1 Directional Derivatives and Differentiability

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1 Wednesday, January 18, Directional Derivatives and Differentiability Let E R N, let f : E R and let x 0 E. Given a direction v R N, let L be the line through x 0 in the direction v, that is, L := { x R N : x = x 0 + tv, t R }, and assume that x 0 is an accumulation point of the set E L. The directional derivative of f at x 0 in the direction v is defined as v (x f (x 0 + tv f (x 0 0 := lim, t 0 t provided the limit exists in R. In the special case in which v = e i, the directional derivative e i (x 0, if it exists, is called the partial derivative of f with respect to x i and is denoted x i (x 0 or f xi (x 0 or D i f (x 0. Remark 1 Let F := {t R : x 0 + tv E} R. If we consider the function of one variable g (t := f (x 0 + tv, t F, then v (x 0, when it exists, is simply the derivative of g at t = 0, v (x 0 = g (0. If v (x 0 exists, then g is differentiable in t = 0 and so it is continuous at t = 0. Thus, the function f restricted to the line L is continuous at x 0. Remark 2 The previous definition continues to hold if in place of R N one takes a normed space V, so that f : E R where E V. In view of the previous remark, one would be tempted to say that if the directional derivatives at x 0 exist and are finite in every direction, then f is continuous at x 0. This is false in general, as the following example shows. Example 3 Let f (x, y := { 1 if y = x 2, x 0, 0 otherwise. Given a direction v = (v 1, v 2, the line L through 0 in the direction v intersects the parabola y = x 2 only in 0 and in at most one point. Hence, if t is very small, f (0 + tv 1, 0 + tv 2 = 0. It follows that f (0 + tv 1, 0 + tv 2 f (0, (0, 0 = lim = lim = 0. v t 0 t t 0 t However, f is not continuous in 0, since f ( x, x 2 = 1 1 as x 0, while f (x, 0 = 0 0 as x 0. 1

2 Exercise 4 Let f (x, y := { x 2 y x 4 +y 2 if (x, y (0, 0, 0 if (x, y = (0, 0. Find all directional derivatives of f at 0 and prove that f is not continuous at 0. The previous examples show that in dimension N 2 partial derivatives do not give the same kind of results as in the case N = 1. To solve this problem, we introduce a stronger notion of derivative, namely, the notion of differentiability. We recall that a function T : R N R is linear if for all x, y R N and T (x + y = T (x + T (y T (sx = st (x for all s R and x R N. Write x = N x ie i. Then by the linearity of T, ( N N T (x = T x i e i = x i T (e i. Define b := (T (e 1,..., T (e N R N. Then the previous calculation shows that T (x = b x for all x R N. Definition 5 Let E R N, let f : E R, and let x 0 E be an accumulation point of E. The function f is differentiable at x 0 if there exists a linear function T : R N R (depending on f and x 0 such that f (x f (x 0 T (x x 0 lim = 0. x x 0 x x 0 provided the limit exists. The function T, if it exists, is called the differential of f at x 0 and is denoted df (x 0 or df x0. Exercise 6 Prove that if N = 1, then f is differentiable at x 0 if and only there exists the derivative f (x 0 R. Remark 7 The previous definition continues to hold if in place of R N one takes a normed space V, so that f : E R where E V. In this case, however, we require T : V R to be linear and continuous. Note that in R N every linear function is continuous. Friday, January 20,

3 Example 8 Let V = {f : [ 1, 1] R : f is differentiable in [ 1, 1]}. Note that V is a normed space with the norm f := max x [ 1,1] f (x. Consider the linear function T : V R defined by T (f := f (0. Then T is linear. To prove that T is not continuous, consider f n (x := 1 n sin ( n 2 x. Then f n 0 1 n 0 but so that f n (x = n cos ( n 2 x T (f n = f n (0 = n and so T is not continuous, since T (f n T (0 = 0. The next theorem shows that differentiability in dimension N 2 plays the same role of the derivative in dimension N = 1. Theorem 9 Let E R N, let f : E R, and let x 0 E be an accumulation point of E. If f is differentiable at x 0, then f is continuous at x 0. Proof. Let T be the differential of f at x 0. Define R (x := f (x f (x 0 T (x x 0. Note that by the definition of differentiability Then lim x x 0 R (x x x 0 = 0. f (x f (x 0 = T (x x 0 + R (x = b (x x 0 + R (x. Hence, by Cauchy s inequality for x E, x x 0, 0 f (x f (x 0 b (x x 0 + R (x b x x 0 + R (x ( R (x = x x 0 b + 0 ( b + 0 x x 0 as x x 0. It follows by the squeeze theorem that f is continuous at x 0. Next we study the relation between directional derivatives and differentiability. The next theorem gives a formula for the vector b used in the previous proof and hence determines T. Here we need x 0 to be an interior point of E. 3

4 Theorem 10 Let E R N, let f : E R be differentiable at some point x 0 E. Then (i all the directional derivatives of f at x 0 exist and (ii for every direction v, v (x 0 = df x0 (v, v (x 0 = N x i (x 0 v i. (1 Proof. Since x 0 is an interior point, there exists B (x 0, r E. Let v R N be a direction and x = x 0 + tv. Note that for t < r, we have that x x 0 = x 0 + tv x 0 = tv = t v = t 1 < r and so x 0 + tv B (x 0, r E. Moreover, x x 0 as t 0 and so, since f is differentiable at x 0, f (x f (x 0 T (x x 0 f (x 0 + tv f (x 0 T (x 0 + tv x 0 0 = lim = lim x x 0 x x 0 t 0 x 0 + tv x 0 f (x 0 + tv f (x 0 tt (v = lim. t 0 t Since t t is bounded by one, it follows that f (x 0 + tv f (x 0 tt (v lim = lim t 0 t But then t t 0 t f (x 0 + tv f (x 0 tt (v t f (x 0 + tv f (x 0 tt (v f (x 0 + tv f (x 0 0 = lim = lim T (v, t 0 t t 0 t = 0. which shows that there exists v (x 0 = T (v. Part (ii follows from the linearity of T. Indeed, writing v = N v ie i, by the linearity of T, v (x 0 = T (v = T ( N N N v i e i = v i T (e i = x i (x 0 v i. Remark 11 If in the previous theorem x 0 is not an interior point but for some direction v R N, the point x 0 is an accumulation point of the set E L, where L is the line through x 0 in the direction v, then as in the first part of the proof we can show that there exists the directional derivative v (x 0 and v (x 0 = df x0 (v. 4

5 If all the partial derivatives of f at x 0 exist, the vector ( (x 0,..., (x 0 R N x 1 x N is called the gradient of f at x 0 and is denoted by f (x 0 or grad f (x 0 or Df (x 0. Note that part (ii of the previous theorem shows that df x0 (v = T (v = f (x 0 v = N x i (x 0 v i. (2 for all directions v. Hence, only at interior points of E, to check differentiability it is enough to prove that Exercise 12 Let f (x f (x 0 f (x 0 (x x 0 lim = 0. (3 x x 0 x x 0 f (x, y := { x 2 y x 2 +y 2 if (x, y (0, 0, 0 if (x, y = (0, 0. Prove that f is continuous at 0, that all directional derivatives of f at 0 exist but that the formula fails. (0, 0 = v x (0, 0 v 1 + y (0, 0 v 2 Exercise 13 Let f (x, y := { x 2 y x 2 +y 4 if (x, y (0, 0, 0 if (x, y = (0, 0. Find all directional derivatives of f at 0. Study the continuity and the differentiability of f at 0. Theorems 9 and 10 give necessary conditions for the differentiability of f at x 0. Let s prove that these conditions are not, however, suffi cient. Example 14 Let f (x, y := { x if y = x 2, x 0, 0 otherwise. Given a direction v = (v 1, v 2, the line L through 0 in the direction v intersects the parabola y = x 2 only in 0 and in at most one point. Hence, if t is very small, f (0 + tv 1, 0 + tv 2 = 0. 5

