Problem 22. Let I : C ([0, 1]; R) Ñ R be deﬁned by ż1 I(f ) = f (x)2 dx . 0 Show that I is diﬀerentiable at every “point” f P C ([0, 1]; R). Hint: Figure out what (DI)(f ) is by computing I(f + h) ´ I(f ), where h P C ([0, 1]; R) is a “small” continuous function. Remark. A map from a space of functions such as C ([0, 1]; R) to a scalar ﬁeld such as R or C is usually called a functional. The derivative of a functional I is usually denoted by δI instead of DI. Proof. For each f P C ([0, 1]; R), deﬁne Lf (h) = 2 ż1 f (x)h(x)dx. 0 claim: Lf P B(C ([0, 1]; R), R). Proof of claim: It is trivial that Lf P L (C ([0, 1]; R), R). Let h P C ([0, 1]; R). Then ˇ ˇ ˇLf (h)ˇ ď 2 ż1 ˇ ˇˇ ˇ ˇf (x)ˇˇh(x)ˇdx ď 2}f }8 }h}8 ; 0 thus ˇ ˇ }Lf }B(C ([0,1];R),R) = sup ˇLf (h)ˇ ď 2}f }8 ă 8 . }h}8 =1 ˇ ˇ ˇI(f + h) ´ I(f ) ´ Lf (h)ˇ = 0. Claim: lim }h}8 Ñ0 }h}8 Proof of claim: Since ż ] ˇ ˇ ˇ ˇ 1 [( )2 ˇ 2 ˇI(f + h) ´ I(f ) ´ Lf (h)ˇ = ˇˇ f (x) + h(x) ´ f (x) ´ 2f (x)h(x) dxˇ 0 ˇż 1 ˇ ˇ ˇ = ˇ h(x)2 dxˇ ď }h}28 , 0 by the sandwich lemma we conclude that ˇ ˇ ˇI(f + h) ´ I(f ) ´ Lf (h)ˇ }h}28 ď lim = 0. 0 ď lim }h}8 Ñ0 }h}8 }h}8 Ñ0 }h}8 Therefore, I is diﬀerentiable at f , and (DI)(f )(h) = Lf (h). ˝ Problem 30. Let f : R Ñ R be diﬀerentiable. Assume that for all x P R, 0 ď f 1 (x) ď f (x). Show that g(x) = e´x f (x) is decreasing. If f vanishes at some point, conclude that f is zero. Proof. To see that g is decreasing, we compute the derivative of g and ﬁnd that g 1 (x) = ´e´x f (x) + e´x f 1 (x) = e´x (f 1 (x) ´ f (x)) ď 0 ; thus g is a decreasing function. Now suppose that f (c) = 0 for some c P R. 1. Since g is decreasing, g(x) ď g(c) = 0 for all x ě c; thus f (x) = ex g(x) = 0 for all x ě c. 2. Since f 1 (x) ě 0, f is an increasing function, thus f (x) ď f (c) = 0 for all x ď c. Since f is assumed to be non-negative, we must have f (x) = 0 for all x ď c. Combining 1 and 2, we conclude that f (x) = 0 for all x P R. ˝ Problem 32. 1. If f : A Ď Rn Ñ Rm and g : B Ď Rm Ñ Rℓ are twice diﬀerentiable and f (A) Ď B, then for x0 P A, u, v P Rn , show that D2 (g ˝ f )(x0 )(u, v) ( ) ( ) = (D2 g)(f (x0 )) (Df )(x0 )(u), Df (x0 )(v) + (Dg)(f (x0 )) (D2 f )(x0 )(u, v) . 2. If p : Rn Ñ Rm is a linear map plus some constant; that is, p(x) = Lx + c for some L P B(Rn , Rm ), and f : A Ď Rm Ñ Rs is k-times diﬀerentiable, prove that ( )( ) Dk (f ˝ p)(x0 )(u(1) , ¨ ¨ ¨ , u(k) ) = (Dk f ) p(x0 ) (Dp)(x0 )(u(1) ), ¨ ¨ ¨ , (Dp)(x0 )(u(k) . Proof. 