Fréchet derivatives and Gâteaux derivatives

Transcription

1 Fréche derivaives and Gâeaux derivaives Jordan Bell Deparmen of Mahemaics, Universiy of Torono April 3, Inroducion In his noe all vecor spaces are real. If X and Y are normed spaces, we denoe by B(X, Y ) he se of bounded linear maps X Y, and wrie B(X) = B(X, X). B(X, Y ) is a normed space wih he operaor norm. 2 Remainders If X and Y are normed spaces, le o(x, Y ) be he se of all maps r : X Y for which here is some map α : X Y saisfying: r(x) = x α(x) for all x X, α(0) = 0, α is coninuous a 0. Following Peno, 1 we call elemens of o(x, Y ) remainders. I is immediae ha o(x, Y ) is a vecor space. If X and Y are normed spaces, if f : X Y is a funcion, and if x 0 X, we say ha f is sable a x 0 if here is some ɛ > 0 and some c > 0 such ha x x 0 ɛ implies ha f(x x 0 ) c x x 0. If T : X Y is a bounded linear map, hen T x T x for all x X, and hus a bounded linear map is sable a 0. The following lemma shows ha he composiion of a remainder wih a funcion ha is sable a 0 is a remainder. 2 Lemma 1. Le X, Y be normed spaces and le r o(x, Y ). If W is a normed space and f : W X is sable a 0, hen r f o(w, Y ). If Z is a normed space and g : Y Z is sable a 0, hen g r o(x, Z). 1 Jean-Paul Peno, Calculus Wihou Derivaives, p. 133, Jean-Paul Peno, Calculus Wihou Derivaives, p. 134, Lemma

2 Proof. r o(x, Y ) means ha here is some α : X Y saisfying r(x) = x α(x) for all x X, ha akes he value 0 a 0, and ha is coninuous a 0. As f is sable a 0, here is some ɛ > 0 and some c > 0 for which w ɛ implies ha f(w) c w. Define β : W Y by { f(w) w β(w) = α(f(w)) w 0 0 w = 0, for which we have (r f)(w) = w β(w), w W. If w ɛ, hen β(w) c α(f(w)). Bu f(w) 0 as w 0, and because α is coninuous a 0 we ge ha α(f(w)) α(0) = 0 as w 0. So he above inequaliy gives us β(w) 0 as w 0. As β(0) = 0, he funcion β : W Y is coninuous a 0, and herefore r f is remainder. As g is sable a 0, here is some ɛ > 0 and some c > 0 for which y ɛ implies ha g(y) c y. Define γ : X Z by { g( x α(x)) x x 0 γ(x) = 0 x = 0. For all x X, (g r)(x) = g( x α(x)) = x γ(x). Since α(0) = 0 and α is coninuous a 0, here is some δ > 0 such ha x δ implies ha α(x) ɛ. Therefore, if x δ 1 hen g( x α(x)) c x α(x) c x ɛ, and hence if x δ 1 hen γ(x) cɛ. This shows ha γ(x) 0 as x 0, and since γ(0) = 0 he funcion γ : X Z is coninuous a 0, showing ha g r is a remainder. If Y 1,..., Y n are normed spaces where Y k has norm k, hen (y 1,..., y n ) = max 1 k n y k k is a norm on n k=1 Y k, and one can prove ha he opology induced by his norm is he produc opology. Lemma 2. If X and Y 1,..., Y n are normed spaces, hen a funcion r : X n k=1 Y k is a remainder if and only if each of r k : X Y k are remainders, 1 k n, where r(x) = (r 1 (x),..., r n (x)) for all x X. Proof. Suppose ha here is some funcion α : X n k=1 Y k such ha r(x) = x α(x) for all x X. Wih α(x) = (α 1 (x),..., α n (x)), we have r k (x) = x α k (x), x X. Because α(x) 0 as x 0, for each k we have α k (x) 0 as x 0, which shows ha r k is a remainder. 2