6 It follows that f (0 + tv 1, 0 + tv 2 f (0, (0, 0 = lim = lim = 0. v t 0 t t 0 t Thus formula (1 holds. Moreover, f is continuous in 0, since f (x, y x 0 as (x, y (0, 0. We claim that f is not differentiable at (0, 0. To negate differentiability, in view of the previous theorem and since (0, 0 is an interior point, we need to prove that the quotient f (x, y f (0, 0 f (0, 0 (x, y (x (y 0 2 does not tend to zero as (x, y (0, 0. Take y = x 2. Then f ( x, x 2 f (0, 0 f (0, 0 (x, x 2 = x 0 0 ( x, x 2 (x (x (x (x Exercise 15 Let f : E R be Lipschitz and let x 0 E. = x x2 + x = x 4 x 1 + x 0. 2 (i Assume that all the directional derivatives of f at x 0 exist and that v (x 0 = N x i (x 0 v i for every direction v. Prove that f is differentiable at x 0. (ii Assume that all the partial derivatives of f at x 0 exist, that the directional derivatives v (x 0 exist for all v S, where S is dense in the unit sphere, and that v (x 0 = N x i (x 0 v i for every direction v S. Prove that f is differentiable at x 0. Monday, January 23, 2012 The next theorem gives an important suffi cient condition for differentiability at a point x 0. Theorem 16 Let E R N, let f : E R, let x 0 E. Assume that there exists r > 0 such that B (x 0, r E and the partial derivatives x j, j = 1,..., N, exist for every x B (x 0, r and are continuous at x 0. Then f is differentiable at x 0. Proof. Let x B (x 0, r. Write x = (x 1,..., x N and x 0 = (y 1,..., y N. Then f (x f (x 0 = (f (x 1,..., x N f (y 1, x 2,..., x N + + (f (y 1,..., y N 1, x N f (y 1,..., y N. By the mean value theorem applied to the function of one variable f (, x 2,..., x N, f (x 1,..., x N f (y 1, x 2,..., x N = x 1 (z 1 (x 1 y 1, 6

7 where z 1 := (θ 1 x 1 + (1 θ 1 y 1, x 2,..., x N for some θ 1 (0, 1. Note that Similarly, for i = 2,..., N, z 1 x 0 x x 0. f (y 1,..., y i 1, x i,..., x N f (y 1,..., y i 1, y i,..., x N = x i (z i (x i y i, where z i := (y 1,..., y i 1, θ i x i + (1 θ i y i, x i+1,..., x N for some θ i (0, 1 and z i x 0 x x 0. Write Then f (x f (x 0 f (x 0 (x x 0 N ( = (z i (x 0 (x i y i. x i x i f (x f (x 0 f (x 0 (x x 0 x x 0 Since xi yi x x 0 1, we have that 0 f (x f (x 0 f (x 0 (x x 0 x x 0 N (z i (x 0 x i y i x i x i x x 0. N (z i (x 0 x i x i. (4 Using the fact that z i x 0 x x 0 0 as x x 0, together with the continuity of x i at x 0, gives 0 f (x f (x 0 f (x 0 (x x 0 x x 0 N (z i (x 0 x i x i 0, which implies the differentiability of f at x 0. The previous theorem can be significantly improved. Indeed, we have the following result. Theorem 17 Let E R N, let f : E R, let x 0 E, and let i {1,..., N}. Assume that there exists r > 0 such that B (x 0, r E and for all j i and for all x B (x 0, r the partial derivative x j exists at x and is continuous at x 0. Assume also that x i (x 0 exists. Then f is differentiable at x 0. 7

8 Proof. Without loss of generality, we may assume that i = N. Reasoning as before we have that f (x f (x 0 f (x 0 (x x 0 N 1 ( = (z i (x 0 (x i y i x i x i ( f (y1,..., y N 1, x N f (y 1,..., y N + x N y N Hence, as before, N 1 0 f (x f (x 0 f (x 0 (x x 0 x x 0 + f (y 1,..., y N 1, y N + (x N y N f (y 1,..., y N x N y N Since t := x N y N 0 as x x 0, we have that (x 0 (x N y N. x N (z i (x 0 x i x i (5 (x 0 x N. f (y 1,..., y N 1, y N + (x N y N f (y 1,..., y N x N y N = f (y 1,..., y N 1, y N + t f (y 1,..., y N (x 0, t x N and so the right-hand side of (5 goes to zero as x x 0. The next exercise shows that the previous conditions are suffi cient but not necessary for differentiability. Example 18 Let f (x, y := { ( x 2 + y 2 sin 1 x+y if (x, y (0, 0 or x + y 0, 0 otherwise. Let s prove that f is differentiable at (0, 0 but the partial derivatives are not continuous at (0, 0. We have ( f (x, 0 f (0, 0 x sin 1 x+0 = 0 = x sin 1 x 0 x 0 x 0 as x 0, since sin 1 x Hence, is bounded. Hence, x (0, 0 = 0. Similarly, y (0, 0 = 0. f (x, 0 f (0, 0 x (0, 0 (x 0 y (0, 0 (y 0 = (x (y 0 2 = (x 2 +y 2 sin 1 x+y x2 +y 2 if x + y 0, 0 otherwise. { x2 + y 2 sin 1 x+y if x + y 0, 0 otherwise. 0 8

9 as (x, y (0, 0 since sin 1 x+y is bounded and x 2 + y 2 0. Hence, f is diff erentiable at (0, 0. On the other hand, if x + y 0, x (x, y = (2x + 0 sin 1 x + y + ( x 2 + y 2 ( cos y (x, y = (0 + 2y sin 1 x + y + ( x 2 + y 2 Taking x = y gives ( cos 1 (x, x = 2x sin x 2x cos 1 2x, 1 (x, x = 2x sin y 2x cos 1 2x ( 1 1 x + y (x + y 2, ( 1 1 x + y (x + y 2. which have no limit as x 0, thus x and y are not continuous at (0, 0. Note that one should also check the existence of the partial derivatives at points such that x + y = 0. Definition 19 Given E R N, we say that Wednesday, January 25, 2012 (i E is disconnected if there exist two nonempty open sets U, V R N such that E U V, E U, E V, E U V =. (ii E is connected if it is not disconnected. We study properties of connected sets. We begin by showing that continuous functions preserve connectedness. Proposition 20 Let F R N and let f : F R M be a continuous function. Then f (E is connected for every connected set E F. Proof. Let E F be a connected set and assume by contradiction that f (E is disconnected. Then there exist two open sets U, V R M such that f (E U V, f (E U, f (E V, f (E U V =. By continuity, f 1 (U and f 1 (V are relatively open, hence, there exist two open sets A and B in R N such that f 1 (U = F A and f 1 (V = F B. Hence, E A B, E A, E B, E A B =, which shows that E is disconnected. 9

