0 such that f (Nδ(p

Definition Let X, Y be metric spaces, E ⊂ X, p ∈ E and f : E → Y. f is said to becontinuous at p if

x∈E, x→plim f (x) = f (p),

⇐⇒ ∀  > 0, ∃ δ > 0 such that if x ∈ N_δ(p) ∩ E, then f (x) ∈ N_(f (p)) ∩ f (E),

⇐⇒ ∀  > 0, ∃ δ > 0 such that f (N_δ(p) ∩ E) ⊂ N_(f (p)) ∩ f (E).

(a) Theorem Let X, Y be metric spaces and f : X → Y. Then f is continuous if and only if f⁻¹(V ) = {x ∈ X | f (x) ∈ V } is open in X for every open set V ⊂ Y.

(b) Theorem Let X, Y be metric spaces and f : X → Y. Then f is continuous if and only if f⁻¹(C) = {x ∈ X | f (x) ∈ C} is closed in X for every closed set C ⊂ Y.

(c) Theorem Let X be a compact metric space, Y be metric spaces. Suppose f : X → Y is a continuous mapping. Then f (X) is compact.

(d) Theorem Let X be a compact metric space, Y be metric spaces. Suppose f : X → Y is a continuous mapping. Then f is uniformly continuous on X, i.e. ∀  > 0, ∃ δ > 0 such that if p, q ∈ X satisfying dX(p, q) < δ then dY(f (p), f (q)) < .

(e) Theorem Let X be a compact metric space, Y be metric spaces. Suppose f : X → Y is a continuous 1 − 1 onto mapping. Then the inverse mapping f⁻¹ : Y → X defined by

For ech y ∈ Y, f⁻¹(y) = x ⇐⇒ ∃! x ∈ X such that f (x) = y is continuous.

Definition Let X be a metric space. Two sets A, B ⊂ X are said to be separated if A ∩ ¯B = ∅ and ¯A ∩ B = ∅.

A set E ⊂ X is said to beconnected if E is not a union of two nonempty separated sets, i.e.

6 ∃ A 6= ∅, B 6= ∅ such that E = A ∪ B, A ∩ ¯B = ∅ and ¯A ∩ B = ∅.

(a) Let E be a connected subset of a metric space X. If E ⊂ F ⊂ ¯E, then F is also connected.

Proof Suppose F is disconnected and F = C ∪ D is a separation of F, i.e.

C 6= ∅, D 6= ∅ and C ∩ ¯D = ∅, ¯C ∩ D = ∅.

Since E is connected, E ⊂ F = C ∪ D and E = E ∩ C
∪ E ∩ D
, either E ∩ C = ∅ or E ∩ D = ∅, i.e. the set E must lie entirely in either C or D. Suppose E ⊂ C. Then ¯E ⊂ ¯C and

D = F ∩ D ⊂ ¯E ∩ D ⊂ ¯C ∩ D = ∅ =⇒ D = ∅.

This contradicts to the fact that D 6= ∅. Hence, F is connected.

(b) Theorem 2.47 A subset E ⊂ R is connected if and only if it is an interval in one of the following forms: (a, b), [a, b), [a, b) or [a, b].

Theorem 4.22 Let X, Y be metric spaces and let f : X → Y be a continuous mapping. If E is a coonected subset of X, then f (E) is connected.

Theorem 4.23 Let f : [a, b] → R be continuous. If f (a) < f (b) and if c is a number such that f (a) < c < f (b), then there exists x ∈ (a, b) such that f (x) = c.

Definition Let f have domain E in Rⁿ and range in R^m, and let p be an interior point of E.

We say that f is differentiable at p if there exists a linear function L : Rⁿ → R^m such that for every  > 0 there exists δ() > 0 such that if x ∈ Rⁿ is any vector satisfying kx − pk ≤ δ(), then x ∈ E and

kf (x) − f (p) − L(x − p)k ≤  kx − pk.

Equivalently, we say that f is differentiable at p if there exists an m × n matrix L : Rⁿ → R^m such that

x→plim

kf (x) − f (p) − L(x − p)k

kx − pk = lim

kx−pk→0

kf (x) − f (p) − L(x − p)k

kx − pk = 0,

i.e., the function f (x) − f (p) − L(x − p) = o(kx − pk), or we say that f (x) − f (p) − L(x − p) is of a little o of kx − pk.

