Math LM (24543) Lectures 02

Transcription

1 Math LM (24543) Lectures 02 Ethan Akin Office: NAC 6/287 Phone: Spring, 2018

2 Contents Continuity, Ross Chapter 3 Uniform Continuity and Compactness Connectedness Real-Valued Functions Sequences and Series of Functions, Ross Chapter 4 Power Series Differentiation, Ross Chapter 5 Mean Value and Inverse Function Theorems Taylor s Theorem Integration, Ross Chapter 6 Integral Properties Fundamental Theorems of Calculus Exponentials and Logarithms

3 Limits and Continuity for Metric Space Functions, Ross Secs. 20, 21 Let (S, d) be a metric space and E S. Remember that a point e S lies in the closure E when there is a sequence {x n } in E with limit e. Equivalently, every open ball centered at e intersects E. Recall that B ɛ (e) = {x S : d(x, e) < ɛ} is the open ball centered at e with radius ɛ > 0. A function f with domain E to a metric space (S, d ) associates to each x E a unique point f (x), the value of f at x, in S. We write this as f : E S. It is important to keep track of the domain. If E 1 E, we can restrict f to E 1, defining f : E 1 S.

4 Theorem: Let f : E S be a function with domain E S mapping to S. Let e E and e S. The following conditions are equivalent and when they hold we say that e is the limit of f (x) as x approaches e, written lim x e f (x) = e. (i) For every sequence {x n } in E with limit e, the sequence {f (x n )} in S has limit e. That is, for {x n E}, x n e implies f (x n ) e. (ii) ( ɛ > 0)( δ > 0)( x E B δ (e)) f (x) B ɛ (e ).

5 Proof: (ii) implies (i): Given a sequence {x n E} with x n e we must show f (x n ) e. That is, given ɛ > 0 we must find an N so that n N implies f (x n ) B ɛ (e ). From (ii) we can choose δ > 0 so that if x E B δ (e), then f (x) B ɛ (e ). Because x n e and δ > 0 there exists N so that n N implies x n B δ (e). But δ has been chosen so that this implies f (x n ) B ɛ (e ). That is, from ɛ we found δ which relates E to S. Then we found the N which works for δ, {x n } and e.

6 For the reverse, we prove the contrapositive: (ii) implies (i) and so we assume ( ɛ > 0)( δ > 0)( x E B δ (e)) f (x) B ɛ (e ). We construct a bad sequence by using δ = 1, 1/2,.... For δ = 1/n there exists x n E B 1/n (e) such that f (x n ) B ɛ (e ). That is, {x n } is a sequence in E with d(x n, e) < 1/n, and so x n e, but d(f (x n ), e ) ɛ > 0 and so {f (x n )} does not have limit e.

7 We can state condition (ii) as: For every open ball B centered at e, there exists an open ball B centered at e such that f (B) B, i.e. f (x) B for every x B. This is useful, because it allows us to extend the result to the case where e or e is ± for S = R. For M R, we refer to the interval (M, + ] as an open ball centered at + and [, M) as an open ball centered at. Our previous definition of s n ± for a sequence {s n } in R simply says that the sequence is eventually in any open ball centered at ±.

8 If e E then the constant sequence x n = e is a sequence in E converging to e and so f (x n ) = f (e) is the constant sequence in S converging to f (e). That is, if e E and lim x e f (x) = e then e = f (e). In that case, we say that the function f is continuous at e. We say that f : E S is a continuous function, or just f is continuous, when it is continuous at every point of E. So f is continuous on E if for every sequence {x n } in E with a limit in E, we have lim n f (x n ) = f (lim n x n ). Or equivalently, ( x E)( ɛ > 0)( δ > 0)( x 1 E B δ (x)) f (x 1 ) B ɛ (f (x)).

9 For E 1 S 1, E 2 S 2 and functions f : E 1 S 2, g : E 2 S 3, assume that f (E 1 ) E 2. That is, if x E 1 then f (x) E 2. We can then define the composition g f : E 1 S 3 by (g f )(x) = g(f (x)). Theorem: For e E 1, if f is continuous at e and g is continuous at f (e) then g f is continuous at e. In particular, if f and g are continuous, then the composition g f is continuous. Proof: If {x n } is a sequence in E 1 with limit e, then by continuity of f at e, {f (x n )} is a sequence in E 2 with limit f (e). By continuity of g at f (e), {g(f (x n ))} is a sequence in S 3 with limit g(f (e)).

10 Since (g f )(x n ) = g(f (x n )) and (g f )(e) = g(f (e)), it follows that g f is continuous at e. Thus, the composition of continuous functions is continuous. As we saw above, f : E S is continuous when ( x E)( ɛ > 0)( δ > 0)( x 1 E B δ (x)) f (x 1 ) B ɛ (f (x)). In general, the choice of δ will depend on the point x as well as on ɛ. However, we call f : E S uniformly continuous when ( ɛ > 0)( δ > 0)( x E)( x 1 E B δ (x)) f (x 1 ) B ɛ (f (x)). Now the δ, which still depends on ɛ, can be chosen, uniformly, to work for all x.

11 For f : S S, A S and B S, we define the image of A: f (A) = {f (x) : x A} S and the pre-image of B: f 1 (B) = {x : f (x) B} S. Exercise A3.1: Assume f : S 1 S 2 is a function. (a) If B 1, B 2 S 2 show that f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ) and f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ). (b) If A 1, A 2 S 1 show that f (A 1 A 2 ) = f (A 1 ) f (A 2 ) and f (A 1 A 2 ) f (A 1 ) f (A 2 ). Give an example of a function f and subsets A 1, A 2 for which the latter inclusion is proper, i.e. is not an equality. (c) If A S 1, B S 2, show that f (A) B iff A f 1 (B).

12 Ross Theorem 21.3: A function f : S S is continuous iff f 1 (U) is open in S whenever U is open in S. A function f : S S is continuous iff f 1 (F ) is closed in S whenever F is open in S. Proof: Assume f is continuous, U is open in S and x f 1 (U), so that f (x) U. Because U is open, there exists ɛ > 0 such that B ɛ (f (x)) U. Because f is continuous at x, there exists δ > 0 such that if x 1 B δ (x), f (x 1 ) B ɛ (x) and so is in U. That is, B δ (x) f 1 (U), an so x is an interior point of f 1 (U). As this is true for every x f 1 (U), it follows that f 1 (U) is open. On the other hand, for ɛ > 0 the ball U = B ɛ (f (x)) is an open set with x f 1 (U). So if f 1 (U) is open, then there exists δ > 0 such that B δ (x) f 1 (U). This says that x 1 B δ (x) implies f (x 1 ) B ɛ (f (x)). Thus, f is continuous at x.

13 Since f 1 (S \ F ) = S \ f 1 (F ), the result for closed sets follows from that from open sets. Ross Theorem 21.4 (i): If a function f : S S is continuous and E S is compact, then f (E) S is compact. Proof: We will prove this two ways. Bolzano-Weierstrass: Let {y n } be a sequence in f (E). Choose x n E with f (x n ) = y n to define a sequence {x n } in E. By the Bolzano-Weierstrass property for E, there is a subsequence {x nk } which converges to a point x E. By continuity of f, {f (x nk ) = y nk } converges to f (x) f (E). That is, {y n } has a subsequence which converges in f (E).