10 Theorem 21 A set C R is connected if and only if it is convex. Proof. Exercise. We now introduce another notion of connectedness, which is simpler to verify. Definition 22 A continuous path is a continuous function ϕ : I R N, where I R is an interval. The set ϕ (I R N is called the range of the path. If I = [a, b], the points ϕ (a and ϕ (b are called endpoints of the path. A set E R N is called pathwise connected if for all x,y E there exists a continuous path with endpoints x and y and range contained in E. Proposition 23 Let E R N be pathwise connected. Then E is connected. Proof. We claim that E is connected. If not, then there exist two nonempty open sets U, V R N such that E U V, E U, E V, E U V =. Let x C U and y C V. By hypothesis there exists a continuous path ϕ : [a, b] R N such that ϕ (a = x, ϕ (b = y and ϕ ([a, b] E. By Proposition 20 and Theorem 21, we have that ϕ ([a, b] is connected. On the other hand, ϕ ([a, b] E U V, x ϕ ([a, b] U, y ϕ ([a, b] V, which is a contradiction. Remark 24 In particular, convex sets and star-shaped sets are connected. Exercise 25 Let E R N be a connected set. Prove that E is connected. The next example and exercise show that in R N a connected set may fail to be pathwise connected, unless the set is open. Exercise 26 Let E R 2 be the set given by E 1 = { (x, y R 2 : x = 0, 1 y 1 }, { E 2 = (x, y R 2 : x > 0, y = sin 1 }, x E = E 1 E 2. Prove that E is connected but not pathwise connected. Definition 27 In the Euclidean space R N, a polygonal path is a continuous path ϕ : [a, b] R N for which there exists a partition a = t 0 < t 1 < < t n = b with the property that ϕ : [t i 1, t i ] R N is affi ne for all i = 1,..., n, that is, for some c i, d i R N. ϕ (t = c i + td i for t [t i 1, t i ], 10

11 Theorem 28 Let O R N be open and connected. Then O is pathwise connected. Proof. Define the sets and U := {x O : there exists a polygonal path with endpoints x and x 0 and range contained in O} V := {x O : there does not exist a polygonal path with endpoints x and x 0 and range contained in O}. We claim that U and V are open. To see this, let x U. Since O is open, there exists B (x, r O. Let s show that B (x, r U. Indeed, let y B (x, r. Since x U, there exists a polygonal path ϕ : [a, b] R N such that ϕ (a = x 0, ϕ (b = x and ϕ ([a, b] O. Define the polygonal path ψ : [a, b + 1] R N as follows { ϕ (t if t [a, b], ψ (t := x + (t b y if t [b, b + 1]. Then ψ (a = x 0, ψ (b + 1 = y. Since ψ ([a, b + 1] = ϕ ([a, b] ψ ([b, b + 1] and ψ ([b, b + 1] is the segment joining x with y, which is contained in B (x, r, we have that ψ ([a, b + 1] is contained in O. Hence, B (x, r U. This shows that U is open. To prove that V is open, let x V. Since O is open, there exists B (x, r O. Let s show that B (x, r V. Indeed, let y B (x, r. Assume by contradiction that y U. Then there exists a polygonal path ϕ : [a, b] R N such that ϕ (a = x 0, ϕ (b = y and ϕ ([a, b] O. Define the polygonal path ψ : [a, b + 1] R N as follows { ϕ (t if t [a, b], ψ (t := y + (t b x if t [b, b + 1]. Then ψ (a = x 0, ψ (b + 1 = x and ψ ([a, b + 1] is contained in O, which shows that x U, which is a contradiction. Hence, B (x, r V. This shows that V is open. Hence, U and V are open and by their definition they are disjoint and O = U V. Moreover, U is nonempty, since x 0 U. Thus, V must be empty, otherwise O would be disconnected. Hence, O = U. It follows that U is pathwise connected. Indeed, if x, y O then there exist two polygonal paths ϕ : [a, b] R N and ψ : [c, d] R N such that ϕ (a = x 0, ϕ (b = x and ϕ ([a, b] O, and ψ (c = x 0, ψ (d = y and ψ ([c, d] O. By gluing these two polygonal paths appropriately, we can construct a polygonal path joining x and y with range contained in O. Exercise 29 Prove that the set R 2 \ Q 2 is connected. Exercise 30 Let E 1, E 2 R N be two connected sets. Prove that if there exists x E 1 E 2, then E 1 E 2 is connected. 11

12 Friday, January 27, 2012 Next we show that if a set is not connected, we can decompose it uniquely into a disjoint union of maximal connected subsets. Proposition 31 Let E R N. Assume that E = α Λ E α, where each E α is a connected set. If α Λ E α is nonempty, then E is connected. Proof. We claim that E is connected. If not, then there exist two nonempty open sets U, V R N such that E U V, E U, E V, E U V =. Since each E α is connected, we must have that either E α U or E α V. On the other hand, if α β, then E α E β is nonempty, while E U V is empty. Thus, all E α either belong to U or to V. This contradicts the fact that E U and that E V. Let E R N. For every x E, let E x be the union of all the connected subsets of E that contain x. Note that E x is nonempty, since {x} is a connected subset. In view of the previous proposition, the set E x is connected. Moreover, if x, y E and x y, then either E x E y = or E x = E y. Indeed, if not, then again by the previous proposition the set E x E y would be connected, contained in E, and would contain x and y, which would contradict the definition of E x and of E y. Thus, we can partition E into a disjoint union of maximal connected subsets, called the connected components of E. Exercise 32 Let C R N be a closed set. Then the connected components of C are closed. Exercise 33 Prove that if U R N is open, then the connected components of U are open. Example 34 In a metric space the connected components of an open set need not be open. Consider the space X = {0} { 1 n : n N} with the metric d (x, y := x y. We claim that the connected component X 0 containing 0 is the singleton {0}. Note that this is not open, since any ball B (0, r contains all 1 n for n > 1 r. To prove the claim, assume by contradiction that X 0 1 contains other element of X, say, m X 0. Consider an irrational number 1 m+1 < r < 1 m and consider the open sets U = (, r X and V = (r, X. They disconnect X 0, which is a contradiction. Next we prove the following theorem. Theorem 35 Let U R N be open and let f : U R be such that for all x U and all i = 1,..., N there exists x i (x = 0. Then f is constant in each connected component of U. 12

13 The proof makes use of the mean value theorem. Theorem 36 (Mean Value Theorem Let x, y R N, with x y, let S be the segment of endpoints x and y, that is, S = {tx + (1 t y : t [0, 1]}, and let f : S R be such that f is continuous in S and there exists the directional derivative v (z for all z S except at most x and y, where v :=. Then there exists θ (0, 1 such that x y x y f (x f (y = (θx + (1 θ y x y. (6 v Lemma 37 Under the hypotheses of the previous theorem, the function g (t := f (tx + (1 t y, t [0, 1], is differentiable for all t (0, 1, with Proof. Fix t 0 (0, 1 and consider f = g (t = (tx + (1 t y x y. v g (t g (t 0 = f (tx + (1 t y f (t 0x + (1 t 0 y t t 0 t t ( 0 t 0 x + (1 t 0 y+ (t t 0 x y x y x y (t t 0 x y = f (t 0x + (1 t 0 y+sv f (t 0 x + (1 t 0 y s f (t 0 x + (1 t 0 y x y, where s := (t t 0 x y. Since s 0 as t t 0, it follows that there exists x y g g (t g (t 0 f (t 0 x + (1 t 0 y+sv f (t 0 x + (1 t 0 y (t 0 = lim = lim x y t t0 t t 0 s 0 s = v (t 0x + (1 t 0 y x y. Proof of the Mean Value Theorem. Consider the function g (t := f (tx + (1 t y, t [0, 1]. Since compositions of continuous functions is continuous, we have that g is continuous. By the previous lemma, we are in a position to apply the mean value theorem to the function g to find θ [0, 1] such that that is, g (1 g (0 = dg (θ (1 0, dt f (x f (y = (θx + (1 θ y x y. v 13

