Solutions to Homework 9

Transcription

1 Solutions to Homework 9 Read the proof of proposition 1.7 on p. 271 (section 7.1). Write a more detailed proof. In particular, state the defintion of uniformly continuous and explain the comment whose sidelengths are less than δ/ n. Solution: Here are some additional details which further explain the content of Proposition 1.7. Definition: Let A R n. A function f : A R is said to be uniformly continuous if for every ɛ > 0 there exists a δ > 0 such that for all x, y A. x y < δ = f(x) f(y) The difference between the above definition and the definition of continuous is subtle. If f is continuous, then for each x X and for each ɛ > 0 there exists a δ > 0 (meaning that δ may depend on both x and ɛ) such that x y < δ = f(x) f(y) for all y A. But if f is uniformly continuous, then for each ɛ > 0 there exists a δ > 0 such that the above condition holds for all x, y A (meaning that δ does not depend on x). A uniformly continous function is obviously continuous. But the converse is not true. For example, if A = (0, 1) and f(x) = 1/x, then f is continuous on A, but it is not uniformly continuous on A. The point is that if x is close to zero, then δ needs to be chosen smaller than if x is not close to zero. (Draw a sketch and think about how small δ needs to be.) The following is a rigorous argument that f is not uniformly continuous: if f(x) = 1/x were uniformly continous on (0, 1), then given ɛ = 1/n, there would exist a δ > 0 such that 1 x 1 y = x y xy < δ xy < ɛ = 1 n for all x, y (0, 1) such that x y < δ. Thus, δ < xy n

2 for all x, y (0, 1) such that x y < δ. Choose y = x + δ/2 so that x y < δ. Substituting into the above inequality and solving for δ one obtains the following: δ < 2x2 2n x, for all x (0, 1). Now let x 0 +. Since the right-hand side of the above inequality approaches zero, δ = 0, which contradicts the assumption that δ > 0. If A is compact, then f is continous if and only if it is uniformly continuous. This is Theorem 1.4 in section 5.1. You should be aware of this result, as it plays an important role in many arguments involving partitions. It will be used in the proof of Proposition 1.7 below. Proposition 1.7 in section 7.1: Let R R n be a rectangle. Suppose f : R mathbbr is continuous. Then f is integrable. Proof (with extra details): To prove f is integrable, we appeal to the convenient criterion (Proposition 1.3). Let ɛ > 0. Since R is compact, by Theorem 1.4 in section 5.1, f is uniformly continous on R. Therefore, there exists a δ > 0 such that x y < δ = f(x) f(y) < ɛ vol(r). Choose a partition P of R such that every subrectangle has a side length less than r = δ/ n. This implies that if x and y belong to the same subrectangle, then x y < r r 2 = r n = δ n n = δ. Therefore, if x and y belong to the same rectangle, then f(x) f(y) < ɛ/vol(r). Since each subrectangle is compact and f is continous, f attains its maximum and minimum values, i.e. the supremum of all values of f(x) such that x belongs to a given rectangle is equal to f(x ) for some x 0 in the rectangle (and similarly, there exists a point x belonging to the subrectangle such that f(x ) is the minimum value. Saying this another way, if M is the supremum of f(x) for x belonging to a subrectangle S and m is the infimum, then there exist x, x S such that M = f(x ) and m = f(x ). Therefore, M m < ɛ/vol(r). It follows that

