In this section, we will introduce our algorithm and prove our main results.
Proof.
Let ν∈Γ. It follows that ∇f(ν)=0 for all ν∈Γ. From (2.5), we deduce that
(3.2)f(yn)=f(yn)-f(ν)≤〈∇f(yn),yn-ν〉.
Thus, by (3.1) and (3.2), we have
(3.3)∥xn+1-ν∥2=∥PC(yn-τnf(yn)∇f(yn)∥∇f(yn)∥2)-ν∥2≤∥yn-τnf(yn)∇f(yn)∥∇f(yn)∥2-ν∥2=∥yn-ν∥2-2τnf(yn)∥∇f(yn)∥2〈∇f(yn),yn-ν〉+τn2f2(yn)∥∇f(yn)∥2≤∥yn-ν∥2-2τnf2(yn)∥∇f(yn)∥2+τn2f2(yn)∥∇f(yn)∥2=∥yn-ν∥2-τn(2-τn)f2(yn)∥∇f(yn)∥2,∥yn-ν∥2=∥αn(u-ν)+(1-αn)(xn-ν)∥2≤αn∥u-ν∥2+(1-αn)∥xn-ν∥2.
It follows that
(3.4)∥xn+1-ν∥2≤αn∥u-ν∥2+(1-αn)∥xn-ν∥2-τn(2-τn)f2(yn)∥∇f(yn)∥2≤αn∥u-ν∥2+(1-αn)∥xn-ν∥2≤max {∥u-ν∥2,∥xn-ν∥2}.
By induction, we deduce
(3.5)∥xn+1-ν∥≤max {∥u-ν∥,∥x0-ν∥}.
Hence, {xn} is bounded.
At the same time, we note that
(3.6)∥yn-ν∥2=∥αn(u-ν)+(1-αn)(xn-ν)∥2≤(1-αn)∥xn-ν∥2+2αn〈u-ν,yn-ν〉.
Therefore,
(3.7)∥xn+1-ν∥2≤(1-αn)∥xn-ν∥2+2αn〈u-ν,yn-ν〉-τn(2-τn)f2(yn)∥∇f(yn)∥2=(1-αn)∥xn-ν∥2+2αn〈u-ν,αn(u-ν)+(1-αn)(xn-ν)〉 -τn(2-τn)f2(yn)∥∇f(yn)∥2=(1-αn)∥xn-ν∥2+2αn2∥u-ν∥2+2αn(1-αn)〈u-ν,xn-ν〉 -τn(2-τn)f2(yn)∥∇f(yn)∥2.
It follows that
(3.8)∥xn+1-ν∥2-∥xn-ν∥2+αn(∥xn-ν∥2-2αn∥u-ν∥2+2αn〈u-ν,xn-ν〉) +τn(2-τn)f2(yn)∥∇f(yn)∥2≤2αn〈u-ν,xn-ν〉.
Next, we will prove that xn→ν. Set ωn=∥xn-ν∥2 for all n≥0. Since αn→0 and inf n τn(2-τn)>0, we may assume without loss of generality that τn(2-τn)≥σ for some σ>0. Thus, we can rewrite (3.8) as
(3.9)ωn+1-ωn+αnUn+σf2(yn)∥∇f(yn)∥2≤2αn〈u-ν,xn-ν〉,
where Un=∥xn-ν∥2-2αn∥u-ν∥2+2αn〈u-ν,xn-ν〉.
Now, we consider two possible cases.
Case 1. Assume that {ωn} is eventually decreasing; that is, there exists N>0 such that {ωn} is decreasing for n≥N. In this case, {ωn} must be convergent and from (3.9) it follows that
(3.10)0≤σf2(yn)∥∇f(yn)∥2≤ωn-ωn+1-αnUn+2αn∥u-ν∥∥xn-ν∥≤ωn-ωn+1+Mαn,
where M>0 is a constant such that sup n{2∥u-ν∥∥xn-ν∥+∥Un∥}≤M. Letting n→∞ in (3.10), we get
(3.11)lim n→∞f(yn)=0.
Since {yn} is bounded, there exists a subsequence {ynk} of {yn} converging weakly to x~∈C. Since, xn-yn→0, we also have {xnk} of {xn} converging weakly to x~∈C. From the weak lower semicontinuity of f, we have
(3.12)0≤f(x~)≤liminf k→∞ f(ynk)=lim n→∞ f(yn)=0.
Hence, f(x~)=0; that is, Ax~∈Q. This indicates that
(3.13)ωw(yn)=ωw(xn)⊂Γ.
Furthermore, by using the property of the projection (c), we deduce
(3.14)limsup n→∞〈u-ν,xn-ν〉=max x~∈ωw(xn)〈u-PΓ(u),x~-PΓ(u)〉≤0.
From (3.8), we obtain
(3.15)ωn+1≤(1-αn)ωn+αn(2αn∥u-ν∥2+2(1-αn)〈u-ν,xn-ν〉).
This together with Lemma 2.1 imply that ωn→0.
Case 2. Assume that {ωn} is not eventually decreasing. That is, there exists an integer n0 such that ωn0≤ωn0+1. Thus, we can define an integer sequence {τn} for all n≥n0 as follows:
(3.16)τ(n)=max {k∈ℕ∣n0≤k≤n,ωk≤ωk+1}.
Clearly, τ(n) is a nondecreasing sequence such that τ(n)→+∞ as n→∞ and
(3.17)ωτ(n)≤ωτ(n)+1,
for all n≥n0. In this case, we derive from (3.10) that
(3.18)σf2(yτ(n))∥∇f(yτ(n))∥2≤Mατ(n)→0.
It follows that
(3.19)lim n→∞f(yτ(n))=0.
This implies that every weak cluster point of {yτ(n)} is in the solution set Γ; that is, ωw(yτ(n))⊂Γ. So, ωw(xτ(n))⊂Γ. On the other hand, we note that
(3.20)∥yτ(n)-xτ(n)∥=ατ(n)∥u-xτ(n)∥→0,∥xτ(n)+1-yτ(n)∥≤ττ(n)f(yτ(n))∥∇f(yτ(n))∥→0.
From which we can deduce that
(3.21)limsup n→∞〈u-ν,xτ(n)-ν〉=max x~∈ωw(xτ(n))〈u-PΓ(u),x~-PΓ(u)〉≤0.
Since ωτ(n)≤ωτ(n)+1, we have from (3.9) that
(3.22)ωτ(n)≤(1-2ατ(n))〈u-ν,xτ(n)-ν〉+2ατ(n)∥u-ν∥2.
Combining (3.21) and (3.22) yields
(3.23)limsup n→∞ ωτ(n)≤0,
and hence
(3.24)lim n→∞ωτ(n)=0.
From (3.15), we have
(3.25)limsup n→∞ ωτ(n)+1≤limsup n→∞ ωτ(n).
Thus,
(3.26)lim n→∞ωτ(n)+1=0.
From Lemma 2.2, we have
(3.27)0≤ωn≤max {ωτ(n),ωτ(n)+1}.
Therefore, ωn→0. That is, xn→ν. This completes the proof.