1 Solutions to selected problems
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1 1 Soluions o seleced problems 1. Le A B R n. Show ha in A in B bu in general bd A bd B. Soluion. Le x in A. Then here is ɛ > 0 such ha B ɛ (x) A B. This shows x in B. If A = [0, 1] and B = [0, 2], hen bd A = 0, 1} bd B = 0, 2}. 2. Le A R n be open and f : A R coninuous wih f(u) > 0. Show here is an open ball B around u such ha f(x) > f(u)/2 for x B. Soluion. Since f is coninuous a u, for ɛ = f(u)/2 here is δ > 0 such ha f(b δ (u)) (f(u) ɛ, f(u) + ɛ). In paricular, f(b δ (u)) > f(u)/2 and we may ake B = B δ (u). 3. Suppose f : R n R is coninuous and f(u) > 0 if u has a leas one raional componen. Prove ha f(u) 0 for all u R n. Soluion. Le u R n be arbirary. Wrie u = (u 1,..., u n ) and le r k } be he rs k digis in he decimal expansion for u 1, and le u k = (r k, u 2,..., u n ). Then u k } u so by coninuiy of f, f(u k )} f(u). Also f(u k ) > 0 for all k since he r k 's are raional. Thus, f(u) Show ha an open ball in R n is bounded. Soluion. Le B r (x) R n be an arbirary ball. We mus show i is conained in a ball B s (0) around 0. Le s = r + x. If y B r (x) hen y y x + x < r + x = s. Thus, B r (x) B s (0). 5. Le f : A R be coninuous wih A R n. If A is bounded, is f(a) bounded? If A is closed, is f(a) closed? Soluion. The answer o boh quesions is no. Consider f(x) = 1/x. Then A = (0, 1) is bounded bu f(a) = (1, ) is no; A = [1, ) is closed bu f(a) = (0, 1] is no. 6. Le f : R n R be coninuous wih f(u) u for every u R n. Prove ha f 1 ([0, 1]) is sequenially compac. Soluion. Since [0, 1] is closed and f is coninuous, f 1 ([0, 1]) is closed. I remains o 1
2 show f 1 ([0, 1]) is also bounded, hence sequenially compac. If x f 1 ([0, 1]), hen 0 f(x) 1 and f(x) x, so in paricular, x 1. Thus, f 1 ([0, 1]) is bounded. 7. Le A R n be sequenially compac and v R n \ A. Prove here is u A such ha u v x v for all x A. (1) Soluion. Le d = inf x A x v. Since d + 1/k is no a lower bound for x v over x A, we may pick u k A such ha d u k v < d + 1/k. Using sequenial compacness, pick u nk } u. By coninuiy of vecor subracion and he norm, u nk v } u v. And by he squeeze heorem in R, u v = d. The poin u is no unique: if A = 1, 1} R and v = 0, hen u = ±1 saisfy (1). 8. A mapping F : R n R m is Lipschiz coninuous if here is K > 0 such ha F (x) F (y) K x y (2) for all x, y R n. Show ha a Lipschiz mapping is uniformly coninuous. Soluion. Suppose (2) holds for F. Le ɛ > 0 and pick δ = K/ɛ. Then F (x) F (x) K x y < Kδ = ɛ whenever x y < δ. 9. Le A be sequenially compac and f : A f(a) coninuous and injecive. Show f 1 is coninuous. Give an example o show he assumpion on A is necessary. Soluion. Le v k } be a sequence in f(a) such ha v k } v, and le u k = f 1 (v k ), u = f 1 (v). Suppose u k } u. Using sequenial compacness of A, pick u nk } w A wih w u. By coninuiy of f, f(u nk )} f(w). As f(u nk )} = v nk } v, we have f(w) = v. Thus, w = f 1 (v) = u, conradicion. To see ha sequenial compacness of A is needed, le x, 0 x < 1 f(x) = 4 x, 2 x 3, f 1 x, 0 x < 1 (x) = 3 x, 1 x 2. Noe ha f is coninuous on A = [0, 1) [2, 3] bu f 1 is no coninuous a 1. 2
