Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Inverse Function Theorem and Implicit Function Theorem Dr. M. Asaduzzaman Professor Dept. of Mathematics University of Rajshahi Email: [email protected] 1 Contents Euclidean Spaces Normed Linear Spaces Metric Induced From Norms Open Sets in ℝn Continuous Functions Directional and Partial Derivatives Derivatives Jacobian and Regular Maps Inverse Function Theorem Implicit Function Theorem 2 Euclidean Spaces Definition: A nonempty set F with addition and multiplication defined in F is called a field if the following conditions hold: i) x+y, xy F, x, y F ii) x+y=y+x and xy=yx x, y F iii) (x+y)+z=x+(y+z) and (xy)z=x(yz) x, y, zF iv) two elements 1 and 0 in F such that x+0=x and x.1=x xF v) xF an element –xF such that x+(-x)=0 and for every nonzero xF an element x-1F such that xx-1=1 vi) x(y+z)=xy+xz, x, y, z F. Example 1. ℝ, the set of all real numbers is a field under usual addition and multiplication. This field is called the real field. Example 2. C, the set of all complex numbers is a field under addition and multiplication of complex numbers. This field is called the complex field. 3 Euclidean Spaces Definition: A nonempty set V with addition and scalar multiplication defined in V is called a vector space over the field F if x, y, zV and , F the following conditions hold: i) x+y, x V ii) x+y=y+x iii) (x+y)+z=x+(y+z) iv) an element 0 in V such that x+0=x v) xV an element –xV such that x+(-x)=0 vi) (x+y)=x+y and (+)x=x+x vii) ()x=(x) and (+)x=x+x viii) 1x=x The elements of V are called vectors and those of F are called scalars. 4 Euclidean Spaces Example: Every field is a vector space over itself. Example: Let n be a positive integer. Then the set ℝn={(x1,x2,,xn): xiℝ, 1≤i ≤n} is a vector space over the real field if we define addition and scalar multiplication component wise. Definition: We define the so-called “inner product “ (or scalar product) of two vectors x=(x1,x2,,xn), y=(y1,y2,,yn) in ℝn by xy=x1y1+x2y2++xnyn and the norm of x by |x|=(x12+x22++xn2)1/2 then the vector space with the above inner product and norm is called n dimensional Euclidean space. We then have following theorem: 5 Euclidean Spaces Theorem: Suppose x, y, zRn, and is a real. Then (a) |x|≥0; (b) |x|=0 iff x=0; (c) |x|=|||x|; (d) |xy|≤|x||y|; (e) |x+y|≤|x|+|y|; (f) |x-z|≤|x-y|+|y-z|. Based on property (a), (b), (c) and (e) of the above theorem we define norm on any vector space over real or complex field. 6 Normed Linear Spaces Definition: Let X be a vector space over the real or complex field K. Then a function ║║: X→ℝ is called a norm in X if it satisfies the following properties: (a) ║x║≥0; (b) ║x║=0 iff x=0; (c) ║x║=||║x║; (d) ║x+y║≤║x║+║y║. A vector space over the real or complex field is called a linear space and a linear space with a norm is called a normed linear space. 7 Normed Linear Spaces Example: Let X =Rn and let ║║: X→ℝ be defined by ║x║= |x1|+|x2|+ +|xn|, where x=(x1,x2,,xn)X. Then X is a normed linear space. Example: Let X =ℝn and let ║║: X→ℝ be defined by ║x║= max{|x1|,|x2|, ,|xn|}, where x=(x1,x2,,xn)X. Then X is a normed linear space. Thus there are many ways to define a norm on the same linear space. 8 Metric Induced from Norms For any normed linear space we have the following theorem: Theorem: Suppose X is a normed linear space . Then for any x, y, zX , we have ║x-y║≤ ║x-z║+ ║y-z║. Thus in a normed linear space X if we define d(x,y)= ║x-y║ then d satisfies all conditions of a metric. This metric is called metric induced from norm. It is clear that different norm on the same linear space induced different metric. 