Selected Solutions. where cl(ω) are the cluster points of Ω and Ω are the set of boundary points.

Transcription

1 Selected Solutions Section 9 (Open and Closed Sets) 9K. Let Ω R p. The original problem asks: Show a point belongs to Ω if and only if it is an interior point. Let nt(ω) be the set of all interior point of Ω. We show Ω = nt(ω) = Ω\ Ω, where Ω are the set of all boundary points of Ω. t follows from this that R p = Ω Ω Ext(Ω), where Ext(Ω) is the set of all exterior points of Ω. Solution. Recall Ω = α G α, where G α is any set contained in Ω. Suppose x Ω. Then x α G α, and so x G α some α. Since G α Ω is open, x is an interior point of Ω. That is, Ω nt(ω). Conversely, suppose x nt(ω). Then r > 0 exists so that B r (x) Ω. Since B r (x) is open, B r (Ω) α G α = Ω, and x Ω. t follows nt(ω) Ω. The first equality follows. By definition of the interior points of Ω, the second equality follows. 9M. The original problem asks: Show a point belongs to Ā if and only if it is either an interior or a boundary point of A. Let Ω R p. We show Ω = Ω Ω = Ω Ω = Ω cl(ω), where cl(ω) are the cluster points of Ω and Ω are the set of boundary points. Solution. Recall Ω = α F α where F α is any closed set containing Ω. To show the first equality, suppose x Ω Ω. Then x must be an exterior point of Ω. That is, r > 0 exists so that B r (x) Ω C, where Ω C = R p \Ω. t follows Ω B r (x) C. Since B r (x) C is a closed set containing Ω, Ω = α F α B r (x) C. This means B r (x) is completely contained in Ω C. Thus x is an exterior point of Ω. That is, x Ω. The contrapositive of this statement is, if x Ω, then x Ω Ω. That is, Ω Ω Ω. Conversely, suppose x Ω Ω. We have to show x F α for all closed sets F α containing Ω. Suppose x Fα C with Ω F α. Then Fα C ΩC. Since Fα C is open, this implies x is an exterior point of Ω. The contrapositive of this is, if x is not an exterior point of Ω, then x F α for all closed sets containing Ω. That is, Ω Ω Ω. The first equality follows. Problem 9K shows Ω = Ω\ Ω. Thus Ω Ω = Ω Ω. The second equality follows. To show the last equality, suppose x Ω Ω. f x Ω, there is noting to show. f x Ω but x Ω, then every neighborhood of x contains a point in Ω and a distinct point, x, not in Ω. That is, x cl(ω) and Ω Ω Ω cl(ω). Conversely, suppose x Ω cl(ω). Again if x Ω, there is nothing to show. f x cl(ω) but x Ω, then every neighborhood of x contains a point in Ω and a point not in Ω, and so x Ω. That is, Ω cl(ω) Ω Ω. The third equality follows. 1

2 Section 11 (Compact Sets) 11C. Prove directly that, if K R p is compact and F K is closed, then F is compact. Solution. Let {G α } be a family of open sets covering F. That is, F α G α. Since F is closed F C is open. Note that K ( α G α ) F C. Since K is compact, K is covered by a finite number of these sets. That is, K ( N i=1 G i) F c for some N N. Since F K, F ( N i=1 G i) F c. However, since F and F C are disjoint, F N i=1 G i, and F is compact. 2