Remarks

(a) If L₁, L₂ are linear functions such that f (x) − f (p) − L₁(x − p) = o(kx − pk) and f (x) − f (p) − L₂(x − p) = o(kx − pk), then L₁ = L₂.

Proof Observe that

x → p ⇐⇒ t → 0

and if L : Rⁿ → R^m is an m × n matrix and let ei denote the unit vector in the positive i−th coordinate of Rⁿ then

L(e_i) = L_m×n·













 0

... 1 ... 0















n×1

=Li1 L_i2 · · · L_in

1×n= the i-th row of L.

Since

0 ≤ lim

x→p

k L1− L2
(x − p)k kx − pk

= lim

x→p

kf (x) − f (p) − L₂(x − p) − f (x) − f (p) − L₁(x − p)
k kx − pk

≤ lim

x→p

kf (x) − f (p) − L₂(x − p)k

kx − pk + lim

x→p

kf (x) − f (p) − L₁(x − p)k kx − pk

= 0,

we get L₁ = L₂ by taking x − p = te_i and the fact that 0 = lim

t→0

k L₁− L₂
(tei)k

kte_ik = k L₁− L₂
(e_i)k which implies that the i-th row of L₁ is equal to the i-th row of L₂.

(b) If L exists at p ∈ E, then it is called the derivative of f at p, denoted by Df (p) or f⁰(p).

Note that the linear function Df (p) : Rⁿ → R^m is defined by

Df (p)(x − p) =



















∂f₁

∂x1

(p) ∂f₁

∂x2

(p) · · · ∂f₁

∂xn

(p)

∂f₂

∂x₁(p) ∂f₂

∂x₂(p) · · · ∂f₂

∂x_n(p)

· · · ·

∂f_m

∂x₁(p) ∂f_m

∂x₂(p) · · · ∂f_m

∂x_n(p)



















m×n











x₁− p₁ x2− p2

... x_n− p_n











n×1

(c) When m = n = 1, f is differentiable at p ∈ (a, b) ⊂ R if there exists L ∈ R such that

x→plim

|f (x) − f (p) − L(x − p)|

|x − p| = lim

x→p

f (x) − f (p) − L(x − p) x − p

= 0

⇐⇒ lim

x→p

f (x) − f (p) − L(x − p)

x − p = 0

⇐⇒ lim

x→p

f (x) − f (p) x − p = L, and the deirivative of f at p is defined to be f⁰(p) = L.

(d) Definition A function f : [a, b] → R is differentiable at x ∈ [a, b] if this limit exists:

f⁰(x) ≡ lim

t→x

f (t) − f (x) t − x .

Questions

(a) If f is continuous on [a, b], is f (x) differentiable? No, consider the absolute value function.

(b) If f is differentiable on [a, b], is f continuous? Yes. Notice:

f (t) − f (x)− = f (t) − f (x)

t − x (t − x) = f⁰(x) · 0 = 0, as t → x.

(c) If f is differentiable on [a, b], is f⁰ continuous? No, consider f (x) = x 4 3 sin(1

x) if x 6= 0 and 0 if x = 0.

Remarks

(a) f⁰ always satisfies the intermediate value property and has no simple discontinuities, if it exists.

(b) Since f⁰ is a limit, f⁰ satisfies sum, product and quotient rules.

Theorems

(a) (f g)⁰ = f⁰g + f g⁰.

Proof Let h = f g. Consider:

h(t) − h(x) = f (t)[g(t) − g(x)] + g(x)[f (t) − f (x)].

Now take t → x after dividing by t − x.

(b) (Mean Value Theorem) If f is continuous on [a, b] and differentiable (a, b) then there exists a point c ∈ (a, b) such that f (b) − f (a) = (b − a)f⁰(c).

(c) (General Mean Value Theorem) If f, g are continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that:

[f (b) − f (a)]g⁰(c) = [g(b) − g(a)]f⁰(c).

Note if g(x) = x, this gives the Mean Value Theorem.

Proof (Idea) Consider h(x) = [f (b) − f (a)]g(x) − [g(b) − g(a)]f (x). Note that h(a) = 0 = h(b). There must be a local min or max between a or b where h⁰(c) = 0. Hence the Generalized Mean Value Theorem follows from Rolle’s Theorem.

Example If f⁰(x) > 0 on (a, b) then f is strictly increasing on (a, b).

