- 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}so it converges by completeness of ℝ^{n}. See Example 11.1.23. Thus there exists f≡ lim_{m→∞}f_{m}. You must show that f is continuous. Consider_{n}. Finally,and since a Cauchy sequence,

< ε whenever m > n for n large enough. Use to show that_{∞}→ 0. - 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. Recall that

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

< δ, then_{∞}< ε. You have to use the definition of a limit in order to establish that something is a limit. - 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 K 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.

- 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. 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.
- 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. Show this next. This is also Problem 22 but here it is again. This is because it is a really nice result. - 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. 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 = Py, 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 9 and Problem 8 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 (If C is a set of open sets whose union contains K,(open cover) then
there are finitely many sets of C whose union contains K.) 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 (A set is compact if every open cover admits a finite subcover.) 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}? - Show that if f is continuous and defined on a compact set K in a metric space, then it is uniformly
continuous. Continuous means continuous at every point. Uniformly continuous means: For every
ε > 0 there exists δ > 0 such that if d< δ, then d< ε. The difference is that δ does not depend on x. Hint: Use the existence of the Lebesgue number in Problem 12 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. Recall the usual definition of continuity. In metric space it is as follows: For,metric spaces, f : D → Y is continuous at x ∈ D means that for all ε > 0 there exists δ > 0 such that if d< δ, then ρ< ε. Continuity on D means continuity at every point of D. - The definition of compactness is that a set K is compact if and only if every open cover (collection of open sets whose union contains K) has a finite subset which is also an open cover. Show that this is equivalent to saying that every open cover consisting of balls has a finite subset which is also an open cover.
- A set K in a metric space is said to be sequentially compact if whenever is a sequence in K, there exists a subsequence which converges to a point of K. Show that if K is compact, then it is sequentially compact. Hint: Explain why if x ∈ K, then there exist an open set B
_{x}containing x which has x_{k}for only finitely many values of k. Then use compactness. This was shown in the chapter, but do your own proof of this part of it. - Show that f : D → Y is continuous at x ∈ D where ,are metric spaces if and only if whenever x
_{n}→ x in D, it follows that f→ f. Recall the usual definition of continuity. f is continuous at x means that for all ε > 0 there exists δ > 0 such that if d< δ, then ρ< ε. Continuity on D means continuity at every point of D. This is in the chapter, but go through the proof and write it down in your own words. - Give an easier proof of the result of Problem 14. Hint: If f is not uniformly continuous, then there
exists ε > 0 and x
_{n},y_{n},d<but d≥ ε. Now use sequential compactness of K to get a contradiction. - This problem will reveal the best kept secret in undergraduate mathematics, the definition of the
derivative of a function of n variables. Let
_{V }be a norm on V and also denote by_{W}a norm on W. Writefor both to save notation. Let U ⊆ V be an open set. Let f : UW be a function having values in W. Then f is differentiable at x ∈ U means that there exists A ∈ℒsuch that for every ε > 0, there exists a δ > 0 such that whenever 0 << δ, it follows thatStated more simply,

Show that A is unique. It is written as Df

= A. This is what is meant by the derivative of f. If V = ℝ^{n}, and W = ℝ^{m}, show that with respect to the usual bases, the matrix of Dfis an m×n matrix whose k^{th}column is. - Let V,W be finite dimensional normed linear spaces and let
_{n=1}^{∞}be a sequence of linear transformations in ℒsuch that sup_{n}< ∞. Show that there exists a subsequenceand A ∈ℒsuch that lim_{n→∞}= 0. - Given an example of a sequence ⊆ℒsuch that the minimum polynomial of each A
_{k}has degree n = dimbut→ 0 and the minimum polynomial of A has degree less than n. Hint: You might want to think in terms of the Jordan form.

Download PDFView PDF