- If you have two norms on F
^{n}or more generally on X a finite dimensional vector space (You can’t escape these. They are encountered as subspaces of F^{n}for example.) and these are equivalent norms in the sense that there are scalars δ,Δ not depending on x such that for all x,(2.20) then the open sets defined in terms of these norms are the same.

- Explain carefully why in ℝ
^{n}, - Explain carefully why if x is a limit point of a set A with respect to then it is also a limit point with respect to
_{1}if the two norms are equivalent as in 2.20. Also show that if x_{n}→ x with respect to, then the same is true with respect to_{1}an equivalent norm satisfying 2.20. - A vector space X with field of scalars either ℝ or ℂ is called a normed linear space if it has a norm
. That is, the following assumptions are satisfied.
(2.21) (2.22) (2.23) A normed linear space is called a Banach space if, in addition to having a norm, it is complete. This means that if

is a Cauchy sequence, then it must converge. Recall that a sequence is a Cauchy sequence if for every ε > 0 there exists N such that if n,m ≥ N, thenThe space of bounded linear transformations ℒ

which are linear transformations mapping X to Y consists of the functions L : X → Y such that L is linear,and

Show that

, called the operator norm is a norm on ℒand thatAlso show that if you have L ∈ℒ

and M ∈ℒfor X,Y,Z normed linear spaces, then - If you have X a normed linear space and Y is a Banach space, show that ℒis a Banach space with respect to the operator norm.
- If X,Y are normed linear spaces, verify that A : X → Y is in ℒif and only if A is continuous at each x ∈ X if and only if A is continuous at 0.
- Verify that Theorem 2.1.17 holds exactly the same if you are in a normed linear space. That is the dist function is still Lipschitz continuous.
- Suppose X is a Banach space and is a sequence of closed sets in X such that B
_{n}⊇ B_{n+1}for all n and no B_{n}is empty. Also suppose that the diameter of B_{n}converges to 0. Recall the diameter is given byThus these sets B

_{n}are nested and diam→ 0. Verify that there is a unique point in the intersection of all these sets. - If X is a Banach space, and Y is the span of finitely many vectors in X, show that Y is closed.
- If X is an infinite dimensional Banach space, show that there exists a sequence
_{n=1}^{∞}such that≤ 1 but for any m≠n,≥ 1∕4. Thus in infinite dimensional Banach spaces, closed and bounded sets are no longer compact as they are in F^{n}. - Generalize Lemma 2.2.8 and Lemma 2.2.9 to the case of a Banach space. That is, show that compact and sequentially compact are the same for a Banach space. Also show that Theorem 2.2.25 holds with no change for a compact subset of a Banach space.
- In the proof of the fundamental theorem of algebra, explain why there exists z
_{0}such that for pa polynomial with complex coefficients, - Explain why a compact set in ℝ has a largest point and a smallest point. Now if f : K → ℝ for K
compact and f continuous, give another proof of the extreme value theorem from using that fis compact.
- A function f : X → ℝ for X a normed linear space is lower semicontinuous if, whenever
x
_{n}→ x,It is upper semicontinuous if, whenever x

_{n}→ x,Explain why, if K is compact and f is upper semicontinuous then f achieves its maximum and if K is compact and f is lower semicontinuous, then f achieves its minimum on K.

- Suppose f
_{n}: S → Y where S is a nonempty subset of X a normed linear space and suppose that Y is a Banach space (complete normed linear space). Generalize the theorem in the chapter to this case: Let f_{n}: S → Y be bounded functions:sup_{x∈S}= C_{n}< ∞. Then there exists bounded f : S → Y such that lim_{n→∞}= 0 if and only ifis uniformly Cauchy. Also show that BCis a Banach space. - Show that no interval ⊆ ℝ can be countable. Hint: First showis not countable. You might do this by noting that every point in this interval can be written as ∑
_{k=1}^{∞}2^{−k}a_{k}where a_{k}is either 0 or 1. Let ℱ be ∪_{n}P. Explain why ℱ is countable. Then let S≡P∖ℱ. Explain why S is uncountable. Let C be all points of the form ∑_{k=1}^{m}2^{−k}a_{k}where a_{k}is 0 or 1. Explain why C is countable. Let J =∖ C. Now let θ : S → J be given by θ= ∑_{k∈S}2^{−k}. Explain why θ is one to one onto J. Ifis countable, show there are onto mappings as indicatedshowing that S is countable after all.