9 Open Sets in Rn We now consider only the Euclidean norm and the metric induced from Euclidean norm. Definition: Let x0ℝn and r>0 be given. We define the set S(x0,r)={xℝn : |x-x0|<r} and call it a neighborhood of x0 of radius r. From the definition it is clear that neighborhood changes with x0 and r. Example: In ℝ1 neighborhoods are open intervals and in ℝ2 neighborhoods are open discs. 10 Open Sets in ℝn r . x0 S(x0,r) O 11 Open Sets in ℝn Definition: Let E be a subset of a Euclidean space ℝn . A point x0E is said to be an interior point of E if there exists a neighborhood S(x0,r) of x0 such that S(x0,r)E. Definition: A subset E of a Euclidean space ℝn is called an open set if every point of E is an interior point of E. Example: For any x0ℝn and for any r>0 the neighborhood S(x0, r) is an open set. Example: Union of any number of neighborhoods is an open set. Example: A finite subset of ℝn cannot be an open set. 12 Continuous functions Definition: Let E be a subset of a Euclidean space ℝn . A point x0 ℝn is said to be an limit point of E if every neighborhood contains infinitely many points of E. Definition: Let E be a subset of ℝn and let :E→ ℝm be a function. Let x0 be a point as well as a limit point of E. Then is said to be continuous at x0 if given there exists a such that |x- x0| whenever |x-x0| and xE. If is continuous at every point of E then is called continuous on E. 13 Continuous functions Example: The function (x,y)=1-x2-y2 is continuous on ℝ2 . Example: The function xy , ( x, y ) (0,0) 2 2 f ( x, y ) x y ( x, y ) (0,0) . 0, is not continuous at (0,0) . 14 Directional and Partial Derivatives Definition: Let be real valued function defined on an open subset E of ℝn and let x0E be given. We define the directional derivative of at x0 in the direction of unit vector u to be f ( x 0 tu ) f ( x 0 ) lim t 0 t provided the limit exists. It will be denoted by Du(x0). 15 Directional and Partial Derivatives Example: Let f ( x, y ) | x 2 y 2 | , x 0 (0,0) . Then directional derivatives of exist only in the four directions (1/2, 1/2). Definition: The partial derivatives of are defined as the directional derivatives in the directions of standard basis vectors e1, e2,, en. The i th partial derivative of at x is f (x). usually denoted by Di(x), f xi (x ) or xi Thus f (x te ) f (x) Di (x) lim t 0 provided the limit exists. i t 16 Directional and Partial Derivatives Problem: Let f ( x, y) ( x 1) y . 2 2 Find the directional derivative at e2 in any direction u. Solution: Let u=(cos , sin ). Then f (e 2 tu) f (e 2 ) Du (e 2 ) lim t 0 t f (t cos ,1 t sin ) f (0,1) lim t 0 t (t cos 1) (1 t sin ) 2 lim t 0 t 2(cos sin ). 2 17 Directional and Partial Derivatives The following example shows that existence of partial derivatives at a point does not imply that directional derivatives exist in all directions. Example: Let (x,y)=(xy) 1/3 and u=(cos , sin). Then f (tu) f (0) Du (0) lim t 0 t (cos sin )1/ 3 lim t 0 t 1/ 3 which exists and equal to 0 if and only if sin cos=0. That is if and only if u=e1, e2, -e1 or –e2. 