- Consider the metric space Cwith the norm≡ max
_{x∈[0,T] }_{∞}. Explain why the maximum exists. Show this is a complete metric space. Hint: If you havea Cauchy sequence in C, then for each x, you havea Cauchy sequence in ℝ^{n}. Recall from the last assignment that this is a complete space. Thus there exists f= lim_{m→∞}f_{m}. You must show that f is continuous. This was in the section on the Ascoli Arzela theorem in more generality if you need an outline of how this goes. Write down the details for this case. Note how f is in bold face. This means it is a function which has values in ℝ^{n}. f=. - For f ∈ C, you define the Riemann integral in the usual way using Riemann sums. Alternatively, you can define it as
Then show that the following limit exists in ℝ

^{n}for each t ∈.You should use the fundamental theorem of calculus from one variable calculus and the definition of the norm to verify this. As a review, in case we don’t get to it in time, for f defined on an interval

and s ∈,means that for all ε > 0, there exists δ > 0 such that if 0 <

< δ, then_{∞}< ε. - A collection of functions ℱ of Cis said to be uniformly equicontinuous if for every ε > 0 there exists δ > 0 such that if f ∈ℱ and< δ, then
_{∞}< ε. Thus the functions are uniformly continuous all at once. The single δ works for every pair t,s closer together than δ and for all functions f ∈ℱ. As an easy case, suppose there exists C such that for all f ∈ℱ,show that ℱ is uniformly equicontinuous. Now suppose G is a collection of functions of C

which is bounded. That is,= max_{t∈[0,T] }_{∞}< M < ∞ for all f ∈G. Then let ℱ denote the functions which are of the formwhere f ∈G. Show that ℱ is uniformly equicontinuous. Hint: This is a really easy problem if you do the right things. Here is the way you should proceed. Remember the triangle inequality from one variable calculus which said that for a < b

≤∫_{a}^{b}ds. ThenReduce to the case just considered using the assumption that these f are bounded.

- Suppose ℱ is a set of functions in Cwhich is uniformly bounded and uniformly equicontinuous as described above. Show it must be totally bounded. Hint: Let ε > 0 be given. Then let 0 = t
_{0}< t_{1}<< t_{p}= T where this is a uniform partition of the intervaland each t_{j}− t_{j−1}= δ∕2 where δ is such that if< δ, then_{∞}< ε∕8 for all f ∈ℱ. If ℱ is not totally bounded, then there exists ε and δ as just described, along with a sequencewhere≥ ε for all k≠m. Take a subsequence p times corresponding to the partition just described to obtain the existence of a subsequence_{m=1}^{∞}such thatis a Cauchy sequence for each of these t_{j}in the partition. Call itto save notation. Now exploit uniform equicontinuity to obtain a contradiction. It goes like this. Pick t. Then it is within δ of some t_{j}. ThenNow note that since there are finitely many t

_{j}, there is a single N such that k,m > N implies the middle term is less than ε∕8 for all t_{j}at once. Get a contradiction from the use of equicontinuity to contradict the assumption that≥ ε for all k≠m. - ↑If A ⊆is totally bounded, show that the closure of A is also totally bounded. In the above problem, explain why the closure of ℱ is compact. This uses the big theorem on compactness. Try and do this on your own, but if you get stuck, it is in the section on Ascoli Arzela theorem. When you have done this problem, you have proved the important part of the Ascoli Arzela theorem in the special case where the functions are defined on an interval. In the next assignment, if I remember, you will use this to prove one of the most important results in the theory of differential equations. This theorem is a really profound result because it gives compactness in a normed linear space which is not finite dimensional. Thus this is a non trivial generalization of the Heine Borel theorem.
- Let V be a vector space with basis . For v ∈ V, denote its coordinate vector as v =where v = ∑
_{k=1}^{n}α_{k}v_{k}. Now defineShow that this is a norm on V .

- Let be a normed linear space. You can let it beif you like. Recallis the usual magnitude of a vector given by
A set A is said to be convex if whenever x,y ∈ A the line segment determined by these points given by tx +

y for t ∈is also in A. Show that every open or closed ball is convex. Remember a closed ball is D≡while the open ball is B≡. This should work just as easily in any normed linear space with any norm. - A vector v is in the convex hull of a nonempty set S if there are finitely many vectors of
S,and nonnegative scalarssuch that
Such a linear combination is called a convex combination. Suppose now that S ⊆ V, a vector space of dimension n. Show that if v =∑

_{k=1}^{m}t_{k}v_{k}is a vector in the convex hull for m > n + 1, then there exist other nonnegative scalarssumming to 1 such thatThus every vector in the convex hull of S can be obtained as a convex combination of at most n + 1 points of S. This incredible result is in Rudin [29]. Convexity is more a geometric property than a topological property. Hint: Consider L : ℝ