14 Heine-Borel: Given U a family of open subsets of S which covers f (E), let f U = {f 1 (U) : U U}. By continuity this is a family of open subsets of S. If x E, then f (x) f (E) and so f (x) U for some U U. So x f 1 (U). This shows that f U is an open cover of E. By the Heine-Borel property for E, there is a finite subcover of E. That is, there exist U 1,..., U k U so that E f 1 (U 1 ) f 1 (U k ). Then for x E, x f 1 (U i ) for some i = 1,..., k and so f (x) U i. That is, {U 1,..., U k } is a finite subcover of f (E).

15 Recall that a function f : S S is one-to-one when f (x 1 ) = f (x 2 ) only when x 1 = x 2. It is onto when f (S) = S. That is, for every y S there exists x S such that f (x) = y. When f : S S is one-to-one and onto, then the inverse function f 1 : S S is defined so that f 1 (y) is the unique x such that f (x) = y. Theorem: If f : S S is a one-to-one, onto continuous function and S is compact, then f 1 : S S is continuous. Proof: If F is a closed subset of S, then F is compact and so f (F ) is compact and so is closed. But f (F ) = (f 1 ) 1 (F ), the pre-image of F by f 1. Since F closed implies that (f 1 ) 1 (F ) is closed, f 1 is continuous. Observe that f (x) = x defines a continuous, one-to-one, onto function from [ 2, 1) [0.1] to [0, 2], but the inverse is not continuous.

16 Ross Theorem 21.4 (ii): If a function f : S S is continuous and E S is compact, then f is uniformly continuous on E. Proof: Given ɛ > 0, there exists, for each x E, δ(x) > 0 so that B δ(x) (x) f 1 (B ɛ/2 (f (x)). Choose a finite subcover for the open cover {B δ(x)/2 (x) : x E}. That is, there exist x 1, x 2,..., x k so that E B δ(x1 )/2(x 1 ) B δ(xk )/2(x k ). Let δ = min(δ(x 1 )/2,..., δ(x k )/2) > 0. If x E, then there exists i such that x B δ(xi )/2(x i ). if z E B δ (x) then d(z, x i ) d(z, x) + d(x, x i ) < δ(x i ). That is, both x and z lie in B δ(xi )(x i ). Hence, d (f (x), f (x i )), d (f (z), f (x i )) < ɛ/2. So d(f (x), f (z)) < ɛ by the Triangle Inequality.

17 Exercise A3.2: Let f 1, f 2,..., f k : E R be real-valued functions on E S. Define f : E R k by f (x) = f 1 (x),..., f k (x). Show that f is continuous at e E iff each f i is continuous at e. HW Exercises: Ross p [R 1 p ]: 21.2, 21.3, 21.8, We will later need Ross Theorem 19.5: If f is a uniformly continuous, real-valued function on an open interval (a, b) with a < b R, then f extends to a continuous function on [a, b]. Proof: If t n, s n a in (a, b) then they are a Cauchy sequences and so by uniform continuity and completeness each {f (t n )}, {f (s n )} converge in R. Also, if u 2n = t n, u 2n 1 = s n then u n a and so {f (u n )} converges to the common limit of {f (t n )} and {f (s n )}. Let f (a) be this common limit.

18 Connectedness, Ross Sec. 22 A disconnection of a subset E of a metric space (S, d) is a pair of open subsets {U 1, U 2 } of S such that (i) {U 1, U 2 } covers E. That is, E U 1 U 2. (ii) U 1 E and U 2 E are disjoint. That is, U 1 U 2 E =. (iii) Neither U 1 E nor U 2 E is empty. We say that E is connected if it has no disconnection. That is, it cannot be thus separated into two pieces. Exercise A3.3: Assume f : S S is a continuous function and E S. Show that if the pair of open subsets {U 1, U 2 } of S is a disconnection of f (E) then {f 1 (U 1 ), f 1 (U 2 )} is a disconnection of E. From this follows

19 Ross Theorem 22.2: If f : S S is a continuous function and E is a connected subset of S, then f (E) is a connected subset of S. Ross Example 22.1: If E R then E is connected iff it is an interval, or, equivalently, iff a < c < b and a, b E imply c E. Proof: If a, b E and c E with a < c < b, then {(, c), (c, + )} is a disconnection of E. Now assume that E is an interval and that {U 1, U 2 } satisfy conditions (ii) and (iii) of a disconnection for E. We will show that {U 1, U 2 } does not cover E. By condition (iii) there exist a U 1 E, b U 2 E. By relabeling if necessary we can assume a < b. Since E is an interval, [a, b] E.

20 There exists ɛ 1 > 0 such that (a ɛ 1, a + ɛ 1 ) U 1 and (b ɛ 1, b + ɛ 1 ) U 2. Let c = sup(u 1 [a, b]) so that a + ɛ 1 c. Because U 2 [a, b] is disjoint from U 1 [a, b], c b ɛ 1. If c were in U 1, then there would exist ɛ 2 < ɛ 1 such that (c ɛ 2, c + ɛ 2 ) U 1 and so (c ɛ 2, c + ɛ 2 ) U 1 [a, b]. Hence, c would not be an upper bound for U 1 [a, b]. If c were in U 2, then there would exist ɛ 2 < ɛ 1 such that (c ɛ 2, c + ɛ 2 ) U 2. Since U 1 [a, b] and U 2 [a, b] are disjoint, (c ɛ 2, c] would be disjoint from U 1 [a, b]. The elements of (c ɛ 2, c) would be upper bounds for U 1 [a, b] contradicting the assumption that c was the least upper bound. Thus, c E \ (U 1 U 2 ).

21 A subset E of a metric space S is called path-connected if for every x, y E there exists a continuous map γ from an interval [a, b] into E with γ(a) = x and γ(b) = y. A subset E of R k is convex if for every x, y E, the line-segment {tx + (1 t)y : t [0, 1]} is contained in E. Clearly, a convex set is path-connected. Exercise A3.4: (a) Prove that a path-connected set is connected. (see Ross Theorem 22.5). (b)prove that if E is a connected subset of S, then its closure E is connected. (Hint: Show that a disconnection of E is a disconnection of E. Be careful about checking (iii)).

22 Limits for Real-Valued Functions, Ross Secs. 17, 18, 19 From the properties of sequential limits, various properties for function limits follows. Theorem (Squeeze Theorem) Let f, h, g : E R with f (x) h(x) g(x) for all x E. If e E and the limits lim x e f (x) and lim x e g(x) exist in R and are equal, then lim x e h(x) exists and agrees with this common value. Ross Example 17.2: Assume that f : [ 1, 1] R with lim x 0 f (x) = 0. If g : [ 1, 1] \ {0} R is defined by g(x) = f (x) sin(1/x) then lim x 0 g(x) = 0.

