In this section, we will prove the strong convergence theorem for a common fixed point of a finite family of relatively nonexpansive mappings in a Banach space by using the hybrid method in mathematical programming. Let us prove a proposition first.

Proof.
We split the proof into seven steps.

Step
1. Show that PF is well defined for every x∈C.

It is easy to know that F(Ti), i=1,2,…,N are closed convex sets and so is F. What is more, F is nonempty by our assumption. Therefore, PF is well defined for every x∈C.

Step
2. Show that Hn and Wn are closed and convex for all n≥0.

From the definition of Wn, it is obvious Wn is closed and convex for each n≥0. By Lemma 6, we also know that Hn is closed and convex for each n≥0.

Step
3. Show that F⊂Hn⋂Wn for all n≥0.

Let u∈F and let n≥0. Then, by the convexity of ∥·∥2, we have
(19)ϕ(u,zn) =ϕ(u,J-1(λn(0)Jxn+∑i=1Nλn(i)JTixn)) =∥u∥2-2〈u,λn(0)Jxn+∑i=1Nλn(i)JTixn〉 +∥λn(0)Jxn+∑i=1Nλn(i)JTixn∥2 ≤∥u∥2-2λn(0)〈u,Jxn〉-2∑i=1Nλn(i)〈u,JTixn〉 +λn(0)∥Jxn∥2+∑i=1Nλn(i)∥JTixn∥2 =λn(0)ϕ(u,xn)+∑i=1Nλn(i)ϕ(u,Tixn) ≤λn(0)ϕ(u,xn)+∑i=1Nλn(i)ϕ(u,xn) =ϕ(u,xn),
and then,
(20)ϕ(u,yn) =ϕ(u,J-1(αnJx+βnJxn+γnJzn)) =∥u∥2-2〈u,αnJx+βnJxn+γnJzn〉 +∥αnJx+βnJxn+γnJzn∥2 ≤∥u∥2-2αn〈u,Jx〉-2βn〈u,Jxn〉-2γn〈u,Jzn〉 +αn∥x∥2+βn∥xn∥2+γn∥zn∥2 =αnϕ(u,x)+βnϕ(u,xn)+γnϕ(u,zn) ≤αnϕ(u,x)+(1-αn)ϕ(u,xn) =ϕ(u,xn)+αn(ϕ(u,x)-ϕ(u,xn)) ≤ϕ(u,xn)+αn(∥x∥2+2〈Jxn-Jx,z〉).
Thus, we have u∈Hn. Therefore, we obtain F⊂Hn for all n≥0.

Next, we prove F⊂Wn for all n≥0. We prove this by induction. For n=0, we have F⊂C=W0. Assume that F⊂Wn. Since xn+1 is the projection of x onto Hn⋂Wn, by Lemma 3, we have
(21)〈xn+1-z, Jx-Jxn+1〉≥0,
for any z∈Hn⋂Wn. As F⊂Hn⋂Wn by the induction assumption,F⊂Wn holds, in particular, for all u∈F. This together with the definition of Wn+1 implies that F⊂Wn+1. Hence, F⊂Hn⋂Wn for all n≥0.

Step
4. Show that ∥xn+1-xn∥→0 as n→∞.

In view of (19) and Lemma 4, we have xn=PWnx, which means that, for any z∈Wn,
(22)ϕ(xn,x)≤ϕ(z,x).
Since xn+1∈Wn and u∈F⊂Wn, we obtain
(23)ϕ(xn,x)≤ϕ(xn+1,x),ϕ(xn,x)≤ϕ(u,x),
for all n≥0. Consequently, limn→∞ϕ(xn,x) exists and {xn} is bounded. By using Lemma 4, we have
(24)ϕ(xn+1,xn)≤ϕ(xn+1,x)-ϕ(xn,x)⟶0,
as n→∞. By using Lemma 2, we obtain ∥xn+1-xn∥→0 as n→∞.

Step
5. Show that ∥xn-zn∥→0 as n→∞.

From xn+1=PHn⋂Wnx∈Hn, we have
(25)ϕ(xn+1,yn)≤ϕ(xn+1,xn) +αn(∥x∥2+2〈Jxn-Jx, xn+1〉)⟶0,
as n→∞. By Lemma 2, we also have ∥xn+1-yn∥→0, and then,
(26)∥xn-yn∥≤∥yn-xn+1∥+∥xn-xn+1∥⟶0,
as n→∞. We observe that
(27)ϕ(zn,xn) =ϕ(zn,yn)+ϕ(yn,xn)+2〈zn-yn, Jyn-Jxn〉 ≤ϕ(zn,yn)+ϕ(yn,xn)+2∥zn-yn∥∥Jyn-Jxn∥,ϕ(zn,yn) =∥zn∥2-2〈zn,αnJx+βnJxn+γnJzn〉 +∥αnJx+βnJxn+γnJzn∥2 ≤αnϕ(zn,x)+βnϕ(zn,xn).
So,
(28)ϕ(zn,xn)≤αnϕ(zn,x)+βnϕ(zn,xn)+ϕ(yn,xn)+2∥zn-yn∥∥Jyn-Jxn∥.
Since limn→∞αn=0, limsupn→∞βn<1, ϕ(yn,xn)→0, and ∥zn-yn∥∥Jyn-Jxn∥→0 as n→∞, we have
(29)ϕ(zn,xn)≤αn1-βnϕ(zn,x)+11-βnϕ(yn,xn) +21-βn∥zn-yn∥∥Jyn-Jxn∥⟶0,
as n→∞. Using Lemma 2, we obtain ∥xn-zn∥→0 as n→∞.