Proof Let a < s < t < b. Then f (s) < f (t). In other words, f (t) − f (s) > 0. But f (t) − f (s) = (t − s)f⁰(c) for some c ∈ (a, b). We know f⁰(c) > 0. So f (t) − f (s) > 0.

Remark Book’s proof uses Rolle’s Theorem which says if h : [a, b] → R has a local max at c ∈ (a, b) and h⁰(c) exists and then h⁰(c) = 0.

L’Hospital’s Rule Suppose f, g : (a, b) → R are differentiable and g⁰(x) 6= 0 for every x ∈ (a, b), where −∞ ≤ a < b ≤ ∞. Suppose

x→alim f⁰(x) g⁰(x) = A.

(a) If lim

x→af (x) = 0 and lim

x→ag(x) = 0, (b) or if lim

x→ag(x) = ∞, then

x→alim f (x) g(x) = A.

Proof

Case 1: −∞ < A < ∞ Since

x→alim f⁰(x)

g⁰(x) = A ⇐⇒ ∀  > 0, ∃ δ > 0 such that if x ∈ (a, a + δ) then A −  < f⁰(x)

g⁰(x) < A + 

⇐⇒ ∃ c = a + δ ∈ (a, b) such that if x ∈ (a, c) then A −  < f⁰(x)

g⁰(x) < A + .

For any x, y satisfying a < x < y < c, the generalized Mean Value Theorem shows that

∃ t ∈ (x, y) such that A −  < f (x) − f (y)

g(x) − g(y) = f⁰(t)

g⁰(t) < A + . (1) (a) If lim

x→af (x) = 0 and lim

x→ag(x) = 0, by letting x → a in (1), we get

A −  ≤ f (y)

g(y) = lim

x→a

f (x) − f (y)

g(x) − g(y) = lim

x→a

f⁰(t)

g⁰(t) ≤ A +  for every y ∈ (a, c)

=⇒ A −  ≤ f (y)

g(y) ≤ A +  for every y ∈ (a, c)

=⇒ lim

y→a

f (y) g(y) = A.

(b) Suppose lim

x→ag(x) = ∞, by keeping y fixed,

∃ c₁ ∈ (a, y) such that if a < x < c₁ then











g(x) > 0, g(x) > g(y) 

g(y) g(x)    

< ,    

f (y) g(x)    

< 

Thus g(x) − g(y)

g(x) > 0 for every a < x < c₁, and

(A − ) ·g(x) − g(y)

g(x) < f (x) − f (y)

g(x) − g(y) · g(x) − g(y)

g(x) < (A + ) ·g(x) − g(y) g(x)

=⇒ A −  − (A − )g(y)

g(x) < f (x)

g(x) −f (y)

g(x) < A +  − (A + )g(y) g(x)

=⇒ A −  − (A − )g(y)

g(x) + f (y)

g(x) < f (x)

g(x) < A +  − (A + )g(y)

g(x) +f (y) g(x)

=⇒ A −  − (A − ) −  < f (x)

g(x) < A +  + (A + ) +  for every x ∈ (a, c₁)

=⇒ lim

x→a

f (x) g(x) = A.

Case 2: A = −∞

Since

x→alim f⁰(x)

g⁰(x) = −∞ ⇐⇒ ∀ q > r ∈ R, ∃ δ > 0 such that if x ∈ (a, a + δ) then f⁰(x)

g⁰(x) < r < q

⇐⇒ ∃ c = a + δ ∈ (a, b) such that if x ∈ (a, c) then f⁰(x)

g⁰(x) < r < q.

For any x, y satisfying a < x < y < c, the generalized Mean Value Theorem shows that

∃ t ∈ (x, y) such that f (x) − f (y)

g(x) − g(y) = f⁰(t)

g⁰(t) < r < q. (2)

(a) If lim

x→af (x) = 0 and lim

x→ag(x) = 0, by letting x → a in (2), we get f (y)

g(y) = lim

x→a

f (x) − f (y) g(x) − g(y) = lim

x→a

f⁰(t)

g⁰(t) ≤ r < q for every y ∈ (a, c)

=⇒ f (y)

g(y) ≤ r < q for every y ∈ (a, c)

=⇒ lim

y→a

f (y)

g(y) ≤ r < q for arbitrary q ∈ R

=⇒ lim

y→a

f (y)

g(y) = −∞.