23 Ross Theorem 17.3, 17.4: f, g : E R are continuous at e E, then f, k f, f + g, f g are continuous at e and if g(x) 0 then f /g is continuous at e. As Ross observes on page 130 ([R 1 page 93), max(f, g) = 1 [(f + g) + f g ] and 2 min(f, g) = 1 [(f + g) f g ] and so max(f, g) and 2 min(f, g) are continuous at e as well. HW Exercises: Ross p [R 1 p ]: 17.4, 17.5, 17.6, 17.7,

24 Ross Theorem 18.1, 18.2: Let f : [a, b] R be a continuous function. (a) There exist x m, x M such that f (x M ) = max{f (x) : x [a, b]}, f (x m ) = min{f (x) : x [a, b]}. In particular, f is a bounded function, i.e. its image f ([a, b]) is a bounded set. (b) (Intermediate Value Theorem): If y is a value between f (a) and f (b) then there exists x (a, b) such that f (x) = y. Proof: [a, b] is a closed, bounded set and so it is compact. Since it is an interval, it is connected. Hence, f ([a, b]) is compact and connected and so is a closed, bounded interval [c, d] with c d in R. So there exist x m, x M such that f (x m ) = c and f (x M ) = d. If f (a) < y < f (b) (or f (b) < y < f (a), then c < y < d and so y [c, d]. So there exists x [a, b] with f (x) = y. Clearly, x a or b.

25 A function f : [a, b] R is strictly increasing when x 1 < x 2 implies f (x 1 ) < f (x 2 ) for all x 1, x 2 [a, b], i.e. f preserves inequalities, and is strictly decreasing when x 1 < x 2 implies f (x 1 ) > f (x 2 ) for all x 1, x 2 [a, b], i.e. f reverses inequalities. In either case, f is one-to-one. Theorem If f : [a, b] R is a continuous one-to-one function, then either f is strictly increasing or f is strictly decreasing. Proof: Assume f (a) < f (b). We show that f is strictly increasing. Suppose instead that there exist x 1 < x 2 in the interval with f (x 1 ) f (x 2 ). If f (x 2 ) f (a) then by the Intermediate Value Theorem, there exists x 3 [x 2, b] with f (x 3 ) = f (a) and so f is not one-to-one. Similarly, if f (x 1 ) f (b), then f is not one-to-one.

26 So we may assume f (a) < f (x 2 ) f (x 1 ) < f (b). Again by the Intermediate Value Theorem, there exists x 3 [a, x 1 ] such that f (x 3 ) = f (x 2 ) and again f is not one-to-one. We similarly show that if f (b) < f (a) then f is strictly decreasing. { x if x [ 1, 0], Exercise A3.5: (a)define f (x) = 1 x if x (0, 1]. Show that f : [ 1, 1] [ 1, 1] is one-to-one, but neither increasing nor decreasing. (b) Show that if f : [a, b] [c, d] is continuous and either strictly increasing or strictly decreasing then the inverse function is continuous.

27 There are some special cases we should look at. Suppose E R and f : E S. If a R and (a, a + ɛ) E then we can restrict f to this interval. If the limit lim x a f (x) exists for the restriction f (a, a + ɛ) then we call it the limit to a from above, written lim x a+ f (x). Similarly, if the limit lim x a f (x) exists for the restriction f (a ɛ, a) then we call it the limit to a from below, written lim x a f (x). The limit of lim x a f (x) on the punctured interval (a ɛ, a + ɛ) \ {a} exists iff both lim x a+ f (x) and lim x a f (x) exist and have the same value. Then the common value is lim x a f (x) on the punctured interval. The limit lim x a f (x) on the whole interval (a ɛ, a + ɛ) also requires that f (a) equals this common value.

28 Pointwise and Uniform Convergence of Sequences of Functions, Ross Secs. 23, 24 We consider real-valued functions defined on a subset E of a metric space (S, d). Let Fun(E) denote the set of all such functions so that Fun(E) = {f : f : E R}. For a sequence {f n } in Fun(E) we say that {f n } converges pointwise to f Fun(E) when for every x E the sequence {f n (x)} in R converges to f (x). We say that {f n } is pointwise Cauchy when for every x E the sequence {f n (x)} is a Cauchy sequence in R. Theorem: If {f n } is a pointwise Cauchy sequence in Fun(E), then it converges pointwise to some function f Fun(E).

29 Proof: A Cauchy sequence of real numbers always has a limit and so we can define f by f (x) = lim f n (x) for each x E. Formally, {f n } converges pointwise to f when ( x E)( ɛ > 0)( N N)( n N) f n (x) f (x) ɛ, and {f n } is pointwise Cauchy when ( x E)( ɛ > 0)( N N)( n, m N) f n (x) f m (x) ɛ. Notice that in each case, the N depends on x as well as on ɛ.

30 Recall that uniform continuity was defined from continuity by interchanging the order of quantifiers requiring that δ be chosen to work for all x. Similarly, we make the following definitions. The sequence {f n } converges uniformly to f when ( ɛ > 0)( N N)( x E)( n N) f n (x) f (x) ɛ, and {f n } is uniformly Cauchy when ( ɛ > 0)( N N)( x E)( n, m N) f n (x) f m (x) ɛ. That is, in each case N can be chosen to work for all x. Clearly, uniform convergence implies pointwise convergence and uniformly Cauchy implies pointwise Cauchy.

31 Lemma: If {f n } is uniformly Cauchy and converges pointwise to f then it converges uniformly to f. Proof: Given ɛ > 0, we can find N so that for all x, n, m N implies f n (x) f m (x) ɛ. Fix x for a moment and observe that lim m f m (x) = f (x) by pointwise convergence. Because t f n (x) t is a continuous function of t we see that lim m f n (x) f m (x) = f n (x) f (x). Because (, ɛ] is a closed set, the inequality f n (x) f m (x) ɛ is preserved in the limit. So for all n N, f n (x) f (x) ɛ. As this is true for every x E, it follows that {f n } converges uniformly to f.

32 Ross Theorem 25.4: If {f n } is a uniformly Cauchy sequence in Fun(E), then it converges uniformly to some function f Fun(E). Proof: Because {f n } is uniformly Cauchy, it is pointwise Cauchy and so converges pointwise to some f. From the above Lemma, it converges uniformly to f. Uniform convergence has nice properties that pointwise convergent does not. For example, let f n : [0, 1] R be defined by f n (x) = x{ n. The sequence {f n } converges pointwise 0 if x < 1, to f with f (x) =. 1 if x = 1.

33 My first day at City College in 1962, I was taking the equivalent of Math 323 (then called Math 13) with Professor Jesse Douglas. He gave us the following homework exercise. Exercise A4.1: For n, m N define f m.n : [0, 1] R by f m,n (x) = [cos(m! πx)] 2n. Define f m (x) = 0 except that f m (x) = 1 when x is a rational number a/b with 0 a b with b N such that b m!. Define f to be the characteristic function of the rationals. That is, f (x) = 0 if x is irrational and = 1 if x is rational. (a) Show that for each fixed m N, f m is the pointwise limit of {f m,n } as n. (b) Show that f is the pointwise limit of {f m } as m. Exercise A4.2: Assume that {f n } on E converges uniformly to f with f n (E) contained in the closed, bounded interval I. Show that f (E) I and if g is a continuous real-valued function on I, then {g f n } converges uniformly to g f. HW Exercises: Ross p [R 1 p ]: 24.2, 24.9, 24.10,

34 Ross Theorem 24.3: Assume the sequence {f n Fun(E)} converges uniformly to f Fun(E). (i) If each f n is continuous at e E, then f is continuous at e. (ii) If each f n is continuous on E, then f is continuous on E. (iii) If each f n is uniformly continuous on E, then f is uniformly continuous on E. (iii): In the proof of (i) choose δ to work for all x and e. Proof: Given ɛ > 0 there exists N so that f N (x) f (x) < ɛ/3 for all x E. (i): Since f N is continuous at e, there exists δ > 0 so that for all x, d(x, e) < δ implies f N (x) f N (e) < ɛ/3. So d(x, e) < δ implies f (x) f (e) f (x) f N (x) + f N (x) f N (e) + f N (e) f (e) < ɛ. (ii): Apply (i) for every e E.