3 Section 12 (Connected Sets) 12B. Suppose C R p is connected and x cl(c). Then C {x} is connected. Solution. As usual we suppose that C = C {x} is disconnected. Then open sets A and B exists so that A C and B C are nonempty, disjoint, and have union C. We show this implies C is disconnected - a contradiction. f x C, then C = C and we immediately conclude C is disconnected and we have our contradiction. Suppose x C. The x is in either A or B. Suppose, without loss of generality, x A. Since A is open and x cl(c), A contains a points in C, and so A C. Since A C and B C are disjoint x B. Thus B C = B C. So far we have Since x A C and x B C, we conclude (A C ) (B C ) = (A C ) (B C) = C. (A C) (B C) = C, and A, B provide the disconnection and the needed contradiction. 12BC. (Not in Book) Suppose {E α } α, where is an index set, are connected for all α and that E α. Then α is connected. α α E Solution. To simplify the notation set E = α E α. Suppose E is not connected. Then open sets A, B exists so that A E and B E are nonempty, disjoint, and have union E. Of course we seek a contradiction. The idea is to show A and B separate E into two components-contradicting the existence of a point common to all E α. Claim #1. With A and B as above, for any α, (A E α ) (B E α ) = E α. Regardless if the E α are connected or not. Proof. For any sets A, B, and E α, (A E α ) (B E α ) E α. f this were a proper subset, then x E α would exist so that x A and x B, but this contradicts x (A E) (B E) = E, and so (A E α ) (B E α ) = E α. Claim #2. With A and B as above, for any α, either A E α = or B E α = but not both. Moreover, for each α, either E α A or E α B but not both. Proof. Here we need that the E α are connected. ndeed, suppose for some index α both A E α and B E α were nonempty. From the previous claim we still have (A E α ) (B E α ) = E α. Moreover, the assumptions on A and B imply that A E α and B E α are disjoint. Thus A and B provide a disconnection of E α - a contradiction. The other statement in the claim now follows. Now weare in alogical bind. Since α E α, thereis a x E α for all α. Thiscontradicts Claim #2, and so no such disconnection of E exists. Thus E is connected. 3

4 12C. The original problem asks: if C R p is connected, show C is connected. We show, assuming C is connected, that any set E satisfying is connected. C E C Solution. From Problem 12B we know C {x} is connected if C is and x is a cluster point of C. So let E α = {C {α} α cl(c)}. Then each E α is connected by Problem 12B. Problem 12BC above applies and any union of the E α are connected. But the set E in the theorem has this form (by Problem 9M). 12H. Show the set is connected. Solution. We first consider the set Ω = {(x,y) 0 < y x 2,x 0} {(0,0)} E = {(x,y) 0 < y < x 2,x > 0}. This set is open so we apply Theorem 12.7 in the book. Specifically, we show E is polygonally connected. Let (x 1,y 1 ) and (x 2,y 2 ) be any two points in E. There are two possibilities: x 1 = x 2. We also assume without loss of generality y 1 < y 2. n this case L = {(x 1,y 1 +t(y 2 y 1 )) 0 t 1} is a line segment joining (x 1,y 1 ) and (x 1,y 2 ). The end points must satisfy, 0 < y 1 < x 2 1 and 0 < y 2 < x 2 1. We need to check y < x2 for all points in L. Under our assumptions, for any 0 t 1, 0 < y 1 y 1 +t(y 2 y 1 ) = y y 2 < x 2 2 = x. That is, y < x 2 for all 0 t 1, and hence, for all points in L. Suppose x 1 x 2. Then, without loss of generality, we may assume x 1 < x 2. n this case we connect (x 1,y 1 ) and (x 2,y 2 ) with two lines. We set L 1 = {(x 1 +t(x 2 x 1 ),y 1 ) 0 t 1}, L 2 = {(x 2,y 1 +t(y 2 y 1 )) 0 t 1}. We note the end point of L 1, (x 2,y 1 ), is the same as the beginning point of L 2. To show L 1 E we have to show y < x 2 for each point in L 1 (i.e. for each 0 t 1). We have y = y 1 < x 2 1 [x 1 +t(x 2 x 1 )] 2 x 2 2, and so y < x 2 for all (x,y) L 1. The treatment of L 2 is exactly as in the first case above. 4