3 if the subrectangles in the partitition are enumerated S 1,..., S k, then U(f, P) L(f, P) = k (M i m i )vol(s i ) < ɛ vol(r) k vol(s i ) = ɛ. Therefore, f is integrable since it satisfies Proposition 1.3 (Convenient Criterion). Section 7.1, exercise #1: Suppose that f(x, y) = 0 if 0 y 1/2 and that f(x, y) = 1 if 1/2 < y 1. Prove that f is integrable on R = [0, 1] [0, 1] and compute R f da. Solution: Let ɛ > 0. Choose a positive integers n such that 1/2n < ɛ. Choose a partion P = P 1 P 2 as follows: let P 1 = {0 = x 0 < 1 2n = x 1 < 2 2n = x 2 < < 2n 2n = 1 = x 2n and let P 2 = {y 0 = 0 < y 1 }. By design, x n = 1/2. If we enurate the subrectangles S 1,..., S 2n, where S i = [x i 1, x i ] [0, 1], then M i = 0 if 1 i (n 1) and M i = 1 if n i 2n. (This is due to the fact that the line y = 1/2 intersects the subrectangle S n.) On the other hand, m i = 0 if 1 i n and m i = 1 if n + 1 i 2n. Therfore, U(f, P) L(f, P) = 2n (M i m i )area(s i ) = (M n m n ) 1 2n = 1 2n < ɛ. Therefore f is integrable by Proposition 1.3. Furthermore, since L(f, P) f da U(f, P) R it follows that R f da = 1/2. (A direct computation shows that L(f, P) = 1/2 and U(f, P) = (n + 1)/2n, and the latter decreases to a limit of 1/2 as n. ) Section 7.1, exercise #2: Show directly that the function f(x, y) = 1 if x = y and f(x, y) = 0 if x y is integrable on R = [0, 1] [0, 1] and compute R f da.

4 Solution: Let ɛ > 0. Choose a positive integer n such that 1/n < ɛ. Let P = P 1 P 2, where P 1 = P 2 = {0 < 1/n < 2/n < < n/n = 1}. The subrectangles can be enumerated by the doubly indexed sequence {S i,j } where i, j = 1,..., n and where S i,j is the subrectangle whose upper right vertex is (i/n, j/n). The line y = x intersects the rectangle S i,j if and only if i = j. Therefore M i,j = m i,j = 0 if i j. And if i = j, then M i,j = 1 and m i,j = 0. (The former is easy to see, and the latter seems clear. A clarification of the latter is that the line y = x can meet at most two vertices of a rectangle, and so there are at least two vertices which do not lie on this line. Therefore there are points in S i,i whose f-values are equal to zero.) A direct computation shows that n U(f, P) L(f, P) = (M i,j m i,j )area(s i,j ) = i,j=1 n (M i,i m i,i ) 1 n 2 = n 1 n 2 < ɛ. Therefore f is integrable by Proposition 1.3. As in the previous problem, the lower and upper sums with respect to P can be computed. The lower sum is equal to zero and the upper sum is equal to 1/n. Since 1/n limits to zero as n, it follows that R f da = 0. Section 7.1, exercise #3: Show directly that the function f(x, y) = 1 if y < x and f(x, y) = 0 if x y is integrable on R = [0, 1] [0, 1] and compute R f da. Solution: The solution is very similar to the previous exercise. Let ɛ > 0 and choose P as in the previous exercise. In this case, M i,j = 1 if j i and m i,j = 1 if j < i; similarly, M i,j = 0 if i < j and m i,j = 0 if i j. It follows that the difference of the upper and lower sums is again less that 1/n. (You ll want to write this out.) Therefore, f is integrable by Proposition 1.3. The lower and upper sums can be computed directly. The lower sum is equal to (n 1)n/2n 2 and the upper sum is equal to n(n + 1)/2n 2 ; both limit to 1/2, and so R f da = 1/2. (To compute these sums you will want to use the following identity: n = n(n + 1). 2