35 Power Series, Ross Secs. 23, 25, 26 If {g n : n = 0, 1,... } is a sequence of real-valued functions on E S, then the partial sums S n = n k=0 g k for n = 0, 1,... is a sequence of real-valued functions. We say that the infinite sum S = k=0 g k exists and the series converges when the sequence of functions {S n } converges at least pointwise. We say that the series converges uniformly when the sequence of partial sums converges uniformly. If the series converges uniformly and each g n is continuous on E then the limit function S is continuous on E. Ross Example 25.5 (Uniform N th Term Test)- If the series converges uniformly on E S, then lim n sup{ g n (x) : x E} = 0.

36 Proof: Observe that the sequences of partial sums {S n } and {S n 1 } both converge uniformly to S = k=0 g k. So the difference g n = S n S n 1 converges uniformly to 0. Alternatively, use the Cauchy criterion as in Ross. Ross Theorem 25.7 (Weierstrass M-Test)- Assume {M k } is a non-negative sequence such that k=0 M k converges. If {g n } is a sequence of real-valued functions on E such that g k (x) M k for all x E and all k = 0, 1,... then k=0 g k converges absolutely and uniformly.

37 Proof: If N n < m then S m (x) S n (x) = m k=n+1 g k (x) m k=n+1 g k (x) m k=n+1 It follows that the sequence {S k } is uniformly Cauchy and so is uniformly convergent. M k. Suppose that {g n } is a sequence of continuous, real-valued functions on E and that E is the union of a sequence {U n } of open subsets of S. If the series g k converges uniformly on each U n, but not necessarily uniformly on all of E, then, nonetheless, the limit function S is continuous on each U n and so is continuous on all of E.

38 For a sequence {a n : n = 0, 1,... } the corresponding power series is n=0 a nx n. That is, it is the series with g n (x) = a n x n. Recall from the notes on Ross Section 14: Lemma: (a) If L > 0 and either lim sup( a n ) 1/n < L or lim sup( a n+1 / a n ) < L, then there exists M > 0 such that for all n N, a n M L n. (b) If l > 0 and either lim sup( a n ) 1/n > l or lim inf( a n+1 / a n ) > l, then there exists m > 0 such that frequently, a n > m l n. Notice that in (a) we said before that it a n < M L n only when n N for some N N. Replacing M by max( a 0, a 1 /L,..., a N /L N, M) if necessary, we can get the inequality for all n N.

39 Ross Theorem 23.1: For the power series n=0 a nx n let α = lim sup( a n ) 1/n and R = 1/α [If α = 0 we set R = and if α = we set R = 0]. (a) For any R 1 < 0 the series converges uniformly on the interval [ R 1, R 1 ] and so it converges to a continuous function on the open interval ( R, R). (b) If x > R then the series diverges. Proof: (a) With R 1 < R, choose L so that R 1 < 1 < R. Then L α < L and so by the Lemma, there exists M > 0 so that for all n, a n < M L n. If x R 1 then n N implies that a n x n < M(R 1 L) n. Since r = R 1 L < 1, the series Mr n is a convergent geometric series. So n=0 a nx n converges uniformly on [ R 1, R 1 ] by the Weierstrass M-Test.

40 (b) If x > R, choose l so that x > 1 > R. Then α > l and l so by the Lemma, there exists m > 0 so that for infinitely many n, a n > m l n. Hence, a n x n > m( x l) n infinitely often and x l > 1. Since a n x n > 1 infinitely often, then N th Term Test implies that the series diverges. The reciprocal R of lim sup( a n ) 1/n is called the radius of convergence of the series. If the limit lim( a n+1 / a n ) exists then its reciprocal is the radius of convergence as well via a similar argument. So if lim( a n+1 / a n ) exists, it must equal lim sup( a n ) 1/n.

41 We will later show that we can differentiate and integrate a power series term by term. For the moment we observe Ross Lemma 26.3 If n=0 a nx n has radius of convergence, then na n x n 1 = (k + 1)a k+1 x k, and n=1 n=0 k=0 a n n + 1 x n+1 = have radius of convergence R. k=1 a k 1 k x k Proof: Multiplying and dividing by x 0 does not affect convergence. The result follows because lim n 1/n = 1 and similarly lim(n + 1) 1/n = 1.

42 We will also be concerned with power series centered at a point c R where c is not 0. These are of the form n=0 a n(x c) n. If we use the change of variable X = x c then this becomes the ordinary power series n=0 a nx n. If this has radius of convergence R and so converges when X lies in the interval ( R, R), then n=0 a n(x c) n converges for x in the interval (c R, c + R), centered at c and with radius R. HW Exercises: Ross p. 192 [R 1 p ]: 23.1a,c,e,g, Ross p. 207 [R 1 p. 144]: 25.6, 25.7, 25.9,

43 Basic Properties of the Derivative, Ross Sec. 28 There are different notations for the derivative. If Q is a quantity that is changing in time, then the change in Q, written Q is the difference between the new value and the old: Q = Q new Q old so that Q new = Q old + Q. If y = f (x) then the slope is y x = f (x new) f (x old ). x new x old For the derivative we take the limit as the new value of x = x new approaches the old x old = a. We can write h = x = x a so that x = a + h and take the limit as h approaches 0.

44 f f (x) f (a) f (a + h) f (a) (a) = lim = lim. x a x a h 0 h We assume that the domain of f is an interval E which contains a, but notice that the difference quotient is defined only on E \ {a}. We say that f is differentiable at a, or that f is defined at a, when this limit exists. Notice that when a is an end-point of the interval E, then this is a left-hand or right-hand limit. When f is differentiable at every point of E then then the derivative function f is defined on E. We write y = f (x) = dy for the derivative on E. The function f is dx called continuously differentiable (or C 1 ) on E when f is defined and continuous on E.

45 Notice that f is differentiable at x = a with derivative f (a) = L exactly when the difference f (a + h) f (a) L h = o(h). To say that a function is o(h) o(h) is to say that the limit lim h 0 = 0. This certainly implies h that lim h 0 o(h) = 0. So if f is differentiable at a, then It follows that lim f (a + h) f (a) = lim L h + o(h) = 0. h 0 h 0 Ross Theorem 28.2: If a function f on an interval E is differentiable at a E, then it is continuous at a. We can also write f (x) f (a) = f (a)(x a) + o(x a). It follows that if C > f (a), then for x close enough to a f (x) f (a) x a < C.

