We say that the function f obtains a maximum value provided that there. We say that the function f obtains a minimum value provided that there
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1 Math 311 W08 Day 10 Section 3.2 Extreme Value Theorem (It s EXTREME!) 1. Definition: For a function f: D R we define the image of the function to be the set f(d) = {y y = f(x) for some x in D} We say that the function f obtains a maximum value provided that there exists x 0 in D such that f(x) f(x 0 ) for all x in D. In this case, x 0 is called the maximizer of f. We say that the function f obtains a minimum value provided that there exists x 0 in D such that f(x 0 ) f(x) for all x in D. In this case, x 0 is called the minimizer of f. So, a function has a maximum value if f(d) has a maximum element and a minimum value if f(d) has a minimum element. 2. Examples: a. The function f: (0, 1) R given by f(x) = x does not have a maximum or minimum value. b. The function f: [0, 1] R given by f(x) = x has a maximum (1) and minimum (0) value. c. The function f: (0, 1) R given by f(x) = 5 does have a maximum and minimum value (both 5). d. The function f: [0, 1] R given by f(0) = 7 and f(x) = 1/x otherwise, does not have a maximum value, but does have a minimum value (1).
2 3. We will prove that a continuous function defined on a closed bounded interval always has a maximum value. To do this, we will first show that the image of the function is bounded above, and then show that the supremum of the image is actually a function value (in the image). 4. Lemming The image of a continuous function on a closed bounded interval f: [a, b] R is bounded above. In other words, there is a real number M such that f(x) M for all x in [a, b]. Proof. Suppose not. Then for each natural number n, there is an x n in [a, b] such that f(x n ) > n. By the Sequential Compactness Theorem, {x n } has a subsequence, { x nk }, that converges to a point in [a, b], call it x 0. Then the image of this subsequence, f( x nk ) must converge to f(x 0 ) since f is continuous. But f( x nk )> n k > k for each natural number k. Thus the sequence f( x nk ) is convergent AND unbounded. I don t think so! Thus the image of the function must be bounded above.
3 5. EXTREME Value Theorem: A continuous function on a closed bounded interval ( f: [a, b] R ) obtains both a maximum and minimum value. Proof: By the previous Lemming, we know that f([a, b]) is bounded above (and it is clearly nonempty) so it has a supremum, c. Now for each natural number n, the number c 1/n is not an upper bound. Thus there is a number in f([a, b]), call it f(x n ) such that c 1/n < f(x n ) c. Clearly the sequence f(x n ) converges to c. And{ x n } is a sequence in [a, b], so by the Sequential Compactness Theorem, it must have a subsequence, x nk { }, that converges to an element in [a, b], call it x 0. Then since f is continuous, the image of this subsequence, f( x nk ), must converge to must converge to f(x 0 ). But since f( x nk ) is a subsequence of a sequence that converges to c, f( x nk ) must converge to c. Thus c = f(x 0 ). So the supremum is a function value WOO HOO! Now note that since f is continuous, so is f. Since f has a maximum (by above) f has a minimum at the same point.
4 3.3 The Intermediate Value Theorem 1. The Nested Interval Theorem (Theorem 2.29). For each natural number n, let a n and b n be numbers such that a n < b n, and consider the sequence of intervals I n = [a n, b n ]. If I n+1 I n and lim n!" [b n # a n ] = 0, then there is exactly one point that belongs to each interval I n for all n and both of the sequences { a n } and { b n } converge to this point. Proof: The sequence { a n } of left endpoints is an increasing sequence that is clearly bounded above by b 1. So it converges (by MCT) to its sup (call it a). Similarly the sequence { b n } of right endpoints is an decreasing sequence that is clearly bounded below by a 1. So it converges to its inf (call it b). Then by the difference property, lim n!" [b n # a n ] = b a (so b = a since this limit is assumed to be 0). Thus both sequences of endpoints converge to the same point. Since this point is the sup of the left endpoints, it is bigger than or equal to all the left endpoints. Since it is the inf of the right endpoints, it is smaller than or equal to all the right endpoints. Thus it is in each interval I n.
5 2. The Intermediate Value Theorem. Suppose f: [a, b] R is continuous and let c be a number that is strictly between f(a) and f(b). Then there exists a point x 0 in the open interval (a, b) at which f(x 0 ) = c. Proof Outline: Assume that f(a) < c < f(b). Let a 1 = a and b 1 = b. Split the interval [a 1, b 1 ]in half. Plug the midpoint, m 1, into the function. If the result is bigger than c, define [a 2, b 2 ] to be [a 1, m 1 ] and if the result is smaller than c, define [a 2, b 2 ] to be [m 1, b 1 ]. (If it is ever equal to c then you are done!) Then do it again and again and again You have made a sequence of nested intervals whose lengths are converging to zero. Thus by the nested interval theorem there is a point, x 0, to which both sequences of endpoints converge. Plugging these sequences into the function, we get two image sequences that converge to f(x 0 ). But since f(a n ) < c, for each index n, we know that f(x 0 ) c, and similarly, f(b n ) c. Thus f(x 0 ) = c. Extra Credit Opportunity! There is a gap (something left unproven) in the proof in the book (and I did not patch it in this outline). I ll add 3 points to your Exam 2 score if you can find it.
6 Class Work for Wednesday 1. Why is every polynomial a continuous function? 2. Prove that f(x) = 2x 3 + 3x 7 has a maximum, minimum, and root in the interval [-5, 10]. 3. Use the sequential definition of continuity to prove that f(x) = 3x 7 is continuous. 4. Use the ε-δ to prove that f(x) = 3x 7 is continuous. 5. Suppose that S is a set of real numbers that is not sequentially compact. Prove that either 1) there is an unbounded sequence of numbers in S, or 2) there is a sequence of numbers in S that converges to a point not in S. 6. Let a and b be real numbers with a < b. a. Find a continuous function on (a, b) with an image that is unbounded below. b. Find a continuous function on (a, b) that has a bounded (from below) image but does not have a minimum value. c. Find a continuous function on (a, b) that has both a maximum and a minimum value.
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