1. Bounded linear maps. A linear map T : E F of real Banach

Transcription

1 DIFFERENTIABLE MAPS 1. Bounded linear maps. A linear map T : E F of real Banach spaces E, F is bounded if M > 0 so that for all v E: T v M v. If v r T v C for some positive constants r, C, then T is bounded: for any v E \ {0} we have T ( rv v ) C, so T v C r v. The vector space L(E, F ) of bounded linear maps is usually endowed with the operator norm: T = sup{ T v ; v 1}. Exercise 1. Show that L(E, F ) is complete in this norm, if E, F are Banach spaces. If F = R, we have the space of bounded linear functionals on E, denoted E (dual space). 2. Differentiable maps. f : U F (U E; E, F Banach) is differentiable at x 0 U if there exists a linear map T : E F satisfying: ( ɛ > 0)( δ > 0) h δ f(x 0 + h) f(x 0 ) T [h] ɛ h. The map T is unique, denoted df(x 0 ) and called differential (or derivative ) of f at x 0. Another notation for this is: (pronounced little oh of h ). f(x 0 + h) f(x 0 ) = df(x 0 )[h] + o(h) Remarks. 1. If f is continuous at x 0, it is easy to see that when f is differentiable at x 0 we have df(x 0 ) L(E, F ): the differential is a bounded linear map. Conversely, if f is differentiable at x 0 and df(x 0 ) is a bounded linear map (which is always the case for finite-dimensional E, F ), then f is continuous at x 0. (This follows easily from the definition.) 2. If f is differentiable at x 0, all directional derivatives v f(x 0 ) exist, and depend linearly on v: setting h = tv in the definition and dividing by t, we have for v 0: ( ɛ > 0)( δ > 0) t < This shows v f(x 0 ) = df(x 0 )[v]. δ v f(x 0 + tv) f(x 0 ) df(x 0 )[v] ɛ v. t 1

2 Example 1. Any bounded bilinear vector-valued map f : V W F is differentiable at every point (a, b) V W, with derivative: df(a, b)[v, w] = f(a, w) + f(v, b). (Bounded means f(v, w) F M v V w W for some M > 0. This is always the case if V, W, F are finite-dimensional. ) To see this, note that by bilinearity: f(a+v, b+w) = f(a, b)+f(a, w)+f(v, b)+f(v, w), f(v, w) M v V w W M (v, w) 2, where we take the sup norm in V W : (v, w) V W = max{ v V, w W }. Example 2. Let U = GL(n) L(R n ) be the set of invertible linear operators. Then the inversion map f(a) = A 1 is differentiable in U, with derivative: df(a)[v ] = A 1 V A 1, V L(R n ). Lemma. Let A L(R n ) be invertible. Let c = (2 A 1 ) 1 (operator norm). Then for all V L(R n ) with V c, A + V is invertible with (A + V ) 1 1/c. Proof. x A 1 Ax, so Ax 2c x. Thus V c (A + V )x Ax V x c x. Thus V c implies A + V is invertible and x = (A + V )(A + V ) 1 x c (A + V ) 1 x, so (A + V ) 1 1/c. Remark. This shows (without using the determinant) that the set of invertible linear operators is open in finite dimensions. In infinite dimensions (general Banach spaces) the set GL(E) L(E) of bounded linear operators which are invertible (with bounded inverse) is also open in the space of bounded linear operators, but this requires more theory. Now write: (A + V ) 1 A 1 = A 1 V A 1 + r(v ). Multiplying on the left by A + V we find: r(v ) = (A + V ) 1 (V A 1 ) 2. Thus r(v ) (A + V ) 1 A 1 2 V 2. This implies lim V 0 r(v )/ V = 0. Remark. In fact the inversion map f(a) = A 1 is C. 3. Mean value theorems. 2

