Math 312 Spring08 Day 13

Transcription

1 HW 3: Due Monday May 19 th # 2, 3, 6, # 1, 3, 6 Math 312 Spring08 Day Continuous Functions and Mappings. 1. Definition. Let A be a subset of R n i. A mapping F: A R n is said to be continuous at the point u in A provided that whenever a sequence {u k } in A converges to u, the image sequence {F (u k )} converges to F(u). ii. A mapping F: A R n is said to be continuous provided that it is continuous at every point in its domain. 2. Proposition 11.1 For each index i with 1 i n, the ith component projection function p i : R n R is continuous. Proof: Follows directly from the Componentwise Convergence Criterion (Do you see how?)

2 3. Theorem 11.3 Let A be a subset of R n that contains the point u and suppose that the functions h: A R and g: A R are both continuous at u. i. For any real numbers α and β, the function α h + βg: A R is continuous at u. ii. Also the product function hg: A R is continuous at u. iii. Furthermore if g(v) 0 for all v in A, then the quotient function h/g: A R is continuous at u. Proof: Just like in the old days, these follow directly from the sum, product, and quotient properties for sequences of real numbers. 4. Theorem 11.5 Let A be a subset of R n that contains the point u. Suppose that the mapping G: A R m is continuous at the point u. Let B be a subset of R m with G(A) B and suppose that the mapping H: B R k is continuous at the point G(u). Then the composition H!G: A R k is continuous at u. Proof: Let {u k } be a sequence in A that converges to the point u

3 5. Definition. Given a mapping F: A R m, where A is a subset of R n, and an index i with 1 i m, we define the function F i : A R, to be the composition of F with the ith component projection function. We call the function F i the ith component function of F. Thus F(u) = (F 1 (u),, F m (u)) 6. Theorem 11.9 (The Component Continuity Criterion) Let A be a subset of R n that contains the point u. The mapping F = (F 1,, F m ) : A R m is continuous at u if and only if each of its component functions F i : A R is continuous at u. Proof: Follows directly from the Componentwise Convergence Criterion 7. Corollary Let A be a subset of R n that contains the point u and suppose that the mappings G: A R m and H: A R m are both continuous at the point u. Then for any real numbers α and β, the mapping α H + βg: A R m is continuous at u. Proof: Follows directly from (you guessed it!) the Componentwise Convergence Criterion (and Theorem 11.3).

4 8. Theorem Let A be a subset of R n that contains the point u. The following assertions about F: A R m are equivalent: i. The mapping F: A R m is continuous at the point u; that is, for a sequence {u k } in A, lim dist(f(u k ), F(u)) = 0 if lim dist(u k, u) = 0. k!" k!" ii. For every ε > 0, there exists a δ > 0 such that for every v in A, dist(f(u), F(v)) < ε if dist(u, v) < δ. Proof: Let F: A R m. Suppose that ii) is not true. Then there exists an ε > 0 for which no adequate δ > 0 exists. Then in particular, for every k in N, there is a point v k in A with the property that dist(u, v k ) < 1/k but dist(f(u), F(v k )) ε. First note that the sequence {v k } in A clearly converges to u, so lim k!" dist(v k, u) = 0. However, since for each index k, dist(f(v k ), F(u)) ε, it is just as clear that lim k!" dist(f(v k ), F(u)) # 0. Thus i) is not true. So by contrapositive, we have shown that i) ii). Now suppose that ii) is true and let {u k } be a sequence in A for which lim k!" dist(u k, u) = 0. Let ε > 0. By ii) there exists a δ > 0 such that for every v in A, dist(u, v) < δ dist(f(u), F(v)) < ε. Now choose K such that k K dist(u k, u) <!. Then k K dist(f(u k ), F(u)) < ε. Therefore lim k!" dist(f(u k ), F(u)) = 0. Thus i) is true. So we have proven that i) and ii) are equivalent.

5 Theorem Let O be an open subset R n of and consider the mapping F: O R m. The following are equivalent. i. The mapping F: O R m is continuous. ii. F -1 (V) is an open subset of R n whenever V is an open subset of R m. Note: F -1 (V) = {u in O: F(u) is in V}, there is NO assumption that F is invertible. Proof: Saved for Wednesday Sequential Compactness, Extreme Values, and Uniform Continuity 1. Definition: Let A be a subset of R n. Then A is said to be sequentially compact provided that every sequence in A has a subsequence that converges to a point in A. 2. Definition: A subset A of R n is said to be bounded provided that there is a number M such that u! M for all points u in A.

6 3. Theorem (Balzano-Weierstrass Theorem) A subset of R n is sequentially compact if and only if it is closed and bounded in R n. Proof: Suppose the set A of R n is sequentially compact. Then A is closed since given any sequence in A that converges, this sequence must have a subsequence that converges to a point in A (so the limit of the whole sequence is a point in A). Now suppose that A is not bounded. Then for any natural number, k, there is a point u k in A that has a norm bigger than k. But then the sequence {u k }has a subsequence {u ki } that converges to some point u in A. But then the sequence of real numbers u ki converges to u. This is a contradiction since the sequence u ki is unbounded. Now suppose that the subset A of R n is closed and bounded. Let {u k } be a sequence in A. Since the sequence {u k } is bounded, it has a convergent subsequence (generalization using induction of the fact that a bounded sequence of real numbers has a convergent subsequence). Now since A is closed, this subsequence must converge to an element of A. Thus A is sequentially compact. 4. Theorem Let A be a subset of R n and suppose that the mapping F: A R m is continuous. If the domain A is sequentially compact, then the image F(A) is also sequentially compact.

7 Proof: To do on Wednesday 5. Lemma Every nonempty sequentially compact subset of R has a smallest and largest member. Proof: Let A be a nonempty sequentially compact subset of R. Since A is bounded, it has a greatest lower bound (call it a) and a least upper bound (call it b). Now for each k in N, there exists an element a k in A such that a a k < a +1/k and an element b k in A such that b -1/k < b k b. Note that the sequence {a k } converges to a, and the sequence {b k } converges to b. These sequences are both in A, so since A is closed, a and b are in A as well! Theorem (The Extreme Value Theorem) Let A be a nonempty sequentially compact subset of R n and suppose that the function f:a R is continuous. Then the function f:a R attains a smallest and largest value. Proof. f(a) is sequentially compact by Theorem But then by Lemma 11.21, since f(a) lives in R, it has a smallest and largest element (thus f attains a smallest and largest value). 6. Definition. A nonempty subset A of R n is said to have the Extreme Value Property provided that every continuous function f: A R has a maximum and a minimum functional value.

8 7. Theorem A nonempty subset of R n has the Extreme Value Property if and only if it is sequentially compact.