1 Solutions to selected problems

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1 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 B = [0, 2], then bd A = 0, 1} bd B = 0, 2}. 2. Let A R n be open and f : A R continuous with f(u) > 0. Show there is an open ball B around u such that f(x) > f(u)/2 for x B. Solution. Since f is continuous at u, for ɛ = f(u)/2 there is δ > 0 such that f(b δ (u)) (f(u) ɛ, f(u) + ɛ). In particular, f(b δ (u)) > f(u)/2 and we may take B = B δ (u). 3. Suppose f : R n R is continuous and f(u) > 0 if u has at least one rational component. Prove that f(u) 0 for all u R n. Solution. Let u R n be arbitrary. Write u = (u 1,..., u n ) and let r k } be the rst k digits in the decimal expansion for u 1, and let u k = (r k, u 2,..., u n ). Then u k } u so by continuity of f, f(u k )} f(u). Also f(u k ) > 0 for all k since the r k 's are rational. Thus, f(u) Show that an open ball in R n is bounded. Solution. Let B r (x) R n be an arbitrary ball. We must show it is contained in a ball B s (0) around 0. Let s = r + x. If y B r (x) then y y x + x < r + x = s. Thus, B r (x) B s (0). 5. Let f : A R be continuous with A R n. If A is bounded, is f(a) bounded? If A is closed, is f(a) closed? Solution. The answer to both questions is no. Consider f(x) = 1/x. Then A = (0, 1) is bounded but f(a) = (1, ) is not; A = [1, ) is closed but f(a) = (0, 1] is not. 6. Let f : R n R be continuous with f(u) u for every u R n. Prove that f 1 ([0, 1]) is sequentially compact. Solution. Since [0, 1] is closed and f is continuous, f 1 ([0, 1]) is closed. It remains to 1

2 show f 1 ([0, 1]) is also bounded, hence sequentially compact. If x f 1 ([0, 1]), then 0 f(x) 1 and f(x) x, so in particular, x 1. Thus, f 1 ([0, 1]) is bounded. 7. Let A R n be sequentially compact and v R n \ A. Prove there is u A such that u v x v for all x A. (1) Solution. Let d = inf x A x v. Since d + 1/k is not a lower bound for x v over x A, we may pick u k A such that d u k v < d + 1/k. Using sequential compactness, pick u nk } u. By continuity of vector subraction and the norm, u nk v } u v. And by the squeeze theorem in R, u v = d. The point u is not unique: if A = 1, 1} R and v = 0, then u = ±1 satisfy (1). 8. A mapping F : R n R m is Lipschitz continuous if there is K > 0 such that F (x) F (y) K x y (2) for all x, y R n. Show that a Lipschitz mapping is uniformly continuous. Solution. Suppose (2) holds for F. Let ɛ > 0 and pick δ = K/ɛ. Then F (x) F (x) K x y < Kδ = ɛ whenever x y < δ. 9. Let A be sequentially compact and f : A f(a) continuous and injective. Show f 1 is continuous. Give an example to show the assumption on A is necessary. Solution. Let v k } be a sequence in f(a) such that v k } v, and let u k = f 1 (v k ), u = f 1 (v). Suppose u k } u. Using sequential compactness of A, pick u nk } w A with w u. By continuity 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, contradiction. To see that sequential compactness of A is needed, let x, 0 x < 1 f(x) = 4 x, 2 x 3, f 1 x, 0 x < 1 (x) = 3 x, 1 x 2. Note that f is continuous on A = [0, 1) [2, 3] but f 1 is not continuous at 1. 2