3 3.1 Real-valued functions. Let f : U R be continuous in the open set U E. Let a, b U, and suppose the line segment [a, b] U; parametrize it by a t = a + t(b a), t [0, 1]. Let v = b a E, and suppose the directional derivative v f(a t ) exists for each t (0, 1). Then there exists a θ (0, 1) so that: f(b) f(a) = v f(a θ ). This follows from the usual Mean Value Theorem for real-valued functions of one real variable, applied to the function φ(t) = f(a t ), t [0, 1]. It follows from the definition of directional derivative that φ (t) = v f(a t ). 3.2 Vector-valued functions of one real variable. Let f : [a, b] F be continuous, and differentiable on (a, b) (where F is a Banach space and [a, b] R.) Let φ : [a, b] R be continuous in [a, b], differentiable in (a, b). Then: f (t) φ (t) t (a, b) f(b) f(a) φ(b) φ(a). In particular, considering φ(t) = Mt we have: f (t) M t (a, b) f(b) f(a) M(b a). (Note that f (t) L(R; F ) is naturally identified with a vector f (t) F, the tangent vector.) Proof. Pick an arbitrary ɛ > 0, and define: G = {c [a, b]; ( t [a, c]) f(t) f(a) φ(t) φ(a) + ɛ(t a).} Clearly G is a non-empty closed sub-interval [a, γ] [a, b]. We want to show γ = b. By contradiction, assume γ < b. By differentiability of f and φ at γ, we may find δ (γ, b) so that, for t [γ, δ]: f(t) f(γ) (t γ)f (γ) ɛ 2 (t γ); φ(t) φ(γ) (t γ)φ (γ) ɛ 2 (t γ). This implies, for t [γ, δ] (using f (γ) φ (γ)): f(t) f(a) f(t) f(γ) + f(γ) f(a) ɛ 2 (t γ) + (t γ)φ (γ) + φ(γ) φ(a) + ɛ(γ a) ɛ(t γ) + φ(t) φ(γ) + φ(γ) φ(a) + ɛ(γ a) = φ(t) φ(a) + ɛ(t a), 3

4 contradicting the maximality of γ. This contradiction shows G = [a, b], and letting ɛ 0 concludes the proof. Corollary 1. Mean Value Inequality. Let U E be open, f : U F continuous, a, b U with the line segment [a, b] U; assume f is differentiable at each point of the segment [a, b]. Then: f(b) f(a) ( sup df(x) ) b a. x [a,b] Proof. Let a t = a + t(b a), t [0, 1] be the standard parametrization of [a, b]; consider g : [0, 1] F, g(t) = f(a t ). Then g (t) = df(a t )[b a], so g (t) M, where M := (sup x [a,b] df(x) ) b a. So the corollary follows from the theorem, noting that f(a) = g(0), f(b) = g(1). Corollary 2. First-order Taylor formula with remainder. With the same notation as in Corollary 1, assume now f is differentiable in a neighborhood V U of the segment [a, b]. Then for each x 0 V : f(b) f(a) df(x 0 )[b a] ( sup df(x) df(x 0 ) ) b a. x [a,b] Proof. Apply the theorem to the function g(t) = f(a t ) df(x 0 )[t(b a)], t [0, 1], with derivative g (t) = df(a t )[b a] df(x 0 )[b a]. Note: g (t) M := ( sup df(x) df(x 0 ) ) b a and g(0) = f(a), g(1) = f(b) df(x 0 )[b a]. x [a,b] Remark 1. Note that in the case x 0 = a we could let g(t) = f(a + tv) t v f(a), g (t) = v f(a t ) v f(a) for v = b a, so (assuming only the existence of this directional derivative at each point of the segment [a, b]) we could take: M = sup v f(a t ) v f(a). t [0,1] If the directional derivative v f(x) is continuous in x [a, b], since v f(x) is homogeneous with respect to v ( λv f(x) = λ v f(x)), M is bounded by a small number times b a for b a small, and we have: f(b) f(a) v f(a) ɛ b a for b a small (v = b a) 4