(b) Suppose lim

x→ag(x) = ∞, by keeping y fixed, choose  > 0 such that r + |r| +  < q ⇐⇒

0 <  < q − r 1 + |r|,

∃ c₁ ∈ (a, y) such that if a < x < c₁ then











g(x) > 0, g(x) > g(y) 

g(y) g(x)    

< ,    

f (y) g(x)    

< 

Thus g(x) − g(y)

g(x) > 0 for every a < x < c₁, and

f (x) − f (y)

g(x) − g(y) · g(x) − g(y)

g(x) < r · g(x) − g(y) g(x)

=⇒ f (x)

g(x) −f (y)

g(x) < r −rg(y) g(x)

=⇒ f (x)

g(x) < r −rg(y)

g(x) +f (y) g(x)

=⇒ f (x)

g(x) < r + |r| +  < q for every x ∈ (a, c₁)

=⇒ lim

x→a

f (x)

g(x) ≤ q for arbitrary q ∈ R

=⇒ lim

x→a

f (x)

g(x) = −∞.

Case 3: A = ∞ Since

x→alim f⁰(x)

g⁰(x) = ∞ ⇐⇒ ∀ p < r ∈ R, ∃ δ > 0 such that if x ∈ (a, a + δ) then f⁰(x)

g⁰(x) > r > p

⇐⇒ ∃ c = a + δ ∈ (a, b) such that if x ∈ (a, c) then f⁰(x)

g⁰(x) > r > p.

For any x, y satisfying a < x < y < c, the generalized Mean Value Theorem shows that

∃ t ∈ (x, y) such that f (x) − f (y)

g(x) − g(y) = f⁰(t)

g⁰(t) > r > p. (3)

(a) If lim

x→af (x) = 0 and lim

x→ag(x) = 0, by letting x → a in (3), we get f (y)

g(y) = lim

x→a

f (x) − f (y) g(x) − g(y) = lim

x→a

f⁰(t)

g⁰(t) ≥ r > p for every y ∈ (a, c)

=⇒ f (y)

g(y) ≥ r > p for every y ∈ (a, c)

=⇒ lim

y→a

f (y)

g(y) ≥ r > p for arbitrary p ∈ R

=⇒ lim

y→a

f (y) g(y) = ∞.

(b) Suppose lim

x→ag(x) = ∞, by keeping y fixed, choose  > 0 such that r − |r| −  > p ⇐⇒

0 <  < r − p 1 + |r|,

∃ c₁ ∈ (a, y) such that if a < x < c₁ then











g(x) > 0, g(x) > g(y) 

g(y) g(x)    

< ,    

f (y) g(x)    

< 

Thus g(x) − g(y)

g(x) > 0 for every a < x < c₁, and

f (x) − f (y)

g(x) − g(y) ·g(x) − g(y)

g(x) > r ·g(x) − g(y) g(x)

=⇒ f (x)

g(x) − f (y)

g(x) > r − rg(y) g(x)

=⇒ f (x)

g(x) > r − rg(y)

g(x) + f (y) g(x)

=⇒ f (x)

g(x) > r − |r| −  > p for every x ∈ (a, c₁)

=⇒ lim

x→a

f (x)

g(x) ≥ p for arbitrary p ∈ R

=⇒ lim

x→a

f (x) g(x) = ∞.

Question Let f : R → R be differentiable to any order (i.e., f is class C^∞ or smooth). How can we approximate f near the point x = a?

Let P₀(x) = f (a). Can we find a polynomial P₁(x) such that P₁(a) = f (a) and moreover P₁⁰(a) = f⁰(a). In other words, they agree to first order, et cetera. Then let:

P₁(x) = f (a) + f⁰(a)(x − a),

P₁(x) = f (a) + f⁰(a)(x − a) + f⁰⁰(a)(x − a)²

2 .

Definition The n-th order Taylor polynomial of f (x) centered at x = a is given by:

P_n(a, x) = P_n(x) =

n

X

k=0

f^(k)(a)(x − a)^k

k! .

Taylor’s Theorem For C^∞, f : R → R, then:

f (x) = P_n(x) + f⁽ⁿ⁺¹⁾(c)(x − c)⁽ⁿ⁺¹⁾ (n + 1)!

for some c between x and a for a particular x.

Remarks

(a) The error is E_n(x) = f (x) − P_n(x).