3 Suppose ha each r k is a remainder. Thus, for each k here is a funcion α k : X Y k saisfying r k (x) = x α k (x) for all x X and α k (x) 0 as x 0. Then he funcion α : X n k=1 Y k defined by α(x) = (α 1 (x),..., α n (x)) saisfies r(x) = x α(x). Because α k (x) 0 as x 0 for each of he finiely many k, 1 k n, we have α(x) 0 as x 0. 3 Definiion and uniqueness of Fréche derivaive Suppose ha X and Y are normed spaces, ha U is an open subse of X, and ha x 0 U. A funcion f : U Y is said o be Fréche differeniable a x 0 if here is some L B(X, Y ) and some r o(x, Y ) such ha f(x) = f(x 0 ) + L(x x 0 ) + r(x x 0 ), x U. (1) Suppose here are bounded linear maps L 1, L 2 and remainders r 1, r 2 ha saisfy he above. Wriing r 1 (x) = x α 1 (x) and r 2 (x) = x α 2 (x) for all x X, we have L 1 (x x 0 )+ x x 0 α 1 (x x 0 ) = L 2 (x x 0 )+ x x 0 α 2 (x x 0 ), x U, i.e., L 1 (x x 0 ) L 2 (x x 0 ) = x x 0 (α 2 (x x 0 ) α 1 (x x 0 )), x U. For x X, here is some h > 0 such ha for all h we have x 0 + x U, and hen L 1 (x) L 2 (x) = x (α 2 (x) α 1 (x)), hence, for 0 < h, L 1 (x) L 2 (x) = x (α 2 (x) α 1 (x)). Bu α 2 (x) α 1 (x) 0 as 0, which implies ha L 1 (x) L 2 (x) = 0. As his is rue for all x X, we have L 1 = L 2 and hen r 1 = r 2. If f is Fréche differeniable a x 0, he bounded linear map L in (1) is called he Fréche derivaive of f a x 0, and we define Df(x 0 ) = L. Thus, f(x) = f(x 0 ) + Df(x 0 )(x x 0 ) + r(x x 0 ), x U. If U 0 is he se of hose poins in U a which f is Fréche differeniable, hen Df : U 0 B(X, Y ). Suppose ha X and Y are normed spaces and ha U is an open subse of X. We denoe by C 1 (U, Y ) he se of funcions f : U Y ha are Fréche differeniable a each poin in U and for which he funcion Df : U B(X, Y ) is coninuous. We say ha an elemen of C 1 (U, Y ) is coninuously differeniable. We denoe by C 2 (U, Y ) hose elemens f of C 1 (U, Y ) such ha Df C 1 (U, B(X, Y )); 3

4 ha is, C 2 (U, Y ) are hose f C 1 (U, Y ) such ha he funcion Df : U B(X, Y ) is Fréche differeniable a each poin in U and such ha he funcion D(Df) : U B(X, B(X, Y )) is coninuous. 3 The following heorem characerizes coninuously differeniable funcions R n R m. 4 Theorem 3. Suppose ha f : R n R m is Fréche differeniable a each poin in R n, and wrie f = (f 1,..., f m ). f C 1 (R n, R m ) if and only if for each 1 i m and 1 j n he funcion is coninuous. f i x j : R n R 4 Properies of he Fréche derivaive If f : X Y is Fréche differeniable a x 0, hen because a bounded linear map is coninuous and in paricular coninuous a 0, and because a remainder is coninuous a 0, we ge ha f is coninuous a x 0. We now prove ha Fréche differeniaion a a poin is linear. Lemma 4 (Lineariy). Le X and Y be normed spaces, le U be an open subse of X and le x 0 U. If f 1, f 2 : U Y are boh Fréche differeniable a x 0 and if α R, hen αf 1 + f 2 is Fréche differeniable a x 0 and D(αf 1 + f 2 )(x 0 ) = αdf 1 (x 0 ) + Df 2 (x 0 ). Proof. There are remainders r 1, r 2 o(x, Y ) such ha and Then for all x U, f 1 (x) = f 1 (x 0 ) + Df 1 (x 0 )(x x 0 ) + r 1 (x x 0 ), x U, f 2 (x) = f 2 (x 0 ) + Df 2 (x 0 )(x x 0 ) + r 2 (x x 0 ), x U. (αf 1 + f 2 )(x) (αf 1 + f 2 )(x 0 ) = αf 1 (x) αf 1 (x 0 ) + f 2 (x) f 2 (x 0 ) and αr 1 + r 2 o(x, Y ). = αdf 1 (x 0 )(x x 0 ) + αr 1 (x x 0 ) +Df 2 (x 0 )(x x 0 ) + r 2 (x x 0 ) = (αdf 1 (x 0 ) + Df 2 (x 0 ))(x x 0 ) +(αr 1 + r 2 )(x x 0 ), 3 See Henri Caran, Differenial Calculus, p. 58, 5.1, and Jean Dieudonné, Foundaions of Modern Analysis, enlarged and correced prining, p. 179, Chaper VIII, Henri Caran, Differenial Calculus, p. 36,