14 Remark 38 If f is defined on a larger domain E and all points of S except at most x and y are interior points of E, then by (1, we have that N v (θx + (1 θ y = (θx + (1 θ y (x i y i x i x y, i=i and so we can rewrite (6 in the form f (x f (y = N i=i x i (θx + (1 θ y (x i y i. We now turn to the proof of Theorem 35. Proof of Theorem 35. Since all the partial derivatives are zero, in particular they are continuous. It follows from Theorem 16 that f is differentiable for all x U. Step 1: Let x U. Since U is open, there exists B (x, r U. We claim that f is constant in B (x, r. Fix y B (x, r. By the mean value theorem and Remark 38, there exists θ (0, 1 such that f (x f (y = N i=i x i (θx + (1 θ y (x i y i = 0. This proves the claim. Step 2: Let U α be a connected component of U. By the previous exercise, U α is open. Next fix x 0 U α, let c := f (x 0, and define the set A := {x U α : f (x = c}. We claim that A is open. Indeed, if x A, then by the previous step there exists B (x, r U α such that f is constant in B (x, r. Hence, f = c in B (x, r, which implies that B (x, r A. Thus every point of A is an interior point and so A is open. Moreover, A is nonempty since x 0 A. Next consider the set A 1 = {x U α : f (x c} = f 1 ((, c (c,. Since (, c (c, is open, f is continuous, and U α is open, we have that f 1 ((, c (c, is open. Thus, we can write U α = A A 1, where A and A 1 are open and disjoint. Since A is nonempty and U α is connected, necessarily A 1 must be empty, since otherwise we would negate the fact that U α is connected. Hence, U α = A, that is, f (x = c for all x U α. Monday, January 30,

15 2 Higher Order Derivatives Let E R N, let f : E R and let x 0 E. Let i {1,..., N} and assume that there exists the partial derivatives x i (x for all x E. If j {1,..., N} and x 0 is an accumulation point of E L, where L is the line through x 0 in the direction e j, then we can consider the partial derivative of the function x i with respect to x j, that is, ( 2 f =:. x j x i x j x i Note that in general the order in which we take derivatives is important. Example 39 Let If y 0, then f (x, y := (x, y = x x = y3 x 2 + y 2, { y 2 arctan x y if y 0, 0 if y = 0. ( y 2 arctan x y = y 2 1 ( 1 + x y 2 x ( x y and y (x, y = ( y 2 arctan x y y = 2y arctan x y xy2 x 2 + y 2, = 2y arctan x y + 1 y2 ( 1 + x y 2 y ( x y while at points (x 0, 0 we have: x (x f (x 0 + t, 0 f (x 0, , 0 = lim = lim = 0, t 0 t t 0 t y (x f (x 0, 0 + t f (x 0, 0 t 2 arctan x0 0, 0 = lim = lim t 0 t 0 t t 0 t = lim t 0 t arctan x 0 t = 0, where we have used the fact that arctan x0 t { (x, y = x y 3 x 2 +y 2 if y 0, 0 if y = 0, (x, y = y is bounded and t 0. Thus, { 2y arctan x y xy2 x 2 +y if y 0, 2 0 if y = 0. 15

16 To find 2 f y x (0, 0, we calculate while Hence, 2 f y x (0, 0 = y = lim t 0 2 f (0, 0 = x y x 2 f x y (0, 0 2 f y x (0, 0. Exercise 40 Let Prove that ( x (0, 0 = lim x t 0 t 3 0+t 0 2 = lim 1 = 1, t t 0 ( y (0, 0 = lim y t = lim = lim 0 = 0. t 0 t t 0 f (x, y := { 2 f x y (0, 0 2 f y x (0, 0. (0, 0 + t x (0, 0 t (0 + t, 0 y (0, 0 t x 3 y xy 3 x 2 +y 2 if (x, y (0, 0, 0 if (x, y = (0, 0. We present an improved version of the Schwartz theorem. Theorem 41 (Schwartz Let E R N, let f : E R, let x 0 E, and let i, j {1,..., N}. Assume that there exists r > 0 such that B (x 0, r E and for all x B (x 0, r, the partial derivatives x i (x, x j (x, and 2 f x j x i (x exist. Assume also that is continuous at x 0. Then there exists 2 f x i x j (x 0 and 2 f x j x i 2 f (x 0 = 2 f (x 0. x i x j x j x i Lemma 42 Let A : (( r, r \ {0} (( r, r \ {0} R. Assume that the double limit lim (s,t (0,0 A (s, t exists in R and that the limit lim t 0 A (s, t exists in R for all s ( r, r\{0}. Then the iterated limit lim s 0 lim t 0 A (s, t exists and lim lim A (s, t = lim A (s, t. s 0 t 0 (s,t (0,0 Proof. Let l = lim (s,t (0,0 A (s, t. Then for every ε > 0 there exists δ = δ ((0, 0, ε > 0 such that A (s, t l ε for all (s, t (( r, r \ {0} (( r, r \ {0}, with s t 0 2 δ. 16

17 Fix s ( δ 2, δ 2 \ {0}. Then for all t ( δ 2, δ 2 \ {0}, A (s, t l ε and so letting t 0 in the previous inequality (and using the fact that the limit lim t 0 A (s, t exists, we get lim A (s, t l ε t 0 for all s ( δ 2, 2 δ \{0}. But this implies that there exists lims 0 lim t 0 A (s, t = l. Proof of Theorem 41. Step 1: Assume that N = 2. Let t, s < r 2. Then the points (x 0 + s, y 0, (x 0 + s, y 0 + t, and (x 0, y 0 + t belong to B ((x 0, y 0, r. Define A (s, t := f (x 0 + s, y 0 + t f (x 0 + s, y 0 f (x 0, y 0 + t + f (x 0, y 0, st g (x := f (x, y 0 + t f (x, y 0. By the mean value theorem A (s, t = g (x 0 + s g (x 0 st = g (ξ t = x (ξ t, y 0 + t x (ξ t, y 0 t where ξ is between x 0 and x 0 + t. Fix t and consider the function By the mean value theorem, h (y := x (ξ t, y. h (b h (a = h (c (b a = 2 f y x (ξ t, c (b a for some c between a and b. Taking b = t and a = 0, we get x (ξ t, y 0 + t x (ξ t, y 0 = 2 f y x (ξ t, η t t where η t is between y 0 and y 0 + t. Hence, A (s, t = 2 f y x (ξ t, η t 2 f y x (x 0, y 0, where we have used the fact that (ξ, η (x 0, y 0 as (s, t (0, 0 together with the continuity of 2 f y x at (x 0, y 0. Note that this shows that there exists the limit lim A (s, t = 2 f (s,t (0,0 y x (x 0, y 0. 17

18 On the other hand, for all s 0, lim A (s, t = 1 [ f t 0 s lim (x0 + s, y 0 + t f (x 0 + s, y 0 t 0 t y = (x 0 + s, y 0 y (x 0, y 0. s Hence, we are in a position to apply the previous lemma to obtain 2 f y x (x 0, y 0 = lim A (s, t = lim (s,t (0,0 lim s 0 t 0 A (s, t f (x ] 0, y 0 + t f (x 0, y 0 t y = lim (x 0 + s, y 0 y (x 0, y 0 = 2 f s 0 s x y (x 0, y 0. Step 2: In the case N 2 let x = (x 1,..., x N, x 0 = (c 1,..., c N. Assume that 1 < i < j < N (the cases i = 1 and j = N are similar and apply Step 1 to the function of two variables F (x i, x j := f (c 1,..., c i 1, x i, c i+1,..., c j 1, x j, c j+1,..., c N Wednesday, February 1, 2012 Next we prove Taylor s formula in higher dimensions. We set N 0 := N {0}. A multi-index α is a vector α = (α 1,..., α N (N 0 N. The length of a multiindex is defined as α := α α N. Given a multi-index α, the partial derivative α x α α x α := α x α1 1, xα N N where x = (x 1,..., x N. If α = 0, we set 0 f x 0 := f. Example 43 If N = 3 and α = (2, 1, 0, then (2,1,0 (x, y, z = 3 (2,1,0 x 2 y. Given a multi-index α and x R N, we set is defined as α! := α 1! α N!, x α := x α1 1 xα N N. If α = 0, we set x 0 := 1. Using this notation, we can extend the binomial theorem. 18