5 Theorem 12.7 applies and E is connected. Next we apply Problem 12C. We may add as many cluster points of E as we like and the resulting set is still connected. Thus E r = {(x,y) 0 < y x 2,x 0} {0,0)} is connected. By symmetry (or following the same arguments) the set E l = {(x,y) 0 < y x 2,x 0} {0,0)} is connected. Finally, we apply Problem 12BC (made up problem above). We have E l E r = {0,0} and so Ω = E l E r is connected. To show Ω is not polygonally connected consider the point (0,0) and any other point (x,y), x > 0 in Ω. Any line segment through the origin must have the form y = mx for some m > 0. (f m 0, we see the line segment is not in Ω.) For such a line segment to be in Ω we need y = mx x 2 for x > 0. This requires x(x m) > 0 for x > 0. This inequality is seen to be false for any 0 < x < m, and so the line y = mx exits the parabola for all positive slopes m. Thus Ω is not polygonally connected. 5

6 Section 21 (Linear Functions) 12L. Suppose f L(R p,r q ), the space of linear functions. Define the operator norm to be Show that (a) f pq is a norm, (b) f(x) f pq x for all x R p. f pq = sup f(x). x 1 Solution. We need to check Definition 8.5 on Page 54. Since f(x) 0 for all x R p, f pq 0 and (i) is verified. f f pq = 0, then f(x) = 0 for all x 1. Let z R p and not zero. Then z z has norm one and so ( ) 0 = z f = 1 z z f(z). That is, f(z) = 0 for all z R p, and so f = 0. This verifies (ii) in Definition 8.5. Note, for a R, af pq = sup af(x) = sup a f(x) = a sup f(x) = a f pq. x 1 x 1 x 1 This verifies (iii). Finally note f +g pq = sup f(x)+g(x) x 1 sup ( f(x) + g(x) ) x 1 sup f(x) + sup g(x) x 1 x 1 = f pq + g pq, and f pq is a norm. x To prove (b), note that for any nonzero x R, x has norm one. By definition of supremum ( ) x f f pq. x By the linear properties of f this is the same as f(x) f pq x. This is (b). 6

7 12L. Suppose f L(R p,r q ), the space of linear functions. Set E 1 = { f(x) x 1}, E 2 = { f(x) x = 1}, E 3 = {M f(x) M x, x R p }. Prove supe 1 = supe 2 = infe 3. Solution. These sets are nonempty and, since q f(x) i=1 j=1 p c 2 ij x, each set is bounded above and bounded below by zero. Thus all three have a supremum and an infimum. We set f pq = supe 1, α = supe 2, and β = infe 3. We claim f pq = α = β. To see f pq = α, we have, since E 2 E 1, α f pq. However, ( ) 1 sup f(x) sup f(x) = sup x x 1 x 1 x f = sup f(x). x 1 x x =1 This implies f pq α, and so f pq = α. To see α = β, we note E 3 is the same as E 3 = {M f(x) M, x = 1} by the linearity properties of f. By the definition of supremum we have α = β. Proof #2 Here is a proof not involving the set E 2. Since, by the previous problem, f(x) f pq x, we must have β f pq. Let M E 3. Then, for any x 1, we have f(x) M x M, and so M is an upper bound for E 1. By definition of supremum f pq M. This means f pq is a lower bound for the set E 3. Hence, f pq β, and so β = f pq. 21O. Give an example of linear map in L(R p,r q ) such that q p f pq < c 2 ij. i=1 j=1 Solution. We claim (as an exercise) f pq is equal to the square root of the double sum when p = 1 or q = 1 or p = q = 1. Suppose p 2 and q 2. Set c 11 = 1 and c 22 = 1 and c ij = 0 otherwise. Then f pq = 1 while the square root of the double sum is 2. 7