5 assuming only existence and continuity of the directional derivative v f at points of the segment [a, b]. (See below.) Remark 2. Specializing to the case x 0 = a, we get a first order Taylor formula with remainder (for f at the point a:) f(a+h) = f(a)+df(a)[h]+r a (h), R a (h) ( sup df(x) df(a) ) h x [a,a+h] (for all h E small enough that a + h U, assuming f differentiable and continuous in U.) For example, if we know that df satisfies a Lipschitz condition in U (with constant L), we have: R a (h) L h 2, a lower order error (quadratic) for h small. Exercise 2. Let f : U R n be differentiable in the convex open set U R m. Show that: f(x) f(y) sup x y x y = sup df(z). z U Hint. For the direction where you might need a hint, let u R m be a unit vector, and consider: ηdf(z)[u] = f(x) f(z) r z (x), x = z + ηu, η R. Divide by η and let η = x z tend to zero, using the estimate for r z (x) given by the definition of differential. 4. Partial derivatives and functions of class C k. Consider first f : R 2 R, with coordinates (x, y) in R 2. Simple examples show that just the existence of partial derivatives x f(a, b), y f(a, b) at a given point (a, b) R 2 does not even imply continuity at (a, b), let alone differentiability (in the vector sense) at (a, b). However, this changes if we assume x f, y f are defined in a whole open neighborhood U R 2 of (a, b), and define continuous (real-valued) functions in U. More generally, we have the important result: Theorem. With U R m open, let f : U R have all its partial derivatives xi f defined and continuous in U. Then f is differentiable in U, df : U L(R m, R) is continuous and we have: df(x)[h] = m xi f(x)h i, h = (h 1,..., h m ) R m. i=1 5

6 (In particular, since h f(x) = df(x)[h], we see that directional derivatives depend linearly on h.) Conversely, if f is differentiable in U with df : U L(R m, R) continuous, then all partials xi f are defined and continuous in U, and this expression holds. This result follows directly from a more abstract one, dealing with a general notion of partial derivative. Let E 1, E 2, F be Banach spaces and consider f : U 1 U 2 F continuous, with U i open in E i (so U = U 1 U 2 is open in the product space E 1 E 2 ). Given a point (a, b) U 1 U 2, we have continuous partial maps : x 1 U 1 f(x 1, b) F ; x 2 U 2 f(a, x 2 ) F. If the first map is differentiable at a U 1, we call its differential at (a, b) the first partial derivative of f at (a, b), denoted d 1 f(a, b) L(E 1 ; F ); and similarly for the second map (if differentiable at b U 2 ), with differential denoted d 2 f(a, b) L(E 2 ; F ). Remark on norms: we take in E = E 1 E 2 the norm (h, k) E = max{ h E1, k E2 }, and in the sequel omit the subscripts in the norms. Theorem. Assume d 1 f : U L(E 1 ; F ) and d 2 f : U L(E 2 ; F ) exist and are continuous in an open set U = U 1 U 2 E 1 E 2. Then f is differentiable in U and df : U L(E 1 E 2 ; F ) is continuous. We have, for (a, b) U 1 U 2 and (h, k) E 1 E 2 : df(a, b)[(h, k)] = d 1 f(a, b)[h] + d 2 f(a, b)[k]. Conversely, if f is differentiable in U with df continuous from U to L(E; F ), we have that d i f are defined and continuous in U (for i = 1, 2), and this expression holds in U. Proof. Fix (a, b) U and let ɛ > 0 be given. From the definition of partial derivative at (a, b), we have that there exists an r > 0 so that if h r: f(a + h, b) f(a, b) d 1 f(a, b)[h] ɛ h. We seek an estimate of the form: f(a + h, b + k) f(a, b) d 1 f(a, b)[h] d 2 f(a, b)[k] ɛ (h, k), and to start off we break up the increment into two steps: f(a + h, b + k) f(a, b) d 1 f(a, b)[h] d 2 f(a, b)[k] 6