5 Section 7.1, exercise #4: Show directly that the function f(x, y) = 1 if x {1/2, 1/3, 1/4,... } and f(x, y) = 0 otherwise is integrable on R = [0, 1] [0, 1] and compute R f da. Strategy: We again appeal to the convenient criterion (Proposition 1.3). The idea is to choose a partition which cuts R into thin vertical strips which are very narrow if the strip contains the line x = 1/k for some k and which are as wide as possible if the strip does not contain such a line. This is not possible to achieve for all k, but it can be achieved for finitely many such k. And fortunately, the remaining such vertical lines, although infinite in number, accumulate in a region of smaller and smaller area. Solution: Let ɛ > 0. Choose a positive integer n such that 3/n < ɛ. Let P = P 1 P 2 where P 1 = {0 < 1 n + 1 n 2 < 1 n 1 1 n 2 < 1 n n 2 < 1 n 2 1 n 2 < < 1 k + 1 n 2 < 1 k 1 1 n 2 < 1 k n 2 < 1 k 2 1 n 2 < < n 2 < n 2 < 1 1 n 2 < 1} and let P 2 = {0 < 1}. The first subrectangle intersects the verticle line x = 1/n, and so M 1 = 1. The second subrectangle does not intersect any vertical line of the form x = 1/m for some positive integer m, and so M 2 = 0. Every other subrectangle has M = 1 followed by M = 0. The subrectangles having M = 1 have a width of 2/n with the exception of the first and last rectangle. Thus, U(f, P) = ( 1 n + 1 n 2 ) + 2 n n n 2 = ( 1 n + 1 n 2 ) + (n 2) 2 n n 2 = 1 n + (n 1) 2 n 2 < 3 n < ɛ Since L(f, P) = 0, f satisfies the criterion of Proposition 1.3 and so f is integrable. The value of R f da is equal to zero since the upper sum above is less than 3/n which tends to zero as n. Section 7.1, exercise #5: Let R R n be a rectangle. Suppose that f : R R is nonnegative, continuous, and positive at some point a R.

6 Prove that R f dv > 0. Show that the result is false if f is not assumed to be continuous. The solution below is short on details because the details will obscure the argument. Try to read it through once or twice and then try to identify which statements might require some justification. Can you justify these statments? Solution: Let W be the open interval (f(a)/2, ). Since f(a) > 0 and since f is continuous, the set U = f 1 (W ) is open and nonempty and contains a. Let V be the interior of R. The set V U is open and nonempty; it is nonempty since either a V or a belongs to the frontier of R in which case every open set containing a must intersect the interior of V. Let b V U and choose a closed square S with center b and with edges of length δ > 0 such that S V U. By design, f(x) > f(a)/2 for every x S. Let g(x) = f(a)/2 for all x S and let g(x) = 0 for all x / S. This function is integrable (since g is continous at every point of R except at points belonging to the frontier of S, and the frontier of S has volume zero so Proposition 1.8 applies) and f(x) g(x) for all x R. By Proposition 1.11, f dv g dv = 1 2 f(a)δ2 > 0. R R The result is false if f is not assumed to be continous. For example, let f be the function in exercise 2 of section 7.1. Section 7.1, exercise #6: Let R n be a region and suppose f : R is continuous. (a) If m and M are, respectively, the minimum and maximum values of f, prove that m vol() f dv M vol(). Solution: Let g(x) = m for all x R n. Then g is integrable on. By Proposition 1.11 and Proposition 1.5, f dv g dv = m 1 dv = m vol().

7 The inequality involving M is proved in a similar manner. (b) (Mean value theorem for integrals) Suppose that is (path) connected. Prove that there is a point c such that f dv = f(c) vol(). Solution: By part (a), the value of the integral lies between m vol() and M vol(). And the image of f is the interval [m, M] R. Since f is continous, the function f attains every value in between m and M by the intermediate value theorem, and so, in particular, f attains the value of f dv. The intermediate value theorem is not directly applicable since the domain is not a closed bounded interval in R. The following argument fills this gap in the above argument. Suppose that f(a) = m and f(b) = M, where a, b. Let g : [0, 1] be a path joining a to b. The function (f g) : [0, 1] R is continuous. Since f(g(0)) = m and f(g(1)) = M, by the intermediate value theorem there exists a t [0, 1] such that f(g(t)) = f dv. Let c = g(t).