3 10. Le f : A R m be coninuous and A sequenially compac. Show f is uniformly coninuous. Soluion. Le u k }, v k } be sequences in A such ha u k v k } 0. Suppose f(u k ) f(v k ) } 0. Then along a subsequence, f(u nk ) f(v nk ) } c > 0. (Why?) Since A is sequenially compac, we can pick sub-subsequences u mnk } u A and v mnk } v A, and u = v since u k v k } 0. By coninuiy of f, f(u mnk )} f(u) and f(v mnk )} f(v) = f(u). Thus, f(u mnk ) f(v mnk ) } 0, conradicion. 11. We say u R n is a a limi poin of A R n if here is a sequence in A \ u} ha converges o u. Prove ha u is a limi poin of A if and only if every open ball around u conains inniely many poins of A \ u}. Soluion. Le u be a limi poin of A and δ > 0. Le u k } in A\u} be such ha u k } u, and pick N such ha k N implies u k B δ (A) \ u}. Noice u k : k N} mus be an innie se: if i were nie wih elemens v 1,..., v n, hen for ɛ = min 1 i n v i u we would have u k / B ɛ (u) for every k N. 12. Le A R n and le f be he characerisic funcion of A. Show ha f is coninuous a u if and only if u / bd A. Can one make an analogous saemen abou lim x u f(x)? Soluion. Suppose u / bd A. Then if u A, here is δ > 0 such ha B δ (u) A. Given any ɛ > 0, if x u < δ hen x A and so f(x) f(u) = 1 1 = 0 < ɛ. If u / A, here is δ > 0 such ha B δ (u) R n \ A and an analogous argumen holds. Conversely, if u bd A, hen for each k N here is v k, w k B 1/k (u) such ha v k A, w k R n \ A. Then v k } u and w k } u, bu f(v k )} 1, f(w k )} 0. An analogous saemen does no hold for limis. I is rue ha if u / bd(a) hen lim x u f(x) exiss in he argumen above, jus replace x u < δ wih 0 < x u < δ. However, he converse is false: if A = 0} hen 0 bd(a) ye lim x 0 f(x) = 0 exiss. (In he above argumen, v k and w k could equal u.) 13. Le f : A R m wih u A a limi poin of A R n. Show ha if f does no have a limi a u, hen f is no coninuous a u. Soluion. We prove he conraposiive. Suppose f is coninuous a u. Then lim x u f(x) = f(u). To see his, le ɛ > 0 and pick δ > 0 such ha f(x) f(u) < ɛ whenever x u < δ and x A; hen in paricular, f(x) f(u) < ɛ whenever 0 < x u < δ and x A. 3
4 14. Le f : A R m wih u A a limi poin of A R n. Show ha f is coninuous a u if and only if lim x u f(x) = f(u). Soluion. We only need o observe ha for any ɛ > 0, he saemens and f(x) f(u) < ɛ whenever x A and 0 < x u < δ f(x) f(u) < ɛ whenever x A and x u < δ are equivalen, since f(x) f(u) = 0 < ɛ when x u = Le A R n and suppose 0 is a limi poin of A. Suppose f : A R is such ha f(x) c x 2 for all x A, where c > 0 is consan. Suppose g : A R is such ha lim x 0 g(x)/ x 2 = 0. Prove here is r > 0 such ha f(x) g(x) (c/2) x 2 for x A wih 0 < x < r. Soluion. We use ɛ = c/2 in he deniion ɛ-δ deniion of lim x 0 g(x)/ x 2 = 0. Pick δ > 0 so ha if x A wih 0 < x < δ, hen g(x)/ x 2 = g(x) / x 2 < c/2 and so f(x) g(x) f(x) g(x) c x 2 (c/2) x 2 = (c/2) x Le f(x, y) = xy 2 x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0). Show f is coninuous a (0, 0) and has direcional derivaives a (0, 0) in every direcion, bu is no diereniable a (0, 0). Soluion. To see f is coninuous a (0, 0): for (x, y) (0, 0), f(x, y) f(0, 0) = xy 2 x 2 + y 0 2 = x x 0 as (x, y) 0. x y 2 (See problem 14 above.) For he direcional derivaives D (a,b) f(0, 0): f((0, 0) + (a, b)) f(0, 0) lim 0 0 f(a, b) Now noe ha D 1 f(0, 0) = 0, D 2 f(0, 0) = 0 and 0 2 ab 2 2 a b 2 = ab2 a 2 + b 2. f((0, 0) + (x, y)) f(0, 0) 0x 0y (x, y) = xy 2 x 2 +y 2 0x 0y x2 + y 2 = xy 2. (3) (x 2 + y 2 ) 3/2 4
