Recall the setting and notation of section2.3.1, where (X,τ) is a topological vector
space. Throughout this section, unless otherwise stated, convergence for sequences inXwill be with respect to theτtopology and convergence for sequences in count- able Cartesian products of X will be in the product topology of the τ topology on X.
We will use x to refer to elements of XN. We can then use (xn)∞n=0 to denote a sequence{x0,. . .,xn,. . .}, where(xn)∞n=0 ∈(XN)N.
14
See Williams (1991)9.7i) and note the trivialσ-algebra is a subset ofFt+1.
15
See Williams (1991),9.9.
16
See section5in Çinlar (2011).
17
RemarkA.2.1. LetX=Q
i∈FXidenote a Cartesian product of topological spaces. Let
πi: X → Xi denote the projection map defined as πi(x) = xi for each i ∈ F. Recall
each projection map will be a continuous function on X when X has the product topology (see section 2.14 by Aliprantis and Border (2006)). Also recall (section 1.8
by Tao (2013)) the image of a (sequentially) compact set under a continuous function
is (sequentially) compact. Accordingly, if a set C with C ⊂ X is (sequentially) compact in the product topology, thenπi(C) will be (sequentially) compact.
Finally, let the function φt: Gt+1(x) → R+ denote (xi)ti=+01 7→ ρt(xt,xt+1) for each t
and letU(x) := P∞t=0βtρt(xt,xt+1).
Lemma A.2.2. Let Assumption2.3.2hold and letxsatisfyx∈ S0. If(xn)∞n=0is a sequence
withxn ∈ G(x)for each nandU(xn) →BforB > infρ, then there exists a sub-sequence
(xnk)∞
k=0 such that for allt∈ N:
lim k→∞ρt(x nk t ,x nk t+1)→θt
whereθt ∈ Rfor eacht andθt>infρfor at-least onet.
Proof. By Assumption2.3.2, for eacht andn,
mt≥ρt(xnt,xtn+1) ≥infρ (A.2.14)
Accordingly, for eachn,(ρt(xnt,xnt+1))∞t=0will belong to the set
Q∞
t=0[infρ,mt]. Note
each[infρ,mt]will be sequentially compact in the order topology (see Kelley (1975),
p149 and Tao (2016), p43) on R∪{−∞,∞}. Then, by the sequential Tychonoff’s
Theorem (see Proposition 1.8.12 by Tao (2010)), Q∞
in the product topology of the order topology. There then exists a sub-sequence of (xn)∞n=0, (xnk)∞ k=0, such that (ρ(x nk t ,x nk
t+1))∞k=0 converges for each t. Let θt :=
lim k→∞ρ(x nk t ,x nk t+1)and note: B= lim k→∞ ∞ X t=0 βtρt xntk,x nk t+1 ≤ ∞ X t=0 lim k→∞β t ρt xntk,x nk t+1 = ∞ X t=0 βtθt (A.2.15)
Since (A.2.14) holds, and P∞
t=0βtmt <∞ by Assumption2.3.2, we can pass limits
through in the second equality using Fatou’s Lemma for series (fact A.1.5 in the
appendix).
Finally, recall infρ ∈ {−∞,0}. Then if B > infρ, then the above means there is at least one θt >infρ.
Lemma A.2.3. Let x satisfy x ∈ S0. If(xn)∞n=0 is a sequence with xn ∈ G(x)for each n
and for somet:
ρt(xnt,xnt+1)→θt
withθt>infρ, then there exists >infρandN∈ Nsuch that for alln > N, (xni)ti+=01∈
UCφt().
Proof. There exists ι such that := θt−ι satisfies > infρ. For N large enough
and any n > N, ρt(xnt,xnt+1) ∈ [,θt+ι], implying ρt(xnt,xnt+1) ≥ and (xni) t+1 i=0 ∈
UCφt().
Lemma A.2.4. Let assumptions2.3.1- 2.3.3hold and letx satisfy x ∈ S0. If (xn)∞n=0 is a
sequence such thatxn ∈ G(x)for eachn∈ NandU(xn)→BwhereB > infρ, then there existsxwithx∈ G(x)such thatB≤U(x)<∞.
Proof. Letxsatisfyx ∈S0and let(xn)∞n=0be a sequence such thatxn ∈ G(x)for each
n and U(xn) → B where B > infρ. By Lemma A.2.2 there exists a sub-sequence (xnj)∞
j=0 such that for each t ∈ N, θt := lim j→∞ρt(x
nj
t ,x nj
t+1) > infρ for at-least one t.
