{d\left(x_1,T{x}_2\right)}{2}\right\}\right)\right)\\ \le \psi \left(\xi \left(\text{max}\left\{d\left(x_1,x_2\right),d\left(x_2,T{x}_2\right),\frac{d\left(x_1,T{x}_2\right)d\left(x_2,T{x}_2\right)}{2}\right\}\right)\right)\\ =\psi (\xi (\max \{d(x_1,x_2),d(x_2,T{x}_2)\})).\end{array}$$(9)

From the above inequality, it follows that and thus $$\max \{d(x_1,x_2),d(x_2,T{x}_2)\}=d(x_1,x_2)$$(10)

and thus $$0<\xi (d(x_2,T{x}_2))\le \psi (\xi (d(x_1,x_2))).$$(11)

For q₁ > 1, by using Lemma 1.6, there exists x₃ ∈ Tx₂ such that $$\xi (d(x_2,x_3))<q_1\xi (d(x_2,T{x}_2)).$$(12)

If x₂ = x₃ or x₃ ∈ Tx₃, then x₃ is a fixed point of T and hence we have noting to prove. Now, we may assume that x₂ ≠ x₃ and x₃ ∉ Tx₃. From (11) and (12), we get $$0<\xi (d(x_2,x_3))<q_1\psi (\xi (d(x_1,x_2)))=\psi (q\psi (\xi (d(x_0,x_1)))).$$(13)

Since ψ is a strictly increasing function, we obtain $$0<\psi (\xi (d(x_2,x_3)))<{\psi }^2(q\psi (\xi (d(x_0,x_1)))).$$(14)

By continuing this process, we can construct a sequence {x_n} in X such that x_n ≠ x_n+1 ∈ Tx_n, $$a(x_n,x_{n+1})\ge 1$$(15)

and $$0<\xi (d(x_{n+1},x_{n+2}))<{\psi }^n(q\psi (\xi (d(x_0,x_1))))$$(16)

for all n ∈ ℕ∪{0}.

Next, we prove that {x_n} is a Cauchy sequence in X. Let m, n ∈ ℕ. Without loss of generality, we may assume that m > n. From the triangle inequality and (16), we obtain $$\xi (d(x_m,x_n))\le {\displaystyle \sum }_{i=n}^{m-1}\xi (d(x_i,x_{i+1}))<{\displaystyle \sum }_{i=n}^{m-1}{\psi }^{i-1}(q\psi (\xi (d(x_0,x_1)))).$$

Using the property of ψ, we get $\underset{n,m\to \infty }{\lim }\xi (d(x_m,\text{}{x}_n))=\text{}0.$. By the continuity of ξ and (ξ3), we have $$\underset{n,m\to \infty }{\lim }d(x_m{x}_n)=0$$(17)

This shows that {x_n} is a Cauchy sequence in (X, d). Since (X, d) is a complete metric space, there exists x* ∈ X such that x_n → x* as n → ∞, that is, $\underset{n\to \infty }{\lim }d(x_{n,}{x}^{*})=0$. From the continuity of T, we obtain $$\underset{n\to \infty }{\lim }H(T{x}_n,Tx*)=0.$$(18)

Further, we have $$d(x*,Tx*)=\underset{n\to \infty }{\lim }d(x_{n+1},Tx*)\le \underset{n\to \infty }{\lim }H(T{x}_n,Tx*)=0.$$

By the closedness of Tx*, we get x* ∈ Tx*. Therefore, x* is a fixed point of T. This completes the proof. □

Following the proof of Theorem 2.2, we can construct a Cauchy sequence {x_n} in X such that x_n → x* as n → ∞ and $$\alpha (x_n,x_{n+1})\ge 1$$(19)

for all n ∈ ℕ. From the condition $(S_3^{\prime })$, we get $$\xi \left(d\left(x_{n+1},T{x}^{*}\right)\right)\le \psi \left(\xi \left(\max \left\{d\left(x_n,x^{*}\right),d\left(x_n,T{x}_n\right),d\left(x^{*},T{x}^{*}\right),\frac{d\left(x_n,T{x}^{*}\right)+d\left(x^{*},T{x}_n\right)}{2}\right\}\right)\right)$$(20)

for all n ∈ ℕ. Suppose that d(x*, Tx*) > 0 and let $\epsilon :=\frac{d(x^{*},T{x}^{*})}{2}$. Since x_n → x* as n → ∞, we can find N₁ ∈ ℕ such that $$d(x*,x_n)<\frac{d(x*,Tx*)}{2}$$(21)

for all n ≥ N₁. Furthermore, we obtain $$d(x*,T{x}_n)\le d(x*,x_{n+1})<\frac{d(x*,Tx*)}{2}$$(22)

for all n ≥ N₁. Also, since {x_n} is a Cauchy sequence, there exists N₂ ∈ ℕ such that $$d(x_n,T{x}_n)\le d(x_n,x_{n+1})<\frac{d(x*,Tx*)}{2}$$(23)

for all n ≥ N₂. Since d(x_n, Tx*) → d(x*, Tx*) as n → ∞, it follows that there exists N₃ ∈ ℕ such that $$d(x_n,Tx*)<\frac{3d(x*,Tx*)}{2}$$(24)

for all n ≥ N₃. Using (21)-(24), it follows that $$\max \left\{d(x_n,x*),d(x_n,T{x}_n),d(x*,Tx*),\frac{d(x_n,Tx*)+d(x*,T{x}_n}{2}\right\}=d(x*,Tx*)$$(25)

for all n ≥ N := max{N₁, N₂, N₃}. For each n ≥ N, from (20) and the triangle inequality, it follows that $$\begin{array}{l}\xi \left(d\left(x^{*},T{x}^{*}\right)\right)\\ \le \xi \left(d\left(x^{*},x_{n+1}\right)\right)+\xi \left(d\left(x_{n+1},T{x}^{*}\right)\right)\\ \le \xi \left(d\left(x^{*},x_{n+1}\right)\right)+\psi \left(\xi \left(\max \left\{d\left(x_n,x^{*}\right),d\left(x_n,T{x}_n\right),d\left(x^{*},T{x}^{*}\right),\frac{d\left(x_n,T{x}^{*}\right)+d\left(x^{*},T{x}_n\right)}{2}\right\}\right)\right)\\ =\xi \left(d\left(x^{*},x_{n+1}\right)\right)+\psi \left(\xi \left(d\left(x^{*},T{x}^{*}\right)\right)\right).\end{array}$$

Letting n → ∞ in the above inequality, we get $$\xi (d(x*,Tx*))\le \psi (\xi (d(x*,Tx*))).$$

This implies that ξ(d(x*,Tx*)) = 0, which is a contradiction. Therefore, d(x*,Tx*)) = 0, that is, x* ∈ Tx*. This completes the proof. □