19 Theorem 44 (Multinomial Theorem Let x = (x 1,..., x N R N and let n N. Then (x x N n = α m ulti-index, α =n n! α! xα. Proof. Exercise. Given an open set U R N, for every nonnegative integer m N 0, we denote by C m (U the space of all functions that are continuous together with their partial derivatives up to order m. We set C (U := C m (U. Theorem 45 (Taylor s Formula Let U R N be an open set, let f C m (U, m N, and let x 0 U. Then for every x U, f (x = α m ulti-index, 0 α m m=0 1 α f α! x α (x 0 (x x 0 α + R m (x, where lim x x 0 R m (x x x 0 m = 0. Proof. Step 1: Since x 0 U and U is open, there exists B (x 0, r U. We prove Taylor s formula with Lagrange s reminder, that is, f (x = 0 α <m 1 α f α! x α (x 0 (x x 0 α + α =m 1 α f α! x α (x 0 + c (x x 0 (x x 0 α, (7 for all x B (x 0, r and where c (0, 1. Fix x B (x 0, r, let h := x x 0 and consider the function g (t := f (x 0 + th defined for t [0, 1]. By Lemma 37 and Remark 38, we have that dg N dt (t = (x 0 + th h i = (h f (x 0 + th x i with for all t [0, 1]. By repeated applications of Lemma 37 and Remark 38, we get that d (n g dt n (t = (h n f (x 0 + th for all n = 1,..., m, where (h n means that we apply the operator h = h h N x 1 x N 19

20 n times to f. By the multinomial theorem, and the fact that for functions in C m partial derivatives commute, ( n (h n = h h N x 1 x N n! α = hα α! x α, and so d (n g dt n (t = α multi-index, α =n α multi-index, α =n Proof. Using Taylor s formula for g, we get m 1 g (1 = g (0 + n=1 1 d (n g n! dt n (0 (1 0n + 1 m! for some c (0, 1. Substituting, we obtain 1 α f f (x 0 + h = α! x α (x 0 h α + 0 α =m 1 n! α! hα α f x α (x 0 + th. α =m Friday, February 03, 2012 d (m g (c (1 0m dtm 1 α f α! x α (x 0 + ch h α. (8 Step 2: We add and subtract α =m 1 α f α! x (x α 0 (x x 0 α in (7, to get f (x = 1 α f α! x α (x 0 (x x 0 α and set R m (x := 0 α m + α =m α =m 1 α! (x x 0 α ( α f x α (x 0 + c (x x 0 α f x α (x 0, ( 1 α! (x x 0 α α f x α (x 0 + c (x x 0 α f x α (x 0. Fix ε > 0 and find δ = δ (x 0, ε > 0 such that α f x α (x α f x α (x 0 ε for all α = m and all x U with x x 0 δ. Then, since (x x 0 α x x 0 m, we have that R m (x x x 0 m 1 α f α! x α (x 0 + c (x x 0 α f x α (x 0 α =m ε α =m 1 α!, 20

21 R which shows that lim m(x x x0 x x = 0. 0 m Example 46 Let s calculate the limit (1 + x y 1 lim (x,y (0,0 x2 + y. 2 By substituting we get 0 0. Consider the function f (x, y = (1 + x y 1 = e log(1+xy 1 = e y log(1+x 1, which is defined in the set U := { (x, y R 2 : 1 + x > 0 }. The function f is of class C. Let s use Taylor s formula of order m = 1 at (0, 0, f (x, y = f (0, 0 + ( (0, 0 (x 0 + x y (0, 0 (y 0 + o x2 + y 2. We have and so ( (x, y = e y log(1+x 1 = e y log(1+x 1 y x x 1 + x, y (x, y = ( e y log(1+x 1 = e y log(1+x y log (1 + x, y ( f (x, y = (x (y 0 + o x2 + y 2, which means that ( (1 + x y 1 o x2 + y 2 lim = lim = 0. (x,y (0,0 x2 + y 2 (x,y (0,0 x2 + y 2 Note that if we had to calculate the limit (1 + x y 1 lim (x,y (0,0 x 2 + y 2, then we would need Taylor s formula of order m = 2 at (0, 0, f (x, y = f (0, 0 + (0, 0 (x 0 + (0, 0 (y 0 x y f (2, 0! x 2 (0, 0 (x f (0, 0 (x 0 (y 0 (1, 1! x y f (0, 2! y 2 (0, 0 (y 02 + o ( x 2 + y 2. Another simpler method would be to use the Taylor s formulas for e t and log (1 + s. Exercise 47 Calculate the limit lim (x,y (2,0 1 cos [(x 2 y] ]. log [1 + (x y 2 21

22 3 Local Minima and Maxima Definition 48 Let f : E R, where E R N, and let x 0 E. We say that (i f attains a local minimum at x 0 if there exists r > 0 such that f (x f (x 0 for all x E B (x 0, r, (ii f attains a global minimum at x 0 if f (x f (x 0 for all x E, (iii f attains a local maximum at x 0 if there exists r > 0 such that f (x f (x 0 for all x E B (x 0, r, (iv f attains a global maximum at x 0 if f (x f (x 0 for all x E. Theorem 49 Let f : E R, where E R N. Assume that f attains a local minimum (or maximum at some point x 0 E. If there exists a direction v and δ > 0 such that the set {x 0 + tv : t ( δ, δ} E (9 and if there exists v (x 0, then necessarily, v (x 0 = 0. In particular, if x 0 is an interior point of E and f is differentiable at x 0, then all partial derivatives and directional derivatives of f at x 0 are zero. Proof. Exercise. Remark 50 In view of the previous theorem, when looking for local minima and maxima, we have to search among the following: Interior points at which f is diff erentiable and f (x = 0, these are called critical points. Note that if f (x 0 = 0, the function f may not attain a local minimum or maximum at x 0. Indeed, consider the function f (x = x 3. Then f (0 = 0, but f is strictly increasing, and so f does not attain a local minimum or maximum at 0. Interior points at which f is not diff erentiable. The function f (x = x attains a global minimum at x = 0, but f is not differentiable at x = 0. Boundary points. To find suffi cient conditions for a critical point to be a point of local minimum or local maximu, we study the second order derivatives of f. Definition 51 Let f : E R, where E R N, and let x 0 E. The Hessian matrix of f at x 0 is the N N matrix 2 f (x x f 1 x N x 1 (x 0 H f (x 0 : = whenever it is defined. =. 2 f x 1 x N (x 0 ( 2 f x i x j (x 0 N, i,j=1. 2 f (x x 2 0 N 22