(b) When n = 0,

f (x) = f (a) + f⁰(c)(x − a)

for some c between a and x at a particular x. Notice that this is equivalent to the Mean-Value Theorem. Thus Taylor’s Theorem is a generalization of Mean-Value Theorem.

Proof Let k ∈ R be defined by

f (b) − P(n−1)(a, b) = k(b − a)ⁿ

n! . (4)

Our goal is to show k = f⁽ⁿ⁾(c) for some c ∈ (a, b). Let:

g(x) = f (b) − P_(n−1)(x, b) − k(b − x)ⁿ n! .

Then g(a) = 0 using (1). But it also holds that g(b) = 0 because P (b, b) = f (b). Then there exists c ∈ (a, b) such that g⁰(c) = 0. Notice that:

g⁰(x = −P_(n−1)⁰ (x, b) + k(b − x)ⁿ⁻¹ (n − 1)!

Notice that:

P₍n − 1)(x, b) =

n−1

X

k=0

f^(k)(x)(b − x)^k k!

Then it follows that:

P_(n−1)⁰ (x, b) = f⁰(x) − f⁰(x) + f⁰⁰(x)(b − x) − f⁰⁰(x)(x − b) + · · · + f⁽ⁿ⁾(x)(b − x)⁽ⁿ⁻¹⁾ (n − 1)!

At x = c, then

0 = g⁰(c) = [−fⁿ(c) + k](b − c)ⁿ⁻¹ n − 1! . Since the term on the right is not zero, our conclusion follows.

Question How about derivatives of complex functions?

Example Let f : R → C given by f (x) = e^ix = cos x + i sin x. Consider f on [0, 2π]. Notice f (0) = 1 = f (2π). So f (2π) − f (0) = 0. The mean-value theorem would imply there is a point in (0, 2π) with derivative zero. But f⁰(x) = ie^ix and moreover |f⁰(x)| = 1. Thus the mean-value theorem does not hold for complex differentiation. Similarly, L’hˆopital’s rule does not hold.

Theorem Let f : [a, b] → R^k be continuous and differentiable on (a, b). Then exists c ∈ (a, b) such that |f (b) − f (a)| ≤ (b − a)|f⁰(c)|.

Inverse Function Theorem Suppose that

U ⊆ Rⁿ is open, f : U → Rⁿ is C¹, x₀ ∈ U,

dfx0 : Rⁿ → Rⁿ is invertible.

Then there exists

a neighborhood V of x₀ in U a neighborhood W of f (x₀) in Rⁿ x₀ ∈ U,

dfx0 : Rⁿ → Rⁿ is invertible.

such that f has a C¹ inverse g = f⁻¹ : W → V such that

f (g(y)) = y ∀ y ∈ W and g(f (x)) = x ∀ x ∈ V.

Moreover

 ∂g_i

∂y_j(y)



1≤i,j≤n

= dgy = (df_g(y))⁻¹ = ∂f_i

∂x_j(g(y))

−1 1≤i,j≤n

∀ y ∈ W and g is smooth whenever f is smooth.

Remarks

(a) The theorem says that a continuously differentiable function f between regions in Rⁿ is locally invertible near points where its differential is invertible,

i.e.

W = {y = (y1(x), . . . , yn(x)) = (f1(x), . . . , fn(x)) = f (x) | x ∈ V }, V = {x = (x₁(y), . . . , x_n(y)) = (g₁(y), . . . , g_n(y)) = g(y) | y ∈ W }, and each coordinate function is continuously differentiable.

(b) Let 0 < a < 1, define f : R → R by

f (x) =







ax + x²sin1

x if x 6= 0

0 if x = 0

.

Compute f⁰(x) for all x ∈ R. Show that f⁰(0) > 0, yet f is not one-to-one in any neighbor- hood of 0.

Hint Note that

(i) there exists a sequence of points {x_n → 0} at which f⁰(x_n) = 0, and f⁰⁰(x_n) 6= 0, (ii) if f⁰(p) = 0, f⁰⁰(p) 6= 0, then f cannot be one-to-one near p.

Remark This example shows that in the Inverse Function Theorem, the hypothesis that f is C¹ cannot be weaken to the hypothesis that f is differentiable.

(c) Define f : R² → R² by

f (x, y) = (e^xcos y, e^xsin y).

Show that f is C¹ and df_(x,y) is invertible for all (x, y) ∈ R² and yet f is not one-to-one function globally. Why doesn’t this contradict the Inverse Function Theorem?