5 The following lemma gives an alernae characerizaion of a funcion being Fréche differeniable a a poin. 5 Lemma 5. Suppose ha X and Y are normed space, ha U is an open subse of X, and ha x 0 U. A funcion f : U Y is Fréche differeniable a x 0 if and only if here is some funcion F : U B(X, Y ) ha is coninuous a x 0 and for which f(x) f(x 0 ) = F (x)(x x 0 ), x U. Proof. Suppose ha here is a funcion F : U B(X, Y ) ha is coninuous a x 0 and ha saisfies f(x) f(x 0 ) = F (x)(x x 0 ) for all x U. Then, for x U, f(x) f(x 0 ) = F (x)(x x 0 ) F (x 0 )(x x 0 ) + F (x 0 )(x x 0 ) = F (x 0 )(x x 0 ) + r(x x 0 ), where r : X Y is defined by { (F (x + x 0 ) F (x 0 ))(x) x + x 0 U r(x) = 0 x + x 0 U. We furher define (F (x+x 0) F (x 0))(x) x x + x 0 U, x 0 α(x) = 0 x + x 0 U 0 x = 0, wih which r(x) = x α(x) for all x X. To prove ha r is a remainder i suffices o prove ha α(x) 0 as x 0. Le ɛ > 0. Tha F : U B(X, Y ) is coninuous a x 0 ells us ha here is some δ > 0 for which x < δ implies ha F (x + x 0 ) F (x 0 ) < ɛ and hence (F (x + x 0 ) F (x 0 ))(x) F (x + x 0 ) F (x 0 ) x < ɛ x. Therefore, if x < δ hen α(x) < ɛ, which esablishes ha r is a remainder and herefore ha f is Fréche differeniable a x 0, wih Fréche derivaive Df(x 0 ) = F (x 0 ). Suppose ha f is Fréche differeniable a x 0 : here is some r o(x, Y ) such ha f(x) = f(x 0 ) + Df(x 0 )(x x 0 ) + r(x x 0 ), x U, where Df(x 0 ) B(X, Y ). As r is a remainder, here is some α : X Y saisfying r(x) = x α(x) for all x X, and such ha α(0) = 0 and α(x) 0 as x 0. For each x X, by he Hahn-Banach exension heorem 6 here is some λ x X such ha λ x x = x and λ x v v for all v X. Thus, r(x) = (λ x x)α(x), x X. 5 Jean-Paul Peno, Calculus Wihou Derivaives, p. 136, Lemma Waler Rudin, Funcional Analysis, second ed., p. 59, Corollary o Theorem