18 Derivatives In order to define derivaties of a function whose domain is an open subset of ℝn, let us take another look at the familiar case n=1, and let us see how to interpret the derivative in that case in a way which will naturally extend to n1. If is a real function with domain (a,b)ℝ1 and if x(a,b), then (x) is usually defined to be the real number f ( x h) f ( x ) lim , h 0 h provided the limit exists. Thus (1) x+hxxh rh where the remainder r(h) is small in the sense that r ( h) lim 0. h 0 h 19 Derivatives Thus (1) expresses the difference x+hx as the sum of the linear function that takes h to xh, plus a small remainder. We can therefore regard the derivative of at x, not as a real number but as the linear operator on ℝ1 that takes h to xh. Let us next consider a function f that maps (a,b)ℝ1 into ℝm . In that case, fx was defined to be that vector y Rm for which f ( x h) f ( x ) lim h 0 h y 0. We can again rewrite this in the form f ( x h) f ( x) hy r (h) (2) where r(h)/h→0 as h→0. The main term on the RHS of (2) is again a linear function of h. 20 Derivatives Each yℝm induces a linear transformation of ℝ1 into ℝm, by associating to each h ℝ1 the vector hy ℝm. This identification of ℝm with L(ℝ1, ℝm) allow us to regard f(x) as a member of L(ℝ1, ℝm). Thus, if f is a differential mapping of (a,b)ℝ1 into ℝm, and if x(a,b), then f(x) is the linear transformation of ℝ1 into ℝm that satisfies (3) f ( x h) f ( x ) f ' ( x ) h lim 0, h 0 h or, equivalently, (4) lim h 0 f ( x h) f ( x ) f ' ( x ) h h 0. 21 Derivatives Motivating from the above fact, we define derivative of a function from ℝn into ℝm. Definition: Suppose E is an open set in ℝn, f maps E into ℝm, and xE. If there exists a linear transformation A of Rn into ℝm such that (5) lim h 0 f (x h) f (x) Ah h 0, then we say that f is differentiable at x, and we write (5) f(x)=A. If f is differentiable at every xE , we say that f is differentiable in E 22 Derivatives The following theorem proves that if a function f is differentiable at x then f(x) is unique. Theorem: Suppose E is an open set in ℝn, f maps E into ℝm, and xE. If there exists a linear transformation A1 and A2 of ℝn into ℝm such that lim h 0 f (x h) f (x) Ah h 0, holds with A=A1 and with A=A2. Then A1=A2. 23 Derivatives We see that if a function f:E→ ℝm is differentiable at a point xE then f(x)L(ℝn, ℝm). Since every mn matrix with real entries are in L(ℝn, ℝm) and every member of L(ℝn, ℝm) has a matrix representation with respect to given bases of ℝn and ℝm. In order to make everything simple we will always consider the standard bases. Thus when we see f(x), we shall think of f(x) as an mn matrix which is the matrix of f(x) with respect to standard bases. 24 Derivatives The following theorem extends chain rule of derivatives of single variable. Theorem: Suppose E is an open set in ℝn, f maps E into ℝm, f is differentiable at x0E, g maps an open set containing f(E) into ℝk, and g is differentiable at f(x0). Then the mapping F of E into Rk defined by F(x)=g(f(x)) is differentiable at x0, and F(x0)=g(f(x0))f(x0). x0 f f(x0) F g g(f(x0)) 25 Derivatives Suppose E is an open set in ℝn, f maps E into ℝm. Let {e1, e2,…., en} and {u1, u2,…., um} be the standard bases of ℝn and ℝm respectively. Let the component functions of f be 1, 2,…., m. Then for xE, f(x)=(1(x), 2(x), … , m(x)) and each i is a real valued function on E. We denote partial derivative of i with respect to xj by the notation Dji. These partial derivatives are called partial derivatives of f. If we consider i as row index and j as column index then partial derivatives of f form an mn matrix. 26 Derivatives The following theorem shows that if f is differentiable at a point x then its partial derivatives exist at x, and they determine f(x) completely. Theorem: Suppose f maps an open set E ℝn into ℝm, and f is differentiable at point xE. Then the partial derivatives (Dji)(x) exist, and the matrix of f(x) is ( D1 f1 )( x) ( Dn f1 )( x) .......................................... . ( D1 f m )( x) ( Dn f m )( x) 27 Derivatives The following theorem shows that differentiability at a point x is sufficient for a function to be continuous at the point. Theorem: Suppose f maps an open set E ℝn into ℝm, and f is differentiable at point x0E. Then f is continuous at x0. 