8 Section 22 (Global Properties of Continuous Functions) 22F. A set Ω R p is disconnected if and only if there exists a continuous function f : Ω R such that f(ω) = {0,1}. Solution. Suppose Ω is disconnected. Then open sets A and B provide a disconnection of Ω. Set { 0 x A Ω f(x) = 1 x B Ω. An application of Theorem 22.1, the global continuity theorem, shows f is continuous. Alternatively, let V be any neighborhood of 0. For any x A Ω, we have f(x) V. A similar statement holds for any neighborhood of 1. Since Ω = (A Ω) (B Ω), f is continuous on Ω. Conversely, suppose f is continuous and f(ω) = {0, 1}. We could apply the contrapositive of Theorem 22.3-the preservation of connectedness. Since {0, 1} is readily seen to be disconnected, Ω must be disconnected. Alternatively, by the global continuity theorem, f(( 1/2, 1/2)) = A Ω for some open set A. Similarly, f((1/2,3/2)) = B Ω for some open B. By the definition of a function A Ω and B Ω must be disjoint. Thus A and B form a disconnection of Ω. 8

9 Section 39 (The Derivative) 39H. Let f : R 2 R 2 be given by F(x,y) = (x 2 +x 2 sin 1x,y ) x 0, = (0,y) x = 0. Show that D 1 F exists at every point and that D 2 F exists and is continuous in a neighborhood of (0,0). Show that F is Fréchet differentiable at (0,0). Solution. Since F is vector valued, we need only check each component. Set F = (F 1,F 2 ). We calculate D 1 F 1 (0,0) = lim t 0 F 1 (t,0) F 1 (0,0) t = lim t 0 t+tsin(1/t) = 0. By the algebra of differentiable functions and the one-dimensional chain rule, D 1 F exists for all (x,y) R 2. We easily see D 1 F 2 = 0, and D 2 F 2 = 1 for all x,y. Thus both are continuous on R 2. We do not have enough to apply Theorem However since the Gâtaux derivatives exists, we must have, if the Fréchet derivative exists, D u F(0,0) = DF(0,0)(u). The above calculations show that, if the Fréchet exists, it must represented by the 2 2 matrix ( ) 0 0 DF(0,0) =. 0 1 Now we just check F(u 1,u 2 ) F(0,0) DF(0,0)(u) = (u 2 1 (1+sin(1/u 1)),0) u 2 1 (u 1,u 2 ) 2. Thus, we take δ = ǫ in the definition of the derivative and we see F is Fréchet differentiable at (0,0). 9

10 39T. Let Ω R p and f : Ω R be such that D 1 f,d 2 f,...d p f exist and are bounded on some neighborhood of c Ω. Show f is continuous at c. Solution. Here is a proof when p = 2. Applying the one-dimensional mean-value theorem, we find in the given neighborhood, f(x,y) f(c 1,c 2 ) f(x,y) f(x 1,y) + f(c 1,y) f(c 1,c 2 ) = f x (c 1,y)(x c 1 ) + f y (c 1,c 2)(y c 2 ) M 1 x c 1 +M 2 y c 2 2max{M 1,M 2 } x c, where c 1 is between x and c 1, c 2 is between y and c 2, and M 1, M 2 are the bounds on the partial derivatives. This shows f is continuous at c.. 39U. Let Ω R 2 and f : Ω R. Suppose D 1 f exists and is continuous on a neighborhood of c and that D 2 f exists at c Ω. Show f is Fréchet differentiable at c. Solution. The quantity we must show is small is f(x,y) f(c 1,c 2 ) D 1 f(c)(x c 1 ) D 2 f(c)(y,c 2 ) f(x,y) f(c 1,y) D 1 f(c)(x c 1 ) + f(c 1,y) f(c 1,c 2 ) D 2 f(c)(y c 2 ). On the first term on the right we apply the mean-value theorem to find ( ) f(x,y) f(c 1,y) D 1 f(c)(x c 1 ) D 1 f(c 1,y) D 1f(c 1,c 2 ) (x c 1 ) D 1 f(c 1,y) D 1f(c 1,c 2 ) x c 1, where c 1 (y) is between x and c 1. Given any ǫ > 0, the continuity of D 1 implies the existence of a δ 1 > 0 such that for all x c < δ 1, D 1 f(c 1,y) D 1f(c 1,c 2 ) < ǫ/2. To deal with the second term above, set u = e 2, t = y c 2 in the definition of D 2 f(c). Then δ 2 > 0 exists so that f(c+tu) f(c) D 2 f(c) t < ǫ/2 for all 0 < t < δ 2. This is the same as (even if t < 0), f(c 1,y) f(c 1,c 2 ) D 2 f(c) y c 2 < ǫ/2 for y c 2 < δ 2. Set δ = min{δ 1,δ 2 }. Returning to our first inequality, for x c < δ, f(x,y) f(c 1,c 2 ) D 1 f(c)(x c 1 ) D 2 f(c)(y,c 2 ) < ǫ 2 x c 1 + ǫ 2 y c 2 ǫ 2 2 x c < ǫ x c and the result follows. 10