6 Define F : U B(X, Y ) by i.e. for x U and v X, Then for x U, F (x) = Df(x 0 ) + (λ x x0 )α(x x 0 ), F (x)(v) = Df(x 0 )(v) + (λ x x0 v)α(x x 0 ) Y. r(x x 0 ) = (λ x x0 (x x 0 ))α(x x 0 ) = F (x)(x x 0 ) Df(x 0 )(x x 0 ), and hence f(x) = f(x 0 ) + F (x)(x x 0 ), x U. To complee he proof i suffices o prove ha F is coninuous a x 0. Bu boh λ 0 = 0 and α(0) = 0 so F (x 0 ) = Df(x 0 ), and for x U and v X, (F (x) F (x 0 ))(v) = (λ x x0 v)α(x x 0 ) = λ x x0 v α(x x 0 ) v α(x x 0 ), so F (x) F (x 0 ) α(x x 0 ). From his and he fac ha α(0) = 0 and α(x) 0 as x 0 we ge ha F is coninuous a x 0, compleing he proof. We now prove he chain rule for Fréche derivaives. 7 Theorem 6 (Chain rule). Suppose ha X, Y, Z are normed spaces and ha U and V are open subses of X and Y respecively. If f : U Y saisfies f(u) V and is Fréche differeniable a x 0 and if g : V Z is Fréche differeniable a f(x 0 ), hen g f : U Z is Fréche differeniable a x 0, and is Fréche derivaive a x 0 is D(g f)(x 0 ) = Dg(f(x 0 )) Df(x 0 ). Proof. Wrie y 0 = f(x 0 ), L 1 = Df(x 0 ), and L 2 = Dg(y 0 ). Because f is Fréche differeniable a x 0, here is some r 1 o(x, Y ) such ha f(x) = f(x 0 ) + L 1 (x x 0 ) + r 1 (x x 0 ), x U, and because g is Fréche differeniable a y 0 here is some r 2 o(y, Z) such ha g(y) = g(y 0 ) + L 2 (y y 0 ) + r 2 (y y 0 ), y V. For all x U we have f(x) V, and using he above formulas, g(f(x)) = g(y 0 ) + L 2 (f(x) y 0 ) + r 2 (f(x) y 0 ) ) ( ) = g(y 0 ) + L 2 (L 1 (x x 0 ) + r 1 (x x 0 ) + r 2 L 1 (x x 0 ) + r 1 (x x 0 ) ) = g(y 0 ) + L 2 (L 1 (x x 0 )) + L 2 (r 1 (x x 0 )) + r 2 (L 1 (x x 0 ) + r 1 (x x 0 ). 7 Jean-Paul Peno, Calculus Wihou Derivaives, p. 136, Theorem

7 Define r 3 : X Z by r 3 (x) = r 2 (L 1 x + r 1 (x)), and fix any c > L 1. Wriing r 1 (x) = x α 1 (x), he fac ha α(0) = 0 and ha α is coninuous a 0 gives us ha here is some δ > 0 such ha if x < δ hen α(x) < c L 1, and hence if x < δ hen r 1 (x) (c L 1 ) x. Then, x < δ implies ha L 1 x + r 1 (x) L 1 x + r 1 (x) L 1 x + (c L 1 ) x = c x. This shows ha x L 1 x + r 1 (x) is sable a 0 and so by Lemma 1 ha r 3 o(x, Z). Then, r : X Z defined by r = L 1 r 1 + r 3 is a sum of wo remainders and so is iself a remainder, and we have g f(x) = g f(x 0 ) + L 2 L 1 (x x 0 ) + r(x x 0 ), x U. Bu L 1 B(X, Y ) and L 2 B(Y, Z), so L 2 L 1 B(X, Z). This shows ha g f is Fréche differeniable a x 0 and ha is Fréche derivaive a x 0 is L 2 L 1 = Dg(y 0 ) Df(x 0 ) = Dg(f(x 0 )) Df(x 0 ). The following is he produc rule for Fréche derivaives. By f 1 f 2 we mean he funcion x f 1 (x)f 2 (x). Theorem 7 (Produc rule). Suppose ha X is a normed space, ha U is an open subse of X, ha f 1, f 2 : U R are funcions, and ha x 0 U. If f 1 and f 2 are boh Fréche differeniable a x 0, hen f 1 f 2 is Fréche differeniable a x 0, and is Fréche derivaive a x 0 is D(f 1 f 2 )(x 0 ) = f 2 (x 0 )Df 1 (x 0 ) + f 1 (x 0 )Df 2 (x 0 ). Proof. There are r 1, r 2 o(x, R) wih which f 1 (x) = f 1 (x 0 ) + Df 1 (x 0 )(x x 0 ) + r 1 (x x 0 ), x U and f 2 (x) = f 2 (x 0 ) + Df 2 (x 0 )(x x 0 ) + r 2 (x x 0 ), x U. Muliplying he above wo formulas, f 1 (x)f 2 (x) = f 1 (x 0 )f 2 (x 0 ) + f 2 (x 0 )Df 1 (x 0 )(x x 0 ) + f 1 (x 0 )Df 2 (x 0 )(x x 0 ) +Df 1 (x 0 )(x x 0 )Df 2 (x 0 )(x x 0 ) + r 1 (x x 0 )r 2 (x x 0 ) +f 1 (x 0 )r 2 (x x 0 ) + r 2 (x x 0 )Df 1 (x 0 )(x x 0 ) +f 2 (x 0 )r 1 (x x 0 ) + r 1 (x x 0 )Df 2 (x 0 )(x x 0 ). Define r : X R by r(x) = Df 1 (x 0 )xdf 2 (x 0 )x + r 1 (x)r 2 (x) + f 1 (x 0 )r 2 (x) + r 2 (x)df 1 (x 0 )x +f 2 (x 0 )r 1 (x) + r 1 (x)df 2 (x 0 )x, 7