^{m}→ V × ℝ defined byExplain why ker

≠. This will involve observing that ℝ^{m}has higher dimension that V × ℝ. Thus L cannot be one to one because one to one functions take linearly independent sets to linearly independent sets and you can’t have a linearly independent set with more than n + 1 vectors in V × ℝ. Next, letting a ∈ ker∖and λ ∈ ℝ, note that λa ∈ker. Thus for all λ ∈ ℝ,Now vary λ till some t

_{k}+ λa_{k}= 0 for some a_{k}≠0. You can assume each t_{k}> 0 since otherwise, there is nothing to show. This is a really nice result because it can be used to show that the convex hull of a compact set is also compact. You might try to show this if you feel like it. - Show that the usual norm in F
^{n}given bysatisfies the following identities, the first of them being the parallelogram identity and the second being the polarization identity.

^{n}. - Let K be a nonempty closed and convex set in an inner product space which is complete. For example, F
^{n}or any other finite dimensional inner product space. Let yK and letLet

be a minimizing sequence. That isExplain why such a minimizing sequence exists. Next explain the following using the parallelogram identity in the above problem as follows.

Hence

is a Cauchy sequence and converges to some x ∈ X. Explain why x ∈ K and= λ. Thus there exists a closest point in K to y. Next show that there is only one closest point. Hint: To do this, suppose there are two x_{1},x_{2}and considerusing the parallelogram law to show that this average works better than either of the two points which is a contradiction unless they are really the same point. This theorem is of enormous significance. - Let K be a closed convex nonempty set in a complete inner product space (Hilbert space) and let y ∈ H. Denote the closest point to y by Px. Show that Px is characterized as being the solution to the following variational inequality
for all z ∈ K. That is, show that x = Py if and only if Re

≤ 0 for all z ∈ K. Hint: Let x ∈ K. Then, due to convexity, a generic thing in K is of the form x + t,t ∈for every z ∈ K. ThenIf x = Px, then the minimum value of this on the left occurs when t = 0. Function defined on

has its minimum at t = 0. What does it say about the derivative of this function at t = 0? Next consider the case that for some x the inequality Re≤ 0. Explain why this shows x = Py. - Using Problem 11 and Problem 10 show the projection map, P onto a closed convex
subset is Lipschitz continuous with Lipschitz constant 1. That is
- Suppose, in an inner product space, you know Re. Show that you also know Im. That is, give a formula for Imin terms of Re. Hint:
Now consider matching real and imaginary parts.

- Suppose K is a compact subset of a metric space. Also let C be an open cover of K. Show that there exists δ > 0 such that for all x ∈ K, Bis contained in a single set of C. This number is called a Lebesgue number. Hint: For each x ∈ K, there exists Bsuch that this ball is contained in a set of C. Now consider the balls
_{x∈K}. Finitely many of these cover K._{i=1}^{n}Now consider what happens if you let δ ≤ min. Explain why this works. You might draw a picture to help get the idea. - Suppose C is a set of compact sets in a metric space and suppose that the intersection of every finite subset of C is nonempty. This is called the finite intersection property. Show that ∩C, the intersection of all sets of C is nonempty. This particular result is enormously important. Hint: You could let U denote the set. If ∩C is empty, then its complement is ∪U = X. Picking K ∈C, it follows that U is an open cover of K. Therefore, you would need to haveis a cover of K. In other words,
Now what does this say about the intersection of K with these K

_{i}? - If is a compact metric space and f : X → Y is continuous whereis another metric space, show that if f is continuous on X, then it is uniformly continuous. Recall that this means that if ε > 0 is given, then there exists δ > 0 such that if d< δ, then ρ< ε. Compare with the definition of continuity. Hint: If this is not so, then there exists ε > 0 and x
_{n},_{n}such that d< 1∕n but ρ≥ ε. Now use compactness to get a contradiction. - Prove the above problem using another approach. Use the existence of the Lebesgue
number in Problem 14 to prove continuity on a compact set K implies uniform
continuity on this set. Hint: Consider C≡. This is an open cover of X. Let δ be a Lebesgue number for this open cover. Suppose d< δ. Then both x,are in Band so both are in f
^{−1}. Hence ρ<and ρ<. Now consider the triangle inequality. - Let h > 0 be given and let f∈ ℝ
^{n}for each x ∈ ℝ^{n}. Also let→ fbe continuous andLet x

_{h}be a solution to the followingwhere x

_{h}≡ x_{0}if s−h ≤ 0. Explain why there exists a solution. Hint: Consider the intervals,and so forth. Next explain why these functions_{h>0}are equicontinuous and uniformly bounded. Now use the result of Problem 5 to argue that there exists a subsequence, still denoted by x_{h}such thatin C

as discussed in Problem 4. Use what you learned about the Riemann integral in single variable advanced calculus to explain why you can pass to a limit and conclude thatHint:

^{′}= f, x= x_{0}. When you have done this, you will have proved the celebrated Peano existence theorem from ordinary differential equations.