5 When x y and y 0 he limi of (3) is 2 3/2 0, so f is no diereniable a (0, 0). 17. Dene f : R 2 R by f(x, y) = xy x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0). Show he parial derivaives of f are no coninuous a (0, 0). Soluion. For (x, y) (0, 0), we can calculae parial derivaives as usual (i.e., wihou resoring o he deniion): 1 D 1 f(x, y) = y3 x 2 y (x 2 + y 2 ) 2, D 2f(x, y) = x3 xy 2 (x 2 + y 2 ) 2. In paricular, D 1 f(0, y) = y 1, D 1 f(x, 0) = x 1 do no have limis as y 0, x 0. Thus, hey are no coninuous a (0, 0). (See problem 13 above.) 18. Dene g : R 2 R by g(x, y) = Is g coninuously diereniable? x 2 y 4 x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0). Soluion. For (x, y) (0, 0) we can calculae parial derivaives as usual: D 1 g(x, y) = 2xy6 (x 2 + y 2 ), D 2g(x, y) = 4x4 y 3 + 2x 2 y 5. 2 (x 2 + y 2 ) 2 Moreover, from he limi deniion of parial derivaives, g(, 0) g(0, 0) D 1 g(0, 0) 0 g(0, 0) g(0, ) D 2 g(0, 0) Since D 1 g(x, y) and D 2 g(x, y) are raional funcions, hey are coninuous excep a heir asympoes (x, y) = (0, 0). Thus, o esablish coninuiy of D 1 g and D 2 g we need only 1 D i f D ei f f x i = 0, = 0. 5
6 check ha lim (x,y) (0,0) D i g(x, y) = 0 for i = 1, 2. For (x, y) (0, 0), D 1 g(x, y) 0 = 2xy 6 x 4 + 2x 2 y 2 + y 4 = 2xy 2 x 4 + 2x x y2 0 as (x, y) (0, 0) y 4 y 2 D 2 g(x, y) 0 = 4x 4 y 3 + 2x 2 y 5 x 4 + 2x 2 y 2 + y 4 = 4x 2 y + 2y 3 x y2 2x2 y + y 3 0 as (x, y) (0, 0). y 2 x 2 This shows D 1 g(x, y) and D 2 g(x, y) are coninuous; hus, g is coninuously diereniable. 19. Suppose g : R 2 R has he propery g(x, y) x 2 + y 2 for all (x, y) R 2. Show g has parial derivaives wih respec o boh x and y a (0, 0). Soluion. Our assumpion on g forces g(0, 0) = 0 and hus g(, 0) g(0, 0) 2 =, g(0, ) g(0, 0) 2 =. Taking limis as 0 in he above expressions, we nd ha D 1 g(0, 0) = D 2 g(0, 0) = Suppose f : R 2 R has rs-order parial derivaives and D 1 f(x, y) = D 2 f(x, y) = 0 for all (x, y) R 2. Show ha f c for some c R; ha is, f is a consan funcion. Soluion. Fix y 0 R and consider g : R R dened by g(x) = f(x, y 0 ). Suppose g is nonconsan. Then g(a) g(b) for some a < b, and by MVT here is r (a, b) such ha D 1 f(r, y 0 ) = g (r) = [g(b) g(a)]/(b a) 0, conradicion. Thus, g is consan, say g c. Le (x, y) R 2. By repeaing he argumen above, we nd ha he funcion h : R R dened by h(z) = f(x, z) is consan. Since h(y 0 ) = g(x) we mus have h c, and in paricular f(x, y) = c. Since (x, y) R 2 was arbirary, we can conclude f c. 21. Given φ, ψ : R 2 R, a funcion f : R 2 R is called a poenial funcion for φ, ψ if D 1 f(x, y) = φ(x, y) and D 2 f(x, y) = ψ(x, y) for all (x, y) R 2. Show ha when a poenial funcion exiss for φ, ψ, i is unique up o an addiive consan. Then show ha if here is a poenial funcion for φ, ψ and φ, ψ are coninuously diereniable, hen D 1 ψ(x, y) = D 2 φ(x, y) for all (x, y) R 2. (4) 6