8 for which we have, for x U, f 1 (x)f 2 (x) = f 1 (x 0 )f 2 (x 0 )+f 2 (x 0 )Df 1 (x 0 )(x x 0 )+f 1 (x 0 )Df 2 (x 0 )(x x 0 )+r(x x 0 ). Therefore, o prove he claim i suffices o prove ha r o(x, R). Define α : X R by α(0) = 0 and α(x) = Df1(x0)xDf2(x0)x x for x 0. For x 0, α(x) = Df 1(x 0 )x Df 2 (x 0 )x x Df 1(x 0 ) x Df 2 (x 0 ) x x = Df 1 (x 0 ) Df 2 (x 0 ) x. Thus α(x) 0 as x 0, showing ha he firs erm in he expression for r belongs o o(x, R). Likewise, each of he oher five erms in he expression for r belongs o o(x, R), and hence r o(x, R), compleing he proof. 5 Dual spaces If X is a normed space, we denoe by X he se of bounded linear maps X R, i.e. X = B(X, R). X is iself a normed space wih he operaor norm. If X is a normed space, he dual pairing, : X X R is x, ψ = ψ(x), x X, ψ X. If U is an open subse of X and if a funcion f : U R is Fréche differeniable a x 0 U, hen Df(x 0 ) is a bounded linear map X R, and so belongs o X. If U 0 are hose poins in U a which f : U R is Fréche differeniable, hen Df : U 0 X. In he case ha X is a Hilber space wih inner produc,, he Riesz represenaion heorem shows ha R : X X defined by R(x)(y) = y, x is an isomeric isomorphism. If f : U R is Fréche differeniable a x 0 U, hen we define f(x 0 ) = R 1 (Df(x 0 )), and call f(x 0 ) X he gradien of f a x 0. Wih U 0 denoing he se of hose poins in U a which f is Fréche differeniable, f : U 0 X. (To define he gradien we merely used ha R is a bijecion, bu o prove properies of he gradien one uses ha R is an isomeric isomorphism.) Example. Le X be a Hilber space, A B(X), v X, and define f(x) = Ax, x x, v, x X. 8

9 For all x 0, x X we have, because he inner produc of a real Hilber space is symmeric, f(x) f(x 0 ) = Ax, x x, v Ax 0, x 0 + x 0, v = Ax, x Ax 0, x + Ax 0, x Ax 0, x 0 x x 0, v = A(x x 0 ), x + Ax 0, x x 0 x x 0, v = x x 0, A x + x x 0, Ax 0 x x 0, v = x x 0, A x + Ax 0 v = x x 0, A x A x 0 + A x 0 + Ax 0 v = x x 0, (A + A)x 0 v + x x 0, A (x x 0 ). Wih Df(x 0 )(x x 0 ) = x x 0, (A + A)x 0 v, or Df(x 0 )(x) = x, (A + A)x 0 v, we have ha f is Fréche differeniable a each x 0 X. Furhermore, is gradien a x 0 is f(x 0 ) = (A + A)x 0 v. For each x 0 X, he funcion f : X R is Fréche differeniable a x 0, and hus Df : X X, and we can ask a wha poins Df has a Fréche derivaive. For x 0, x, y X, (Df(x) Df(x 0 ))(y) = y, (A + A)x v y, (A + A)x 0 v = y, (A + A)(x x 0 ). For D(Df)(x 0 )(x x 0 )(y) = y, (A + A)(x x 0 ), in oher words wih D 2 f(x 0 )(x)(y) = D(Df)(x 0 )(x)(y) = y, (A + A)x, we have ha Df is Fréche differeniable a each x 0 X. Thus D 2 f : X B(X, X ). Because D 2 f(x 0 ) does no depend on x 0, i is Fréche differeniable a each poin in X, wih D 3 f(x 0 ) = 0 for all x 0 X. Here D 3 f : X B(X, B(X, X )). 6 Gâeaux derivaives Le X and Y be normed spaces, le U be an open subse of X, le f : U Y be a funcion, and le x 0 U. If here is some T B(X, Y ) such ha for all v X we have f(x 0 + v) f(x 0 ) lim = T v, (2) 0 hen we say ha f is Gâeaux differeniable a x 0 and call T he Gâeaux derivaive of f a x 0. 8 I is apparen ha here is a mos one T B(X, Y ) ha 8 Our definiion of he Gâeaux derivaive follows Jean-Paul Peno, Calculus Wihou Derivaives, p. 127, Definiion

