Fix a point a = ( a 1 , β¦ , a n ) β U a=(a_1,\dots,a_n)\in U a = ( a 1 β , β¦ , a n β ) β U f f f a a a Differentiability at a Point and Jacobian Matrix for Maps Between Euclidean Spaces . Write f = ( f 1 , β¦ , f m ) f=(f_1,\dots,f_m) f = ( f 1 β , β¦ , f m β )
Because U U U open , there exists Ο > 0 \rho>0 Ο > 0 x β R n x\in\mathbb{R}^n x β R n
β i = 1 n ( x i β a i ) 2 < Ο 2 , \sum_{i=1}^n (x_i-a_i)^2<\rho^2, i = 1 β n β ( x i β β a i β ) 2 < Ο 2 , one has x β U x\in U x β U
Let Ξ΅ > 0 \varepsilon>0 Ξ΅ > 0
Ξ· = Ξ΅ m n . \eta=\frac{\varepsilon}{mn}. Ξ· = mn Ξ΅ β . For each pair of indices j β { 1 , β¦ , m } j\in\{1,\dots,m\} j β { 1 , β¦ , m } r β { 1 , β¦ , n } r\in\{1,\dots,n\} r β { 1 , β¦ , n }
x β¦ β f j β x r ( x ) x\mapsto \frac{\partial f_j}{\partial x_r}(x) x β¦ β x r β β f j β β ( x ) is continuous at a a a C^1 Map on an Open Subset of Euclidean Space . Hence there exists Ξ΄ j r > 0 \delta_{jr}>0 Ξ΄ j r β > 0 x β U x\in U x β U
β i = 1 n ( x i β a i ) 2 < Ξ΄ j r 2 , \sum_{i=1}^n (x_i-a_i)^2<\delta_{jr}^2, i = 1 β n β ( x i β β a i β ) 2 < Ξ΄ j r 2 β , one has
β£ β f j β x r ( x ) β β f j β x r ( a ) β£ < Ξ· . \left|\frac{\partial f_j}{\partial x_r}(x)-\frac{\partial f_j}{\partial x_r}(a)\right|<\eta. β β x r β β f j β β ( x ) β β x r β β f j β β ( a ) β < Ξ· . Let
Ξ΄ = min β‘ ( Ο , Ξ΄ 11 , β¦ , Ξ΄ m n ) . \delta=\min\Bigl(\rho,\delta_{11},\dots,\delta_{mn}\Bigr). Ξ΄ = min ( Ο , Ξ΄ 11 β , β¦ , Ξ΄ mn β ) . Now let h = ( h 1 , β¦ , h n ) β R n h=(h_1,\dots,h_n)\in\mathbb{R}^n h = ( h 1 β , β¦ , h n β ) β R n
0 < β i = 1 n h i 2 < Ξ΄ 2 . 0<\sum_{i=1}^n h_i^2<\delta^2. 0 < i = 1 β n β h i 2 β < Ξ΄ 2 . For each r β { 0 , 1 , β¦ , n } r\in\{0,1,\dots,n\} r β { 0 , 1 , β¦ , n }
x ( r ) = a + ( h 1 , β¦ , h r , 0 , β¦ , 0 ) . x^{(r)}=a+(h_1,\dots,h_r,0,\dots,0). x ( r ) = a + ( h 1 β , β¦ , h r β , 0 , β¦ , 0 ) . Then x ( 0 ) = a x^{(0)}=a x ( 0 ) = a x ( n ) = a + h x^{(n)}=a+h x ( n ) = a + h r r r
β i = 1 n ( x i ( r ) β a i ) 2 = β i = 1 r h i 2 β€ β i = 1 n h i 2 < Ξ΄ 2 β€ Ο 2 , \sum_{i=1}^n (x_i^{(r)}-a_i)^2=\sum_{i=1}^r h_i^2\le \sum_{i=1}^n h_i^2<\delta^2\le \rho^2, i = 1 β n β ( x i ( r ) β β a i β ) 2 = i = 1 β r β h i 2 β β€ i = 1 β n β h i 2 β < Ξ΄ 2 β€ Ο 2 , so every point x ( r ) x^{(r)} x ( r ) U U U
Fix a coordinate index j β { 1 , β¦ , m } j\in\{1,\dots,m\} j β { 1 , β¦ , m }