28 Derivatives The following example shows that existence of partial derivatives at a point is not enough for a function to be differentiable or even continuous. Example: Let 2 xy2 , ( x, y ) (0,0) f ( x, y ) x 2 y 4 0 , ( x, y ) (0,0), then both partial derivatives x and y exist at (0,0) but the function is not continuous at (0,0). 29 Derivatives Definition: A differentiable mapping f of an open set Eℝn into Rm is said to be continuously differentiable in E if f is a continuous mapping of E into L(ℝn , ℝm ). More explicitly, it is required that to every corresponds a such that f ' ( y) f ' ( x) If yE and |x-y|. If this is so, we also say that f is a 𝒞- mapping, or that f𝒞(E). 30 Derivatives The following theorem shows that existence of continuous partial derivatives at a point for a function is a sufficient condition to be differentiable at the point. Theorem: Suppose f maps an open set E ℝn into ℝn. Then f𝒞(E) if and only if the partial derivatives Dji exists and are continuous on E for 1≤i ≤m, 1 ≤j ≤n. 31 Derivatives The following example shows that continuity of partial derivatives at a point is not a necessary condition for a function to be differentiable there. Example: Let ( x y) 2 sin ( x1 y ) , x y f ( x, y) 0, x y, then (x and y exist for all (x,y)) is differentiable at every point on the line x=y but the functions x and y are not continuous on the line x=y. 32 Derivatives The existence of a derivative for a real valued function of one variable is a fact of considerable interest. Geometrically it says that a tangent line exists. However, the fact that a real valued function of several variables has partial derivatives is not in much interest. For, one thing, the existence of directional derivatives in the directions of standard basis vectors e1, e2,…., en does not imply that directional derivatives in all direction exist. Moreover, the function need not have a tangent hyper plane even if there is a derivative in every direction. It can be shown that most of the basic properties of differentiable functions of one variable remain true for differentiable functions of several variables. 33 Derivatives Now we will see the geometrical similarities: Theorem: Suppose a real function defined on (a,b)ℝ1 is differentiable at a point x0(a,b). Then the tangent line to the curve y=(x) has the equation y=(x0)+(x0)(x-x0). y (x0,f(x0)) o a x0 y=f(x) b x 34 Derivatives Theorem: Suppose a real function defined on an open set ERn is differentiable in E and x0E is a point on the hyper surface (x)=c . Then the tangent hyper plane to the hyper surface z=(x) at the point (x0,(x0)) n has the equation z f ( x 0 ) Di f ( x 0 )( xi x0 i ). i 1 z z0 f x ( x0 , y0 )( x x0 ) f y ( x0 , y0 )( y y0 ). E 35 Derivatives Theorem: Suppose a real function defined on an open set Eℝ2 is differentiable in E and (x0,y0)E be a point on the curve (x,y)=c. Then the gradient vector (x, y) of at (x0,y0) represents the direction of the normal to the curve (x,y)=c at (x0,y0). y f(x,y)=c (x0,y0) o x 36 Derivatives Theorem: Suppose a real function defined on an open set Eℝn is differentiable at a point x0E. Then the gradient vector (D1, D2, … , Dn) of at x0 represents the direction of the normal to the hyper surface (x)=c at x0. x0 37 Jacobian Suppose f maps an open set E ℝn into itself, and x0E. Then f is continuous at x0. Let f=(1, …. ,n), so that 