46 The usual properties of the derivative: linearity, the product rule and the quotient rule are listed in Ross Theorem Notice that for real-valued functions f, g on E, (f + g) = f + g and cf = c f, but (f g) ( f ) ( g). For the product, one uses f (a + h)g(a + h) f (a)g(a) = f (a + h)g(a + h) f (a + h)g(a) + f (a + h)g(a) f (a)g(a) = f (a + h)[g(a + h) g(a)] + [f (a + h) f (a)]g(a). For the chain rule Ross Theorem 28.4: If f is differentiable at a E 1 and g is differentiable at f (a) E 2 where E 1 and E 2 are intervals with f (E 1 ) E 2, then g f is differentiable at a with (g f ) (a) = g (f (a)) f (a).

47 Proof: If for x a close to a, f (x) f (a), then (g f )(x) (g f )(a) x a = (g f )(x) (g f )(a) f f (x) f (a) (x) f (a). x a As x a, f (x) f (a) because differentiability of f at a implies continuity of f at a. So the limit on the right exists and equals g (f (a)) f (a). If there is a sequence x n a such that f (x n ) = f (a), then 0 = lim f (x n) f (a) x n a = f (a). Since f (a) = 0, we complete the proof by showing that (g f ) (a) exists and equals 0.

48 Fix L > g (f (a)) and let ɛ > 0 = f (a). There exists δ 1 > 0 so that y f (a) < δ 1 implies g(y) g(f (a)) L y f (a). There exists δ 2 > 0 so that x a < δ 2 implies f (x) f (a) < ɛ x a. Choose δ 2 small enough that ɛδ 2 < δ 1 so that f (x) f (a) < δ 1. Then g(f (x)) g(f (a)) L f (x) f (a) < Lɛ x a. Since Lɛ can be chosen arbitrarily small, we have g(f (x)) g(f (a)) lim x a = 0. x a

49 Mean Value Theorem Recall the result from elementary calculus Ross Theorem 29.1 Assume f is defined on an open interval containing x 0 and f has a local maximum or minimum at x 0. If f is differentiable at x 0, then f (x 0 ) = 0. Proof: Assume ɛ > 0 so that for x (x 0 ɛ, x 0 + ɛ), f (x 0 ) f (x) (local maximum case). That is, f (x) f (x 0 ) 0. f f (x (x 0 ) = lim 0 +h) f (x 0 ) h 0. If h > 0 then the difference h quotient is non-negative and if h < 0 then it is non-positive. Hence, the limit is 0 and is 0. That is, it is 0.

50 Ross Theorem 29.2 (Rolle s Theorem) - Assume f is a continuous function on [a, b], with a < b, which is differentiable on (a, b). If f (a) = f (b), then there exists x 0 in (a, b) such that f (x 0 ) = 0. Proof: By compactness, f achieves its maximum and minimum values. So there exist x 0, x 1 in [a, b] such that f (x 0 ) f (x) f (x 1 ) for all x [a, b]. If both x 0 and x 1 are end-points then since f (a) = f (b), f is a constant function and so f (x) = 0 for all x [a, b]. Otherwise, either x 0 or x 1 is in (a, b) and so f = 0 at that point.

51 Ross Theorem 29.3 (Mean Value Theorem) - Assume f is a continuous function on [a, b], with a < b, which is differentiable on (a, b). There exists x 0 in (a, b) such that f (b) f (a) = f (x 0 )(b a). Proof: Apply Rolle s Theorem to f (b) f (a) g(x) = f (x) [f (a) + (x a)], that is, to f (x) L(x) b a where L is the equation of the line through (b, f (b)) and f (b) f (a) (a, f (a)) which has slope. b a The Mean Value Theorem (= MVT) lies behind many standard results. Ross Definition 29.6: A real-valued function f on an interval I is increasing (or strictly increasing) when x 1 < x 2 in I implies f (x 1 ) f (x 2 ) (or implies f (x 1 ) < f (x 2 )). It is decreasing (or strictly decreasing) when x 1 < x 2 in I implies f (x 1 ) f (x 2 ) (or implies f (x 1 ) > f (x 2 )).

52 Ross Corollaries 29.4, 29.7: For a differentiable real-valued function f on an interval I. f is increasing iff f (x) 0 for all x I. f is decreasing iff f (x) 0 for all x I. f is constant iff f (x) = 0 for all x I. Proof: If f is increasing, then the difference quotient f (x) f (a) 0 for all x, a in I with x a. So the limit x a f (a) 0. If f is not increasing, then there exist a < b in I with f (b) f (a) f (a) > f (b) and so < 0. By the MVT there exists x b a in the interval (a, b) I such that f f (b) f (a) (x) = < 0. b a The proofs of the remaining results are similar. The same argument shows that if f (x) > 0 for all x I then f is strictly increasing, but this time the converse is not true. Think of f (x) = x 3.

53 We showed above that a one-to-one continuous function f on an interval I is either strictly increasing or strictly decreasing and the inverse function is continuous. So with J = f (I ), the inverse function g = f 1 is defined and continuous on the interval J. So if y = f (x) then x = g(y). If x x 0 in I then y = f (x) y 0 = f (x 0 ) in J and conversely if y y 0 in J then x = g(y) x 0 = g(y 0 ) in I. Ross Theorem 29.9 (Inverse Function Theorem) - Assume f is a one-to-one continuous function on an open interval I with g = f 1 the continuous inverse function on the open interval J = f (I ). If f is differentiable at x 0 I and f (x 0 ) 0, then g is differentiable at y 0 = f (x 0 ) and g (y 0 ) = 1/f (x 0 ). Proof: For x x 0 in I, y = f (x) y 0 = f (x 0 ) in J and so 1 is the limit of x x 0 as x x f (x 0 ) f (x) f (x 0 ) 0.

54 x x 0 f (x) f (x 0 ) = x x 0 y y 0 = g(y) g(y 0) y y 0. Since y y 0 iff x x 0, it follows that 1 g(y) g(y 0 ) y y 0 as y y 0. f (x 0 ) is the limit of An important application of the MVT is an estimate of the amount that a function stretches distance. Theorem: Let f be a C 1 function on a closed, bounded interval I. Let L = sup{ f (x) : x I }. There exists e I such that L = f (e) and L is a Lipschitz Constant for f on I. That is, for all x 1, x 2 I, f (x 1 ) f (x 2 ) L x 1 x 2.

55 Proof: Because f is assumed to be continuous on the compact interval I, it achieves its maximum value L. If x 1 < x 2 I then by the MVT there exists c (x 1, x 2 ) I such that f (x 1) f (x 2 ) x 1 x 2 = f (c). So f (x 1 ) f (x 2 ) = f (c) x 1 x 2 L x 1 x 2. The uniform limit of a sequence of differentiable functions need not be differentiable and even if it is, the derivatives need not converge to derivative of the limit. 1 Exercise A5.1: (a) Show that if f n (x) = + x n 2, then each f n is C 1 and {f n } converges uniformly on [ 1, 1] to f (x) = x. Show that x is not differentiable at 0. (b) Show that if f n (x) = cos(n2 x), then {f n n } converges uniformly to 0, but the derivative sequence does not converge.