7 Soluion. Suppose f and g are wo poenial funcions for φ, ψ and le h = f g. Then D i h(x, y) = D i f(x, y) D i g(x, y) = 0 for all (x, y) R 2, i = 1, 2, so by Problem 20, h is consan. Thus, f and g dier by a consan. The saemen in (4) follows from he assumpion D 1 f and D 2 f are coninuously diereniable and Theorem Dene f : R 2 R by f(x, y) = x 3 y xy 3 x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0). Show ha D 1 f(0, y) = y for all y R and D 2 f(x, 0) = x for all x R. Conclude ha D 2 D 1 f(0, 0) = 1 bu D 1 D 2 f(0, 0) = 1. Soluion. Away from (0, 0) we can calculae parial derivaives of f as usual, i.e., wihou resoring o he limi deniion. Thus, for (x, y) (0, 0), Thus, for y 0 and x 0, D 1 f(x, y) = (x2 + y 2 )(3x 2 y y 3 ) (x 3 y xy 3 )(2x) (x 2 + y 2 ) 2, D 2 f(x, y) = (x2 + y 2 )(x 3 3xy 2 ) (x 3 y xy 3 )2y (x 2 + y 2 ) 2. D 1 f(0, y) = y2 ( y 3 ) (y 2 ) 2 = y, D 2 f(x, 0) = x2 x 3 (x 2 ) 2 = x, and, from he limi deniion of parial derivaives, D 1 f(0, 0) = 0 = D 2 f(0, 0). (Check his! See Problem 18 for a similar calculaion.) Thus, D 2 D 1 f(0, y) = 1 and D 1 D 2 f(x, 0) = 1 for all x, y R. In paricular, D 2 D 1 f(0, 0) = 1 bu D 1 D 2 f(0, 0) = Le A R 2 be an open se conaining (x 0, y 0 ). Prove ha here is r > 0 such ha (x, y) A whenever x x 0 < 2r and y y 0 < 2r. Soluion. Pick ɛ > 0 such ha B ɛ (x 0, y 0 ) A. If 0 < r < ɛ/(2 2), hen (x, y) (x 0, y 0 ) = (x x 0 ) 2 + (y y 0 ) 2 < (2r) 2 + (2r) 2 = 8r 2 = 2 2r < ɛ whenever x x 0 < 2r and y y 0 < 2r. 7
8 24. Suppose f : R n R and g : R R are coninuously diereniable. Find a formula for (g f)(x) in erms of f(x) and g (f(x)). Soluion. By he deniion of parial derivaive and he mean value heorem, D i (g f)(x) 0 g(f(x + e i )) g(f(x)) g (z ) f(x + e i) f(x), 0 where z is on he line segmen beween f(x) and f(x + e i ). Noice lim 0 g (z ) = g (f(x)) due o coninuiy of f and g, and D i f(x) : 0 [f(x + e i ) f(x)]/. Thus, ha is, (g f)(x) = f(x) g (f(x)). D i (g f)(x) = g (f(x))d i f(x), i = 1,..., n, 25. Suppose f : R n R is such ha D v f(x) exiss. Prove ha D cv f(x) = cd v f(x) for any nonzero c R. Soluion. This follows from he compuaion f(x + cv) f(x) D cv f(x) 0 0 c f(x + cv) f(x) c = c lim s 0 f(x + sv) f(x) s = cd v f(x). 26. Suppose f : R n R has rs-order parial derivaives and ha x R n is a local minimizer for f, ha is, here is ɛ > 0 such ha Prove ha f(x) = 0. Soluion. Due o he assumpion (5), for i = 1,..., n, f(x + h) f(x) for h B ɛ (0). (5) f(x + e i ) f(x) D ei f(x) 0, 0 f(x e i ) f(x) D ei f(x) 0, 0 and by problem 25, D ei f(x) = D ei f(x) 0. Thus D ei f(x) = 0, and so f(x) = Consider f(x, y, z) = xyz + x 2 + y 2. 8
9 Find θ (0, 1) such ha f(1, 1, 1) f(0, 0, 0) = D 1 f(θ, θ, θ) + D 2 f(θ, θ, θ) + D 3 f(θ, θ, θ). Soluion. Noe ha D 1 f(x, y, z) = yz + 2x, D 2 f(x, y, z) = xz + 2y, D 3 f(x, y, z) = xy. and f(1, 1, 1) f(0, 0, 0) = 3. Thus, we solve 3θ 2 + 4θ = 3 o ge θ = Dene f : R 2 R by f(x, y) = x x 2 +y 2, y if y 0, 0, if y = 0. a) Prove f is no coninuous a (0, 0). b) Prove f has direcional derivaives in all direcions a (0, 0). c) Prove for any c R here is p such ha Does his conradic Corollary 13.18? p = 1 and D p f(0, 0) = c. Soluion. a) Noe ha f(x, y) 1 when y = x2 1 x 2 and x 0. Approaching (0, 0) along his curve shows f is no coninuous a zero, since f(0, 0) = 0 1. b) When b 0, Also, D (a,b) f(0, 0) 0 f(a, b) f(0, 0) f(a, 0) f(0, 0) D (a,0) f(0, 0) 0 a 2 a b 2 0 b Thus, f has direcional derivaives in every direcion a (0, 0). c) We show such p exiss by solving he equaions a 2 a b + 1 = c, 2 a2 + b 2 = 1... a = = a = 0. a 2 b c 1 + c 2, b = c 2. This does no conradic he corollary since f is no coninuously diereniable. (See also Problem 33 below.) 9