- Using the above problem as needed, let B be a countable set of real numbers. Say B =
_{n=1}^{∞}. LetLet g

≡∑_{k=1}^{∞}2^{−k}f_{k}. Explain why g is continuous on ℝ ∖ B and discontinuous on B. Note that B could be the rational numbers. - Let
For f a continuous function defined on

, extend it to have f= ffor x > 1 and f= ffor x < −1. ConsiderThis involves elementary calculus and change of variables. Show that p

_{n}is a polynomial and that p_{n}converges uniformly to f on. This is the way Weierstrass originally proved the famous approximation theorem. - In fact the Bernstein polynomials apply for, f having values in a normed linear space and a similar result will hold. Give such a generalization.
- Consider the following
(2.24) where f :

^{n}→ X a normed linear space. Show→ 0 as min→∞. Hint: Do consider Lemma 2.8.1 first and you may see how to do this. - Theorem 2.2.41 gave an example of a function which is everywhere continuous and nowhere
differentiable. The first examples of this sort were given by Weierstrass in 1872 who gave an
example involving an infinite series in which each term had all derivatives everywere
and yet the uniform limit had no derivative anywhere. Using the example of Theorem
2.2.41, give an example of an infinite series of functions, each term being a polynomial
defined on , ∑
_{k=1}^{∞}p_{k}= ffor which it makes absolutely no sense to write f^{′}= ∑_{k=1}^{∞}p_{k}^{′}because f^{′}fails to exist at any point. In other words, you cannot differentiate an infinite series term by term. The derivative of a sum is not the sum of the derivatives when dealing with an infinite “sum”. Also show that if you have any differentiable function g and ε > 0, there exists a nowhere differentiable function h such that< ε. This is in stark contrast with what will be presented in complex analysis in which, thanks to the Cauchy integral formula, uniform convergence of differentiable functions does lead to a differentiable function. Hint: Use Weierstrass approximation theorem and telescoping series to get the example of a series which can’t be differentiated term by term. - Consider ℝ ∖. Show this is not connected.
- Show S ≡∪is connected but not arcwise connected.
- Let A be an m × n matrix. Then A
^{∗}, called the adjoint matrix, is obtained from A by taking the transpose and then the conjugate. For example,Formally,

_{ij}= A_{ji}. Show that=and=. The inner product is described in the chapter. Recall≡∑_{j}x_{j}y_{j}. - Let X be a subspace of F
^{m}having dimension d and let y ∈ F^{m}. Show that x ∈ X is closest to y in the Euclidean normout of all vectors in X if and only if= 0 for all u ∈ X. Next show there exists such a closest point and it equals ∑_{j=1}^{d}u_{j}for_{j=1}^{d}an orthonormal basis for X. - Let A : F
^{n}→ F^{m}be an m × n matrix. (Note how it is being considered as a linear transformation.) Show Im≡is a subspace of F^{m}. If y ∈ F^{m}is given, show that there exists x such that y −Ax is as small as possible (Ax is the point of Imclosest to y) and it is a solution to the least squares equationHint: You might want to use Problem 24.

- 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}. By definition, an inner product space is just a vector space which has an inner product. - 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 29 and Problem 28 show the projection map, P onto a closed convex subset is
Lipschitz continuous with Lipschitz constant 1. That is
- Let A : ℝ
^{n}→ ℝ^{n}be continuous and let f ∈ ℝ^{n}. Also letdenote the standard inner product in ℝ^{n}. Letting K be a closed and bounded and convex set, show that there exists x ∈ K such that for all y ∈ K,Hint: Show that this is the same as saying P

= x for some x ∈ K where here P is the projection map discussed above. Now use the Brouwer fixed point theorem. This little observation is called Browder’s lemma. It is a fundamental result in nonlinear analysis. - ↑In the above problem, suppose that you have a coercivity result which is
Show that if you have this, then you don’t need to assume the convex closed set is bounded. In case K = ℝ

^{n}, and this coercivity holds, show that A maps onto ℝ^{n}.

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