Math 140B – Homework 2. Due Friday, April 14.

Problem 1. (Mean value property. Friday, April 7.) Solve Exercise 4 and 5, Rudin, Chapter 5.

Problem 2. (Mean value property. Friday, April 7.) A f : R → R is said to be Lipschitz if there exists a constant M > 0 such that for all x, y ∈ R we have |f(x) − f(y)| ≤ M|x − y|.

(i) Show that any Lipschitz function is uniformly continuous over R. (ii) Show that if f is differentiable and f 0 is bounded, then f is Lipschitz. In partic- ular, conclude that f(x) = sin x is uniformly continuous over R. (iii) Give an example of a differentiable function which is not Lipschitz. You may try f(x) = x2. (iv) Give an example of a function which is Lipschitz but not differentiable. You may try f(x) = |x|. Problem 3. (Mean value property. Friday, April 7.) For any

f :[−1, 1] → R such that f is three times differentiable on (−1, 1), and f(−1) = 0, f(0) = 0, f(1) = 1, f 0(0) = 0, 000 there must exist x0 ∈ (−1, 1) such that f (x0) = 3. This is a modification of Rudin, Problem 17, Chapter 5.

1 3 2 Hint: You may want to consider the auxiliary function g(x) = f(x) − 2 (x + x ). Compute g(−1), g(0), g(1), g0(0). Then apply the mean value property a number of times to the function g and conclude.

Problem 4. (Rolle’s theorem for vector-valued functions. Friday, April 7.)

Let ~ k ~ f : I → R , f = (f1, . . . , fk) be a vector-valued function defined over an interval I. We say that f~ is differentiable if each fi is differentiable and we set ~0 0 0 f (x) = (f1(x), . . . , fk(x)).

2 Show that Rolle’s theorem fails for vector-valued functions. Specifically, let f~ : R → R be the differentiable function f~(x) = (cos x, sin x). Show that f~(0) = f~(2π), but that there is no c ∈ (0, 2π) such that f~0(c) = 0. 1 2

Remark: The can be salvaged if we replace equality by inequality. In other words, there exists c ∈ (a, b) such that

||f~(b) − f~(a)|| ≤ ||f~0(c)||(b − a).

As usual, ||~x|| denotes the length of a vector. This is proven in Rudin, Theorem 5.19, and you can read the proof if you wish.

Problem 5. (L’Hopital rule. Monday, April 10.) Solve Exercises 9 and 11, Rudin, Chapter 5.

Problem 6. (Taylor’s theorem. Wednesday, April 12.) Use Taylor’s theorem for the π function f(x) = cos x to prove that for all x ∈ [0, 2 ] we have x2 x2 x4 1 − ≤ cos x ≤ 1 − + . 2! 2! 4!

Extra credit (not required). (Taylor . Real analytic functions. Wednesday, April 12.) Let f : R → R be infinitely many times differentiable. We say the function f is of class Cω (or that f is a real analytic function) if for all a ∈ R, the function f equals its Taylor expansion f 00(a) f (n)(a) f(x) = f(a) + f 0(a)(x − a) + (x − a)2 + ... + (x − a)n + .... 2! n! for all x in a sufficiently small neighborhood of a. In , we sometimes incorrectly learn that every function which is infinitely many times differentiable equals its , but this is not the case as we will show below. Consider the function ( exp 1  if x < 0 f(x) = x . 0 if x ≥ 0 First calculate the first of f: (i.1) Show that f 0(0) = 0, directly from the definition. (i.2) Write down a formula for f 0(x) which is valid for all values of x. Now calculate the . (ii.1) Consider the formula you found in (i.2). First, from this formula, show that f 00(0) = 0 by using the definition of the derivative. (ii.2) Next, differentiate f 0 and write down a formula for f 00(x) which is valid for all x. Now onto higher order . 3

(iii) Do the same for third derivatives. You should be able to prove first that f 000(0) = 0 from the definition. Write down an expression for the third derivative. Do you notice any patterns in the general formula? (iv) Continuing in this fashion, show that f (n)(0) = 0 for all n. You will need to use induction on n and formulate a hypothesis about the form of f (n)(x) for all n, which you will then need to prove by induction. (v) Explain that (iv) implies that f is not real analytic i.e. it does not equal its Taylor series at a = 0 in any neighborhood of a.