- Show the Brouwer fixed point theorem is equivalent to the nonexistence of a
continuous retraction onto the boundary of B.
- Using the Jordan separation theorem, prove the invariance of domain theorem
n ≥ 2. Thus an open ball goes to some open. Hint: You might consider
Band show f maps the inside to one of two components of ℝ
^{n}∖ f. etc. - Give a version of Proposition 21.6.5 which is valid for the case where n = 1.
- It was shown that if f is locally one to one and continuous, f : ℝ
^{n}→ ℝ^{n}, andthen f maps ℝ

^{n}onto ℝ^{n}. Suppose you have f : ℝ^{m}→ ℝ^{n}where f is one to one and lim_{}→∞= ∞. Show that f cannot be onto. - Can there exists a one to one onto continuous map, f which takes the unit interval
to the unit disk B? Hint: Think in terms of invariance of domain.
- Let m < n and let B
_{m}be the ball in ℝ^{m}and B_{n}be the ball in ℝ^{n}. Show that there is no one to one continuous map from B_{m}to B_{n}. Hint: It is like the above problem. - Consider the unit disk,
and the annulus

Is it possible there exists a one to one onto continuous map f such that f

= A? Thus D has no holes and A is really like D but with one hole punched out. Can you generalize to different numbers of holes? Hint: Consider the invariance of domain theorem. The interior of D would need to be mapped to the interior of A. Where do the points of the boundary of A come from? Consider Theorem 6.9.3. - Suppose C is a compact set in ℝ
^{n}which has empty interior and f : C → Γ ⊆ ℝ^{n}is one to one onto and continuous with continuous inverse. Could Γ have nonempty interior? Show also that if f is one to one and onto Γ then if it is continuous, so is f^{−1}. - Let K be a nonempty closed and convex subset of ℝ
^{n}. Recall K is convex means that if x,y ∈ K, then for all t ∈,tx +y ∈ K. Show that if x ∈ ℝ^{n}there exists a unique z ∈ K such thatThis z will be denoted as Px. Hint: First note you do not know K is compact. Establish the parallelogram identity if you have not already done so,

Then let

be a minimizing sequence,Now using convexity, explain why

and then use this to argue

is a Cauchy sequence. Then if z_{i}works for i = 1,2, consider∕2 to get a contradiction. - In Problem 9 show that Px satisfies the following variational inequality.
for all y ∈ K. Then show that

≤. Hint: For the first part note that if y ∈ K, the function t →^{2}achieves its minimum onat t = 0. For the second part,Explain why

and then use a some manipulations and the Cauchy Schwarz inequality to get the desired inequality. Thus P is called a retraction onto K.

- Establish the Brouwer fixed point theorem for any convex compact set in ℝ
^{n}. Hint: If K is a compact and convex set, let R be large enough that the closed ball, D⊇ K. Let P be the projection onto K as in Problem 10 above. If f is a continuous map from K to K, consider f∘P. You want to show f has a fixed point in K. - Suppose D is a set which is homeomorphic to B. This means there exists a continuous one to one map, h such that h= D such that h
^{−1}is also one to one. Show that if f is a continuous function which maps D to D then f has a fixed point. Now show that it suffices to say that h is one to one and continuous. In this case the continuity of h^{−1}is automatic. Sets which have the property that continuous functions taking the set to itself have at least one fixed point are said to have the fixed point property. Work Problem 7 using this notion of fixed point property. What about a solid ball and a donut? Could these be homeomorphic? - Suppose Ω is any open bounded subset of ℝ
^{n}which contains 0 and that f : Ω → ℝ^{n}is continuous with the property thatfor all x ∈ ∂Ω. Show that then there exists x ∈Ω such that f

= 0. Give a similar result in the case where the above inequality is replaced with ≤. Hint: You might consider the function - Suppose Ω is an open set in ℝ
^{n}containing 0 and suppose that f : Ω → ℝ^{n}is continuous and≤for all x ∈ ∂Ω. Show f has a fixed point in Ω. Hint: Consider h≡ t+x for t ∈. If t = 1 and some x ∈ ∂Ω is sent to 0, then you are done. Suppose therefore, that no fixed point exists on ∂Ω. Consider t < 1 and use the given inequality. - Let Ω be an open bounded subset of ℝ
^{n}and let f,g : Ω → ℝ^{n}both be continuous such thatfor all x ∈ ∂Ω. Show that then

Show that if there exists x ∈ f

^{−1}, then there exists x ∈^{−1}. Hint: You might consider h≡f+ tand argue 0hfor t ∈. - Suppose f : ℝ
^{n}→ ℝ^{n}is continuous and satisfiesShow that f must map ℝ

^{n}onto ℝ^{n}. Hint: First show f is one to one. Then use invariance of domain. Next show, using the inequality, that the points not in fmust form an open set because if y is such a point, then there can be no sequenceconverging to it. Finally recall that ℝ^{n}is connected.It is obvious that f is one to one. This follows from the inequality. If U are the points not in the image of f, then U must be open because if not, then for y one of these points, there would be a sequence f