23 Remark 52 If the hypotheses of Schwartz s theorem are satisfied for all i, j = 1,..., N, then 2 f (x 0 = 2 f (x 0, x i x j x j x i which means that the Hessian matrix H f (x 0 is symmetric. Given an N N matrix H, the characteristic polynomial of H is the polynomial p (t := det (ti N H, t R. Theorem 53 Let H be an N N matrix. If H is symmetric, then all roots of the characteristic polynomial are real. Theorem 54 Given a polynomial of the form p (t = t N + a N 1 t N 1 + a N 2 t N a 1 t + a 0, t R, where the coeffi cients a i are real for every i = 0,..., N 1, assume that all roots of p are real. Then (i all roots of p are positive if and only if the coeffi cients alternate sign, that is, a N 1 < 0, a N 2 > 0, a N 3 < 0, etc. (ii all roots of p are negative if and only a i > 0 for every i = 0,..., N 1. Monday, February 06, 2012 Proof of Theorem 53. We begin by observing that if x and y R N, then (Hx y = x (Hy. Indeed, we have ( N N N N (Hx y = H ij x j y i = x j H ji y i = x (Hy. j=1 Step 1: We claim that there cannot be more than N distinct eigenvalues. To see this it is enough to show that if x and y are eigenvectors corresponding to different eigenvalues λ and µ, then x and y are orthogonal. Indeed, we have and so λ (x y = (λx y = (Hx y = x (Hy = x (µy = µ (x y, j=1 (λ µ (x y = 0. Since λ µ, it follows that x y = 0. Step 2: We prove the existence of one real eigenvalue λ 1. Consider the function f (x := (Hx x N N j=1 x 2 = H ijx i x j x x2 N 23

24 defined for all x R N, x 0. The expression (Hx x is called the Rayleigh x 2 quotient. Note that f is of class C and that ( x x 2 N ( N N j=1 H N N ij (δ i,k x j + x i δ j,k j=1 H ijx i x j 2x k (x = x k = (x x2 N 2 ( x xn ( 2 N j=1 H kjx j + N H ikx i ( N j=1 N H ijx i x j 2x k (x x2 N 2 (10 = 2 ( x xn 2 ( N j=1 H N N kjx j j=1 H ijx i x j 2x k = 2 (Hx k x 2 2(Hx x x 4 x k (x x2 N 2 for all k = 1,..., N, where in the third equality we have used the fact that H is symmetric. Since f is a continuous function and B (0, 1 is compact, by the Weierstrass theorem, there exists max x B(0,1 f (x = f (x 1. Hence, f (x f (x 1 for all x B (0, 1. If now x R N, x 0, then belongs to B (0, 1, and so ( x f (x = f f (x 1, x where we have used the fact that f (tx = f (x for t > 0. This shows that x 1 is a point of absolute maximum for f in R N \ {0}. Since x 1 is an interior point and f is of class C, it follows from Theorem 49 that x 1 is a critical point of f, that is, f (x 1 = 0 for all k = 1,..., N. It follows by (10 and the fact that x 1 = 1 that or, equivalently, that where Hx 1 ((Hx 1 x 1 x 1 = 0, Hx 1 = λ 1 x 1, λ 1 := (Hx 1 x 1. This shows that λ 1 is a real eigenvalue of H and x 1 is the corresponding eigenvector. Note that λ 1 = f (x 1 = max {(Hx x : x = 1}. x x 24

25 Step 3: Let 1 n N 1 and assume by induction that real eigenvalues λ 1,..., λ n of H have been found with corresponding orthonormal eigenvectors x 1,..., x n. Note that the eigenvalues may not be distinct, but x i x j = δ i,j for all i, j = 1,..., n. Consider the set Y := { x R N : x x i = 0 for all i = 1,..., n }. Note that Y is a subspace of R N of dimension N k. Since Y B (0, 1 is compact, by the Weierstrass theorem, there exists max f (x = f (x n+1. x Y B(0,1 Hence, f (x f (x n+1 for all x Y B (0, 1. If now x Y, x 0, then belongs to Y B (0, 1, and so x x f (x = f ( x f (x n+1, x where we have used the fact that f (tx = f (x for t > 0. In particular, if v Y is a direction, the for every t R, f (x n+1 + tv f (x n+1, which shows that the function g (t := f (x n+1 + tv has a maximum at t = 0. Hence, 0 = g (0 = v (x n+1. Using Theorems 16 and 10, we get that f (x n+1 v = 0 for all vectors v Y of norm one, and, in turn, for all v Y. By (10 and the fact that x n+1 = 1 it follows that 2 (Hx n+1 ((Hx n+1 x n+1 x n+1 v = 0 (11 for v Y. On the other hand, Hx n+1 belongs to Y. Indeed, for all i = 1,..., n (Hx n+1 x i = x n+1 (Hx i = x n+1 (λ i x i = λ i (x n+1 x i = 0, where we have used the fact that x n+1 Y. Thus the vector f (x n+1 = 2 (Hx n+1 ((Hx n+1 x n+1 x n+1 belongs to Y and so it is orthogonal to itself by (11. It follows that 0 = f (x n+1 = 2 (Hx n+1 ((Hx n+1 x n+1 x n+1, so that Hx n+1 = λ n+1 x n+1, 25

26 where λ n+1 := (Hx n+1 x n+1. This proves that λ n+1 is an eigenvalue and x n+1 a corresponding eigenvector. Note that λ n+1 may coincide with some of the previous eigenvalues, but since x n+1 belongs to Y, x n+1 is distinct from all the previous eigenvectors x 1,..., x n. Moreover, λ n+1 = f (x n+1 = max {(Hx x : x Y B (0, 1}. This induction argument shows that there exist N distinct orthonormal eigenvectors x 1,..., x N. Thus, the eigenvectors form an orthonormal basis in R N. Step 4: Since for any t R and i = 1,..., N, (ti N H x i = (t λ i x i, it follows that (see the following exercise, det (ti N H = (t λ 1 (t λ N. Exercise 55 Let H be an N N symmetric matrix and let x 1,..., x N R N be orthonormal eigenvectors corresponding to eigenvalues λ 1,..., λ N R (possibly repeated. Let A be the N N matrix whose columns are x T 1,..., x T N. (i Prove that A T A = I N. (ii Prove that det A = ±1. (iii Prove that HA = ( λ 1 x T 1 λ N x T N. (iv Prove that det H = λ 1... λ N. Wednesday, February 08, 2012 The next theorem gives necessary and suffi cient conditions for a point to be of local minimum or maximum. Theorem 56 Let U R N be open, let f : U R be of class C 2 (U and let x 0 U be a critical point of f. (i If H f (x 0 is positive definite, then f attains a local minimum at x 0, (ii if f attains a local minimum at x 0, then H f (x 0 is positive semidefinite, (iii if H f (x 0 is negative definite, then f attains a local maximum at x 0, (iv if f attains a local maximum at x 0, then H f (x 0 is negative semidefinite. 26

27 Proof. (i Assume that H f (x 0 is positive definite. Then by Remark??, N N j=1 2 f x j x i (x 0 v i v j m v 2 (12 for all v R N and for some m > 0. We now apply Taylor s formula of order two to obtain f (x = f (x 0 + f (x 0 (x x 0 + = f (x N N j=1 α =2 1 α f α! x α (x 0 (x x 0 α + R 2 (x 2 f x j x i (x 0 (x x 0 i (x x 0 j + R 2 (x, where we have used the fact that x 0 is a critical point and where lim x x 0 R 2 (x x x 0 2 = 0. Using the definition of limit with ε = m 2, we can find δ > 0 such that R 2 (x x x 0 2 m 2 for all x E with x x 0 δ. Using this property and (12, we get ( f (x f (x 0 + m x x R 2 (x = f (x 0 + x x 0 2 m + R 2 (x x x 0 2 ( f (x 0 + x x 0 2 m m = f (x 0 + x x 0 2 m 2 2 > f (x 0 for all x E with 0 < x x 0 δ. This shows that f attains a (strict local minimum at x 0. (ii Since x 0 U and U is open, there exists B (x 0, r U. Let v R N and consider the function g (t := f (x 0 + tv, t ( r, r. Since g (t = f (x 0 + tv f (x 0 = g (0 for all t ( r, r, g attains a local minimum at t = 0. By two applications of Lemma 37 and Remark 38, we have that for t ( r, r, g (t = g (t = N N j=1 x i (x 0 + tv v i, N 2 f x j x i (x 0 + tv v i v j. 27