11 Section 40 (The Chain Rule and MVT) 40H. Let f,g : Ω R q and c Ω with f and g differentiable at c. Set h(x) = f(x) g(x). Show h is differentiable and for all u R p. Solution. We have to estimate Dh(c)u = (Df(c)(u)) g(c)+(dg(c)(u)) f(c) f(c+u) g(c+u) f(c) g(c) ( ) f(c) Dg(c)(u)+g(c) Df(c)(u). By adding and subtracting appropriate terms we may write this as the sum of the two terms ( ) ( ) f(c+u) g(c+u) g(c) Dg(c)(u) + f(c+u) f(c) Dg(c)(u) and ( ) g(c) f(c+u) f(c) Df(c)(u). Let ǫ > 0. We have to estimate each term in these two terms. We do them in order. Since f is Fréchet differentiable at c, it is continuous at c. By setting ǫ = 1 in the definition of continuity, we can conclude f is bounded in a neighborhood of c. Thus δ 1 > 0, M 1 > 0 exist so that f(c + u) M for all u < δ 1. Since g is Fréchet differentiable at c, δ 2 > 0 exists so that g(c+u) g(c) Dg(c)(u) < ǫ u for all u < δ 2. Continuity of f implies δ 3 > 0 exists so that f(c + u) f(c) < ǫ for all u < δ 3. Moreover, Dg(c)(u) Dg(c) pq u. Finally, δ 4 > 0 exists so that f(c + u) f(c) Df(c)(u) < ǫ u for all u < δ 4. Set δ = min{δ 1,δ 2,δ 3,δ 4 }. Then, for u < δ, f(c+u) g(c+u) f(c) g(c) ( f(c) Dg(c)(u)+g(c) Df(c)(u)) Mǫ u +ǫ Dg(c) pq u + g(c) ǫ u = ǫ(m + Dg(c) pq + g(c) ) u. The result follows. 11

12 Section 41 (Mapping Theorems and mplicit Functions) 41J. Let h : R R be given by h(x) = { x+2x 2 sin 1 x x 0 0 x = 0. Show h does not belong to C 1 (R) and that h is not injective on any neighborhood of 0. However, h is surjective on a neighborhood of 0 and Dh(0) is invertible. Solution. One calculates using the definition of a derivative that h (0) = 1. For x 0 h is easy to calculate and we find h (x) = 1+4xsin 1 x 2cos 1 x. We see h is not continuous at the origin since h (x) does not have a limit as x approaches the origin. To see h is not injective on any neighborhood of the origin let B ǫ (0) be any ǫ-neighborhood of the origin. Choose n so large that 1 π 2 +2πn < 1 2πn < ǫ. Note that ) while h ( 1 π 2 +2πn = 1+ 4 π 2 +2πn > 0 ( ) 1 h = 1 2 < 0. 2πn Since h is continuous for x > 0, the intermediate-value theorem applies x 0 exists between these two points so that h (x 0 ) = 0. Together these indicate x 0 is a relative minimum. This implies h is not injective near x 0, and hence, h is not injective on any neighborhood of the origin. Next we show a neighborhood of the origin exists where h is surjective. Note that, for x 1, sin(1/x) > 0, and so h(x) x for such x. Similarly, for x 1, sin(1/x) < 0, and so h(x) x for x 1. By the intermediate-value theorem h takes on all values between [ 1,1] and so, h is onto the neighborhood ( 1, 1). 41L. Let L L(R p,r p ). Prove (a) Suppose L 0 is invertible and L L 0 is sufficiently small. Show L is invertible. (b) Again suppose L 0 is invertible. Show that the map L L 1 is continuous in a neighborhood of L 0. (c) f F : Ω R p with Ω R p open, f C 1 (Ω), and, for some c Ω, Df(c) is invertible, show Df(x) is invertible in some neighborhood of c. Solution. Suppose L 1 0 exists. Recall that a linear operator in invertible if and only if m 0 > 0 exists so that x m 0 L 0 (x). 12