7 = {f(a+h, b) f(a, b) d 1 f(a, b)[h]}+{f(a+h, b+k) f(a+h, b) d 2 f(a, b)[k]}, where we already have an estimate for the first part. From the 1st order Taylor formula with remainder (Corollary 2 above) we have that (for k small enough, say k r, so that b + k U 2 ): f(a+h, b+k) f(a+h, b) d 2 f(a+h, b)[k] ( sup z k d 2 f(a+h, b+z) d 2 f(a+h, b) ) k. By continuity of d 2 f at (a, b), the expression on the right is bounded above by ɛ k, if h r 1 and k r 1, for some r 1 = r 1 (ɛ, a, b) > 0, r 1 < r. To use this estimate, we further decompose the expression corresponding to the increment in the second variable into two parts: f(a + h, b + k) f(a + h, b) d 2 f(a, b)[k] = {f(a+h, b+k) f(a+h, b) d 2 f(a+h, b)[k]}+{d 2 f(a+h, b)[k] d 2 f(a, b)[k]}, where the first term has just been estimated, while the second is bounded by ɛ k if h r 1, again by continuity of d 2 f at (a, b). We conclude that, if (h, k) = max{ h, k } r 1 : f(a+h, b+k) f(a, b) d 1 f(a, b)[h] d 2 f(a, b)[k] 3ɛ max{ h, k } = 3ɛ (h, k). This shows f is differentiable at (a, b), with differential given by the expression in the theorem. Given this expression, the continuity of df in U follows from that of d 1 f and d 2 f. Conversely, assume f is differentiable at (a, b) U. Then it follows directly from the definitions that the partial derivatives are given by: d 1 f(a, b)[h] = df(a, b)[(h, 0)], d 2 f(a, b) = df(a, b)[(0, k)], and since (h, k) = (h, 0) + (0, k) the desired expression for df follows. Continuity of d 1 f and d 2 f follows immediately from that of df, given these expressions. Corollary. Let f : U R n, U R m open. The following are equivalent: (i) all partial derivatives xj f i are defined and continuous in U; (ii) f is differentiable in U, and df : U L(R m ; R n ) is continuous; (iii) For each v R m, the directional derivative v f is defined and continuous in U (with values in R n ), and depends linearly on v. Proof. (i) (ii) follows from the theorem; (ii) (iii) since v f(x) = df(x)[v] where f is differentiable. (i) is a special case of (iii). 7

8 Remark. (Gateaux and Fréchet derivatives.) It is natural to wonder if one can give a direct, coordinate-independent proof of the equivalence of (ii) and (iii) in the corollary, valid also in the infinite-dimensional case. We have the following classical definition: Definition: f : U E F is Gateaux differentiable at x 0 if there is a map v Df(x 0, v) from E to F such that: lim f(x 0 + tv) f(x 0 ) tdf(x 0, v) = 0. t 0 (That is, if all directional derivatives exist at x 0, but don t necessarily depend linearly on v.) Fréchet differentiable means differentiable as defined earlier. We have the following classical theorem: Theorem. If f is Fréchet differentiable at x 0 it is Gateaux differentiable at x 0 (and Df(x 0, v) depends linearly on v). Conversely if f is Gateaux differentiable in a neighborhood U of x 0 with Df(x, v) depending linearly on v and continuously on x (that is, Df C(U; L(E; F ))), then f is Fréchet differentiable in U with df continuous, and df(x)[v] = Df(x, v) (x U, v E.) (This can be proved starting from Remark 1 in the previous section. Can you do it?) Proposition. Let f : U R m R have a local min (or local max) at a U. If f is differentiable at a, df(a)[v] = 0, v R m. Proof. Indeed if df(a)[v] = v f(a) > 0 (say) for some v, we have f(a + tv) > f(a) for t > 0 small enough (so f can t have a local max at a), and f(a + tv) < f(a) for t < 0 ( t small), so f can t have a local min at a. Definition. Let U R m be open, f : U R n. We say f is of class C 1 in U, denoted f C 1 (U; R n ), if all partial derivatives xi f(x) are defined in U, and are continuous functions from U to R n. Equivalently, f is differentiable at each x U, and the differential df is continuous from U to L(R m, R n ). (This second definition also makes sense for C 1 (U; F ), where E, F are Banach spaces and U E is open.) 5. The Chain Rule. (Differential of a composition.) Let E, F, G be Banach spaces, U E and V F open subsets, f : U F, g : V G continuous maps, x 0 U and y 0 V with f(x 0 ) = y 0. In particular, the composition h = g f is defined and continuous in a neighborhood of x 0, and takes values in G. 8