Re-label (xnj)∞
j=0 to(xn)∞n=0, and let P^ denote the subset ofNsuch thatt ∈ P^ if and
only ifθt>infρ. The set P^ will be non-empty, but could be finite or infinite.
To prove part1of the lemma, consider first the case whenP^ is infinite and then the
case whenP^ is finite.
Suppose P^ is infinite and consider any t ∈ N. There will exist k > t such that
ck > infρ. By Lemma A.2.3, there exists N and > infρ such that for all n > N, (xni)ki=+01∈ UCφk().
By Assumption 2.3.1, UCφ
k() will be sequentially compact in the product topol- ogy. The space πt(UCφk()) will also be sequentially compact by the argument in Remark A.2.1. LetΞt:={x0
1,. . .,xNt }∪πt(UCφk()). Since{x
0
1,. . .,xNt }is sequentially
compact,Ξtwill be sequentially compact. Moreover, notexnt ∈Ξt for eachn∈ N.
Since t was arbitrary, we can construct a Ξt as above for every t ∈ N. Now let
Ξ := Qt∈NΞt. Using the Sequential Tychonoff Theorem (Proposition 1.8.12 by Tao
(2010)), Ξwill be sequentially compact. Since for each t, xn
t ∈Ξt for eachn, xn ∈ Ξ
for eachn. There then exists a sub-sequence(xnj)∞
j=0converging to x, withx ∈Ξ.
We now confirm x ∈ G(x) by showing xt+1 ∈ Γt(xt) for all t ∈ N. Pick any t ∈ N,
there will be ak satisfyingk > tsuch thatck >infρ. By Lemma A.2.3, there exists > infρ and Jsuch that for all j > Jwe have (xnij)ki=+01 ∈ UCφk(). By Assumption
2.3.1, UCφ
k() will be sequentially compact, moreover, UCφk() ⊂ G
k+1(x) by the
definition of UCφk() at (2.22). As such, the sub-sequence (xnj
i ) k+1
(xi)ki=+01, with (xi)ki=+01 ∈ Gk+1(x), allowing us to conclude xt+1 ∈ Γt(xt). Since thet
was arbitrary,xt+1∈ Γt(xt)for eacht ∈Nand x∈ G(x).
Now assume P^ is finite;P^ will have a maximum element, which we now callk. By Lemma A.2.3, there exists > infρ and N ∈ N such that (xn
t)kt=+01 ∈ UCφk() for each n > N. By Assumption 2.3.1, UCφ
k() will be sequentially compact in the product topology. As such, there exists a sub-sequence (xnj)∞
j=0 such that (x nj
t )∞j=0
for each t ≤k+1. Define (xt)∞t=0 by setting xt= lim j→∞x
nj
t fort ≤k+1 and picking
any xt+1 ∈Γt(xt)fort ≥k+1.
To confirm (xt)∞t=0 is feasible, let us check xt+1 ∈ Γt(xt) for each t. Once again,
note by definition, UCφk() ⊂ Gk+1(x). Since UC
φk() is sequentially compact,
(xt)kt=+01 ∈ G(x) and xt+1 ∈ Γt(xt) for all t satisfying t ≤ k. On the other hand, if
t > k, by construction,xt+1 ∈Γt(xt), confirming(xt)∞t=0 ∈ G(x). Re-label (xnj)∞ j=0 to(x n)∞ n=0. By Assumption2.3.2, we have: ρt(xnt,xnt+1) ≤mt
for eacht and n, whereP∞t=0βtmt<∞. Whence Fatou’s Lemma18 gives:
B=lim sup n→∞ ∞ X t=0 βtρt(xnt,xnt+1)≤ ∞ X t=0 lim sup n→∞ βtρt(xnt,xtn+1)<∞ (A.2.16) 18
See fact A.1.5in the appendix and letΩ=Z+andµbe the counting measure. Also see Equation
Upper-semicontinuity ofρt(Assumption2.3.3) and the growth condition (Assump-
tion 2.3.2) imply:19
lim sup
n→∞
ρt(xnt,xtn+1)≤ρt(xt,xt+1)≤mt, t∈ N (A.2.17)
To complete the proof, combine (A.2.17) with (A.2.16) and conclude:
B≤ ∞
X t=0
βtρt(xt,xt+1) =U(x)<∞
Proof of theorem 2.3.1. Fixx ∈S0. IfU(x) =infρfor allx∈ G(x), then our solution
will be anyx∈ G(x).
