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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I recommend putting all the object level points in spoilers and including metadata outside of the spoilers, like so: "I think I've solved problem #5, here's my solution <spoilers>" or "I'd like help with problem #3, here's what I understand <spoilers>" so that people can choose what to read.
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'.