13 We have We require L 0 L op 1/(2m 0 ). Then x m 0 L 0 (x) ) m 0 ( (L 0 L)(x) + L(x) m 0 L 0 L op x +m 0 L(x). x 2m 0 L(x), and Part (a) follows. To prove Part (b) we set f(l) = L 1. We need to show f is continuous near L 0. We know from Part (a) that, if L B 1 (L 0 ), it is invertible. Let L 1 and L 2 be two such operators, and 2m 0 let y R p. We need to show we can make L 1 1 (y) L 1 2 (y) small for L 1 L 2 op small enough. Set x 1 = L 1 1 y and x 2 = L 1 2 y so that L(x 1) = L 2 (x 2 ). Since L 1 is invertible, we have, for some m 1 > 0, This implies x 1 x 2 m 1 L 1 (x 1 x 2 ) m 1 L 1 (x 1 ) L 1 (x 2 )+L 2 (x 2 ) L 2 (x 2 ) m 1 (L 1 L 2 )(x 2 ) m 1 L 1 L 2 op x 2 m 1 L 1 L 2 op L 1 2 op y. L 1 1 L 1 2 op m 1 L 1 2 op L 1 L 2 op. This shows that, in fact, the map f(l) = L 1 is uniformly continuous on B 1 (L 0 ). 2m 0 For Part (c) we can apply Part (a). We just need to show we can make Df(x) Df(c) op small as we need. Since f C 1 (Ω) we may compute the Fréchet derivatives using partial derivatives. So Df(x) Df(c) 2 op is bounded by the square of the sums of terms like D i f j (x) D i f j (c) which can be made small by the continuity of the derivatives. 41S Suppose f : R 2 R is C 1 (R 2 ). Show that f is not injective on any open set in R 2. Solution Let Ω be any open set and suppose D 1 f(c) 0 for some c Ω. Set F : R 2 R 2 to be The Fréchet derivative is F(x,y) = (f(x,y),y). DF(x) = ( fx f y 0 1 Our assumptions imply DF(c) is surjective. Thus the surjective mapping theorem applies and F is onto in a neighborhood of F(c). Let (u,v 1 ) and (u,v 2 ) with v 1 v 2 and both points in a neighborhood of F(c) where F is guaranteed to be surjective by the surjective mapping theorem. ). 13