9 Assume f is differentiable at x 0 and g is differentiable at y 0, with df(x 0 ) L(E; F ), df(y 0 ) L(F ; G). Then h is differentiable at x 0, with differential: d(g f)(x 0 ) = dg(y 0 ) df(x 0 ) L(E; G). Proof. Given 0 < ɛ < 1, we may find r > 0 so that if s r and t r: f(x 0 + s) = f(x 0 ) + df(x 0 )[s] + o 1 (s), g(y 0 + t) = g(y 0 ) + dg(y 0 )[t] + o 2 (t), where o 1 (s) ɛ s, o 2 (t) ɛ t. We also have: df(x 0 )[s] a s for some a > 0, so df(x 0 )[s]+o 1 (s) (a+1) s if s r. Thus if we have s r/(a + 1) we may let df(x 0 )[s] + o 1 (s) = t and we have the estimate: Thus with h = g f: o 2 (df(x 0 ) + o 1 (s)) ɛ(a + 1) s. h(x 0 + s) = g(f(x 0 + s)) = g(y 0 + df(x 0 )[s] + o 1 (s)) = g(y 0 ) + dg(y 0 )[df(x 0 )[s] + o 1 (s)] + o 2 (df(x 0 )[s] + o 1 (s)) = h(x 0 ) + dg(y 0 ) df(x 0 )[s] + dg(y 0 )[o 1 (s)] + o 2 (df(x 0 )[s] + o 1 (s)), and we just need to show the last two terms can be combined into a remainder term o 3 (s). Using dg(y 0 )[o 1 (s)] b o 1 (s) bɛ s (for some b > 0), we have: o 3 (s) ɛ(a + b + 1) s, for s min{r, r/(a + 1)}. This concludes the proof. Classical notation. For f : U R m R n differentiable, given by f(x) = (f 1 (x),..., f n (x)) with f i : U R, in the standard bases of R m, R n the differential df(x) L(R m, R n ) is given by an n m matrix, the Jacobian matrix Jf(x): [Jf(x)] ij = xj f i (x), j = 1,..., m, i = 1,..., n. In the Chain Rule situation, we also have g : V R n R p, g(y) = (g 1 (y),..., g p (y)), y V. The Jacobian matrix of g, Jg, is of type p n, and the Chain Rule takes the form (for x 0 U): J(g f)(x 0 ) = Jg(f(x 0 ))Jf(x 0 ) M p m (matrix product), 9

10 xj (g i f)(x 0 ) = n yk g i (f(x 0 )) xj f k (x 0 ), i = 1,..., p, j = 1,..., m. k=1 From this we immediately have: Corollary 1. If f C 1 (U; R n ) and g C 1 (V ; R p ) with U R m, V R n and f(u) V, then g f C 1 (U; R p ). Corollary 2. If f : U R m V R n is differentiable at x 0 U and admits an inverse mapping g : V U, differentiable at y 0 = f(x 0 ), then df(x 0 ) L(R n, R m ) is an isomorphism, with inverse dg(y 0 ) L(R n, R m ). In particular, m = n. Proof. From g f = id U and f g = id V it follows via the Chain Rule that dg(y 0 )df(x 0 ) = I m and df(x 0 )dg(y 0 ) = Im, hence dg(y 0 ) = (df(x 0 )) 1. Definition. f : U V is a diffeomorphism if it is a differentiable bijection, with differentiable inverse. Remark. x x 3 is a simple example of a differentiable homeomorphism which is not a diffeomorphism. It is a non-trivial theorem in Topology that U R n and V R m are not homeomorphic If n m. But we already know they can t be diffeomorphic. Let f : U V be a diffeomorphism between open subsets U, V in R m, with inverse G : V U. Then if f C 1 (U; V ), it follows that also g C 1 (V ; U) (and similarly for C k, for any 1 k.) To see this, let Inv : GL(R n ) L(R m ; R m ) be the inversion map, which is smooth (C k for each k 1). Then the map dg : V L(R m ; R m ) can be written as the composition of three maps: So we have: dg = Inv df g, df : U GL(R m ). Corollary 3. Let f : U V be a bijection of class C k (k 1) between open sets U, V R m. If the inverse map f 1 : V U is differentiable, then f 1 C k (V ; R m ). (In this case we say f is a C k diffeomorphism from U to V.) Exercise 3. Let f : U R n and g : U R p be differentiable (U R m open). Define h : U \ {x; f(x) = 0} R p by h(x) = g(x) f(x) (euclidean norm in R p ). Compute dh(x)[v], for each v R m. 10