3 10. Let f : A R m be continuous and A sequentially compact. Show f is uniformly continuous. Solution. Let u k }, v k } be sequences in A such that 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 sequentially compact, we can pick sub-subsequences u mnk } u A and v mnk } v A, and u = v since u k v k } 0. By continuity of f, f(u mnk )} f(u) and f(v mnk )} f(v) = f(u). Thus, f(u mnk ) f(v mnk ) } 0, contradiction. 11. We say u R n is a a limit point of A R n if there is a sequence in A \ u} that converges to u. Prove that u is a limit point of A if and only if every open ball around u contains innitely many points of A \ u}. Solution. Let u be a limit point of A and δ > 0. Let u k } in A\u} be such that u k } u, and pick N such that k N implies u k B δ (A) \ u}. Notice u k : k N} must be an innite set: if it were nite with elements v 1,..., v n, then for ɛ = min 1 i n v i u we would have u k / B ɛ (u) for every k N. 12. Let A R n and let f be the characteristic function of A. Show that f is continuous at u if and only if u / bd A. Can one make an analogous statement about lim x u f(x)? Solution. Suppose u / bd A. Then if u A, there is δ > 0 such that B δ (u) A. Given any ɛ > 0, if x u < δ then x A and so f(x) f(u) = 1 1 = 0 < ɛ. If u / A, there is δ > 0 such that B δ (u) R n \ A and an analogous argument holds. Conversely, if u bd A, then for each k N there is v k, w k B 1/k (u) such that v k A, w k R n \ A. Then v k } u and w k } u, but f(v k )} 1, f(w k )} 0. An analogous statement does not hold for limits. It is true that if u / bd(a) then lim x u f(x) exists in the argument above, just replace x u < δ with 0 < x u < δ. However, the converse is false: if A = 0} then 0 bd(a) yet lim x 0 f(x) = 0 exists. (In the above argument, v k and w k could equal u.) 13. Let f : A R m with u A a limit point of A R n. Show that if f does not have a limit at u, then f is not continuous at u. Solution. We prove the contrapositive. Suppose f is continuous at u. Then lim x u f(x) = f(u). To see this, let ɛ > 0 and pick δ > 0 such that f(x) f(u) < ɛ whenever x u < δ and x A; then in particular, f(x) f(u) < ɛ whenever 0 < x u < δ and x A. 3

4 14. Let f : A R m with u A a limit point of A R n. Show that f is continuous at u if and only if lim x u f(x) = f(u). Solution. We only need to observe that for any ɛ > 0, the statements 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 equivalent, since f(x) f(u) = 0 < ɛ when x u = Let A R n and suppose 0 is a limit point of A. Suppose f : A R is such that f(x) c x 2 for all x A, where c > 0 is constant. Suppose g : A R is such that lim x 0 g(x)/ x 2 = 0. Prove there is r > 0 such that f(x) g(x) (c/2) x 2 for x A with 0 < x < r. Solution. We use ɛ = c/2 in the denition ɛ-δ denition of lim x 0 g(x)/ x 2 = 0. Pick δ > 0 so that if x A with 0 < x < δ, then 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 Let f(x, y) = xy 2 x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0). Show f is continuous at (0, 0) and has directional derivatives at (0, 0) in every direction, but is not dierentiable at (0, 0). Solution. To see f is continuous at (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 the directional derivatives D (a,b) f(0, 0): f((0, 0) + t(a, b)) f(0, 0) lim t = lim f(ta, tb) t = lim t 2 ab 2 t 2 a 2 + t 2 b 2 = ab2 a 2 + b 2. To see f is not dierentiable at (0, 0), suppose to the contrary L : R 2 R is linear and f((0, 0) + (x, y)) f(0, 0) L(x, y) lim (x,y) 0 (x, y) 4 = 0. (3)

5 We may write L(x, y) = ax + by for some scalars a, b, and so f((0, 0) + (x, y)) f(0, 0) L(x, y) (x, y) = xy 2 x 2 +y 2 ax by x2 + y 2. (4) The limit of (4) as x 0 with y 0 is a; its limit as y 0 with x 0 is b. Thus, a = b = 0. But then the limit of (4) as x 0 with y x is (1/2 a b)/ 2 0, contradiction. 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 the partial derivatives of f are not continuous at (0, 0). Solution. For (x, y) (0, 0), we can calculate partial derivatives as usual (i.e., without resorting to the denition): 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 particular, D 1 f(0, y) = y 1, D 1 f(x, 0) = x 1 do not have limits as y 0, x 0. Thus, they are not continuous at (0, 0). (See problem 13 above.) 18. Dene g : R 2 R by g(x, y) = Is g continuously dierentiable? x 2 y 4 x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0). Solution. For (x, y) (0, 0) we can calculate partial derivatives 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 the limit denition of partial derivatives, g(t, 0) g(0, 0) D 1 g(0, 0) = lim t g(0, 0) g(0, t) D 2 g(0, 0) = lim t 1 D i f D ei f f x i 5 = lim 0 0 t = lim 0 0 t = 0, = 0.