56 Theorem: Let {f n } be a sequence of differentiable real-valued functions on an interval I. If {f n } converges uniformly to f and {f n} converges uniformly to g, then f is differentiable with f = g. Proof: For x a in I, we apply the MVT to f m f n for n, m N to get a z I such that (f m (x) f m (a)) (f n (x) f n (a)) = (f m(z) f n(z))(x a). Let ɛ > 0. Since {f n} is uniformly Cauchy, there exists N N such that n, m N implies f m(x) f n(x) ɛ for all x I. Letting m so that f m(x) g(x) we have g(x) f n(x) ɛ for all x I. So for m, n N, (f m (x) f m (a)) (f n (x) f n (a)) ɛ x a. Again let m and so we have,

57 (i) g(x) f N (x) ɛ for all x I. (ii) (f (x) f (a)) (f N (x) f N (a)) ɛ x a for all x, a I. Now for the function f N and the point a I there exists δ > 0 so that (iii) f N (x) f N (a) f N (a)(x a) ɛ for all x such that x a < δ Putting these three together we have for x a < δ that f (x) f (a) g(a)(x a) 3ɛ x a. f (x) f (a) That is, lim x a x a = g(a).

58 From this result we can prove directly Ross Theorem 26.5, that we can differentiate a power series term-by-term. Also we will later reprove a slightly weaker version of the above theorem the way Ross does Theorem 26.5, That is, by using integration and the Fundamental Theorem of Calculus. Exercise A5.2: Prove the Power Rule for Rational Exponents r: If f (x) = x r then f is differentiable with f (x) = rx r 1. (a) Use the Product Rule and Mathematical Induction to prove it for r = n N. (b) Use the Quotient Rule to prove it for r = n. (c) Use the Inverse Function Theorem to prove it for r = 1/n. (d) Use the Chain Rule to prove it for r = m/n. HW Exercises: Ross p [R 1 p. 160]: 28.4, 28.6, 28.7, Ross p [R 1 p. 167]: 29.2, 29.5, 29.7, 29.11

59 Taylor s Theorem If f (x) is the limit of the series k=0 a k(x c) k on the interval (c R, c + R) then f (c) = a 0, f (c) = a 1, f (c) = 2a 2,..., f (k) (c) = k!a k and so a k = f (k) (c)/k!. If a function has derivatives of all orders at c then we call f (k) (c) k=0 (x c) k the Taylor series of f centered at c. (The k! series centered at 0 is usually called the MacLaurin series). The remainder R n is defined by: R n (x) = f (x) n 1 k=0 f (k) (c) (x c) k. k!

60 Ross Theorem 31.3 (Taylor s Theorem)- Let f be defined on (a, b) and a < c < b and the first n derivatives of f are defined on (a, b). For x c in (a, b) there exists y between x and c such that R n (x) = f (n) (y) (x c) n. n! Proof: With x fixed, define g(t) = n 1 k=0 f (k) (c) (t c) k + M k! n! (t c)n f (t), where M is chosen so that g(x) = 0. Observe that g (k) (c) = 0 for k = 0, 1,..., n 1. Inductively, with x 0 = x use Rolle s Theorem to get x k between x k 1 and c so that g (k) (x k ) = 0 for k = 1,..., n. Since g (n) (t) = M f (n) (t), it follows that with y = x n, M = f (n) (y).

61 If A n = sup{ f (n) (y) : y ( R + c, c + R)} and A n n=0 R n n! converges then R n 0 uniformly on ( R + c, c + R) and so the Taylor series converges to f uniformly. The Taylor series may converge but to the wrong function. We note first that for any n N, lim x 0 e 1/x2 /x n = 0. Proof: For x < 1, 0 e 1/x2 /x n e 1/x2 /x 2n. Letting t = 1/x 2 we see that lim x 0 e 1/x2 /x 2n = lim t e t t n and this is 0 by l Hospital s (or l Hôpital s) Rule.

62 Define g n (x) = { e 1/x2 /x n if x 0 0 if x = 0. The product and chain rules imply that g n(x) = 2g n+3 (x) ng n+1 (x) for x 0. At x = 0, the limit of the difference quotient is 0. Hence, the formula holds at x = 0 as well. It then follows by induction that for g = g 0, each g (k) is a linear combination of the g n s. In particular, g is infinitely differentiable with g (n) (0) = 0 for all n. So the Taylor series for g with c = 0 has a n = 0 for all n. It converges but to 0 rather than to g.

63 The Riemann Integral, Ross Sec. 32 We are going to consider functions on an interval [a, b] with a < b in R. A partition P of [a, b] is a finite set of points of [a, b] which include the end-points, so that with the points listed in order P = {a = t 0 < t 1 < < t n = b}. A partition cuts [a, b] into subintervals which intersect only at the end-points I(P) = {I 1,..., I n } with I k = [t k 1, t k ] for k = 1,..., n. We let I k = t k t k 1, the length of interval I k. A partition P 1 refines P when P P 1. That is, the cut-points of P are all included among the cut-points of P 1. Then each interval I of I(P) is itself partitioned by the points of P 1 I. Thus, we have for each I I(P) I = {J : J I(P 1 ) and J I }, I = { J : J I(P 1 ) and J I }

64 If P 1 and P 2 are partitions, then P 1 P 2 is a common refinement of both P 1 and P 2. The trivial partition is P 0 = {a < b} with I(P 0 ) = {[a, b]}. Every partition is a refinement of the trivial partition. We consider f a bounded, real-valued function on the interval [a, b]. For a S [a, b] we define M(f, S) = sup{f (x) : x S}, m(f, S) = inf{f (x) : x S}. Using this we define the upper and lower Darboux sums associated with a partition P = {t 0,..., t n }:

65 U(f, P) = {M(f, I ) I : I I(P)} = L(f, P) = {m(f, I ) I : I I(P)} = n M(f, I k ) I k, k=1 n m(f, I k ) I k. k=1 Theorem: Let f be a bounded, real-valued function on the interval [a, b]. (a) If P 1 is a refinement of P, then L(f, P) L(f, P 1 ) U(f, P 1 ) U(f, P). In particular, m(f, [a, b])(b a) L(f, P) U(f, P) M(f, [a, b])(b a). (b) For any partitions P 1 and P 2, L(f, P 1 ) U(f, P 2 ).

66 Proof: (a) Since each interval of I(P) is a union of intervals of I(P 1 ), we have, with I I(P), J I(P 1 ). U(f, P 1 ) = M(f, J) J M(f, I ) J I J I I J I M(f, I ) J = M(f, I ) I = U(f, P). I J I I Similarly, L(f, P 1 ) L(f, P) because m(f, J) m(f, I ) when J I. Clearly, L(f, P) U(f, P) for any partition P. For the last result of (a) we use that P refines the trivial partition. (b) Use the common refinement: L(f, P 1 ) L(f, P 1 P 2 ) U(f, P 1 P 2 ) U(f, P 2 ).