14 Thus (x 1,y 1 ) and (x 2,y 2 ) exist in a neighborhood of c such that F(x 1,y 1 ) = (u,v 1 ) and F(x 2,y 2 ) = (u,v 2 ). That is, F(x 1,y 1 ) = (u,v 1 ) = (f(x 1,y 2 ),y 1 ), F(x 2,y 2 ) = (u,v 2 ) = (f(x 1,y 2 ),y 2 ). t follows f(x 1,y 1 ) = f(x 2,y 2 ) while (x 1,y 1 ) (x 2,y 2 ) and f is not injective. f D 1 f = 0 on some open set of R 2, then applying the mean-value theorem along the x axis we see f does not depend on x and again f is not injective. 41S Supposeg : R R 2 is C 1 (R). Show that if c R, thentherestriction of g toany neighborhood of c is not a surjective map onto a neighborhood of g(c). Solution Let g(x) = (g 1 (x),g 2 (x)) and suppose D 1 g 1 (c) 0. Set G : R 2 R 2 to be G(x,y) = (g 1 (x),g 2 (x)+y). The Fréchet derivative is DG(x,y) = ( g 1 (x) 0 g 2 (x) 1 Our assumptions imply DG(c, 0) is injective. Thus the injective mapping theorem applies and we see G is injective in a neighborhood of (c,0) to a neighborhood of G(c,0). The proof g is not surjective is by contradiction. Suppose g is onto in a neighborhood of g(c) and G is injective on this neighborhood. Let us denote the neighborhood B r (g(c)). Let (u,v) B r (g(c)). We can perturb v some and still remain in this neighborhood. That is, v+η B r (g(c)) for some real η 0. Then x 1 exists so that g(x 1 ) = (u,v) or G(x 1,0) = (u,v) and G(x 1,η) = (u,v +η). Since we are assuming g is a surjective, a x 2 exists so that g(x 2 ) = (u,v +η). That is, G(x 1,η) = G(x 2,0). This implies G is not injective - a contradiction, and so g must not be surjective. f g 1 (c) = 0 but g 2 (c) 0, we can change G(x,y) = (g 1+y,g 2 ) and reach the same conclusions. Thus suppose D 1 g(c) = 0. ). 14

15 Section 43 (The ntegral in R p ) 43L. Let Ω R p be bounded and, J be two cells such that Ω J. Suppose f : Ω R is bounded and f = f if x Ω and zero otherwise. Show that in integral of f over exists if and only if the integral of f J over J exists, in which case the integrals are equal. Solution. Suppose f is integrable over. Let ǫ > 0. Then a partition of exists so that U(P,f ) L(P,f ) < ǫ. We need to construct an appropriate partition for J. Suppose = [a 1,b 1 ] [a 2,b 2 ] [a p,b p ], J = [a 1,b 1 ] [a 2,b 2 ] [a p,b p ] By definition the partition P is constructed by taking cross products of the partitions of [a k,b k ] for 1 k p. Since J, we have a k a k < b k b k. n this case we simply add the two points a k and b k to the partition for [a k,b k ] and for k = 1,...,p and call this partition Q. We denote the cells in Q by {J i } N i=1. Let A = {i J i Ω = } and B = {i J i Ω }. Then U(Q,f J ) = N M i (f J )c(j i ) i=1 = A M i (f J )c(j i )+ B M i (f J )c(j i ) = 0+ B M i (f )c(j i ) = A M i (f )c(j i )+ B M i (f )c(j i ) = U(P,f ), where A are the indices where cells of P are disjoint from Ω. A similar calculation holds for the lower Darboux sums, and we conclude U(Q,f J ) L(Q,f J ) = U(P,f ) L(P,f ) < ǫ. t follows f J is integrable. To see the integrals are the same, we have, under our assumptions, ˆ ˆ U(P,f ) L(P,f ) = U(P,f ) f + f L(P,f ) < ǫ, and so, Thus ˆ J ˆ f J f = ˆ U(P,f ) ˆ ˆ J J < 2ǫ. f < ǫ. ˆ f J U(Q,f J ) + U(Q,f J ) f ˆ f J U(Q,f J ) + U(P,f ) f 15

