Proof.
The proof is divided into four steps.
(I) We first prove that limn→∞∥xn-p∥ exists for any p∈Γ.
Since p∈Γ, p∈C≔⋂i=1NF(Si) and Ap∈Q≔⋂i=1NF(Ti). It follows from (9) that
(10)∥xn+1-p∥2=∥un-p+αn(Sn(mod N)un-un)∥2=∥un-p∥2+2αn〈un-p,Sn(mod N)un-un〉+αn2∥un-Sn(mod N)un∥2.
Because Si is ϱi-strictly pseudononspreading, for each v∈H1, we have
(11)∥Sn(mod N)un-Sn(mod N)v∥2 ≤∥un-v∥2+ϱ∥un-Sn(mod N)un-(v-Sn(mod N)v)∥2 +2〈un-Sn(mod N)un,v-Sn(mod N)v〉.
Taking v=p, we have
(12)∥Sn(mod N)un-p∥2≤∥un-p∥2+ϱ∥un-Sn(mod N)un∥2.
This implies that
(13)∥Sn(mod N)un-p∥2 =∥Sn(mod N)un-un+un-p∥2 =∥Sn(mod N)un-un∥2+2〈Sn(mod N)un-un,un-p〉 +∥un-p∥2 ≤∥un-p∥2+ϱ∥un-Sn(mod N)un∥2.
Thus it yields that
(14)2αn〈Sn(mod N)un-un,un-p〉 ≤αn(ϱ-1)∥un-Sn(mod N)un∥2.
Substituting (14) into (10) and simplifying, we have
(15)∥xn+1-p∥2≤∥un-p∥2+αn(ϱ-1)∥un-Sn(mod N)un∥2+αn2∥un-Sn(mod N)un∥2=∥un-p∥2-αn(1-ϱ-αn)∥un-Sn(mod N)un∥2.
On the other hand,
(16)∥un-p∥2 =∥xn-p+γA*(Tn(mod N)-I)Axn∥2 =∥xn-p∥2+2γ〈xn-p,A*(Tn(mod N)-I)Axn〉 +γ2∥A*(Tn(mod N)-I)Axn∥2 =∥xn-p∥2+2γ〈xn-p,A*(Tn(mod N)-I)Axn〉 +γ2〈A*(Tn(mod N)-I)Axn,A*(Tn(mod N)-I)Axn〉 =∥xn-p∥2+2γ〈xn-p,A*(Tn(mod N)-I)Axn〉 +γ2〈AA*(Tn(mod N)-I)Axn,(Tn(mod N)-I)Axn〉 ≤∥xn-p∥2+2γ〈xn-p,A*(Tn(mod N)-I)Axn〉 +γ2∥A∥2∥(Tn(mod N)-I)Axn∥2.
Since Ti is κi-strictly pseudononspreading and noting that Ap∈⋂i=1NF(Ti), we have
(17)∥Tn(mod N)Axn-Ap∥2 =∥Tn(mod N)Axn-Axn+Axn-Ap∥2 =∥Tn(mod N)Axn-Axn∥2+∥Axn-Ap∥2 +2〈Tn(mod N)Axn-Axn,Axn-Ap〉 ≤∥Axn-Ap∥2+κ∥Tn(mod N)Axn-Axn∥2.
This leads to
(18)〈Tn(mod N)Axn-Axn,Axn-Ap〉 ≤κ-12∥Tn(mod N)Axn-Axn∥2.
By (18), we have
(19)〈Tn(mod N)Axn-Axn,Tn(mod N)Axn-Ap〉=〈Tn(mod N)Axn-Axn,Tn(mod N)Axn+Axn-Axn-Ap〉=∥(Tn(mod N)-I)Axn∥2+〈Tn(mod N)Axn-Axn,Axn-Ap〉≤∥(Tn(mod N)-I)Axn∥2+κ-12∥(Tn(mod N)-I)Axn∥2=κ+12∥(Tn(mod N)-I)Axn∥2.
It follows from (19) that
(20)2γ〈xn-p,A*(Tn(mod N)-I)Axn〉 =2γ〈A(xn-p),(Tn(mod N)-I)Axn〉 =2γ〈A(xn-p)+(Tn(mod N)-I)Axn -(Tn(mod N)-I)Axn,(Tn(mod N)-I)Axn〉 =2γ〈Tn(mod N)Axn-Ap,(Tn(mod N)-I)Axn〉 -2γ∥(Tn(mod N)-I)Axn∥2 ≤[γ(1+κ)-2γ]∥(Tn(mod N)-I)Axn∥2 =[γ(κ-1)]∥(Tn(mod N)-I)Axn∥2.