67 Define the upper and lower Darboux integrals: U(f ) = inf P U(f, P), L(f ) = sup L(f, P). P Since all the lower sums are smaller than all the upper sums we have L(f ) U(f ) by Exercise 4.8. We say that f is integrable when U(f ) = L(f ) in which case the common value is the integral which we denote by b a f. Clearly, f is integrable iff for every ɛ there exist partitions P 1, P 2 so that U(f, P 1 ) L(f, P 2 ) < ɛ. Replacing P 1 and P 2 by P = P 1 P 2 we obtain Ross Theorem 32.5: f is integrable iff there exists a partition P such that U(f, P) L(f, P) < ɛ. Observe that L( f, P) = M(f, P) and so L( f ) = U(f ). In particular, if f is integrable then f is integrable and b f = b f. a a

68 For a partition P we define the mesh of P to be the maximum length of the intervals of I(P). That is, mesh(p) = max{ I : I I(P)} = max{t k t k 1 : k = 1,..., n}. Ross Theorem 32.7: For every ɛ > 0, there exists δ > 0 such that mesh(p) < δ implies U(f, P) U(f ) < ɛ and L(f ) L(f, P) < ɛ. Proof: First choose a partition P 0 = {u 0 < < u m } so that U(f, P) U(f ) < ɛ/2. Let K = M(f, [a, b]) m(f, [a, b]) + 1 so that K > 0. Let δ = ɛ/(2km). Assume that mesh(p) < δ. let P k = P {u 0, u k,..., u m }. Then P 1 is the common refinement P P 0 and so U(f, P 1 ) < U(f ) + ɛ/2. We show inductively that U(f, P k ) < U(f ) + ɛ( k 2m ).

69 Moving from P k to P k+1 we are removing u k unless u k P. If u k P then P k+1 = P k and so U(f, P k+1 ) = U(f, P k ). Now assume u k P and let I I(P k+1 ) be the interval which contains u k. Thus, I is the union of the two intervals I 1, I 2 of I(P k ) which have u k as an end-point. U(f, P k+1 ) U(f, P k ) = M(f, I ) I [M(f, I 1 ) I 1 + M(f, I 2 ) I 2 ] = [M(f, I ) M(f, I 1 )] I 1 + [M(f, I ) M(f, I 2 )] I 2. Since M(f, I ) = max(m(f, I 1 ), M(f, I 2 )) at least one of these terms is 0 and the other is bounded by K mesh(p k ). Since P k is a refinement of P, the mesh of P k is bounded by m. So the difference is bounded by K δ < ɛ/2m. By induction U(f, P k ) < U(f ) + ɛ( k 2m ). Since P m = P, it follows that U(f, P) < U(f ) + ɛ.

70 The proof for L(f ) is similar. Alternatively, we can apply the U(f ) result to f. It follows that if f is integrable, then for every ɛ > 0 there exists δ > 0 so that mesh(p) < δ implies that L(f, P) ( b f ɛ, b f ] and U(f, P) [ b f, b f + ɛ). a a a a If f is not integrable, then for ɛ = U(f ) L(f ) > 0, U(f, P) L(f, P) ɛ for every partition P. We define a Riemann sum for a partition P by choosing a point in every interval I I(P). If P = {t 0 < < t n } we define Π(P) = {x = x 1,..., x n R n : x k I k for k = 1,..., k}. Notice that x k I k when t k 1 x k t k.

71 For each partition P and x Π(P) the Riemann sum R(f, P, x ) is given by R(f, P, x ) = n f (xk ) I k. k=1 Clearly, for any x, we have m(f, I k ) f (x k ) M(f, I k) and so L(f, P) R(f, P, x ) U(f, P). On the other hand, U(f, P) = sup{r(f, P, x ) : x Π(P)}, L(f, P) = inf{r(f, P, x ) : x Π(P)}. Proof: Let ɛ > 0. For U(f, P), choose x so that f (x k ) M(f, I k) ɛ/(b a) and for L(f, P), choose x so that f (x k ) m(f, I k) + ɛ/(b a) for k = 1,..., n.

72 Theorem: (a) If f is integrable, then for every ɛ > 0 there exists δ > 0 so that mesh(p) < δ implies b R(f, P, x ) < ɛ for every x Π(P). a f (b) If f is not integrable, and 0 < ɛ < U(f ) L(f ) then for every partition P there exist x 1, x 2 such that R(f, P, x 1) R(f, P, x 2) > ɛ. Proof: (a) If L(f, P) ( b a f ɛ, b f + ɛ), then R(f, P, x ) ( b f ɛ, b f + ɛ). a a U(f, P) [ b f, b a a a f ] and (b) Let ɛ 0 = [U(f ) L(f ) ɛ]/2. Choose x 1, x 2 so that R(f, P, x 1) > U(f, P) ɛ 0 and R(f, P, x 2) < L(f, P) + ɛ 0.

73 Integral Properties, Ross Sec. 33 Ross Theorem 33.1: If f is a monotone function on [a, b], then f is integrable. Proof: Assume that f is increasing. For any interval [c, d] [a, b], the M(f, [c, d]) = f (d), m(f, [c, d]) = f (c). Choose P which subdivides [a, b] into n intervals of equal length. Thus, t k = a + k (b a) and so I n k = 1 (b a). for n every k. U(f, P) L(f, P) = n (f (t k ) f (t k 1 )) ( 1 (b a)) n k=1 because the sum telescopes. = (f (b) f (a)) ( 1 (b a)), n

74 Given ɛ > 0 we can choose n large enough that ɛ > (f (b) f (a)) ( 1 (b a)). n If f is decreasing, we can use an analogous argument. Alternatively, apply the result to f. Ross Theorem 33.1: If f is a continuous function on [a, b], then f is integrable. Proof: Given ɛ > 0, there exists, by uniform continuity, a δ > 0 so that x 1 x 2 < δ implies f (x 1 ) f (x 2 ) ɛ/(b a). For any interval [c, d] [a, b] with [c, d] = d c < δ, we have M(f, [c, d]) m(f, [c, d]) ɛ/(b a).

75 Let P be a partition mesh(p) < δ. U(f, P) L(f, P) = n (M(f, I k ) m(f, I k )) I k k=1 (ɛ/(b a)) n I k = ɛ. k=1 If c, k R with c > 0 then M(cf, S) = c M(f, S), m(cf, S) = c m(f, S) and M(f + k, S) = M(f, S) + k, m(f + k, S) = m(f, S) + k for any S [a, b]. So for any partition P we have U(cf, P) = c U(f, P), L(cf, P) = c L(f, P), U(f + k, P) = U(f ) + k(b a), L(f + k, P) = L(f ) + k(b a).

76 We have already observed that U( f, P) = L(f, P). If f, g are bounded functions on [a, b], then M(f + g, S) M(f, S) + M(g, S), m(f + g, S) m(f, S) + m(g, S). So for any partition P L(f, P)+L(g, P) L(f +g, P) U(f +g, P) U(f, P)+U(g, P). So we obtain Ross Theorem 33.3: If f and g are integrable and c R then cf and f + g are integrable with b cf = c b f and a a b (f + g) = b f + b g. a a a Theorem: If g is integrable with b g = 0, then a U(f + g) = U(f ) and L(f + g) = L(g).

77 Proof: U(f ) U(f + g) + U( g) and U(f + g) U(f ) + U(g). Similarly, L(f ) L(f + g) L(f ). Theorem If g(x) equals 0 for a finite set of points in [a, b] then g is integrable with b a g = 0. Proof: Let P 0 be a partition such that f (x) = 0 at the points not in P 0. Let K = M(f, [a.b]) m(f, [a, b]). If P is a partition which refines P 0 then M(f, P) m(f, P) 2Kmesh(P). Thus, if f is integrable and we change its values on a finite set of points, the new function ˆf is integrable and with the same integral because ˆf = f + g where g = ˆf f is 0 except at finitely many points.