11 Hint. Show first (using the Chain Rule) that if φ(x) = f(x) = f(x), f(x) has differential given by (at points x where f(x) 0): dφ(x)[v] = 1 df(x)[v], f(x). f(x) (Using v, w to denote the standard inner product in R m.) 6. From infinitesimal to local (strong differentiability). In this section we are interested in results that, starting from properties of the differential df(a) at a point a U, allow conclusions about the mapping behavior of f, in a small ball B r (a). Theorem. Let f : U R m, U R m open, be differentiable at a U. If the differential df(a) GL(m), then there exists δ > 0 so that 0 < x a < δ f(x) f(a). Lemma. Let A GL(n), Then Ax c x for all x R n, where c = 1/ A 1. More generally, if A L(R m ; R n ) is injective (and hence an isomorphism onto a subspace E R n ), we have the same inequality, where c is the reciprocal of the norm of A 1 : E R m. Proof. x = A 1 Ax A 1 Ax = (1/c) Ax. Proof of theorem. Since f is differentiable at a: f(x) = f(a) + df(a)[x a] + r a (x) x a, where lim x a r a (x) = 0. Let c = df(a) 1, choose δ > 0 so that x a δ r a (x) c/2. Thus for all x U with 0 < x a < δ: f(x) f(a) df(a)[x a] r a (x) x a c x a c 2 x a = c x a > 0, 2 so f(x) f(a). Remark. The lemma and the theorem clearly hold in the infinite-dimensional case, assuming df(a) GL(E), the space of T L(E) with bounded inverse. Exercise 4. Show that if f : U R n, U R m open, is differentiable at a U and df(a) L(R m, R n ) is injective, then we may find c > 0, δ > 0 so that x a δ f(x) f(a) c x a. In particular, for x B δ (a): x a f(x) f(a). Hint: see the lemma at the start of this section, and the proof of the theorem just given. 11

12 Proposition. If f : U R m R n is differentiable at a U and df(a) L(R m, R n ) is surjective, then in any ball with center a there exist points x with f(x) > f(a), and also (unless f(a) = 0) points x with f(x) < f(a). Proof. If the first statement were false, a U would be a point of local maximum for the function φ(x) = f(x) (euclidean norm). Its derivative at a is dφ(a)[v] = 1 f(a) df(a)[v], f(a), so if dφ(a) = 0 there is no vector v R m with df(a)[v] = f(a), contradicting the hypothesis (note we must have f(a) 0, otherwise f would vanish in a whole neighborhood of a, forcing df(a) = 0.) The second statement is proved analogously. Given these results, it is natural to ask: (1) If f is differentiable at a, with df(a) injective, is f injective in a neighborhood of a? (2) If f is differentiable at a, with df(a) surjective, is f(a) an interior point of f(u), for some neighborhood U of a? The answer to both questions is no. There are examples of functions f : R 2 R 2, differentiable at 0 with df(0) = Id, but not injective in any neighborhood of 0, and with f(0) not in the interior of f(u), for any neighborhood U of 0. Example. Let f(x) = x 2 + x2 sin 1 x, f(0) = 0. Then f (0) = 1/2 but f is not injective in any neighborhood of zero. To see this, consider f (x) = x sin 1 x cos 1 x (for x 0.) In any interval (0, δ) with δ > 0 small, f changes sign infinitely often, so f cannot be monotone in any such interval, hence not injective either.(exercise.) Example. Let f(x, y) = (x, y), unless x > 0 and 0 < y < x 2, in which case we set f(x, y) = (x, x 2 ). This function is discontinuous at the points (x, 0) with x > 0, but is continuous at all other points, including the origin, and it is differentiable at the origin, with df(0) = Id. Exercise 5. Verify the claim about differentiability at the origin for this example. Since f maps the vertical line segment [(x, 0), (x, x 2 )] to the point (x, x 2 ) (for any x > 0), it is not injective in any neighborhood of the origin. Also no point (x, y) with x > 0, 0 < y < x 2, is in the image of f, so f(0) = 0 is not interior to any f(u), for any neighborhood U of the origin. Remark. It is possible to modify this example so as to obtain a continuous map f : R 2 R 2 with df(0) = id, which is not injective in any neighborhood of 0. However if f : U R m is continuous in U (a U R m 12