→ y. Then by the inequality,is a Cauchy sequence and so it converges to x. Thus f= lim_{n→∞}f= y. Now by invariance of domain, fis open. However, ℝ^{n}is connected and so in fact, U is empty. - Let f : ℂ → ℂ where ℂ is the field of complex numbers. Thus f has a real and
imaginary part. Letting z = x + iy,
Recall that the norm in ℂ is given by

=and this is the usual norm in ℝ^{2}for the ordered pair. Thus complex valued functions defined on ℂ can be considered as ℝ^{2}valued functions defined on some subset of ℝ^{2}. Such a complex function is said to be analytic if the usual definition holds. That isIn other words,

(21.10.12) at a point z where the derivative exists. Let f

= z^{n}where n is a positive integer. Thus z^{n}= p+ iqfor p,q suitable polynomials in x and y. Show this function is analytic. Next show that for an analytic function and u and v the real and imaginary parts, the Cauchy Riemann equations hold.In terms of mappings show 21.10.12 has the form

^{T}and h is given by h_{1}+ ih_{2}. Thus the determinant of the above matrix is always nonnegative. Letting B_{r}denote the ball B= Bshowwhere f

= z^{n}. In terms of mappings on ℝ^{2},Thus show

Hint: You might consider

where the a

_{j}are small real distinct numbers and argue that both this function and f are analytic but that 0 is a regular value for g although it is not so for f. However, for each a_{j}small but distinct d= d. - Using Problem 17, prove the fundamental theorem of algebra as follows. Let pbe a nonconstant polynomial of degree n,
Show that for large enough r,

>for all z ∈ ∂B. Now from Problem 15 you can conclude d= d= n where f= a_{n}z^{n}. - The proof of Sard’s lemma made use of the hard Vitali covering theorem. Here is
another way to do something similar. Let U be a bounded open set and let
f : U → ℝ
^{n}be in C^{1}. Let S denote the set of x ∈ U such that Dfhas rank less than n. Thus it is a closed set. Let U_{m}=, a closed set. It suffices to show that for S_{m}≡ U_{m}∩ S, fhas measure zero because f= ∪_{m}fthese sets increasing in m. By definition of differentiability,for each x ∈ U. Explain why the above function of x is measurable. Now by Eggoroff’s theorem, there is measurable set A of measure less than

such that off A, the convergence is uniform. Let C_{k}be a countable union of non overlapping half open rectangles one of which is of the form ∏_{i=1}^{n}(a_{i},b_{i}] such that each has diameter less than 2^{−k}. Consider the half open rectangles which have nonempty intersection with S_{m}∖ A,ℐ_{k}. Then repeat the argument given in the first section of this chapter. Show that for k large enough, the rank condition and uniform convergence above implies that m_{n}is less than ε. Now show that fis contained in a set of measure no more than m^{n}10^{n}= 2 ε. Thus fhas measure no more than 3 ε. Since ε is arbitrary, this establishes the desired conclusion. - Let X be a Banach space and let Ω be a symmetric and bounded open set. Let
F : Ω → X be odd and compact 0. Show using Corollary 21.9.3 that Dis an odd integer.
- Let F be compact. Suppose I − F is one to one on B. Then using similar reasoning to the finite dimensional case, show that there is a δ > 0 such that
- Let F be compact. Suppose I −F is locally one to one on an open set Ω. Show that
maps open sets to open sets. This is a version of invariance of domain.
- Suppose is locally one to one and F is compact. Suppose also that lim
_{∥x∥ →∞}= ∞. Show thatis onto. - As a variation of the above problem, suppose F : X → X is compact
and
Then I − F is onto. Note that I − F is not one to one.

- Suppose F is compact and ≥ α. Show that thenis onto.
- The Jordan curve theorem is: Let C denote the unit circle,
Suppose γ : C → Γ ⊆ ℝ

^{2}is one to one onto and continuous. Then ℝ^{2}∖ Γ consists of two components, a bounded component (called the inside) U_{1}and an unbounded component (called the outside), U_{2}. Also the boundary of each of these two components of ℝ^{2}∖ Γ is Γ and Γ has empty interior. Using the Jordan separation theorem, prove this important result. - This problem is from [?] Recall Theorem 21.9.11. It allowed you to say that
D= 0 provided 0Fand λx≠Fx for all x ∈ ∂Ω,λ ∈. This was for F compact and defined on an infinite dimensional space X. Suppose now that F is compact and F : → X where 0 ∈ Ω an open set in X. Suppose also that F= 0 and that
Show that there is a sequence α

_{n}→ 0 each α_{n}≠0, and for some x_{n}≠0,x_{n}− α_{n}F= 0 . Note that when α = 0, there is only one solution to= 0 , but this says that there are many small α_{n}≠0 for which there is a nonzero solution to= 0 . That is there exist arbitrarily small α_{n}such thatF= 0 . This says that 0 is a bifurcation point for I − αF. Hint: Let α_{n}↓ 0 and pick r_{n}such that for all= r_{n},Thus 0

α_{n}Fand also α_{n}F≠λx for all x ∈ ∂B. Use the theorem to conclude thatand then consider the homotopy I − α

_{n}tF. If it sends no point of ∂Bto 0 then you would have

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