78 Theorem If f is uniformly continuous or monotonic on the open interval (a, b) then f is integrable. Proof: Since f is uniformly continuous on (a, b), the restriction to (a, b) extends to a continuous function f on [a, b] and so f is continuous. As it is a continuous on [a, b], f is integrable and agrees with f except possibly at the end-points. If f is monotonic on (a, b), then since we have assumed f is bounded, we can define f (b) = sup{f (x) : x (a, b)} and f (a) = inf{f (x) : x (a, b)}. Again f is integrable and agrees with f except possibly at the end-points. Observe that for S [a, b] M(f, S) m(f, S) = sup{f (x 1 ) f (x 2 ) : x 1, x 2 S} = sup{ f (x 1 ) f (x 2 ) : x 1, x 2 S}.

79 If f ([a, b]) [c, d] and g on [c, d] is Lipschitz continuous, i.e. for all y 1, y 2 [c, d], g(y 1 ) g(y 2 ) K y 1 y 2 with K 0, then M(g f, S) m(g f, S) K (M(f, S) m(f, S)). Theorem: If f is integrable with f ([a, b]) [c, d] and g on [c, d] is Lipschitz continuous, then g f is integrable. Proof: For any partition P, U(g f, P) L(g f, P) K [U(g f, P) L(g f, P)]. Corollary: If f is integrable, then f and f n are integrable for n N. If there exists ɛ > 0 such that f (x) ɛ for all x [a, b] and f is integrable, then 1/f and f m/n for m, n N are integrable. If f and g are integrable then f g, max(f, g) and min(f, g) are integrable.

80 Proof: Since x 1 x 2 x 1 x 2, it follows that g(x) = x is Lipschitz continuous. If g is differentiable on [c, d] with g bounded on [c, d] then g is Lipschitz on [c, d]. If g(x) = x n then on [ K, K] the derivative is bounded by nk n 1. For g(x) = x m/n or g(x) = 1/x on the interval [ɛ, K] the derivative is bounded by the values on the end-points. Because f g = 1[(f + 4 g)2 (f g) 2 ], max(f, g) = 1 [(f + g) + f g ] and 2 min(f, g) = max( f, g), it follows that these are integrable when f and g are. If I is a subinterval of [a, b] and f is a function on [a, b] we denote by f I the restriction of f to I, that is, the function with domain I which agrees with f.

81 Ross Theorem 33.6 If f is a function on [a, b] and P 0 = {t 0 < < t n } is a partition with I(P 0 ) = {I k : k = 1,..., n}, then U(f ) = n U(f I k ), L(f ) = n L(f I k ). k=1 k=1 In particular, f is integrable iff each f I k is integrable, in which case, b f = n tk a k=1 t k 1 f. Proof: For any partition P which refines P 0 we see that U(f, P) = n U(f I k, P I k ), L(f ) = n L(f I k, P I k ). k=1 k=1 Notice that if P k is a partition of I k for k = 1,..., n, then P = P k is a partition of [a, b] which refine P 0 and has P I k = P k.

82 In particular, U(f ) L(f ) = n k=1 [U(f I k) L(f I k )]. Since each term is non-negative, the sum is zero iff each term is zero. We call f piecewise monotonic or piecewise continuous when there is a partition P 0 = {t 0 < < t n } of [a, b] such that f is monotonic or uniformly continuous on each open interval (t k 1, t k ). So we have Ross Theorem 33.8 If f is piecewise monotonic or piecewise continuous, then it is integrable. If f g on [a, b] then for any S [a, b], M(f, S) M(g, S), m(f, S) m(g, S). It follows that If f g on [a, b] then U(f ) U(g), L(f ) L(g).

83 In particular, we have Ross Theorem 33.4 If f, g are integrable and f g on [a, b], then b f b g. a a Since f f, f it follows that if f is integrable, then b f b f. a a In any case, we have m f M m (b a) L(f ) U(f ) M (b a). Theorem If for ɛ > 0, f (x) g(x) ɛ, then U(f ) U(g), L(f ) L(g) ɛ (b a). Proof: f ɛ g f + ɛ.

84 Ross Theorem 33.4: If f is continuous and non-negative on [a, b] then b f = 0 implies f (x) = 0 for all x [a, b]. a Proof: If f (x 0 ) > 0 then there is an interval I [a, b] with positive length on which f f (x 0 )/2. Let g(x) = f (x 0 )/2 for x I and = 0 otherwise. Clearly f g on [a, b] and so b f b g = f (x 0) I > 0. a a 2 1 We define function f. b a b a f to be the average value of an integrable

85 Ross Theorem 33.9 (Integral Intermediate Value Theorem)- If f is continuous on [a.b] there exists x (a, b) such that f (x) = 1 b f. b a a Proof: The continuous function f achieves its maximum value M and its minimum value m. Since m f M we have m (b a) b f M (b a). By the Intermediate Value a Theorem x exists. Following Ross it can be shown that x can be chosen in (a, b). Theorem: If a sequence of integrable functions {f n } converges uniformly to f on [a, b] then f is integrable and b f = lim b a n f a n.

86 Proof: If f (x) f n (x) ɛ for all x [a, b] and f n M then f M + ɛ and so f is bounded. Furthermore, U(f ) b a f n, L(f ) b a f n ɛ. So { b a f n} converges to both U(f ) and L(f ). HW Exercises: Ross p. 280 [R 1 p. 191]: 32.6, Ross p. 289 [R 1 p ] 33.3, 33.4,

87 Fundamental Theorems of Calculus, Ross Sec. 34 Following Ross we say that a function defined on the open interval (a, b) is integrable on [a, b] if some, and hence every, extension to a function on [a, b] is integrable. Since the extensions differ merely at the end-points, they all have the same integral. Ross Theorem 34.1 (Fundamental Theorem of Calculus I)- If f is a continuous function on [a, b] which is differentiable on (a, b) and if the derivative f is integrable on [a, b] then b a f = f (b) f (a). Proof: Since f is integrable, for every ɛ > 0 there exists δ > 0 so that if mesh(p) < δ then every Riemann sum R(f, P, x) is ɛ close to b a f.

88 We will show that for every partition P = {t 0 < < t n } there exists x Π(P) such that R(f, P, x) = f (b) f (a). Observe first that since t 0 = a, t n = b the sum n k=1 (f (t k) f (t k 1 )) telescopes and equals f (b) f (a). By the MVT applied to the interval [t k 1, t k ] there exists x k (t k 1, t k ) such that f (t k ) f (t k 1 ) = f (x k ) (t k t k 1 ) = f (x k ) I k. The choices x k define x such that f (b) f (a) = R(f, P, x). Ross Theorem 34.2 (Integration By Parts)- If u and v are continuous functions on [a, b] with derivatives u, v on (a, b) which are integrable on [a, b] then b a u v + b uv = u(b)v(b) u(a)v(a). a

89 Proof: Apply the Product Rule (uv) = u v + uv. Ross Theorem 34.3 (Fundamental Theorem of Calculus II)- If f is integrable on [a, b] and F (x) = x f, then F is continuous a on [a, b] (with F (a) = 0). If f is continuous at x 0 [a, b] then F is differentiable at x 0 with F (x 0 ) = f (x 0 ). Proof: If a x < y b then F (y) F (x) = y f. So if x f K on [a, b] then F (y) F (x) = y x f y x f K y x. Thus, F is Lipschitz continuous with Lipschitz constant the bound on f.