This is the third of three sets of fixed point exercises. The first post in this sequence is here, giving context.
Note: Questions 15 form a coherent sequence and questions 610 form a separate coherent sequence. You can jump between the sequences.

Let be a complete metric space. A function is called a contraction if there exists a such that for all , . Show that if is a contraction, then for any , the sequence converges. Show further that it converges exponentially quickly (i.e. the distance between the th term and the limit point is bounded above by for some )

(Banach contraction mapping theorem) Show that if is a complete metric space and is a contraction, then has a unique fixed point.

If we only require that for all , then we say is a weak contraction. Find a complete metric space and a weak contraction with no fixed points.

A function is convex if , for all and . A function is strongly convex if you can subtract a positive parabaloid from it and it is still convex. (i.e. is strongly convex if is convex for some .) Let be a strongly convex smooth function from to , and suppose that the magnitude of the second derivative is bounded. Show that there exists an such that the function given by is a contraction. Conclude that gradient descent with a sufficiently small constant step size converges exponentially quickly on a strongly convex smooth function.

A finite stationary Markov chain is a finite set of states, along with probabilistic rule for transitioning between the states, where represents the space of probability distributions on . Note that the transition rule has no memory, and depends only on the previous state. If for any pair of states , the probability of passing from to in one step is positive, then the Markov chain is ergodic. Given an ergodic finite stationary Markov chain, use the Banach contraction mapping theorem to show that there is a unique distribution over states which is fixed under application of transition rule. Show that, starting from any state , the limit distribution exists and is equal to the stationary distribution.

A function from a partially ordered set to another partially ordered set is called monotonic if implies that . Given a partially ordered set with finitely many elements, and a monotonic function from to itself, show that if or , then is a fixed point of for all .

A complete lattice is a partially ordered set in which each subset of elements has a least upper bound and greatest lower bound. Under the same hypotheses as the previous exercise, extend the notion of for natural numbers to for ordinals , and show that is a fixed point of for all with or and all ( means there is an injection from to , and means there is no such injection).

(KnasterTarski fixed point theorem) Show that the set of fixed points of a monotonic function on a complete lattice themselves form a complete lattice. (Note that since the empty set is always a subset, a complete lattice must be nonempty.)

Show that for any set , forms a complete lattice, and that any injective function from to defines a monotonic function from to . Given injections and , construct a subset of and a subset of of such that and .