By using (15), (16), (19), and (20), we have
(21)∥xn+1-p∥2≤∥xn-p∥2+γ2∥A∥2∥(Tn(mod N)-I)Axn∥2 +[γ(κ-1)]∥(Tn(mod N)-I)Axn∥2 -αn(1-σ-αn)∥un-Sn(mod N)un∥2≤∥xn-p∥2-γ(1-κ-γ∥A∥2)∥(Tn(mod N)-I)Axn∥2 -αn(1-ϱ-αn)∥un-Sn(mod N)un∥2≤∥xn-p∥2.
This shows that limn→∞∥xn-p∥ exists.
(II) We now prove that limn→∞∥un-p∥ exists.
In fact, by (21), we have
(22)[γ(1-κ-γ∥A∥2)]∥(Tn(mod N)-I)Axn∥2 +αn(1-ϱ-αn)∥un-Sn(mod N)un∥2 ≤∥xn-p∥2-∥xn+1-p∥2.
This implies that
(23)limn→∞∥(Tn(mod N)-I)Axn∥=0,(24)limn→∞∥un-Sn(mod N)un∥=0.
By virtue of (16), (23), and (24), it follows that limn→∞∥un-p∥ exists and limn→∞∥xn-p∥=limn→∞∥un-p∥.
(III) Now, we prove that limn→∞∥xn+1-xn∥=0 and limn→∞∥un+1-un∥=0.
In fact, it follows from (9) that
(25)∥xn+1-xn∥=∥(1-αn)un+αnSn(mod N)un-xn∥=∥(1-αn)(xn+γA*(Tn(mod N)-I)Axn) +αnSn(mod N)un-xn∥=∥(1-αn)(γA*(Tn(mod N)-I)Axn) +αn(Sn(mod N)un-xn)∥=∥(1-αn)(γA*(Tn(mod N)-I)Axn) +αn(Sn(mod N)un-un)+αn(un-xn)∥=∥(1-αn)(γA*(Tn(mod N)-I)Axn) +αn(Sn(mod N)un-un) +αnγA*(Tn(mod N)-I)Axn∥=∥γA*(Tn(mod N)-I)Axn+αn(Sn(mod N)un-un)∥.
This together with (23) and (24) leads to limn→∞∥xn+1-xn∥=0.
Similarly, it follows from (9), (23), and (25) that
(26)∥un+1-un∥ =∥xn+1+γA*(Tn+1(mod N)-I)Axn+1 -[xn+γA*(Tn(mod N)-I)Axn]∥ ≤∥xn+1-xn∥+∥γA*(Tn+1(mod N)-I)Axn+1∥ +∥[γA*(Tn(mod N)-I)Axn]∥ ⟶0 (as n→∞).
(IV) Finally, we prove that xn⇀x* and un⇀x*, which is a solution of the MSSFP.
In fact, since {un} is bounded, there exists a subsequence {uni}⊂{un} such that {uni}⇀x*∈H1. Hence, for any positive integer j=1,2,…,N, there exists a subsequence {ni(j)}⊂{ni} with ni(j)(mod N)=j such that {uni(j)}⇀x*. Again, by (24) we know that ∥uiN+j-SjuiN+j∥→0, as n→∞; therefore, we have that
(27)∥uni(j)-Sjuni(j)∥⟶0, as ni(j)⟶∞.
Since Sj is demiclosed at zero, it follows that x∈F(Sj). By the arbitrariness of j=1,2,…,N, we have
(28)x*∈C≔⋂i=1NF(Si).
Moreover, from (9) and (24), we have xni=uni-γA*(Tni(mod N)-I)Axni⇀x*. Since A is a bounded linear operator, it follows that Axni⇀Ax*. For any positive integer k=1,2,…,N, there exists a subsequence {xni(k)}⊂{xni} with ni(k)(mod N)=k such that Axni(k)⇀Ax* and ∥Axni(k)-TkAxni(k)∥→0. Since Tk is demiclosed at zero, we have Ax*∈F(Tk). By the arbitrariness of k∈{1,2,…,N}, it follows that Ax*∈Q≔⋂k=1NF(Tk). This together with x*∈C shows that x*∈Γ; that is, x* is a solution to the MSSFP.
Now, we prove that xn⇀x* and un⇀x*.
Suppose on the contrary that there exists another subsequence {unl}⊂{un} such that {unl}⇀y*∈Γ with y*≠x*. Consequently, by virtue of the existence of limn→∞∥xn-p∥ and the Opial property of Hilbert space, we have
(29)lim infni→∞∥uni-x*∥<lim infni→∞∥uni-y*∥=lim infn→∞∥un-y*∥=lim infnj→∞∥unj-y*∥<lim infnj→∞∥unj-x*∥=lim infn→∞∥un-x*∥=lim infni→∞∥uni-x*∥.
This is a contradiction. Therefore un⇀x*. By (9) and (24), we have
(30)xn=un-γA*(Tn(mod N)n-I)Axn⇀x*.
Therefore, the conclusion follows.
This completes the proof of Theorem 12.