1. First of all, we show that g ˝ f is twice diﬀerentiable. Since g and f are both diﬀerentiable, the chain rule implies that g ˝ f is diﬀerentiable and (( ) ) D(g ˝ f )(x) = (Dg)(f (x))(Df )(x) = (Dg) ˝ f (Df ) (x) . Since g and f are twice diﬀerentiable, Dg and Df are diﬀerentiable. By the chain rule ( ) again, (Dg) ˝ f is diﬀerentiable; thus the product rule implies that (Dg) ˝ f (Df ) is diﬀerentiable. Therefore, g ˝ f is twice diﬀerentiable. Now by Proposition 6.69 in 共筆, we have D2 (g ˝ f )(x0 )(u, v) = n ÿ B 2 (g ˝ f ) (x0 )ui vj . Bx Bx j i i,j=1 By the chain rule, m [ ÿ B(g ˝ f ) Bg B 2 (g ˝ f ) B ˇˇ B ˇˇ Bfk ] (x0 ) = (x) = (f (x)) (x) ˇ ˇ Bxj Bxi Bxj x=x0 Bxi Bxj x=x0 k=1 Byk Bxi m [ Bg ÿ B ˇˇ Bfk ] = (f (x)) (x) ˇ Bxj x=x0 Byk Bxi k=1 = m ÿ m m ÿ ÿ B2g Bfℓ Bfk Bg B 2 fk (f (x0 )) (x0 ) (x0 ) + (f (x0 )) (x0 ) ; By By Bx Bx By Bx Bx ℓ k j i k j i k=1 ℓ=1 k=1 thus D2 (g ˝ f )(x0 )(u, v) n [ ÿ m m ] ÿ ÿ B2g Bg Bfℓ Bfk B 2 fk = (f (x0 )) (x0 ) (x0 ) + (f (x0 )) (x0 ) ui vj Byℓ Byk Bxj Bxi Byk Bxj Bxi i,j=1 k,ℓ=1 k=1 n n m (ÿ )( ÿ ) ÿ Bfℓ Bfk B2g (f (x0 )) (x0 )vj (x0 )ui = Byℓ Byk Bxj Bxi j=1 i=1 k,ℓ=1 m n ( ÿ ) ÿ Bg B 2 fk + (f (x0 )) (x0 )ui vj . Byk Bxj Bxi i,j=1 k=1 ( ) Letting (Df )(x0 )(w) r denote the r-th component of (Df )(x0 )(w), we obtain that D2 (g ˝ f )(x0 )(u, v) m ÿ ( )( ) B2g = (f (x0 )) (Df )(x0 )(v) ℓ (Df )(x0 )(u) k Byℓ Byk k,ℓ=1 m ÿ ( ) Bg (f (x0 )) D2 f )(x0 )(u, v) k Byk k=1 ( ) ( ) = (D2 g)(f (x0 )) (Df )(x0 )u, (Df )(x0 )v + (Dg)(f (x0 )) (D2 f )(x0 )(u, v) . + 2. The validity of the desired equality for the case k = 1 is the chain rule. Suppose that the desired holds for k = K. Then for k = K + 1, by Corollary 6.70 in 共筆 we obtain that DK+1 (f ˝ p)(x0 )(u(1) , ¨ ¨ ¨ , u(K+1) ) = n ÿ j=1 (K+1) uj B ˇˇ (DK (f ˝ p))(x)(u(1) , ¨ ¨ ¨ , u(K) ) ˇ Bxj x=x0 ˇ ( )( ) (K+1) B ˇ = uj (Dk f ) p(x) (Dp)(x)(u(1) ), ¨ ¨ ¨ , (Dp)(x)(u(k) . ˇ Bxj x=x0 j=1 n ÿ Noting that (Dp)(x)(ur ) = Lu(r) (which is independent of x), by Proposition 6.69 in 共筆 we ﬁnd that ( )( ) (DK f ) p(x) (Dp)(x)(u(1) ), ¨ ¨ ¨ , (Dp)(x)(u(K) n ÿ BK f (p(x))(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK , = ByjK ¨ ¨ ¨ Byj1 j ,¨¨¨ ,j =1 1 K where (Lur )s denotes the s-th component of the vector Lu(r) . As a consequence, ( )( ) B ˇˇ (DK f ) p(x) (Dp)(x)(u(1) ), ¨ ¨ ¨ , (Dp)(x)(u(K) ˇ Bxj x=x0 j=1 n m ˇ ÿ ÿ BK f (K+1) B ˇ = uj (p(x))(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK ˇ Bxj x=x0 j ,¨¨¨ ,j =1 ByjK ¨ ¨ ¨ Byj1 j=1 n ÿ (K+1) uj 1 = n ÿ (K+1) uj j=1 = m ÿ j1 ,¨¨¨ ,jK ,jK+1 m ÿ j1 ,¨¨¨ ,jK ,jK+1 = m ÿ j1 ,¨¨¨ ,jK ,jK+1 K Bpj B K+1 f (p(x0 )) K+1 (x0 )(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK ByjK+1 ¨ ¨ ¨ Byj1 Bxj =1 ) (ÿ B K+1 f (K+1) BpjK+1 uj (x0 ) (Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK (p(x0 )) ByjK+1 ¨ ¨ ¨ Byj1 Bxj j=1 =1 n ( ) B K+1 f (p(x0 )) (Dp)(x0 )u(K+1) j (Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK K+1 ByjK+1 ¨ ¨ ¨ Byj1 =1 m ÿ B K+1 f (p(x0 ))(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK (Lu(K+1) )jK+1 By ¨ ¨ ¨ By jK+1 j1 j1 ,¨¨¨ ,jK ,jK+1 =1 ( )( ) = (DK+1 f ) p(x0 ) (Dp)(x0 )(u(1) ), ¨ ¨ ¨ , (Dp)(x0 )(u(K+1) = which shows the validity of the desired equality for the case k = K + 1. ˝ Problem 34. Let f : Rn Ñ Rm be diﬀerentiable, and Df is a constant map in B(Rn , Rm ); that is, (Df )(x1 )(u) = (Df )(x2 )(u) for all x1 , x2 P Rn and u P Rn . Show that f is a linear term plus a constant and that the linear part of f is the constant value of Df . Proof. Since Df is a constant map, Df is continuous; thus f P C 1 . Therefore, the Taylor Theorem implies that f (x) = f (0) + (Df )(c)(x ´ 0) for some c on the line segment joining x and 0. Let L = (Df )(x). Then f (x) = f (0) + L(x ´ 0) = Lx + f (0) . ˝ Problem 38. Prove Corollary 7.5; that is, show that if U Ď Rn is open, f : U Ñ Rn is of class C 1 , and (Df )(x) is invertible for all x P U, then f (W) is open for every open set W Ď U. Proof. Let W Ď U be an open set. For each x P W, (Df )(x) is invertible; thus the inverse function theorem implies that there exists δx ą 0 such that ( ) ( ) (a) D(x, δx ) Ď W; (b) f D(x, δx ) is open; (c) f : D(x, δ) Ñ f D(x, δx ) is one-to-one and onto. Ť Since W = D(x, δx ), xPW ď ( ) f (W) = f D(x, δx ) xPW is the union of inﬁnitely many open sets; thus f (W) is open. ˝ Problem 40. Let f : R2 Ñ R be of class C 1 , and for some (a, b) P R2 , f (a, b) = 0 and fy (a, b) ‰ 0. Show that there exist open neighborhoods U of a and V of b such that every y P V corresponds to a unique x P U such that f (x, y) = 0. In other words, there exists a function y = y(x) such that y(a) = b and f (x, y(x)) = 0 for all x P U. Proof. Let z = (x, y) and w = (u, v), where x, y, u, v P R. Deﬁne w = F (z), where F is ( ) given by F (x, y) = x, f (x, y) . Then F : D Ñ R2 , and [ ] 1 0 [ ] (DF )(z) = . fx (x, y) fy (x, y) We note that the Jacobian of F at (a, b) is fy (a, b) ‰ 0, so the inverse function theorem ( ) implies that there exists open neighborhoods O Ď R2 of (a, b) and W Ď R2 of a, f (a, b) = (a, 0) such that (a) F : O Ñ W is one-to-one and onto; (b) the inverse function F ´1 : W Ñ O is of class C r ; ( ) ( )´1 (c) (DF ´1 ) x, f (x, y) = (DF )(x, y) . W.L.O.G. we can assume that O = U ˆ V, where U Ď R and V Ď R are open, and a P U, b P V. ( ) Write F ´1 (u, v) = φ(u, v), ψ(u, v) , where φ : W Ñ U and ψ : W Ñ V. Then ( ) ( ) (u, v) = F φ(u, v), ψ(u, v) = φ(u, v), f (u, ψ(u, v)) ( ) which implies that φ(u, v) = u and v = f (u, ψ(u, v)). Let y(x) = ψ(x, 0). Then u, f (u) P ( ) U ˆ V is the unique point satisfying f u, y(u) = 0 if u P U. Therefore, f : U Ñ V, and ( ) f x, y(x) = 0 @x P U . ( ) ( ) Since G(a, b) = (a, 0) = G a, f (a) , (a, b), a, f (a) P O, and G : O Ñ W is one-to-one, we must have b = f (a). ˝

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