1, …. ,n are real valued functions with domain E. We suppose all first order partial derivatives (Dji)(x0) exist for 1≤i,j≤n. In this case we say that f possesses first order partial derivatives at x0. The determinant ( D1 f1 )( x 0 ) ( Dn f1 )( x 0 ) .......................................... ( D1 f n )( x 0 ) ( Dn f n )( x 0 ) is called the Jacobian of the function f at the point x0 and is denoted by Jf(x0). Another common notation of Jacobian is ( f1 , f 2 , , f n ) . ( x10 , x20 , , xn 0 ) 38 Jacobian Definition: A function f is said be one-to-one, or invertible, if f(x1) and f(x2) are distinct whenever x1 and x2 are distinct points of Df. Example: The linear map f(x,y)=(x-y, x+y) is one-to-one in the domain Df=ℝ2. Example: The map f(x,y)=(x2-y2, 2xy) is not one-to-one in the domain Df=ℝ2, since f(x,y)=f(-x,-y). Definition: A function f may fail to be one-to-one, but be oneto-one on a subset S of Df. By this we mean that f(x1) and f(x2) are distinct whenever x1 and x2 are distinct points of S. In this case, f is not invertible, but if fS is defined on S by fS(x)=f(x), xS, and left undefined for xS, then fS is invertible. We say that fS is the restriction of f to S. 39 Jacobian Definition: If a function f is one-to-one on a neighborhood of x0, we say that f is locally invertible at x0. Example: The map f(x,y)=(ex cos y, ex sin y) is not one-toone in the domain Df=ℝ2, since f(x,y)=f(x,y+2). But this map is one-to-one on S={(x,y): x , ≤ y 2 }. This map is locally invertible on the entire plane. Remark: The above example proves that locally invertible at every point of the domain of a function does not imply that the function is invertible. 40 Jacobian Definition: A function f : ℝn → ℝn is said to be regular on an open set S if f is one-to-one and continuously differentiable on S, and Jf(x)0 if xS. Example: If f(x,y)=(x-y, x+y), then 1 1 Jf ( x, y ) 2, for all (x,y)ℝ2. 1 1 So f is regular on ℝ2. Example: If f(x,y)=(x+y, 2x+2y), then 1 1 Jf ( x, y ) 0, for all (x,y)ℝ2. 2 2 So f is not regular on any open subset of ℝ2. 41 Jacobian Example: If f(x,y)=(ex cos y, ex sin y), then Jf ( x, y) e x cos y e x sin y e x sin y e x cos y e 2 x ( 0), for all (x,y)ℝ2. So f is regular on any open set S on which f is one-to-one. Remark: The above example proves that non zero Jacobian in the entire plane does not prove that the function is one-toone in the entire plane. 42 Inverse function theorem The inverse function theorem states, roughly speaking, that a continuously differentiable mapping f is invertible in a neighborhood of any point x at which the linear transformation f(x) is invertible. 43 Inverse function theorem Theorem (Inverse function theorem): Suppose f is a continuously differentiable mapping of an open set Eℝn into ℝn, f(a) is invertible for some aE, and b=f(a). Then (a) there exists open set U and V in ℝn such that aU, bV, f is one-to-one on U, and f(U)=V; (b) if g is the inverse of f [ which exists by (a)], defined in V by g(f(x)=x (xU), then g is continuously differentiable in V and g(y)={f(g(y))}-1 (yV). 44 Hypothesis: (i) Eℝn is open. (ii) f : E→ ℝn . (iii) f𝒞(E) . (iv) a E such that Jf(a)0. (v) b=f(a). Conclusion: (i) open sets U and V in ℝn such that aUE, bVf(E) , f is one-to-one and f(U)=V . (ii) if g is the inverse of f defined in V then g𝒞(V) and g(y)={f(g(y))}-1 (yV). b=f(a) a f . U g . V=f(U) f(E) E Rn Rn 45 2 x y ( z 1) 1 Example: Let a=(1,2,1) and 2 f ( x, y, z ) y z ( x 1) 1 . z x ( y 2) 2 3 1 2 z 2 1 f ' (a) 2 x 2 1 1 , 1 2y 4 1 1 1 0 Jf (a) 0 1 1 2. so 1 0 1 Then In this case, it is difficult to describe U of find g=fU-1 explicitly; however we know that f(U) is a nbd. of b=f(a)=(4,2,5), that g(b)=a=(1,2,1), and that 46 1 1 1 0 1 1 1 1 1 g ' (b) [f ' (a)] 0 1 1 1 1 1. 