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The Implicit and Inverse Function Theorems Notes to Supplement Chapter 13 The Implicit and Inverse Function Theorems Notes to supplement Chapter 13. Remark: These notes are still in draft form. Examples will be added to Section 5. If you see any errors, please let me know. 1. Notation Let X and Y be two normed linear spaces, and let f : X ! Y be a function de¯ned in some neighborhood of the origin of X. We say that f(x) = o(jjxjjn) if jjf(x)jjY lim n = 0: x!0 jjxjjX Analogously, we say that f(x) = O(jjxjjn) if there exists some number C, and some neighborhood G of the origin in X such that n jjf(x)jjY · C jjxjjX ; 8x 2 G: 2. Differentiation on Banach Space Recall that a function f : R ! R is \di®erentiable" at a point x0 if there exists a number a such that (1) f(x) = f(x0) + a (x ¡ x0) + o(jx ¡ x0j): We usually write a = f 0(x). The right hand side of (1) is a linear approximation of f, valid near x0. This de¯nition can straight-forwardly be generalized to functions between two Banach spaces. De¯nition 1. Let X and Y be Banach spaces, let f be a map from X to Y , and let x0 be a point in X. We say that f is di®erentiable at x0 if there exists a map A 2 B(X; Y ) such that f(x) = f(x0) + A (x ¡ x0) + o(jjx ¡ x0jj): 0 The number A is called the \Fr¶echet Derivative" of f at x0. We write A = f (x) = df = D f = fx. Note that the de¯nition makes sense even if f is de¯ned only in a neighborhood of x0 (it does not need to be de¯ned on all of x). It follows directly from the de¯nition that if f is di®erentiable at x0, then it is also continuous as x0. The function f 0(x) is not a map from X to Y , it is a map from X to B(X; Y ). n m Example 1: Let f be a function from R to R . Let fi denote the component functions of f so that f = (f1; f2; : : : ; fm). If the partial derivatives @fi fi;j = @xj n all exist at some x0 2 R , then f is di®erentiable at x0 and 2 3 f1;1(x0) f1;2(x0) ¢ ¢ ¢ f1;n(x0) 6 f (x ) f (x ) ¢ ¢ ¢ f (x ) 7 0 6 2;1 0 2;2 0 2;n 0 7 f (x0) = 6 . 7 : 4 . 5 fm;1(x0) fm;2(x0) ¢ ¢ ¢ fm;n(x0) Example: Let (X; ¹) be a measure space and consider the function f : L3(X; ¹) ! L1(X; ¹): ' 7! '3: In order to see if f is di®erentiable, we need to see if there exists an A 2 B(L3;L1) such that jjf(' + Ã) ¡ f(') ¡ AÃjj (2) lim 1 = 0: jjÃjj3!0 jjÃjj3 We have f(' + Ã) = '3 + 3'2 à + 3'Ã2 + Ã3: Therefore, if f is di®erentiable, we must have Aà = 3'2 Ã. That A is a bounded map is clear since (applying HÄolder'sinequality with p = 3, q = 2=3) Z µZ ¶1=3 µZ ¶2=3 2 3 3 2 jjAÃjj1 = 3 j'j jÃj · 3 jÃj j'j = 3 jjÃjj3 jj'jj3 A similar calculation shows that 2 3 jjf(' + Ã) ¡ f(') ¡ AÃjj1 = jj3 'à + à jj1 2 3 2 3 · 3 jj'à jj1 + jjà jj1 · 3 jj'jj3 jjÃjj3 + jjÃjj3: It follows that (2) holds. Thus f is di®erentiable, and f 0('): à 7! 3 '2 Ã. Example: Set I = [0; 1] and consider the function Z 1 f : C(I) ! R : ' 7! sin('(x)) dx: 0 The function f is di®erentiable at ' if there exists an A 2 B(C(I);R) = C(I)¤ such that jf(' + Ã) ¡ AÃj (3) lim = 0: jjÃjju!0 jjÃjju We ¯nd that Z Z f(' + Ã) = sin(' + Ã) = (sin ' cos à + cos ' sin Ã): 2 When jjÃjju is small, Ã(x) is small for every x, and so cos à = 1 + O(à ), and sin à = à + O(Ã2). An informal calculation then yields Z Z f('+Ã) = ((sin ')(1+O(Ã2))+(cos ')(Ã+O(Ã2))) = f(')+ ((cos ')Ã)+O(Ã2): If f is di®erentiable, we must have Z 1 A : à 7! cos('(x)) Ã(x) dx: 0 It remains to prove that (3) holds. We have Z 1¡ ¢ f(' + Ã) ¡ f(') ¡ Aà = sin(' + Ã) ¡ sin(') ¡ cos(') à dx 0 Z 1 = (sin(')(cos(Ã) ¡ 1) + cos(')(sin(Ã) ¡ Ã)) dx: 0 Using that j1 ¡ cos tj · t2 and j sin t ¡ tj · t2 for all t 2 R, we obtain Z 1 2 2 jf(' + Ã) ¡ f(') ¡ AÃj · (j sin '(x)j + j cos '(x)j) jÃ(x)j dx · 2 