advanced calculus (ii) - ncku

Advanced Calculus (II)

WEN-CHING LIEN

Department of MathematicsNational Cheng Kung University

WEN-CHING LIEN Advanced Calculus (II)

Ch11: Differentiability on Rn

Ch11.6: Inverse Function Theorem

Notation:f : Rn → Rn,The Jacobian of f is

∆f (a) := det(Df (a)).

Lemma (11.38)Let V be open in Rn, f : V → Rn, a ∈ V, and r > 0 be sosmall that Br (a) ⊂ V. Suppose that f is continuous and1-1 on Br (a), and its first-order partial derivatives exist atevery point in Br (a). If ∆f 6= 0 on Br (a), then there is aρ > 0 such that Bρ(f (a)) ⊂ f (Br (a)).

Theorem (11.39)Let V be open and nonempty in Rn, and f : V → Rn becontinuous. If f is 1-1 and has first-order partialderivatives on V , and if ∆f 6= 0 on V , then f−1 iscontinuous on f (V ).

Lemma (11.40)

Let V be open in Rn and f : V → Rn be C1 on V. If∆f (a) 6= 0 for some a ∈ V, then there is an r > 0 such thatBr (a) ⊂ V, f is 1-1 on Br (a), ∆f (x) 6= 0 for all x ∈ Br (a),and

det[∂fi∂xj

]n×n6= 0

for all c1, . . . ,cn ∈ Br (a).

Theorem (11.41 Inverse Function Theorem)

Let V be open in Rn and f : V → Rn be C1 on V. If∆f (a) 6= 0 for some a ∈ V, then there exists an open setW containing a such that(i) f is 1-1 on W,

(ii) f−1 is C1 on f (W ), and

(iii) for each y ∈ f (W ),

D(f−1)(y) = [Df (f−1(y))]−1,

where [ ]−1 represents matrix inversion (see TheoremC.5).

D(f−1)(y) = [Df (f−1(y))]−1,

Remark (11.42)The hypothesis ”∆f 6= 0 in Theorem 11.39 can be relaxed.

Proof.If f (x) = x3, then f : R→ R and its inverse f−1(x) = 3

are continuous on R, but ∆f (0) = f ′(0) = 0

Remark (11.42)The hypothesis ”∆f 6= 0 in Theorem 11.39 can be relaxed.

Proof.If f (x) = x3, then f : R→ R and its inverse f−1(x) = 3

are continuous on R, but ∆f (0) = f ′(0) = 0

Remark (11.43)The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot berelaxed. In fact, if f : Br (a)→ Rn is differentiable at a andits inverse f−1 exists and is differentiable at f (a), then∆f (a) 6= 0.

Proof.Suppose to the contrary that f is differentiable at a but∆f (a) = 0. By Exercise 8, p.338, and the Chain Rule,

I = D(f−1 ◦ f )(a) = D(f−1)(f (a))Df (a).

Taking the determinant of this identity, we have

1 = ∆f−1(f (a))∆f (a) = 0,

a contradiction.

I = D(f−1 ◦ f )(a) = D(f−1)(f (a))Df (a).

1 = ∆f−1(f (a))∆f (a) = 0,

a contradiction.

I = D(f−1 ◦ f )(a) = D(f−1)(f (a))Df (a).

1 = ∆f−1(f (a))∆f (a) = 0,

a contradiction.

I = D(f−1 ◦ f )(a) = D(f−1)(f (a))Df (a).

1 = ∆f−1(f (a))∆f (a) = 0,

a contradiction.

I = D(f−1 ◦ f )(a) = D(f−1)(f (a))Df (a).

1 = ∆f−1(f (a))∆f (a) = 0,

a contradiction.

I = D(f−1 ◦ f )(a) = D(f−1)(f (a))Df (a).

1 = ∆f−1(f (a))∆f (a) = 0,

a contradiction.

Remark (11.44)

The hypothesis ”f is C1 on V ” in Theorem 11.41 cannot berelaxed.

Proof.If f (x) = x + 2x2 sin 1

x , x 6= 0, and f (0) = 0, then f : R→ Ris differentiable on V := (−1,1) and f ′(0) = 1 6= 0.However, since

2(4k − 1)π

(4k + 1)π

(4k − 3)π

)for k ∈ N, f is not 1-1 on any open set that contains 0.Therefore, no open subset of f (V ) can be closen onwhich f−1 exists.

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Remark (11.44)

2(4k − 1)π

(4k + 1)π

(4k − 3)π

Theorem (11.47 The Implicit Function Theorem)Suppose that V is open in Rn+p, and F = (F1, . . . ,Fn) :V → Rn is C1 on V. Suppose further that F (x0, t0) = 0 forsome (x0, t0) ∈ V, where x0 ∈ Rn and t0 ∈ Rp. If

∂(F1, . . . ,Fn)

∂(x1, . . . , xn)(x0, t0) 6= 0,

then there is an open set W ⊂ Rp containing t0 and aunique continuously differentiable function g : W → Rn

such that g(t0) = x0, and F (g(t), t) = 0 for all t ∈W.

Example (11.49)

Prove that there exist function u, v : R4 → R, continuouslydifferentiable on some ball B centered at the point(x , y , z,w) = (2,1,−1,−2), such thatu(2,1,−1,−2) = 4, v(2,1,−1,−2) = 3, and theequations

u2 + v2 + w2 = 29,u2

x2 +v2

y2 +w2

z2 = 17

both hold for all (x , y , z,w) in B.

Proof.Set n = 2, p = 4, and

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

This determinant is nonzero when u = 4, v = 3, x = 2,and y = 1. Therefore, such functions u, v exist by theImplicit Function Theorem.

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

F (u, v , x , y , z,w) = (u2 +v2 +w2−29,u2

x2 +v2

y2 +w2

z2 −17).

Then F (4,3,2,1,−1,−2) = (0,0), and

∂(F1,F2)

∂(u, v)= det

[2u 2v2ux2

]= 4uv

(1y2 −

Thank you.

advanced calculus (ii) - ncku

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