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Chapter 8

# Chapter 8 - H.1 “*ﬂ—wm There are three fundamental...

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Unformatted text preview: H.1 “*ﬂ—wm There are three fundamental theorems concerning a function that is continuous on a closed interval [a,b]. The first is the Intermediate-Value Theorem, which is stated and proved on p. 144 of Apostol. We consider the other two here. We begin with a definition. Definition. If the function f is bounded on the interval [c,d], we define the EREE of f on this interval as follows: Let H M(f) sup {f(x); xe [c,df} I m(f) H inf ff(x);2<é[c,d]} . k Then we define the span of f by the equation span (f) = M(f) - m(f). Theorem. {The small-32am theorem}. Let f he continuous on the closed interval [a,b]. Given g:>O, there is a partition x0 < x14< ... < xn ~_§ the interval '[a,b] such that f is bounded 9g each closed subinterval [xi_l,xi], _9g such that the sgan g: f on each closed subinterval is at most 6. _______—.———_——¢——————————-———_——————_———— roof. The proof of this theorem is a bit tricky, but the theorem is so useful that the effort is justified. One proceeds by what is sometimes called "the method of successive bisections," or less elegantly, "chopping . the interval in half repeatedly"! For purposes of this proof, let us make up some terminology. If f is defined on the interval [c,d], we shall say that f is E—Qleasant on the interval [c,d] if there is some partition of [c,d] such that the span of f on each closed subinterval of the partition is at most E. If there is no such partition, we shall say that f is E—unpleasant on [c,d]: WOur object then is to prove that if f is continuous on [a,b], then f is E—pleasant on [a,b]. We make the following remark: Let c be any number with >a< c<:b. If f is s-pleasant on [agc], and if f is also £~pleasant on [c,b], then f is g-pleasant on all of [a,b]. The proof is easy. One merely takes the apprOpriate partitions of [a,c]y and [c,b] and puts them together to get a partition of [a,b]. This simple fact is all we need. We now prove the theorem. Assume that f is continuous on [a,b], and that f is E—unpleasant on [a,b]. We shall derive a contradiction. First step. ,Let c be the midpoint of [a,b]. Since f is s-unpleasant on [a,b], it must be true that f is E—unpleasant on either [a,c], or.on [c,b], or on both.. Let [al’bl] denote the left half [a,c] of our interval if f is E-ununeasant on [a,c]. Otherwise, let [a 'bl] denote the right half [c,b] of our interval. 1 In either case, f is E-unpleasant on [a1,bl]. General step. Assume that [an’bn] is an interval contained in [a,b] and that f is E—unpleasant on [an,bn]. Let cn be the midpoint of [an,bn]. As before, let i if f is g—unpleasant on this half; otherwise, let an+l’bn+l] denote the left half [an,cn] of [an’bn] ,b ] denote the right half. In either case, f. We now have defined a sequence of intervals [a n+1 n+1 isiunpleasant on [an+l,bn+1 [allbl]r [a21b21,... that is "nested" in the sense that each interval contains all its successors. Furthermore, each interval has half ’the length of the preceding one, and f is ﬁ—unpleasant on each of them. It follows by an easy induction proof that the length of the nth interval is bn - an = (b-a)/2n. BecaUSe the intervals are nested, we have Let s be the least upper bound of the numbers ai. Since all the numbers ai belong to the interval [a,b], so does 5. Now we derive a contradiction. We have three cases, according as a< s<:b, or s = a, or s = b. Consider first the case where a<£3<1L Since f is continuous at s, we may choose a neighborhood (s-JT 5+3) of s such that |f(X) - f(S)l < S/2 for all x in this neighborhood of s. Because 3 is the least upper bound of the numbers ai, there must be an n such that - < s §< an- S. Because the ai are increasing, we have < n+1 ‘ an+2 Now let us choose m so large that m > n, and so that (b — a)/2m < 5. Then b -a (5-, sothat b<a +5‘ SS+§. Then m m m m the interval [am'bm] is contained in the interval (3 —5: 5 +5). Therefore the span of f in the interval [am’bm] is at most 5. It follows that f is 6—pleasant on [am,bm], Indeed, ‘for 3&1 partition of [am,bm], the span of f in each subinterval of the partition will be at most 2. The other two cases proceed similarly. For example, suppose s = a. In this case, we have an = a for all n. (This means merely that we chose the left half of the interval at each step of the construction.) Since f is continuous at a, there must be a 5 such that [f(x) — f<a)l< {/2 for a§x<a +5. Choose m so large that (b—a)/2m< 3-, Then b — a < 5: so that b '< a +«;= a + J: Then the m m m m interval [am’bm] is contained in the interval [a, a +¢T)) so the span of f in the interval [am’bm] is at most 5. It follows that f is i—pleasant in [am,bm], as before. The proof when s = b is similar.l3 Here is an important application of this theorem: Theorem. If f is continuous on [a,b], then f ii bounded 93 [a,b] and integrable 92 [a,b]. Proof. Given.