6 Since D 1 g(x, y) and D 2 g(x, y) are rational functions, they are continuous except at their asymptotes (x, y) = (0, 0). Thus, to establish continuity of D 1 g and D 2 g we need only check that 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 completes the proof. 19. Suppose g : R 2 R has the property g(x, y) x 2 + y 2 for all (x, y) R 2. Show g has partial derivatives with respect to both x and y at (0, 0). Solution. Our assumption on g forces g(0, 0) = 0 and thus g(t, 0) g(0, 0) t t 2 t = t, g(0, t) g(0, 0) t t 2 t = t. Taking limits as t 0 in the above expressions, we nd that D 1 g(0, 0) = D 2 g(0, 0) = Suppose f : R 2 R has rst-order partial derivatives and D 1 f(x, y) = D 2 f(x, y) = 0 for all (x, y) R 2. Show that f c for some c R; that is, f is a constant function. Solution. Fix y 0 R and consider g : R R dened by g(x) = f(x, y 0 ). Suppose g is nonconstant. Then g(a) g(b) for some a < b, and by MVT there is r (a, b) such that D 1 f(r, y 0 ) = g (r) = [g(b) g(a)]/(b a) 0, contradiction. Thus, g is constant, say g c. Let (x, y) R 2. By repeating the argument above, we nd that the function h : R R dened by h(z) = f(x, z) is constant. Since h(y 0 ) = g(x) we must have h c, and in particular f(x, y) = c. Since (x, y) R 2 was arbitrary, we can conclude f c. 21. Given φ, ψ : R 2 R, a function f : R 2 R is called a potential function 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 that when a potential function exists for φ, ψ, it is unique up to an additive constant. Then show that if there is a potential function for φ, ψ and φ, ψ are continuously dierentiable, then D 1 ψ(x, y) = D 2 φ(x, y) for all (x, y) R 2. (5) 6

7 Solution. Suppose f and g are two potential functions for φ, ψ and let 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 constant. Thus, f and g dier by a constant. The statement in (5) follows from the assumption D 1 f and D 2 f are continuously dierentiable 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 that D 1 f(0, y) = y for all y R and D 2 f(x, 0) = x for all x R. Conclude that D 2 D 1 f(0, 0) = 1 but D 1 D 2 f(0, 0) = 1. Solution. Away from (0, 0) we can calculate partial derivatives of f as usual, i.e., without resorting to the limit denition. 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 the limit denition of partial derivatives, D 1 f(0, 0) = 0 = D 2 f(0, 0). (Check this! See Problem 18 for a similar calculation.) Thus, D 2 D 1 f(0, y) = 1 and D 1 D 2 f(x, 0) = 1 for all x, y R. In particular, D 2 D 1 f(0, 0) = 1 but D 1 D 2 f(0, 0) = Let A R 2 be an open set containing (x 0, y 0 ). Prove that there is r > 0 such that (x, y) A whenever x x 0 < 2r and y y 0 < 2r. Solution. Pick ɛ > 0 such that B ɛ (x 0, y 0 ) A. If 0 < r < ɛ/(2 2), then (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 continuously dierentiable. Find a formula for (g f)(x) in terms of f(x) and g (f(x)). Solution. By the denition of partial derivative and the mean value theorem, D i (g f)(x) = lim g(f(x + te i )) g(f(x)) t = lim g (z t ) f(x + te i) f(x), t where z t is on the line segment between f(x) and f(x + te i ). Notice lim g (z t ) = g (f(x)) due to continuity of f and g, and D i f(x) := lim [f(x + te i ) f(x)]/t. Thus, that is, (g f)(x) = f(x) g (f(x)). D i (g f)(x) = g (f(x))d i f(x), i = 1,..., n, 8

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