jjÃjju: 0 Therefore (3) holds, and f is di®erentiable. Example: Read Example 13.7 in the textbook. Theorem 1 (Chain rule). Let X, Y , and Z denote Banach spaces. Suppose that the functions f : X ! Y , and g : Y ! Z are di®erentiable. Then g ± f : X ! Z is di®erentiable, and (g ± f)0(x) = g0(f(x)) f 0(x): Note that all properties of the functions in Theorem 1 are local, so it would have been su±cient to assume only that f is di®erentiable in some neighborhood of x, and g is di®erentiable in some neighborhood of f(x). The notion of di®erentiation de¯ned in Def. 1 is the direct generalization of the \Jacobian matrix" of multivariate analysis. We can also de¯ne a generalization of the concept of a directional derivative: De¯nition 2. Let X and Y be Banach spaces and let f denote a function from X to Y . Letting x and u denote elements of X, we de¯ne the directional derivative of f at x, in the direction u, by f(x + t u) ¡ f(x) (Duf)(x) = lim : t!0 t Note that (Duf)(x) is simply an element of Y . Remark 1. In the environment of Banach spaces, the directional derivative is fre- quently called a \G^ateauxderivative". Sometimes, this term is used to denote the vector (Duf)(x), but the text book uses a di®erent terminology. To avoid confusion, we will avoid the term \G^ateauxderivative". Example: Let f be as in Example 1, and let u 2 Rm. Then 2 3 u ¢ rf1(x) 6 7 6 u ¢ rf2(x) 7 (Duf)(x) = 6 . 7 : 4 . 5 u ¢ rfn(x) Note that if X and Y are Banach spaces, and f is a di®erentiable function from X to Y , then f(x + t u) = f(x) + f 0(x)(t u) + o(jjt ujj): Consequently, 0 (Duf)(x) = f (x) u: In other words, every (Fr¶echet) di®erentiable functions has directional derivatives in all directions. The converse is not true. In fact, it is not even true in R2 as the following example shows: 2 2 2 2 Example: Set X = R and Y = R. De¯ne ­ = f(x1; x2) 2 R : x1 < x2 < 2 x1g 2 and set f = ­. Then (Du f)(0) exists for all u 2 R , but f is not even continuous at 0. 3. Partial derivatives Let X, Y , and Z be Banach spaces and let F be a map from X £ Y to Z. Then F is di®erentiable at (^x; y^) 2 X £ Y if and only if there exist maps A 2 B(X; Z) and B 2 B(Y; Z) such that F (x; y) = F (^x; y^) + A (x ¡ x^) + B (y ¡ y^) + o(jjx ¡ x^jjX + jjy ¡ y^jjY ): We call the maps A and B the partial derivatives of F with respect to x and y, A = Fx(^x; y^);B = Fy(^x; y^): Note that A is simply the derivative of the map x 7! F (x; y^), and similarly B is the derivative of the map y 7! F (^x; y). 4. Minimization of functionals (View this section as a large example.) The directional derivative can be used to derive necessary conditions for a sta- 1 tionary point of a function. As an example, set I = [0; 1], X = C0 (I), and let us consider the functional Z 1 (4) I : X ! R : u 7! L(x; u(x); u0(x)) dx; 0 where L = L(x; u; v) is a function that is continuously di®erentiable in each of its arguments. Suppose that u 2 X is a point where I is minimized. Then for any ' 2 X, d 0 = I(u + "') = [D I](u): d" ' In other words, if u is a minimizer, then the directional derivative [D'I](u) must be zero for all ' 2 X. For the particular functional I, we ¯nd that ¯ Z 1 d ¯ £ 0 0 0¤ I(u + "')¯ = Lu(x; u; u ) ' + Lv(x; u; u ) ' dx: d" "=0 0 Performing a partial integration (using that '(0) = '(1) = 0), we obtain Z 1 · ¸ 0 d 0 (5) 0 = Lu(x; u; u ) ¡ Lv(x; u; u ) '(x) dx: 0 dx 1 For (5) to hold for every ' 2 C0 ([0; 1]) we must have d (6) L (x; u(x); u(x)0) ¡ L (x; u(x); u0(x)) = 0; x 2 [0; 1]: u dx v That the (potentially non-linear) ODE (6) holds is a necessary condition that a minimizer u must satisfy. This equation is called the \Euler-Lagrange" equation.
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