(Cantor–Schröder–Bernstein theorem) Given sets and , show that if and , then . ( means there is an injection from to , and means there is a bijection)
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Tomorrow's AI Alignment Forum Sequences post will be "Approvaldirected agents: overview" by Paul Christiano in the sequence Iterated Amplification.
The next post in this sequence will be released on Saturday 24th November, and will be 'Fixed Point Discussion'.
Ex 6:
If at any point fn(b)=fn−1(x), then we're done. So assume that we get a strict increase each time up to n=P. Since there are only P elements in the entire poset, and f is monotone, fn+1(x) has to equal fn(x).
Ex 7:
For a limit ordinal α, define fα(x) as the least upper bound of fn(x) for all n<α. If α>L, then the set fn(x) for n<α is a set of size α that maps into a set of size L by taking the value of the element. Since there are no injections between these sets, there must be two ordinals n<m such thatfn(x)=fm(x). Since f is monotone, that implies that for every ordinal l>n, fl(x)=fn(x)and thus is a fixed point. Since n<α this proves the exercise.
Ex 8:
Starting from x, we can create a fixed point via iteration by taking α>L, and iterating α times as demonstrated in Ex 7. Call this fixed point fx. Suppose there was a fixed point k such that x≤k and k≤fx. Then at some point fn(x)≤fn(k)=k, but fn+1(x)≥fn+1(k)=k, which breaks the monotonicity of f unless k=fx. So fx generated this way is always the smallest fixed point greater than x.
Say we have fixed points xi. Then let x be the least upper bound of xi, and generate a fixed point from fx. So fx will be greater than each element of xi since f is monotone, and is the smallest such fixed point as shown in the above paragraph. So the poset of fixed points is semicomplete with upper bounds.
Now take our fixed points xi again. Now let x be the greatest lower bound of xi, and generate a fixed point fx. Since x≤xi and f is monotonic, fα(x)≤fα(xi)=xi, and so fx is a lower bound of xi. It has to be the greatest such bound because x itself is already the greatest such bound in our poset, and f is monotonic.
Thus the lattice of fixed points has all least upper bounds and all greatest lower bounds, and is thus complete!
#3:
x→x+1x on x≥1 shortens all distances but is strictly monotonic.
#6: (the "show that if" condition follows from the property, the question is likely misstated)
The iteration is so long that it must visit an element twice. We can't have a cycle in the order so the repetition must be immediate.
Thanks, I actually wanted to get rid of the earlier condition that f(x)≥x for all x, and I did that.
Answer to question 1.
Let xi+1=f(xi) for arbitrary x0. Call c=d(x0,x1). Then by induction (i<j)d(xi,xj)≤∑k=j−1k=id(xk,xk+1)≤∑k=j−1k=icqk≤cqi1−q (power series simplification)
Therefore ∀δ>0:∃n∈N∀i>n,j>i:d(i,j)<cqn1−q<δ ie xi is a cauchy sequence. However (X,d) is said to be complete, which by definition means any cauchy sequence is convergent. So xn→y and d(xi,y)≤sup∞j=id(xi,xj)≤cqi1−q So xn converges exponentially quickly
Answer to question 2.
From part 1, as f is continuous, y=limn→∞(f(xn+1))=limn→∞(f(xn))=f(limn→∞(xn))=f(y) So y is a fixed point. Suppose x and y are both fixed points of f(x) a contraction map. Then f(x)=x and f(y)=y so d(f(x),f(y))≤qd(x,y)=qd(f(x),f(y)) therefore d(x,y)=0 so x=y. Thus f has a unique fixed point.
Answer to question 3.
(R,d(x,y)=x−y) is a metric space. Its the real line with normal distance. Let f(x)=√1+x2 . Then f is a contraction map because f is differentiable and f′(x)=x√1+x2has the property ∀x:f′(x)<1. However no fixed point exists as ∀x:f(x)>x. This works because the sequence xi generated from repeated applications of f will tend to infinity, despite successive terms becoming ever closer.