Step 6. Show that ∥xn-Tixn∥→0, i=1,2,…,N.

Since {xn} is bounded and ϕ(p,Tixn)≤ϕ(p,xn), where p∈F, i=1,2,…,N, we also obtain that {Jxn},{JT1xn},…,{JTNxn} are bounded, and hence, there exists r>0 such that {Jxn},{JT1xn},…,{JTNxn}⊂Br(0). Therefore, Proposition 7 can be applied and we observe that
(30)ϕ(p,zn) =∥p∥2-2〈p,λn(0)Jxn+∑i=1Nλn(i)JTixn〉 +∥λn(0)Jxn+∑i=1Nλn(i)JTixn∥2 ≤∥p∥2-2λn(0)〈p,Jxn〉-2∑i=1Nλn(i)〈p,JTixn〉 +λn(0)∥xn∥2+∑i=1Nλn(i)∥Tixn∥2 -1N2g(∑i=1N∑j=1Nλn(i)λn(j)∥JTixn-JTjxn∥ +2∑i=1Nλn(0)λn(i)∥Jxn-JTixn∥) =λn(0)ϕ(p,xn)+∑i=1Nλn(i)ϕ(p,Tixn) -1N2g(∑i=1N∑j=1Nλn(i)λn(j)∥JTixn-JTjxn∥ +2∑i=1Nλn(0)λn(i)∥Jxn-JTixn∥) ≤ϕ(p,xn)-1N2g(∑i=1N∑j=1Nλn(i)λn(j)∥JTixn-JTjxn∥ +2∑i=1Nλn(0)λn(i)∥Jxn-JTixn∥),
where g:[0,∞)→[0,∞) is a continuous strictly increasing convex function with g(0)=0. And
(31)ϕ(p,xn)-ϕ(p,zn) =∥p∥2-2〈p,Jxn〉+∥xn∥2-∥p∥2+2〈p,Jzn〉-∥zn∥2 ≤2∥p∥∥Jzn-Jxn∥+∥xn∥2-∥zn∥2⟶0,
as n→∞. From the properties of the mapping g, we have
(32)limn→∞λn(0)λn(i)∥xn-Tixn∥=0,limn→∞λn(i)λn(j)∥Tixn-Tjxn∥=0,
for all i,j∈{1,2,…,N}. From the condition (d′), we have ∥xn-Tixn∥→0 immediately, as n→∞, i=1,2,…,N; from the condition (d), we can also have ∥xn-Tixn∥→0, as n→∞, i=1,2,…,N. In fact, since liminfn→∞λn(i)λn(j)>0, it follows that
(33)limn→∞∥Tixn-Tjxn∥=0,
for all i,j∈{1,2,…,N}. Next, we note by the convexity of ∥·∥2 and (9) that
(34)ϕ(Tjxn,zn) =∥Tjxn∥2-2〈Tjxn,λn(0)Jxn+∑i=1Nλn(i)JTixn〉 +∥λn(0)Jxn+∑i=1Nλn(i)JTixn∥2 ≤∥Tjxn∥2-2λn(0)〈Tjxn,Jxn〉-2∑i=1Nλn(i)〈Tjxn,JTixn〉 +λn(0)∥xn∥2+∑i=1Nλn(i)∥Tixn∥2 =λn(0)ϕ(Tjxn,xn)+∑i=1Nλn(i)ϕ(Tjxn,Tixn)⟶0,
as n→∞. By Lemma 2, we have limn→∞∥Tixn-zn∥=0 and
(35)∥Tixn-xn∥≤∥Tixn-zn∥+∥xn-zn∥⟶0,
as n→∞ for all i∈{1,2,…,N}.

Step 7. Show that xn→ΠFx, as n→∞.

From the result of Step 6, we know that if {xnk} is a subsequence of {xn} such that {xnk}⇀x^∈C, then x^∈∩i=1NF^(Ti)=∩i=1NF(Ti). Because E is a uniformly convex and uniformly smooth Banach space and {xn} is bounded, so we can assume {xnk} is a subsequence of {xn} such that {xnk}⇀x^∈F and ω=ΠFx. For any n≥1, from xn+1=ΠHn⋂Wnx and ω∈F⊂Hn⋂Wn, we have
(36)ϕ(xn+1,x)≤ϕ(ω,x).
On the other hand, from weakly lower semicontinuity of the norm, we have
(37)ϕ(x^,x) =∥x^∥2-2〈x^,Jx〉+∥x∥2 ≤liminfn→∞(∥xnk∥2-2〈∥xnk∥2,Jx〉+∥x∥2) =liminfn→∞ϕ(xnk,x)≤limsupn→∞ ϕ(xnk,x)≤ϕ(ω,x).
From the definition of ΠFx, we obtain x^=ω, and hence, limn→∞ϕ(xnk,x)=ϕ(ω,x). So, we have limk→∞∥xnk∥=∥ω∥. Using the Kadec-klee property of E, we obtain that {xnk} converges strongly to ΠFx. Since {xnk} is an arbitrary weakly convergent sequence of {xn}, we can conclude that {xn} converges strongly to ΠFx.