16 This implies the two integrals are equal. Conversely, suppose f J is integrable on J. Let ǫ > 0. Then a partition Q of J exists so that U(Q,f J ) L(Q,f J ) < ǫ. This time we need to construct an appropriate partition for. By making a refinement, if necessary, we include the vertices of in the partition Q for J. Set P = Q. Then, calculations almost identical to the previous case show, U(Q,f J ) = U(P,f ), and the result follows as before. 43M. Let Ω R p be bounded and Ω i, i = 1,2 Suppose f : Ω R is bounded and f i = f if x Ω and zero otherwise. Show that in integral of f 1 over 1 exists if and only if the integral of f 2 over 2 exists, in which case the integrals are equal. Solution. Set = 1 2. Then i for i = 1,2. The previous problem show that, if either integral (over i ) exists, the integral over exists and they are equal. Thus if either integral exists, the other exists and they are equal. 43N. Let Ω R p be bounded and f : Ω R be a bounded function. Show that the choice of bounded cell, J, with Ω J does not change the value of J f J. Solution. Suppose J is any cell such that Ω J and J f J exists. f is another cell such that Ω, then Problem 43M shows f exists and equals J f J. Thus the choice of cell does not change the value of the integral. 43S. Suppose R p is a closed cell and that (f n ) is a sequence of real-valued functions which are all integrable. f the sequence converges uniformly on, show f is integrable and ˆ ˆ f n = f. lim n Solution. To show f is integrable copy the MAT 371 version and adapt it to our setting. Here is a way to show the convergence without using any algebra of integration. Let ǫ > 0 be given. Then, uniform convergence implies ǫ/2 < f n (x) f(x) < ǫ/2 for all n N, all x, and some N N. Thus sup (f n (x) f(x)) ǫ/2 < ǫ for such n. We have sup(f n (x)) = sup(f n (x) f(x)+f(x)) sup(f n (x) f(x))+sup(f(x)) < ǫ+sup(f(x)), for n N. Let P be any partition of and J i be an arbitrary cell in P. Then, by increasing N if necessary, we conclude N N exists so that sup(f n (x)) sup(f(x)) < ǫ J i J i 2c() for all n N and every J i. A similar calculation starting with sup (f(x)) implies ǫ 2c() < sup (f n (x)) sup(f(x)) J i J i 16

17 for n N. That is, sup Ji (f n (x)) sup Ji (f(x)) < ǫ/(2c(j)) for all J i and n N. We have U(P,f n ) U(P,f) = M i (f n )c(j i ) M i (f)c(j i ) i i M i (f n )c(j i ) M i (f)c(j i ) i < ǫ 2. We can write this as U(P,f) ǫ 2 < U(P,f n) < U(P,f)+ ǫ 2 for all n N and all partitions P of. Taking the infimum over the partitions, we find ˆ ˆ ˆ f ǫ < f n < f +ǫ. That is, for n N, and the result follows. ˆ ˆ f n f < ǫ, 17

18 Section 44 (Content and the ntegral.) 44S. Let J R 2 and G : J R be such that D 2 D 1 G is continuous on J. Show D 1 D 2 G exists and equals D 2 D 1 G. Solution. The theorem is not true. Here is a counter example. Let D(y) be the Dirichlet function and let F(x,y) be any function in C 2 (J). Set G(x,y) = F(x,y) +D(y). Then D 1 G = D 1 F and D 2 D 1 G = D 2 D 1 F. Moreover, D 2 D 1 G is continuous. However, D 2 G does not exist. Here is a possible fix to the problem. We add the assumption G(0,y) is differentiable for all (0,y) J. Let Ω = [0,x] [0,y]. f we set H(x,y) = D 2 D 1 G(x,y). Then H is continuous on J and so H is integrable. Also both H(,y) and H(x, ) are continuous and hence 1D-integrable for each x or y. Thus we may evaluate the integral of H by an iterated integral. With the help of the Fundamental Theorem of Calculus, we find ˆ x (ˆ y ) F(x,y) := H(s,t) = H(s,t)dt ds Ω 0 ˆ x 0 ( = 0 s G(s,y) ) s G(s,0) ds = G(x,y) G(x,0) G(0,y)+G(0,0). Problem 44R shows D 1 D 2 F = D 2 D 1 F = H. t follows ) D 1 D 2 (G(x,y) G(x,0) G(0,y)+G(0,0) = D 2 D 1 G(x,y). The above calculation shows xg(x,0) exists. With our added assumption, we can calculate the left side of the previous expression, and we conclude D 1 D 2 G(x,y) = D 2 D 1 G(x,y). 18