For Q2, I believe you aren't done:
You have established that there is at most one fixed point, but not that a fixed point exists.
#6:
Assume WLOG f(x)>=xThen by monotonicity, we have x<=f(x)<=f2(x)<=...<=fP(x)If this chain were all strictly greater, than we would have P+1istinct elements. Thus there must be some kuch that fk(x)=fk+1(x)By induction, fn+1(x)=fn(x)=fk(x)or all n>k
#7:
Assume f(x)>=xnd construct a chain similarly to (6), indexed by elements of αIf all inequalities were strict, we would have an injection from αo L.
#8:
Let F be the set of fixed points. Any subset S of F must have a least upper bound xn L. If x is a fixed point, done. Otherwise, consider fα(x) which must be a fixed point by (7). For any q in S, we have f(q)≤x⇒fα(q)≤fα(x)⇒q≤fα(x) Thus fα(x)s an upper bound of S in F. To see that it is the least upper bound, assume we have some other upper bound b of S in F. Then x<=b⇒fα(x)<=fα(b)=b
To get the lower bound, note that we can flip the inequalities in L and still have a complete lattice.
#9:
P(A) clearly forms a lattice where the upper bound of any set of subsets is their union, and the lower bound is the intersection.
To see that injections are monotonic, assume A0⊆A1nd fs an injection. For any function, f(A0)⊆f(A1) If a∉A0nd f(a)∈f(A0)that implies f(a)=f(a′)or some a′∈A0which is impossible since fs injective. Thus fs (strictly) monotonic.
Now h:=g∘fs an injection A→ALet Xe the set of all points not in the image of gand let A′=X∪h(X)∪h2(X)∪...ote that h(A′)=h(X)∪h2(X)∪h3(X)∪...=A′−Xsince no element of Xs in the image of hThen g(B−f(A′))=g(B)−h(A′)=g(B)−(A′−X)=g(B)−A′+g(B)∩X=g(B)−A′On one hand, every element of A not contained in g(B)s in A′y construction, so A−A′⊆g(B) On the other, clearly g(B)⊆Aso g(B)−A′⊆A−A′QED.
#10:
We form two bijections using the sets from (9), one between A' and B', the other between A  A' and B  B'.
Any injection is a bijection between its domain and image. Since B′=f(A′)nd fs an injection, fs a bijection where we can assign each element b′∈B′o the a′∈A′uch that f(a′)=b′Similarly, gs a bijection between B−B′nd A−A′Combining them, we get a bijection on the full sets.
Q1
We wish to show that the terms of xn form a Cauchy sequence, which suffices to demonstrate they converge in a complete space. Take m,n∈N+, and WLOG m<n. Then we know from the definition of contraction that d(xm,xn)≤qm⋅d(x0,xn−m). This converges to 0 as m increases, so the sequence is Cauchy.
It's easy to see that this makes the rate of convergence between terms of the Cauchy sequence exponentially quick. Intuitively that seems like it ought to make the sequence converge to its limit with the same speed, but I don't think that can be made rigorous without more steps.
Q2
Take a sequence {xn=fn(x0)}. This converges to some L. Suppose L was not a fixed point. Then choose an ϵ=(L−f(L))/10 . A sequence {xn} which converges to a limit has, for every ϵ, some N such that ∀n>=N:xn−L<ϵ. Then we know that d(xN,L)<ϵ but d(f(xN),f(L))>ϵ , contradicting the contraction condition. So there is at least one fixed point, L.
Suppose there are two fixed points, f(x)=x, f(y)=y for distinct x and y. If so, d(f(x),f(y))=d(x,y), which again contradicts the contraction condition. So there is at most one fixed point.
Q3
Take as the space {n∈R+:n≥1}, with the usual metric. Define f(x)=x2+1x. This is a weak contraction (toward infinity) and has no fixed points within this space.
Ex1
Let x0:=x, let D:=d(x0,x1), and let k∈N. Then d(xk,xk+1)=d(f(xk−1,f(xk))≤qd(xk−1,xk)⋯≤qkD. For each ϵ∈R+, we find an n∈N such that qn<ϵD, then d(xn,xn+1)≤qnD<ϵ. This proves that (xk)k∈N is a Cauchysequence, which (because (X,d) is complete) means it converges to some point x∗∈X.
Furthermore, given a position n∈N, we have