10 29. Suppose f : R n R is coninuously diereniable and le K = x R n : x = 1}. Show here is a poin x K a which f K aains is smalles value. Now suppose whenever p R n is a uni vecor, f(p), p > 0. Show ha hen x < 1. Soluion. K is sequenially compac and f is coninuous, so f K aains a smalles value by he exreme value heorem (Theorem 11.22). Le p R n wih p = 1. Then f(p), p > 0 and 0 0 f(p p) f(p) f(p), p f(p p) f(p) + f(p), p. 0 This shows ha for > 0 sucienly close o zero, f(p p) f(p) < 0. Thus, he minimum of f K canno be aained a p. (For an ɛ-δ proof of his, le ɛ = f(p), p and pick δ > 0 such ha 1 [f(p p) f(p)] + f(p), p < ɛ whenever 0 < < δ. Then f(p p) f(p) < 0 whenever 0 < < δ; why?) 30. Prove ha sin(2x + 2y) 2x 2y lim = 0. (x,y) (0,0) x2 + y 2 Soluion. Le f(x, y) = sin(2x + 2y). Then D 1 f(x, y) = 2 cos(2x + 2y) = D 2 f(x, y) are coninuous. Thus, f is coninuously diereniable, so since D 1 f(0, 0) = D 2 f(0, 0) = 2, sin(2x + 2y) 2x 2y lim (x,y) (0,0) (x, y) = Suppose f : R 2 R is coninuous and a, b R. Prove Is i rue ha lim [f(x, y) (f(0, 0) + ax + by)] = 0. (6) (x,y) (0,0) f(x, y) [f(0, 0) + ax + by] lim = 0? (7) (x,y) (0,0) x2 + y 2 Soluion. Since f is coninuous, f(x, y) f(0, 0) = 0 as (x, y) (0, 0). Moreover, ax + by 0 as (x, y) (0, 0). This proves (6). Noice (7) is false unless f is also diereniable a (0, 0) wih D 1 f(0, 0) = a, D 2 f(0, 0) = b. 32. Dene f : R 2 R by f(x, y) = x2 + y 2 sin y2, if x 0, x 0, if x = 0. 10
11 Show f is coninuous a (0, 0) and has direcional derivaives in every direcion a (0, 0), bu f is no diereniable a (0, 0) ha is, here is no angen plane o he graph of f a (0, 0). Soluion. Noice f(x, y) f(0, 0) = x 2 + y 2 sin y2 x 2 x 2 + y 2 0 as (x, y) 0. Thus, f is coninuous a (0, 0). Since sin(c) is coninuous for c R, when a 0, f(a, b) f(0, 0) D (a,b) f(0, 0) 0 2 a b 2 sin 2 b 2 0 a a2 + b 2 b2 0 sin a = 0. f(0,b) f(0,0) 0 0 Also, D (0,b) f(0, 0) 0 0 = 0. Thus, f has direcional derivaives in every direcion a (0, 0). Noe ha D 1 f(0, 0) = D 2 f(0, 0) = 0 bu he limi f(x, y) f(0, 0) 0x 0y lim (x,y) (0,0) (x, y) y2 sin (x,y) (0,0) x does no exis (e.g., ake x y 3 and le y 0). Thus, f is no diereniable a (0, 0). 33. Suppose f : R n R is homogeneous, ha is f(0) = 0 and f(x) = f(x) for all R and x R n. Prove ha if f is diereniable, hen f(x) = f(0), x, in paricular, f is linear 2. Thus, if f is homogeneous and no linear, i canno be diereniable. Soluion. Le f be diereniable and homogeneous, and u R n a uni vecor. Then f(u) f(0) f(0), u u 0 + f(u) f(0), u u This shows ha f(u) = f(0), u and hus for R, f(u) = f(u) = f(0), u = f(0), u. = f(u) f(0), u. Any nonzero x R n is a scalar imes a uni vecor, x = x x. Thus, we are done. x 2 f is linear i i is homogeneous and addiive: f(x + y) = f(x) + f(y) for all x, y R n. 11
1 Solutions to selected problems
1 Solutions to selected problems 1. Let A B R n. Show that int A int B but in general bd A bd B. Solution. Let x int A. Then there is ɛ > 0 such that B ɛ (x) A B. This shows x int B. If A = [0, 1] and
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