£>0, choose a partition < < ... < x0 X1 xn of [a,b] such that the span of f on each closed subinterval of the partitionr is at most 5. Define inf Eﬂx) for x < XSXKE, U} H tk = sup {f(x) for Xk-l IA N IA .£< ng Now f is bounded on [a,b]; indeed, f(x) is at most the largest of the numbers t1,...,tn andat least the smallest of the numbers 5 ,...,s . 1 n We define step functions 5 and t by letting their values equal 5k and tk, respectively, on (Xk_1,xk); at the partition points, we set s(xk) = t(xk) = f(xk). Then s(x)\$ f(x) 3 t(x) for all x. Now tk — 5k 5 2 because the span of f on [Xk—I'Xk] is at most i . Therefore 52% T28 = 5’: The Riemann condition applies to Show that f is integrable on [a,b]. [3 Finally, we prove the third big theorem about continuous functions. Theorem. gExtreme—value theorem}. Let f be continuous 93 the closedinterval [a,b].'men there are points x0 and x1 2: [a,b] gygh that for every x in [a,b], we have f(xo) 5 f(x) < f(x1). ~ The number M = f(x1) is called the maximum—value of f on [a,b], and the number m = f(xo) is called the minimum—value of f(x) on [a,b]. Both are called extreme values of -f on [a,b]. Proof. .We show that the point x1 exists; the proof that exists is similar. Xo We know that f is bounded on [a,b], by the previous theorem; define M = sup if(x) for x in [a,b]} . We wish to show that M = f(x1) for some point xl of [a,b]. Suppose this is not true. We derive a contradiction. Then by assumption, we have f(x)<.M for all x in [a,b]. Consider the function 1 g(X) = M __ f(X) ' Since the denominator does not vanish, g is well-defined; because f is continuous, so is g. Therefore g is bounded on [a,b], by the preceding theorem. Choose C so that g(x)_<_ C for x in [a,b]. Then 1 M — f(x) O < 5; Cr so that l/C 5M — Vf(x), or f(x)g M- 1/C, for every x in [a,b]. This contradicts the fact that M is the least upper bound of the values of f(x) on [a,b]. [3 We shall use the extreme—value theorem shortly, when {we prove the fundamental theorems of calculus. H.8 Exercises on the intermediate-value, extreme-value, and small— span theorems. 1. Let f(x) = x + [x] for 0 i X i 3. (a) Draw the graph of f; show f is strictly increasing. (b) Define a function g by the following rule: -If y = f(x) for some x in [0,3], let g(y) equal that x. Because f is strictly increasing, 9 is well—defined. What is the domain of g? 2. Let f(x) = x4 + 2x2 + 1 for o 5 x 5 10. (a) Show f is strictly increasing; what is the domain of its inverse function g? (b) Find an expression for 9, using radicals. ("Radicals" are the symbols /7 3/7 “ , etc.) 5 - 5x4 + 5- for x :12. We will show later 3. Let f(x) = 2x that f is strictly increasing (since its derivative is positive for x > 2). (a) Show that f is unbounded. (Hint: f(x) > x4(2x-5) > 2x - 5.) (b) What is the domain of its inverse function g? [Note: A famous theorem of Modern Algebra states that it‘ is not possible to express 9 in terms of algebraic operations and radicals.] 4. Let f(x) be defined and continuous and strictly increasing for x i O; suppose that f(0) = a. (a) Show that if f' is unbounded, then ‘5 takes on every value greater than a. i (b) If f is bounded, let_ M be the least upper bound of the values of f. Show that f takes on every value between a and M, but does not take on the value M. 8. 1-1.9 Show by example that the conclusion of the intermediate value theorem can fail if f is only continuous on [a,b) and bounded on [a,b}. Show by example that the conclusion of the extreme value theorem can fail if f is only continuous on [a,b) and bounded on [a,b]. Let f(x) = x for 0 < x < 1; let f(l) 5. Show that the conclusion of the small span theOrem fails for the 'function f(x). A function f defined on [a,b] is said to be piecewise— continuous if it is continuous except at finitely many points of [a,b]. Said differently, f is piecewise— continuous if there is some partition of [a,b] such that f is continuous on each open subinterval determined by the partition. (a) Show that if f is bounded on [a,b] and piecewise- continuous on [a,b], then' f is integrable on [a,b]. [Hint: Examine the proof we gave for piecewise—monotonic functions.] (b) Show that f can be piecewise—continuous on [a,b] Without being bounded on [a,b]. Consider the function (-1)[1/X] if x > 0, f(x) = 0 if x = O. Show that f is integrable on [0,1]. Show that f is neither piecewise—monotonic nor piecewise—continuous on [0,1]. 10. Challenge exercise. Define a function f on the [0,1] by setting f(x) = 0 if x is irrational and f(x) = l/n if x is a rational number of the x = m/n, where m and n are positive integers common factors other than 1 ; and f(0) = l (a) Show that f is integrable on [0,1]. H.10 interval ; form having no (b) Show that f is continuous at each irrational and discontinuous at each rational. ...
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Chapter 8 - H.1 “*ﬂ—wm There are three fundamental...

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