28 Hence, g C 2 ( r, r, and so, since g attains a local minimum at t = 0, 0 g (0 = N N j=1 2 f x j x i (x 0 v i v j. Since this is true for every direction v, it follows that H f (x 0 is positive semidefinite. Remark 57 Note that in view of the previous theorem, if at a critical point x 0 the characteristic polynomial of H f (x 0 has one positive root and one negative root, then f does not admit a local minimum or a local maximum at x 0. Theorem 56 shows that if H f (x 0 is positive definite, then f admits a local minimum at x 0. The following exercise shows that we cannot weakened this hypothesis to H f (x 0 positive semidefinite. Example 58 Let f (x, y := x 2 y 4. Let s find the critical points of f. We have ( (x, y = x 2 y 4 = 2x 0 = 0, x x y (x, y = ( x 2 y 4 = 0 4y 3 = 0. y Hence, (0, 0 is the only critical point. Let s find the Hessian matrix at these points. Note that the function is of class C, so we can apply Schwartz s theorem. We have 2 f (x, y = (2x = 2, x2 x 2 f y 2 (x, y = ( 4y 3 = 12y 2, y 2 f y x (x, y = (2x = 0. y Hence, so that H f (0, 0 = ( , ( ( = det λ H 0 1 f (0, 0 ( λ 2 0 = det = (λ 2 (λ λ 0 = (λ 2 λ, 28

29 and so the roots are λ = 2 or λ = 0. Hence, at (0, 0 we cannot have a local maximum. But it could be a local minimum. However, taking f (0, y = y 4, which has a strict maximum at y = 0. This shows that f does not admit a local minimum or a local maximum at (0, 0. 4 Vector-Valued Functions Since we will work with the different euclidean spaces, R M and R N, when needed and to avoid confusion, we will denote by M and N the norms in R M and R N, respectively. We recall that a function T : R N R M is linear if T (x + y = T (x + T (y for all x, y R N and for all s R and x R N. T (sx = st (x Definition 59 Let E R N, let f : E R M, and let x 0 E be an accumulation point of E. The function f is differentiable at x 0 if there exists a linear function T : R N R M (depending on f and x 0 such that f (x f (x 0 T (x x 0 lim = 0. (13 x x 0 x x 0 N provided the limit exists. The function T, if it exists, is called the differential of f at x 0 and is denoted df (x 0 or df x0. Theorem 60 Let E R N, let f : E R M, and let x 0 E be an accumulation point of E. Then f = (f 1,..., f M is differentiable at x 0 if and only if all its components f j, j = 1,..., M, are differentiable at x 0. Moreover, df x0 = ( d (f1 x0,..., d (f M x0. Proof. The proof relies on Exercise?? and is left as an exercise. We study the differentiability of composite functions. Theorem 61 (Chain Rule Let F R M, E R N, let g : F E, g = (g 1,..., g N, and let f : E R. Assume that at some point y 0 F there exist the directional derivatives g1 v (y 0,..., g N v (y 0 for some direction v and f is differentiable at the point g (y 0. Then the composite function f g admits a directional derivative at y 0 in the direction v and (f g v (y 0 = N df g(y0 (e i g i v (y 0. Moreover, if g is differentiable at y 0 and f is differentiable at g (y 0, then f g is differentiable at y 0. 29

30 Proof. Set x 0 := g (y 0. Since f is differentiable at x 0, we can write f (x = f (x 0 + b (x x 0 + R (x, where b := ( df g(y0 (e 1,..., df g(y0 (e N and lim x x 0 R (x x x 0 N = 0. (14 Note that Take x = g (y 0 + tv. Then R (x 0 = 0. f (g (y 0 + tv = f (g (y 0 + b (g (y 0 + tv g (y 0 + R (g (y 0 + tv, and so (f g (y 0 + tv (f g (y 0 t We claim that = f (g (y 0 + tv f (g (y 0 t = b g (y 0 + tv g (y 0 t R (g (y 0 + tv lim = 0. t 0 t If for some t, we have g (y 0 + tv = g (y 0, then (15 + R (g (y 0 + tv. t R (g (y 0 + tv = R (g (y 0 = 0. (16 On the other have, if g (y 0 + tv g (y 0, then R (g (y 0 + tv R (g (y 0 + tv t = g (y 0 + tv g (y 0 t g (y 0 + tv g (y 0 N t t. (17 N Since each function g i admits a finite directional derivative at y 0, it follows that g (y 0 + tv g (y 0 as t 0 (why?. Hence, by (14 lim t 0 R (g (y 0 + tv g (y 0 + tv g (y 0 N = 0, (18 where the limit is taken over all t such that g (y 0 + tv g (y 0. On the other hand, since g(y0+tv g(y0 t g v (y 0, and t t is bounded by one, we have that is bounded, which, together with (17 and (14, implies t t that g(y0+tv g(y0 t N R (g (y 0 + tv lim = 0 t 0 t where the limit is taken over all t such that g (y 0 + tv g (y 0 (here we are using the fact that if a function is bounded and another function goes to zero, 30

31 then their product goes to zero. Together with (16, this implies that the claim holds. Using the claim, it follows that (f g (y 0 + tv (f g (y 0 t = b g (y 0 + tv g (y 0 t b g v (y 0 + 0, which proves the first part of the statement. The second part of the statement is left as an exercise. Remark 62 If x 0 E, then (f g v (y 0 = N x i (g (y 0 g i v (y 0 = f (g (y 0 g v (y 0. Exercise 63 Prove the second part of the theorem. + R (g (y 0 + tv t Example 64 Consider the functions f : R 2 R and g : R 2 R 2 defined by Then f (x 1, x 2 = x 1 x 2 1, g (y 1, y 2 = (y 1 y 2 e y1, y 1 sin (y 1 y 2. (x 1, x 2 = 1x 2 0, x 1 g 1 (y 1, y 2 = 1y 2 e y1, y 1 g 2 (y 1, y 2 = 1 sin (y 1 y 2 + y 1 cos (y 1 y 2 (1y 2, y 1 First method: Consider the composition Then by the chain rule, (f g y 1 while (f g y 2 (x 1, x 2 = x 1 1 0, x 2 g 1 (y 1, y 2 = y 1 1 0, y 2 (f g (y 1, y 2 = f (g 1 (y 1, y 2, g 2 (y 1, y 2. g 2 y 2 (y 1, y 2 = y 1 cos (y 1 y 2 (y 1 1. (y 1, y 2 = x 1 (g 1 (y 1, y 2, g 2 (y 1, y 2 g 1 y 1 (y 1, y 2 + x 2 (g 1 (y 1, y 2, g 2 (y 1, y 2 g 2 y 1 (y 1, y 2 = y 1 sin (y 1 y 2 (1y 2 e y1 + (y 1 y 2 e y1 (1 sin (y 1 y 2 + y 1 cos (y 1 y 2 (1y 2, (y 1, y 2 = x 1 (g 1 (y 1, y 2, g 2 (y 1, y 2 g 1 y 2 (y 1, y 2 + x 2 (g 1 (y 1, y 2, g 2 (y 1, y 2 g 2 y 2 (y 1, y 2 = y 1 sin (y 1 y 2 (y (y 1 y 2 e y1 (y 1 cos (y 1 y 2 (y