Next, suppose at-least one x with x ∈ G(x) satisfies U(x) > infρ. By Assumption
2.3.2, there exists a sequence of real numbers (mt)∞
t=0 such that ρt(xt,xt+1) ≤ mt
for any xinG(x)and
¯ B:= ∞ X t=0 βtmt<∞
Any xwithx ∈ G(x) will satisfy:
U(x) =
∞
X t=0
βtρt(xt,xt+1)≤B¯
Consider the set I := {U(x) |x∈ G(x)}. The set I will be a subset ofR∪{−∞,∞} and so must have a supremum. Let B: = sup Iand note infρ≤B≤B <¯ ∞.
19
If P^ is infinite, then xn
t converges for each t and lim supn→∞ρt(xnt,xnt+1) ≤ ρt(xt,xt+1) by
upper semicontinuity. On the other hand, ifP^is finite, then lim supn→∞ρt(xnt,xnt+1)≤ρt(xt,xt+1)
by upper semicontinuity for each t ≤ k+1 and the inequality holds for t > k+1 since lim supn→∞ρt(xtn,xnt+1) =infρ.
Construct a sequence(xn)∞n=0withxn ∈ G(x)for eachnandU(xn)→Bas follows: for every n ∈ N, take xn such that B−U(xn) < n+11. Such a sequence exists, otherwise for some n, U(x) ≤ B− n1+1 for all x ∈ G(x) and B will not be the supremum of I.
SinceU(xn)→ B, by Lemma A.2.4, there exists x∈ G(x) such that U(x)≥ B. Since Bwas the supremum forI, conclude:
U(x) =B=J(x)<∞
Proof of Proposition 2.3.1. Letx satisfyx ∈ S0. Fix any t∈ Nand anysatisfying
>infρ. We show the upper contour setsUCφt()defined by:
UCφt() ={(xi)it=+01 ∈ Gt+1(x)|ρt(xt,xt+1)≥} (A.2.18)
are sequentially compact. In particular, we first show UCφt() is a sequentially closed subset of Xt+1 and then show UCφt() is contained within a compact and metrizable set.
To show UCφt() is sequentially closed in X
t+1, take any sequence (xn)∞
n=0 with
xn ∈ UCφt() for each n that converges to x = (xi)
t+1
i=0 point-wise. Note x n i ∈ Si
for each i ≤ t+1 and n. Since each Si is sequentially closed (Assumption 2.3.4), xi∈ Si.
By Assumption2.3.5, each Γi has a sequentially closed graph, and thusxi+1 ∈ Γi(xi)
for eachi ≤t+1. Noting the definition ofG(x)t+1 by (2.23), conclude x∈ G(x)t+1.
We now confirm ρt(xt,xt+1) ≥ . By upper semi-continuity of ρt (Assumption
2.3.3), UCρ
(xni)ti=+01 will satisfyρt(xnt,xnt+1)≥and thus {xnt,xnt+1}∈UCρt()for each n. More- over, xt+1 ∈ Γt(xt). Accordingly, {xt,xt+1} ∈ UCρt() and ρt(xt,xt+1) ≥ . We concludex∈ UCφt()and UCφt() is sequentially closed.
SinceE satisfies Assumption 2.3.6, there will exist ¯M such that if(xi)t+1
i=0 ∈ G t+1(x)
andρt(xt,xt+1)≥, thenkxik ≤ M¯ fori ∈ {0,. . .,t+1}. WhenceUCφt()will be a subset of the spaceBM¯ := Πti+=01{xi ∈ Si|kxik ≤M¯ }.
For each i ≤ t+1, the space {xi ∈ Si|kxik ≤ M¯ } will be compact by Alaoglu’s
Theorem.20
Next,L2(Ω,P)is a separable space sinceFis separable.21
As such, since
L2(Ω,P)is reflexive, the spaces{xi∈ Si|kxik ≤ M¯ }are metrizable and sequentially
compact.22
Moreover, by the Sequential Tychonoff’s Theorem,23
the space BM¯ will
be sequentially compact in the product topology (of weak topology on X). By the argument in the preceding paragraph, UCφt() is a sequentially closed subset of
BM¯ , allowing us to concludeUCφt()is sequentially compact.