2 1 0 1 1 1 1 Therefore 1 1 1 1 y1 4 1 g(y ) 2 1 1 1 y2 2 e(y ), 2 1 1 1 1 y3 5 where lim y ( 4 , 2 , 5 ) e( y ) ( y1 4) ( y2 2) ( y3 5) 2 2 2 0. Thus we have approximated g near b=(4,2,5). 47 IImplicit function theorem If is a continuously differentiable real function in the plane, then the equation (x,y)=0 can be solved for y in terms of x in a nbd. of any point (a,b) at which (a,b)=0 and /y0. Similarly x can be solved in terms of y near (a,b) if /x0. For a simple example which illustrates the need for assuming /y0 , consider y (x,y)=x2+y2-1. (x,y)=0 . O x 48 IImplicit function theorem We now consider mappings from ℝn+m to ℝm. It will be convenient to denote points in ℝn+m by (x, u) ( x1 , x2 ,, xn , u1 , u2 ,, um ). To motivate the problem we are interested in, we first ask whether the linear system of m equations in m+n variables a11x1 a12 x2 a1n xn b11u1 b12u2 b1mum 0 a21x1 a22 x2 a2 n xn b21u1 b22u2 b2 mum 0 am1 x1 am 2 x2 amn xn bm1u1 bm 2u.2 bmmum 0 determines u1 , u2 ,, um uniquely in terms of x1 , x2 ,, xn . 49 By rewriting the system in matrix form as (1) where Ax+Bu=0, a11 a A 21 am1 x1 x x 2 , xn a12 a22 am 2 and a1n a2 n , amn b11 b B 21 bm1 b12 b22 am 2 b1m b2 m , bmm u1 u u 2 . un . uniquely for u in We see that the equation (1) can be solved terms of x if the square matrix B is nonsingular. In this case the solution is u=-B-1Ax. 50 For our purpose it is convenient to restate this: If (2) f(x,u)= Ax+Bu, where B is nonsingular, then the system f(x,u)= 0 determines u as a function of x, for all x in ℝn . Notice that f in (2) is a linear transformation from ℝn+m to ℝm. Now the question of whether an arbitrary system f(x,u)= 0 determines u as a function of x is too general to have a useful answer. The implicit function theorem answers this question in an important special case. 51 Notations: We partition the differential matrix of f : ℝn+m →ℝm : f1 x 1 f 2 f ' x1 f m x1 f1 x2 f 2 x2 f m x2 f1 xn f 2 xn f m xn f1 u1 f 2 | u1 | f m | u1 | f1 u2 f 2 u2 f m u2 f1 um f 2 um f m um or f=[fx,fu], where fx denotes the submatrix to the left of the dashed line and fu is to the right. 52 Theorem(Implicit function theorem): Suppose f : ℝn+m →ℝm is continuously differentiable on an open set E ℝn+m containing (x0,u0). Let f (x0,u0)=0, and suppose that fu (x0,u0) is nonsingular. Then there exists a neighborhood M of (x0,u0), contained in E, on which fu (x,u) is nonsingular and a neighborhood N of x0 in ℝn on which a unique continuously ( differentiable map g : ℝn→ℝm is defined, such that g(x0)=u0 and (x,g(x))M and f(x,g(x))=0 if xN. Moreover, g(x)=-[fu(x,g(x)]-1fx(x,g(x), xN. 53 Eℝn+m u f(x,u)=0 ℝm E M u=g(x) * (x0,u0) x0 ℝn * N x 54 Example: Take m=2, n=3, and consider the mapping f=(1,2) of ℝ5 into ℝ2 x1 f ( x , x , u , u , u ) 2 e x2u1 4u2 3 given by 1 1 2 1 2 3 f 2 ( x1 , x2 , u1 , u2 , u3 ) x2 cos x1 6 x1 2u1 u3 . If x0=(0,1) and u0=(3,2,7), then f(x0,u0)=0. With respect to the standard bases, the matrix of f(x0,u0) is 2 3 1 4 0 [f ' (x 0 , u 0 )] . 6 1 2 0 1 Hence 2 3 1 4 0 f x (x 0 , u 0 ) , . fu (x 0 , u 0 ) 6 1 2 0 1 Since fx(x0,u0) is nonsingular, by the implicit function theorem, there exists a continuously differentiable map g defined in a nbd. of u0, such that g(u0)=x0. and f(g(u),u)=0. 55 We can compute g(u0) using the formula g(u0)=-[fx(x0,u0)]-1fu(x0,u0). Since 1 1 3 [f x (x 0 , u 0 )] 20 6 2 1 1 1 3 1 4 0 14 g' (u 0 ) 1 20 6 2 2 0 1 2 1 5 6 5 203 . 1 10 56