10 saisfies (2) for all v X. We wrie f (x 0 ) = T. Thus, f is a map from he se of poins in U a which f is Gâeaux differeniable o B(X, Y ). If V U and f is Gâeaux differeniable a each elemen of V, we say ha f is Gâeaux differeniable on V. Example. Define f : R 2 R by f(x 1, x 2 ) = x4 1 x2 for (x x 6 1, x 2 ) (0, 0) and 1 +x3 2 f(0, 0) = 0. For v = (v 1, v 2 ) R 2 and 0, f(0 + v) f(0) = f(v 1, v 2 ) { 1 = 5 v 4 1 v2 v (0, 0) 6 v1 6+3 v2 3 0 v = (0, 0) = { v 4 1 v2 3 v 6 1 +v3 2 v (0, 0) 0 v = (0, 0). Hence, for any v R 2, we have f(0+v) f(0) 0 as 0. Therefore, f is Gâeaux differeniable a (0, 0) and f (0, 0)v = 0 R for all v R 2, i.e. f (0, 0) = 0. However, for (x 1, x 2 ) (0, 0), f(x 1, x 2 1) = x6 1 x = 1 x6 1 2, from which i follows ha f is no coninuous a (0, 0). We saed in 4 ha if a funcion is Fréche differeniable a a poin hen i is coninuous a ha poin, and so f is no Fréche differeniable a (0, 0). Thus, a funcion ha is Gâeaux differeniable a a poin need no be Fréche differeniable a ha poin. We prove ha being Fréche differeniable a a poin implies being Gâeaux differeniable a he poin, and ha in his case he Gâeaux derivaive is equal o he Fréche derivaive. Theorem 8. Suppose ha X and Y are normed spaces, ha U is an open subse of X, ha f Y U, and ha x 0 U. If f is Fréche differeniable a x 0, hen f is Gâeaux differeniable a x 0 and f (x 0 ) = Df(x 0 ). Proof. Because f is Fréche differeniable a x 0, here is some r o(x, Y ) for which f(x) = f(x 0 ) + Df(x 0 )(x x 0 ) + r(x x 0 ), x U. For v X and nonzero small enough ha x 0 + v U, f(x 0 + v) f(x 0 ) Wriing r(x) = x α(x), f(x 0 + v) f(x 0 ) = Df(x 0)(x 0 + v x 0 ) + r(x 0 + v x 0 ) = Df(x 0) + v α(v) = Df(x 0)v + r(v). = Df(x 0 )v + v α(v). Hence, f(x 0 + v) f(x 0 ) lim = Df(x 0 )v. 0 This holds for all v X, and as Df(x 0 ) B(X, Y ) we ge ha f is Gâeaux differeniable a x 0 and ha f (x 0 ) = Df(x 0 ). 10