13 open) and df(a) L(R m ) is an isomorphism, then f(a) int f(u). (This will be proved later.) These considerations motivate the following definition: Definition. f : U R m R n is strongly differentiable at a U if there exists a linear transformation T L(R m, R n ) and a function r a (x, y), x, y U so that, for all x, y U: f(x) f(y) = T [x y] + r a (x, y), r a (x, y) lim x a,y a x y = 0. In particular, specializing to y = a we see that f is differentiable at a, with T = df(a). It follows easily from the definition that, for any ɛ > 0, we may find r > 0 so that f is Lipschitz in the ball B r (a): f(x) f(y) ( df(a) + ɛ) x y, x, y B r (a). Also, denoting by r a (x) the Taylor remainder (f(x) f(a) = df(a)[x a] + r a (x)) we find that, for x a r, y a r: r a (x) r a (y) = r a (x, y) ɛ x y. Conversely, if r a (x) satisfies a Lipschitz condition with arbitrarily small constant in a sufficiently small ball B r (a), then f is strongly differentiable at a. Theorem. (1) If f : U R n is strongly differentiable at a U R m (open) and df(a) L(R m ; R n ) is injective, then there exist c > 0, r > 0 so that x, y B r (a) f(x) f(y) c x y ; in particular, f is injective in B r (a), and a homeomorphism onto f(b r (a)) R n. (2) Under the same hypotheses on f, if df(a) L(R m, R n ) is surjective, then f(a) int f(u). Proof. (1) We have df(a)v c v for some c > 0, and r a (x, y) (c/2) x y if x, y B r (a), for some r > 0. Hence f(x) f(y) df(a)[x y] r a (x, y) (c/2) x y. So f is bijective from B r (a) to its image, and f 1 is Lispschitz on this image. Hence f is a homeomorphism. (2) The proof will be given later. 13

14 Theorem. Let f : U R n be differentiable in U. If df : U L(R m ; R n ) is continuous at a U, then f is strongly differentiable at a. (In particular this is the case if f C 1 (U; R n )). Conversely, if f is strongly differentiable at a, then df is continuous at a. Proof. The function r a (x) is differentiable in U, with derivative dr a (x) = df(x) df(a), so dr a (a) = 0 and dr is continuous at a. Thus given ɛ > 0 we have dr a (x) ɛ in B r (a), for r sufficiently small. By the mean value inequality this implies f(x) f(y) ɛ x y for x, y in this ball. As observed earlier, this implies f is strongly differentiable at a. For the converse, suppose f is strongly differentiable at a. Combining f(x) f(y) = df(a)[x y] + r a (x, y), f(y) f(x) = df(x)[y x] + r x (y), we find: (df(x) df(a))[y x] = [r a (x, y) + r x (y)]. Given ɛ > 0, we find ρ > 0 and for each x U, 0 < η x < ρ so that: r a (x, y) ɛ y x for x, y B 2ρ (a); y x η x r x (y) ɛ y x. Let x U satisfy x a ρ, and let u R m be a unit vector. Then letting y = x + η x u we have: x, y B 2ρ (a) and y x η x, so η x (df(x) df(a))[u] r a (x, y) + r x (y) 2ɛη x, or (df(x) df(a))[u] 2ɛ, establishing continuity of df at a. Exercise 6. Let f C 1 (U; R n ), U R m open, K U compact. Suppose df L(R m, R n ) is injective for each x K. Then there exist c > 0 and r > 0 so that, for each x K, y U, we have: f(x) f(y) c x y whenever x y < r. Hint: f is strongly differentiable at each x U. 7. Schwarz s theorem. 8. Spaces of differentiable maps. 9. Taylor s theorem. 14