d(xn,x∗)≤∞∑k=nd(xk,xk+1)≤∞∑k=nqkD=qnD∞∑k=0qk=qnD11−q=D⋅qn1−q<D⋅(q1−q)n.
Ex2
Given a any sequence (xk)k∈N in X, it converges to some point x∗, and it's easy to see that x∗ is a fixed point of f. Let y∗ be a fixed point of X. Then, d(f(x∗),f(y∗))=d(x∗,y∗), hence x=y. (Otherwise, this contradicts the fact that f is a contraction.)
Ex3
Choose X⊊R2 as X:=R×N+. Then X is complete because, given any Cauchy sequence (xk)k∈N, it's easy to prove that there is an n∈N such that all but finitely many xk are in the subspace R×{n}. However, the map f:X→X given by f(x,n):=f(x+1n,n+1) has no fixed points since it moves each point by at least 1. (And it's straightforward to verify that f is a weak contraction.)
#1
d(fn(x),fn+1(x))≤qnd(x,f(x))  can show by induction.
∀m>n,d(fn(x),fm(x))≤qn+qn+1+…+qm−1≤qn1−q→n→∞0
Therefore, fn(x) is a Cauchy sequence, and since (X, d) is complete, it must have a limit in X. Suppose y=limn→∞fn(x) . Then d(y,fn+1(x))≤qd(y,fn(x)) , therefore d(y,fn(x))≤qnd(x,y)
#2
Suppose y=limn→∞fn(x) . Let's show that y is a fixed point. Indeed, for any n, d(fn(x),f(y))≤qd(fn+1(x),y) , and if we take the limit in both sides, we get d(y,f(y))≤qd(y,y)=0 .
Let's show uniqueness: suppose x and y are fixed points, then d(x,y)=d(f(x),f(y))≤qd(x,y) , therefore d(x,y) = 0.
#3
X=[1,+∞) , f(x) = x + 1/x.
#4
Suppose f(x)=ϵx2+h(x) , where h is some convex function and ϵ<1/2. Take x,y∈Rn. Since h is convex on segment [x,y], its directional derivative is nondecreasing. Its directional derivative is a projection of gradient of g on the [x,y] line. Therefore, we have ⟨∇h(y),y−x⟩≥⟨∇h(x),y−x⟩ , or ⟨∇h(y)−∇h(x),y−x⟩≥0 . Hence,
x−∇f(x)−y+∇f(y)=x−2ϵx−∇h(x)−y−2ϵy+∇h(y)≤(1−2ϵ)x−y
Therefore, g is a contraction mapping, and from problem 1 it follows that the gradient descent converges exponentially quickly.
#5
Suppose A is an NxN positive matrix, and e is its minimal entry. (Then e < 1/N). Then we can write A = eJ + (1  Ne)Q, where J is a matrix whose entries are all 1, and Q is a matrix whose entries are all nonnegative and the sum of each column is 1 (because the sum of each column is 1 in A and Ne in J). Suppose x and y are probability distributions, i.e. Ndimensional vectors with nonnegative entries whose sum is 1. Then Ax−Ay1=eJ(x−y)+(1−Ne)Q(x−y)1=(1−Ne)Q(x−y)
Denote x+=max(x−y,0) , x−=min(x−y,0) (pointwise max/min). Then x−y=x+−x− ,Q(x−y)1=Qx+−Qx−1≤Qx+1+Qx−1=x+1+x−1=x−y1 ,
so Ax−Ay1≤(1−Ne)x−y1 . The space of all probability distributions with metric induced by .1 norm is a compact subset of Rn, so it is a complete metrics space, therefore, the sequence An(x) converges to a unique fixed point.
#6
Let us assume x≤f(x) (the proof for x≥f(x) is the same). Then, from monotonicity of f, x≤f(x)≤f(f(x))≤… is an ascending chain. This sequence cannot have more that P distinct elements, so an element of this sequence is going to repeat: fm(x)=fn(x),m<n,m<P . Then all the inequalities in fm(x)≤fm+1(x)≤…≤fn(x) must be equalities, so f(fm(x))=fm(x) , fm(x) is a fixed point.
(The second half made me realize how much more comfortable I am with abstract exercises than with regular Analysis à la Ex4.)
Ex6
If f(x)≤x, then all fn(x) are comparable to each other: we have
f(x)≤x⟹f(f(x))≤f(x)⟹f3(x)≤f2(x)
and so on. Furthermore, if n∈N is such that fn(x)≠fn−1(x), then fk(x)≠fk−1(x) for all k∈[n] as well (verify by looking at the contrapositive). Consequently (set n:=P), if fn(x) were not a fixed point of f, then fn+1(x)<fn(x), and hence {x,f(x),f2(x),...,fn+1(x)}⊆P, which means P would have P+2 elements.
If x≤f(x), we get x≤f(x)≤f2(x) and so on, leading to the same argument.
Ex7
Wlog, assume x≤f(x). Set f0(x):=x. Given any nonlimit ordinal β, we find a predecessor α and set fβ(x):=f(fα(x)). Given any limit ordinal ω, we set fω(x):=sup{fα(x)α∈ω}.