11 If X is a vecor space and u, v X, le [u, v] = {(1 )u + v : 0 1}, namely, he line segmen joining u and v. The following is a mean value heorem for Gâeaux derivaives. 9 Theorem 9 (Mean value heorem). Le X and Y be normed spaces, le U be an open subse of X, and le f : U Y be Gâeaux differeniable on U. If u, v U and [u, v] U, hen f(u) f(v) sup f (w) u v. w [u,v] Proof. If f(u) = f(v) hen immediaely he claim is rue. Oherwise, f(v) f(u) 0, and so by he Hahn-Banach exension heorem 10 here is some ψ Y saisfying ψ(f(v) f(u)) = f(v) f(u) and ψ = 1. Define h : [0, 1] R by h() = f((1 )u + v), ψ. For 0 < < 1 and τ 0 saisfying + τ [0, 1], we have h( + τ) h() τ = 1 τ f((1 τ)u + ( + τ)v), ψ 1 f((1 )u + v), ψ τ f((1 )u + v + (v u)τ) f((1 )u + v) =, ψ. τ Because f is Gâeaux differeniable a (1 )u + v, f((1 )u + v + (v u)τ) f((1 )u + v) lim = f ((1 )u + v)(v u), τ 0 τ so because ψ is coninuous, h( + τ) h() lim = f ((1 )u + v)(v u), ψ, τ 0 τ which shows ha h is differeniable a and ha h () = f ((1 )u + v)(v u), ψ. h : [0, 1] R is a composiion of coninuous funcions so i is coninuous. Applying he mean value heorem, here is some θ, 0 < θ < 1, for which h (θ) = h(1) h(0). 9 Anonio Ambrosei and Giovanni Prodi, A Primer of Nonlinear Analysis, p. 13, Theorem Waler Rudin, Funcional Analysis, second ed., p. 59, Corollary. 11

12 On he one hand, h (θ) = f ((1 θ)u + θv)(v u), ψ. On he oher hand, h(1) h(0) = f(v), ψ f(u), ψ = f(v) f(u), ψ = f(v) f(u). Therefore f(v) f(u) = f ((1 θ)u + θv)(v u), ψ ψ f ((1 θ)u + θv)(v u) = f ((1 θ)u + θv)(v u) f ((1 θ)u + θv) v u sup f (w) v u. w [u,v] 7 Aniderivaives Suppose ha X is a Banach space and ha f : [a, b] X be coninuous. Define F : [a, b] X by F (x) = Le x 0 (a, b). For x (a, b), we have F (x) F (x 0 ) = x a f x0 a f = x x a f. x 0 f = f(x 0 )(x x 0 ) + from which i follows ha F is Fréche differeniable a x 0, and ha DF (x 0 )(x x 0 ) = f(x 0 )(x x 0 ). x x 0 (f f(x 0 )), If we idenify f(x 0 ) X wih he map x f(x 0 )x, namely if we say ha X = B(R, X), hen DF (x 0 ) = f(x 0 ). Le X be a normed space, le Y be a Banach space, le U be an open subse of X, and le f C 1 (U, Y ). Suppose ha u, v U saisfy [u, v] U. Wrie I = (0, 1) and define γ : I U by γ() = (1 )u + v. We have and hus by Theorem 6, Dγ() = v u, I, D(f γ)() = Df(γ()) Dγ(), I, ha is, D(f γ)() = Df(γ()) (v u), I, 12

13 i.e. D(f γ)() = Df(γ())(v u), I. If I and + h I, hen D(f γ)( + h) D(f γ)() = Df(γ( + h))(v u) Df(γ())(v u) and hence = (Df(γ( + h)) Df(γ())) (v u), D(f γ)( + h) D(f γ)() Df(γ( + h)) Df(γ()) v u. Because Df : U B(X, Y ) is coninuous, i follows ha D(f γ)( + h) D(f γ)() 0 as h 0, i.e. ha D(f γ) is coninuous a, and hus ha D(f γ) : I B(R, Y ) is coninuous. If we idenify B(R, Y ) wih Y, hen On he one hand, 1 On he oher hand, D(f γ) = D(f γ) : I Y. D(f γ) = (f γ)(1) (f γ)(0) = f(v) f(u). 1 0 ( 1 Df(γ())(v u)d = 0 ) Df((1 )u + v)d (v u); here, Therefore 1 0 ( 1 f(v) f(u) = Df((1 )u + v)d B(X, Y ). 0 ) Df((1 )u + v)d (v u). 13