f j ( a + h ) β f j ( a ) = β r = 1 n ( f j ( x ( r ) ) β f j ( x ( r β 1 ) ) ) . f_j(a+h)-f_j(a)=\sum_{r=1}^n \bigl(f_j(x^{(r)})-f_j(x^{(r-1)})\bigr). f j β ( a + h ) β f j β ( a ) = r = 1 β n β ( f j β ( x ( r ) ) β f j β ( x ( r β 1 ) ) ) . For each r r r Ο r \varphi_r Ο r β 0 0 0 h r h_r h r β
Ο r ( t ) = f j ( a 1 + h 1 , β¦ , a r β 1 + h r β 1 , a r + t , a r + 1 , β¦ , a n ) . \varphi_r(t)=f_j(a_1+h_1,\dots,a_{r-1}+h_{r-1},a_r+t,a_{r+1},\dots,a_n). Ο r β ( t ) = f j β ( a 1 β + h 1 β , β¦ , a r β 1 β + h r β 1 β , a r β + t , a r + 1 β , β¦ , a n β ) . Because the coordinate function f j f_j f j β U U U C 1 C^1 C 1 Ο r \varphi_r Ο r β Continuity at a Point ; hence it is continuous on the closed interval in the sense of Continuity on a Closed Interval . Moreover, at every interior point t t t Ο r \varphi_r Ο r β
Ο r β² ( t ) = β f j β x r ( a 1 + h 1 , β¦ , a r β 1 + h r β 1 , a r + t , a r + 1 , β¦ , a n ) , \varphi_r'(t)=\frac{\partial f_j}{\partial x_r}(a_1+h_1,\dots,a_{r-1}+h_{r-1},a_r+t,a_{r+1},\dots,a_n), Ο r β² β ( t ) = β x r β β f j β β ( a 1 β + h 1 β , β¦ , a r β 1 β + h r β 1 β , a r β + t , a r + 1 β , β¦ , a n β ) , directly from the definition of Partial Derivative of a Coordinate Function . Therefore we may apply Mean Value Theorem in One Dimension . If h r β 0 h_r\ne 0 h r β ξ = 0 c r c_r c r β 0 0 0 h r h_r h r β
f j ( x ( r ) ) β f j ( x ( r β 1 ) ) = Ο r ( h r ) β Ο r ( 0 ) = Ο r β² ( c r ) h r . f_j(x^{(r)})-f_j(x^{(r-1)})
=\varphi_r(h_r)-\varphi_r(0)
=\varphi_r'(c_r)h_r. f j β ( x ( r ) ) β f j β ( x ( r β 1 ) ) = Ο r β ( h r β ) β Ο r β ( 0 ) = Ο r β² β ( c r β ) h r β . If h r = 0 h_r=0 h r β = 0 0 0 0 r r r
ΞΎ ( r ) = ( a 1 + h 1 , β¦ , a r β 1 + h r β 1 , a r + c r , a r + 1 , β¦ , a n ) β U \xi^{(r)}=(a_1+h_1,\dots,a_{r-1}+h_{r-1},a_r+c_r,a_{r+1},\dots,a_n)\in U ΞΎ ( r ) = ( a 1 β + h 1 β , β¦ , a r β 1 β + h r β 1 β , a r β + c r β , a r + 1 β , β¦ , a n β ) β U such that
f j ( x ( r ) ) β f j ( x ( r β 1 ) ) = β f j β x r ( ΞΎ ( r ) ) h r . f_j(x^{(r)})-f_j(x^{(r-1)})=\frac{\partial f_j}{\partial x_r}(\xi^{(r)})h_r. f j β ( x ( r ) ) β f j β ( x ( r β 1 ) ) = β x r β β f j β β ( ΞΎ ( r ) ) h r β . Summing over r r r
f j ( a + h ) β f j ( a ) = β r = 1 n β f j β x r ( ΞΎ ( r ) ) h r . f_j(a+h)-f_j(a)=\sum_{r=1}^n \frac{\partial f_j}{\partial x_r}(\xi^{(r)})h_r. f j β ( a + h ) β f j β ( a ) = r = 1 β n β β x r β β f j β β ( ΞΎ ( r ) ) h r β . Subtracting the linear term defined by the Jacobian matrix at a a a
R j ( h ) : = f j ( a + h ) β f j ( a ) β β r = 1 n β f j β x r ( a ) h r = β r = 1 n ( β f j β x r ( ΞΎ ( r ) ) β β f j β x r ( a ) ) h r . R_j(h):=f_j(a+h)-f_j(a)-\sum_{r=1}^n \frac{\partial f_j}{\partial x_r}(a)h_r