32 Second method: Write the explicit formula for (f g, that is, Then (f g (y 1, y 2 = f (g 1 (y 1, y 2, g 2 (y 1, y 2 = (y 1 y 2 e y1 (y 1 sin (y 1 y 2 1. (f g y 1 (y 1, y 2 = y 1 [(y 1 y 2 e y1 (y 1 sin (y 1 y 2 1] and = (1y 2 e y1 (y 1 sin (y 1 y 2 + (y 1 y 2 e y1 (1 sin (y 1 y 2 + y 1 cos (y 1 y 2 (1y 2 0 (f g y 2 (y 1, y 2 = y 2 [(y 1 y 2 e y1 (y 1 sin (y 1 y 2 1] = (y (y 1 sin (y 1 y 2 + (y 1 y 2 e y1 (y 1 cos (y 1 y 2 (y Next we define the Jacobian of a vectorial function f : E R M. Definition 65 Given a set E R N and a function f : E R M, the Jacobian matrix of f = (f 1,..., f M at some point x 0 E, whenever it exists, is the M N matrix J f (x 0 := It is also denoted f 1 (x 0. f M (x 0 = 1 x 1 (x 0.. M x 1 (x 0 (f 1,..., f M (x 1,..., x N (x 0. 1 x N (x 0.. M x N (x 0 When M = N, J f (x 0 is an N N square matrix and its determinant is called the Jacobian determinant of f at x 0. Thus, ( j det J f (x 0 = det (x 0. x i i,j=1,...,n Remark 66 Let E R N, let f : E R M, and let x 0 E. Assume that f is differentiable at x 0. Then by Theorem 60 all its components f j, j = 1,..., M, are differentiable at x 0 with df x0 = ( d (f 1 x0,..., d (f M x0.. Since x 0 is an interior point, it follows from (2 that for every direction v, Hence, d (f j x0 (v = f j (x 0 v = N j x i (x 0 v i. df x0 (v = ( d (f 1 x0 (v,..., d (f M x0 (v = J f (x 0 v. 32

33 As a corollary of Theorem 61, we have the following result. Corollary 67 Let F R M, E R N, let g : F E, g = (g 1,..., g N, and let f : E R P. Assume that g is differentiable at some point y 0 F and that f is differentiable at the point g (y 0 and that g (y 0 E. Then the composite function f g is differentiable at y 0 and J f g (y 0 = J f (g (y 0 J g (x 0. Definition 68 Given an open set U R N and a function f : U R M, we say that f is of class C m for some nonnegative integer m N 0, if all its components f i, i = 1,..., M, are of class C m. The space of all functions f : U R M of class C m is denoted C ( m U; R M. We set C ( U; R M := C ( m U; R M. 5 Implicit and Inverse Function m=0 Friday, February 10, 2012 Given a function f of two variables (x, y R 2, consider the equation f (x, y = 0. We want to solve for y, that is, we are interested in finding a function y = g (x such that f (x, g (x = 0. We will see under which conditions we can do this. The result is going to be local. In what follows given x R N and y R M and f (x, y, we write (x, y := x 1 x 1 (x, y. M x 1 (x, y 1 x N (x, y. M x N (x, y and (x, y := y 1 y 1 (x, y. M y 1 (x, y 1 y M (x, y. M y M (x, y. Theorem 69 (Implicit Function Let U R N R M be open, let f : U R M, and let (a, b U. Assume that f C m (U for some m N, that f (a, b = 0 and det (a, b 0. y 33

34 Then there exists B N (a, r 0 R N and B M (b, r 1 R M such that B N (a, r 0 B M (b, r 1 U and for every x B N (a, r 0 there exists a unique y B M (b, r 1 such that f (x, y = 0. Moreover the function g : B N (a, r 0 B M (b, r 1 x y satisfies g (a = b, belongs to C m (B N (a, r 0 and ( 1 g x (x = (x, g (x (x, g (x. y x Proof. We present a proof in the case N = M = 1. Step 1: Existence of g. Since y (a, b 0, without loss of generality, we can assume that y (a, b > 0 (the case y (a, b < 0 is similar. Using the fact that is continuous at (a, b, we can find r > 0 such that y and R := [a r, a + r] [b r, b + r] U (x, y > 0 for all (x, y R. y Consider the function h (y := f (a, y, y [b r, b + r]. Since h (y = (a, y > 0 for all y [b r, b + r], y we have that h is strictly increasing. Using the fact that h (b = f (a, b = 0, it follows that 0 > h (b r = f (a, b r, 0 < h (b + r = f (a, b + r. Consider the function k 1 (x := f (x, b r, x [a r, a + r]. Since k 1 (a < 0 and k 1 is continuous at a, there exists 0 < δ 1 < r such that 0 > k 1 (x = f (x, b r for all x (a δ 1, a + δ 1. Similarly, consider the function k 2 (x := f (x, b + r, x [a r, a + r]. Since k 2 (a > 0 and k 2 is continuous at a, there exists 0 < δ 2 < r such that 0 < k 2 (x = f (x, b + r for all x (a δ 2, a + δ 2. Let δ := min {δ 1, δ 2 }. Then for all x (a δ, a + δ, f (x, b r < 0, f (x, b + r > 0. Fix x (a δ, a + δ and consider the function k (y := f (x, y, y [b r, b + r]. Since k (y = (x, y > 0 for all y [b r, b + r], y 34

35 we have that k is strictly increasing. Using the fact that k (b r = f (x, b r < 0 and k (b + r = f (x, b + r > 0, it follows that there exists a unique y (b r, b + r (depending on x such that 0 = k (y = f (x, y. Thus, we have shown that for every x (a δ, a + δ there exists a unique y (b r, b + r depending on x such that f (x, y = 0. We define g (x := y. Step 2: Continuity of g. Fix x 0 (a δ, a + δ. Note that b r < g (x 0 < b + r. Let ε > 0 be so small that b r < g (x 0 ε < g (x 0 < g (x 0 + ε < b + r. Consider the function j (y := f (x 0, y, y [b r, b + r]. Since j (y = y (x 0, y > 0 for all y [b r, b + r], we have that j is strictly increasing. Using the fact that j (g (x 0 = f (x 0, g (x 0 = 0, it follows that f (x 0, g (x 0 ε < 0, f (x 0, g (x 0 + ε > 0. Consider the function j 1 (x := f (x, g (x 0 ε, x (a δ, a + δ. Since j 1 (x 0 < 0 and j 1 is continuous at x 0, there exists 0 < η 1 < δ such that 0 > j 1 (x = f (x, g (x 0 ε for all x (x 0 η 1, x 0 + η 1. Similarly, consider the function j 2 (x := f (x, g (x 0 + ε, x (a δ, a + δ. Since j 2 (x 0 > 0 and j 2 is continuous at x 0, there exists 0 < η 2 < δ such that 0 < j 2 (x = f (x, g (x 0 + ε for all x (x 0 η 2, x 0 + η 2. Let η := min {η 1, η 2 }. Then for all x (a η, a + η, f (x, g (x 0 ε < 0, f (x, g (x 0 + ε > 0. But f (x, g (x = 0 and y [b r, b + r] f (x, y is strictly increasing. It follows that g (x 0 ε < g (x < g (x 0 + ε and so g is continuous at x 0. Step 3: Differentiability of g. Fix x 0 (a δ, a + δ. Consider the segment joining (x, g (x and (x 0, g (x 0, S = {t (x, g (x + (1 t (x 0, g (x 0 : t [0, 1]}. By the mean value theorem there exists θ (0, 1 such that 0 = f (x, g (x f (x 0, g (x 0 = x (θx + (1 θ x 0, θg (x + (1 θ g (x 0 (x x 0 + y (θx + (1 θ x 0, θg (x + (1 θ g (x 0 (g (x g (x 0. 35