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Math 346 Lecture #3 6.3 the General Fréchet Derivative

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Math 346 Lecture #3 6.3 the General Fréchet Derivative Math 346 Lecture #3 6.3 The General Fr´echet Derivative We now extend the notion of the Fr´echet derivative to the Banach space setting. Throughout let (X; k · kX ) and (Y; k · kY ) be Banach spaces, and U an open set in X. 6.3.1 The Fr´echet Derivative Definition 6.3.1. A function f : U ! Y is Fr´echet differentiable at x 2 U if there exists A 2 B(X; Y ) such that kf(x + h) − f(x) − A(h)k lim Y = 0; h!0 khkX and we write Df(x) = A. We say f is Fr´echet differentiable on U if f is Fr´echet differ- entiable at each x 2 U. We often refer to Fr´echet differentiable simply as differentiable. Note. When f is Fr´echet differentiable at x 2 U, the derivative Df(x) is unique. This follows from Proposition 6.2.10 (uniqueness of the derivative for finite dimensional X and Y ) whose proof carries over without change to the general Banach space setting (see Remark 6.3.9). Remark 6.3.2. The existence of the Fr´echet derivative does not change when the norm on X is replace by a topologically equivalent one and/or the norm on Y is replaced by a topologically equivalent one. Example 6.3.3. Any L 2 B(X; Y ) is Fr´echet differentiable with DL(x)(v) = L(v) for all x 2 X and all v 2 X. The proof of this is exactly the same as that given in Example 6.2.5 for finite-dimensional Banach spaces. Example (in lieu of 6.3.4). For X = C([0; 1]; R) with k · kX = k · k1, Y = R with k · kY = j · j, and a fixed polynomial p(t) 2 R[t], the function L : X ! Y defined by Z 1 L(f) = p(t)f(t) dt 0 is a bounded linear functional, bounded because Z 1 Z 1 jL(f)j ≤ jp(t)f(t)j dt ≤ kpk1kfk1 = kpk1kfk1; 0 0 so that jL(f)j kLkX;Y = sup ≤ kpk1; kfk1>0 kfk1 and so by Example 6.3.3, we have for each f 2 X that DL(f)g = L(g) for all g 2 X. Note. When X is an infinite-dimensional Banach space, functions from an open subset of X to Y are often defined by Banach-valued integration. Example (in lieu of 6.3.5). Again with X = C([0; 1]; R) with k · kX = k · k1, Y = R with k · kY = j · j, and a fixed polynomial p(t) 2 R[t], the function Q : X ! R defined by Z 1 Q(f) = p(t)[f(t)]3 dt 0 is not linear (unless p(t) = 0) but Fr´echet differentiable on X. Because DQ(f)g = DgQ(f) if Q is Fr´echet differentiable at f, we guess the form of the Fr´echet derivative DQ(f) by computing the G^ateauxderivative DgQ(f). To this end we have for fixed f 2 X, fixed g 2 X, and r > 0 that Q(f + rg) − Q(f) DgQ(f) = lim r!0 r 1 Z 1 Z 1 = lim p(t)[f(t) + rg(t)]3 dt − p(t)[f(t)]3 dt r!0 r 0 0 1 Z 1 = lim p(t)3rf 2(t)g(t) + 3r2f(t)g2(t) + r3g3(t) dt r!0 r 0 Z 1 = lim p(t)3f 2(t)g(t) + 3rf(t)g2(t) + r2g3(t) dt r!0 0 Z 1 = 3p(t)[f(t)]2g(t) dt; 0 provided we can take the limit inside the integral. (Can you explain why this is justified?) Thus the guess for the Fr´echet derivative of Q at f is the linear transformation B : X ! R defined by Z 1 B(g) = 3p(t)[f(t)]2g(t) dt: 0 Notice that the guess for the Fr´echet derivative of an function defined by an integral is another function defined by an integral. This is typical for functions defined by integra- tion. Now for each fixed f is our guess B for DQ(f) a bounded linear transformation? If we set M = kfk1 and K = kpk1, then for any g 2 X we have Z 1 Z 1 2 2 jB(g)j = 3p(t)[f(t)] g(t) dt ≤ j3p(t)[f(t)] g(t)j dt 0 0 Z 1 Z 1 2 2 2 ≤ 3KM jg(t)j dt ≤ 3KM kgk1 dt = 3KM kgk1; 0 0 so that jB(g)j 2 kBkX;Y = sup : nonzero g 2 X ≤ 3KM : kgk1 Thus our guess B for DQ(f) is a bounded linear transformation. Now to show that our guess B is the Fr´echet derivative of Q at f: jQ(f + h) − Q(f) − B(h)j lim h!0 khk1 Z 1 Z 1 Z 1 1 3 3 2 = lim p(t)[f(t) + h(t)] dt − p(t)[f(t)] dt − 3p(t)[f(t)] h(t) dt h!0 khk1 0 0 0 Z 1 1 2 3 = lim p(t) 3f(t)h (t) + h (t) dt h!0 khk1 0 Z 1 1 2 2 ≤ lim jh (t)j p(t) [f(t)] + h(t) dt h!0 khk1 0 2 Z 1 khk1 2 ≤ lim p(t) [f(t)] + h(t) dt = 0 h!0 khk1 0 because the integral is bounded for small h. Therefore Q is Fr´echet differentiable on X where for each f 2 X we have Z 1 DQ(f)(g) = p(t)[f(t)]2g(t) dt; g 2 X: 0 Definition 6.3.6. For f : U ! Y , if Df : U ! B(X; Y ) given by x ! Df(x) for x 2 U, is continuous, then we say that f is continuously differentiable on U. The set of continuously differentiable functions f : U ! Y is denoted by C1(U; Y ). We will show in the next section that C1(U; Y ) is a vector space. Note. The proof of Theorem 6.2.14 { for f : U ! Rm, with U open in Rn, if all the partial derivatives Difj(x) exist and are continuous for all x 2 U, then f is differentiable at each x 2 U { actually shows that the function x ! Df(x) for x 2 U is continuous because in the standard coordinates each entry of the Jacobian (a matrix representation of Df) is a continuous function on U. Thus the hypotheses of Theorem 6.2.14 imply that f 2 C1(U; Rm). Example. This is from Math 634 { Theory of Ordinary Differential Equations { you are not responsible for knowing or reproducing this example. But it is included to illustrate that the \same" function or operator may be continuously differentiable on one Banach space, i.e., (C([0; 1]; R); k · k1), but not differentiable on 2 another Banach space, i.e., (L ([0; 1]; R); k · k2). For X = C([0; 1]; R) equipped with the 1-norm, we show that the operator F : X ! X defined by F (g)(t) = sin(g(t)) is continuously differentiable on X. For fixed g 2 X, the linear operator B 2 L (X) defined by B(h)(t) = (cos g(t))h(t) is bounded because j(cos g(t))h(t)j ≤ jh(t)j ≤ khk1: The operator F is differentiable with (DF (g)h)(t) = (cos g(t))h(t) for all h 2 X because jF (g + h)(t) − F (g)(t) − (cos g(t))h(t)j = j sin(g(t) + h(t)) − sin g(t) − (cos g(t))h(t)j = j sin g(t) cos h(t) + cos g(t) sin h(t) − sin g(t) − (cos g(t))h(t)j = j(−1 + cos h(t)) sin g(t) + (−h(t) + sin h(t)) cos g(t)j ≤ j(−1 + cos h(t)) sin g(t)j + j(−h(t) + sin h(t)) cos g(t)j = j sin g(t)j j − 1 + cos h(t)j + j cos g(t)j j − h(t) + sin h(t)j ≤ j − 1 + cos h(t)j + j − h(t) + sin h(t)j jh(t )j2 jh(t )j3 = 1 + 2 2 6 khk2 khk3 ≤ 1 + 1 ; 2 6 where we have used the existence of t1; t2 2 [0; 1] for which jh(t )j2 jh(t )j3 j − 1 + cos h(t)j = 1 ; j − h(t) + sin h(t)j = 2 ; 2 6 that follow from Lagrange's Remainder Theorem applied to cos and sin respectively. To show that g ! DF (g) is continuous we start with kDF (g1) − DF (g2)k1 = sup kDF (g1)h − DF (g2)hk1 khk1=1 = sup sup j(cos g1(t))h(t) − (cos g2(t))h(t)j khk1=1 t2[0;1] = sup sup jh(t)j j cos g1(t) − cos g2(t)j: khk1=1 t2[0;1] By the Mean Value Theorem applied to cos whose derivative is bounded in absolute value by 1, we have j cos g1(t) − cos g2(t)j ≤ jg1(t) − g2(t)j: Thus kDF (g1) − DF (g2)k1 ≤ sup sup jh(t)j jg1(t) − g2(t)j khk1=1 t2[0;1] ≤ kg1 − g2k1 which says that DF is Lipschitz continuous on X, and hence DF continuous on X. If we take the \same" operator F on the domain X = L2([0; 1]; R) with its 2-norm, then F (g)(t) = sin g(t) is Lipschitz continuous because Z 1 2 2 kF (g1) − F (g2)k2 = j sin g1(t) − sin g2(t)j dt 0 Z 1 2 ≤ jg1(t) − g2(t)j dt 0 2 = kg1 − g2k2 (where we have used the Mean Value Theorem to get j sin g1(t)−sin g2(t)j ≤ jg1(t)−g2(t)j), but this F is not differentiable at g = 0 (to show this requires a long tedious argument that depends on the properties of Lebesgue integration).
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