Suppose this doesn't define fα(x) for all ordinals α. Then, there is some smallest ordinal α∗ such that fα∗(x) is not defined. This immediately yields a contradiction (regardless of whether α∗ is a limit ordinal or not).
We want this construction to have the properties that f(x)=x⟹fβ(x)=x and that x≤y⟹fβ(x)≤fβ(y). Thus, let x,y∈L and β be an ordinal. If β has a predecessor, the check for both properties are easy. If not, then fβ(x)=sup{fα(x)α∈β} and fβ(y)=sup{fα(y)α∈β}. Then, for the first property, note that the upperbound of a oneelement set is just the element itself. For the second, note that each element in the first set is smaller than some element in the second set, so fβ(y) is an upperbound for the first set, which implies that fβ(x)≤fβ(y) since fβ(x) is the lowest upperbound.
Now, given an ordinal α, our construction defines a function ϕ:α→L. If f(fα(x))≠fα(x), then the chain doesn't become stationary at any earlier point either (to verify, take a smallest α such that [f(fα(x))≠fα(x) but the chain is stationary for smaller ordinals] and derive a contradiction), and hence ϕ is injective, proving that α≤L. (This is the generalized version of the argument from Ex6.)
Ex8
Let f:L→L be monotonic and let L′ be the set of fixed points of f. Then L′ inherits the partial order from L; what needs doing is verify the least upperbound property. So let X⊆L′. Then, X has a least upperbound u in L.
Let ω be some ordinal with ω>L. From the previous exercise, we know that f(fω(u))=fω(u). Choose the smallest α such that f(fα(u))=fα(u). Then, fα(u)∈L′ and x≤u≤fα(u), hence fα(u) is an upperbound of X.
It remains to show that it is the least upperbound. Thus, let u′∈L′ be another upperbound of x. Then, u≤u′ in L, hence fα(u)≤fα(u′)=u′ (apply Ex7).
Ex9
A least upperbound is obtained via ⋃ on all sets, and the greatest lowerbound via ⋂. (Easy checks.) Given any function f:A→B, we trivially have X⊆Y⟹f(X)⊆f(Y); injectivity is not needed.
We define
We need to verify that g(B−f(A′))=A−A′, then A′ and f(A′) are the desired sets.
x∈A(k−1)−A(k)=g(B−f(A(j)))⊆g(B−f(A′))
Ex10
Let A′ be the set constructed in Ex9. Then, we can define a bijection ϕ:A→B via
ϕ:x↦{f(x)x∈A′g−1(x)x∉A′
Ex5 (this is super ugly but I don't think it's worth polishing and it does work. All important ideas are in the first third of the proof, the rest just inelegantly resolves the details.)
We define our metric space as (X,d) where X:=[0,1]d is the set of probability distributions, and d(x,y)=∑dj=1xj−yj. Let x,y∈X and let Δ:=x−y, then d(Ax,Ay) can be computed as
d∑i=1(Ax−Ay)i=d∑i=1(AΔ)i=d∑i=1d∑k=1ak,iδk≤d∑i=1d∑k=1ai,kδk=d∑i=1δi
where the last step holds because multiplying a vector with the statetransition matrix leaves the sum of entries unchanged. (Reasonably easy to verify using that each column of A sums up to 1.)
If x≠y, then Δ has at least one negative entry and the inequality is strict. In that case, let k=argmaxi∈{1,...,d}δi and ℓ=argmini∈{1,...,d}δi. In particular, we have δk>0>δℓ. Note that, when two numbers a,b∈R have different sign, then a+b=a−b and thus a+b−a+b=min(2a,2b). Therefore, the amount that gets canceled out is at least
d∑i=1ai,kδk+ai,ℓδℓ−ai,kδk+ai,ℓδℓ=d∑i=12min(ai,kδk,ai,ℓδℓ)
Let a′ be the smallest entry in A, then we can lowerbound the above as
d∑i=1a′2min(δk,δℓ)=2a′dmin(δk,δℓ)
Wlog, let δk>δℓ. Let K be the sum of all postive entires of Δ, then ∑di=1δi=2K, so the term we want to lowerbound is 2a′dδℓ2K=a′dδℓK. The sum of the negative entries is −K, which means that the one with largest norm among them has norm at least 1d−1K. Thus, the relative decrease is at least a′dKd−1K=a′dd−1>a′
Then, d(x,y)−d(A(x),A(y))d(x,y)≥a′, hence d(A(x),A(y)d(x,y)≤1−a′. This proves that A is a contraction; apply Banach's theorem.