=\sum_{r=1}^n \left(\frac{\partial f_j}{\partial x_r}(\xi^{(r)})-\frac{\partial f_j}{\partial x_r}(a)\right)h_r. R j β ( h ) := f j β ( a + h ) β f j β ( a ) β r = 1 β n β β x r β β f j β β ( a ) h r β = r = 1 β n β ( β x r β β f j β β ( ΞΎ ( r ) ) β β x r β β f j β β ( a ) ) h r β . Now each point ΞΎ ( r ) \xi^{(r)} ΞΎ ( r )
β i = 1 n ( ΞΎ i ( r ) β a i ) 2 β€ β i = 1 n h i 2 < Ξ΄ 2 β€ Ξ΄ j r 2 , \sum_{i=1}^n (\xi_i^{(r)}-a_i)^2\le \sum_{i=1}^n h_i^2<\delta^2\le \delta_{jr}^2, i = 1 β n β ( ΞΎ i ( r ) β β a i β ) 2 β€ i = 1 β n β h i 2 β < Ξ΄ 2 β€ Ξ΄ j r 2 β , so by the choice of Ξ΄ j r \delta_{jr} Ξ΄ j r β
β£ β f j β x r ( ΞΎ ( r ) ) β β f j β x r ( a ) β£ < Ξ· . \left|\frac{\partial f_j}{\partial x_r}(\xi^{(r)})-\frac{\partial f_j}{\partial x_r}(a)\right|<\eta. β β x r β β f j β β ( ΞΎ ( r ) ) β β x r β β f j β β ( a ) β < Ξ· . Hence
β£ R j ( h ) β£ β€ Ξ· β r = 1 n β£ h r β£ . |R_j(h)|\le \eta\sum_{r=1}^n |h_r|. β£ R j β ( h ) β£ β€ Ξ· r = 1 β n β β£ h r β β£. Therefore,
R j ( h ) 2 β€ Ξ· 2 ( β r = 1 n β£ h r β£ ) 2 . R_j(h)^2\le \eta^2\left(\sum_{r=1}^n |h_r|\right)^2. R j β ( h ) 2 β€ Ξ· 2 ( r = 1 β n β β£ h r β β£ ) 2 . Expanding the square and using 2 β£ h r β£ β£ h s β£ β€ h r 2 + h s 2 2|h_r||h_s|\le h_r^2+h_s^2 2β£ h r β β£β£ h s β β£ β€ h r 2 β + h s 2 β
( β r = 1 n β£ h r β£ ) 2 β€ n β r = 1 n h r 2 . \left(\sum_{r=1}^n |h_r|\right)^2\le n\sum_{r=1}^n h_r^2. ( r = 1 β n β β£ h r β β£ ) 2 β€ n r = 1 β n β h r 2 β . Consequently,
R j ( h ) 2 β€ n Ξ· 2 β r = 1 n h r 2 . R_j(h)^2\le n\eta^2\sum_{r=1}^n h_r^2. R j β ( h ) 2 β€ n Ξ· 2 r = 1 β n β h r 2 β . Summing over j = 1 , β¦ , m j=1,\dots,m j = 1 , β¦ , m
β j = 1 m R j ( h ) 2 β€ m n Ξ· 2 β r = 1 n h r 2 . \sum_{j=1}^m R_j(h)^2\le mn\eta^2\sum_{r=1}^n h_r^2. j = 1 β m β R j β ( h ) 2 β€ mn Ξ· 2 r = 1 β n β h r 2 β . Since Ξ· = Ξ΅ / ( m n ) \eta=\varepsilon/(mn) Ξ· = Ξ΅ / ( mn )
m n Ξ· 2 = Ξ΅ 2 m n β€ Ξ΅ 2 , mn\eta^2=\frac{\varepsilon^2}{mn}\le \varepsilon^2, mn Ξ· 2 = mn Ξ΅ 2 β β€ Ξ΅ 2 , because m , n β N m,n\in\mathbb{N} m , n β N m , n β₯ 1 m,n\ge 1 m , n β₯ 1
β j = 1 m R j ( h ) 2 < Ξ΅ 2 β r = 1 n h r 2 . \sum_{j=1}^m R_j(h)^2<\varepsilon^2\sum_{r=1}^n h_r^2. j = 1 β m β R j β ( h ) 2 < Ξ΅ 2 r = 1 β n β h r 2 β . This is exactly the differentiability condition from Differentiability at a Point and Jacobian Matrix for Maps Between Euclidean Spaces . Therefore f f f a a a a β U a\in U a β U f f f U U U