Example Use Inverse Function Theorem to determine whether the system u(x, y, z) = x + xyz

v(x, y, z) = y + xy w(x, y, z) = z + 2x + 3z² can be solved for x, y, z in terms of u, v, w near p = (0, 0, 0).

Solution Set F (x, y, z) = (u, v, w). Then

DF (p) =





u_x u_y u_z vx vy vz

w_x w_y w_z



(p) =





1 + yz xz xy

y 1 + x 0

2 0 1 + 6z



(p) =





1 0 0 0 1 0 2 0 1





and 

1 0 0 0 1 0 2 0 1      

= 1 6= 0.

By the Inverse Function Theorem, the inverse F⁻¹(u, v, w) exists near p = (0, 0, 0), i.e. we can solve x, y, z in terms of u, v, w near p = (0, 0, 0).

Implicit Function Theorem Let

U ⊆ R^m+n≡ R^m× Rⁿ be an open set, f = (f1, . . . , fn) : U → Rⁿ be a C¹ function, (a, b) ∈ U be a point at which f (a, b) = 0, the n × n matrix d_yf |_(a,b)= ∂f_i

∂y_j(a, b)



1≤i,j≤n

is invertible.

Then there exists

a neighborhood V of (a, b) in U, a neighborhood W of a in R^m, and a C¹ function g : W → Rⁿ

such that

{(x, y) ∈ V ⊂ R^m× Rⁿ| f (x, y) = 0}

= {(x, g(x)) | x ∈ W }

= the graph of g over W.

i.e. Letting

S = {(x, y) ∈ R^m× Rⁿ | f (x, y) = 0}

denote the solution set, then S ∩ V is of the dimension m, and it is equal to graph(g), the graph of a C¹ function g, over W . Moreover

dg_x = − (d_yf )⁻¹|_(x,g(x)) · d_xf |_(x,g(x)) and g is smooth whenever f is smooth.

Example Let F : R⁴ → R² be given by

F (x, y, z, w) = (G(x, y, z, w), H(x, y, z, w)) = (y²+ w²− 2xz, y³+ w³+ x³− z³), and let p = (1, −1, 1, 1).

(a) Show that we can solve F (x, y, z, w) = (0, 0) for (x, z) in terms of (y, w) near (−1, 1).

Solution Since

DF (p) =G_x G_y G_z G_w H_x H_y H_z H_w



(p) = −2 −2 −2 2

3 3 −3 3



and 

G_x G_z H_x H_z    

(p) =    

−2 −2 3 −3    

= 12 6= 0,

we can write (x, z) in terms of (y, w) near (−1, 1) by Implicit Function Theorem.

(b) If (x, z) = Φ(y, w) is the solution in part (a), show that DΦ(−1, 1) is given by the matrix

−−2 −2 3 −3

−1

−2 2

3 3



=−1 0

0 1



Solution The Implicit Function Theorem implies that, near p,the solution set {(x, y, z, w) | F (x, y, z, w) = (0, 0)}

can be represented as the graph of (x, z) = Φ(y, w) near (−1, 1).

Hence, we have

∂F

∂y = (0, 0) and ∂F

∂w = (0, 0) near (−1, 1).

Therefore

0 = G_x∂x

∂y + G_y+ G_z∂z

∂y and 0 = G_x∂x

∂w + G_z∂z

∂w + G_w

which implies that

−[G_y, G_w] = [G_x, G_z]







∂x

∂y

∂x

∂w

∂z

∂y

∂z

∂w





. Similarly, we have

−[H_y, H_w] = [H_x, H_z]







∂x

∂y

∂x

∂w

∂z

∂y

∂z

∂w





. Thus, we have

−G_y Gw

H_y H_w



=G_x Gz

H_x H_z









∂x

∂y

∂x

∂w

∂z

∂y

∂z

∂w





, or

DΦ =







∂x

∂y

∂x

∂z ∂w

∂y

∂z

∂w





= −G_x Gz

H_x H_z

−1

G_y Gw

H_y H_w

 .

Hence, DΦ(−1, 1) is given by the matrix

−−2 −2 3 −3

⁻¹

−2 2

3 3



=−1 0

0 1

 .

Remarks

(a) The theorem says that if

f : U ⊂ R^m+n → Rⁿ

is a map of the class C¹(U ), (a, b) is a point in U, and dyf |(a,b) = h∂f_i

∂y_j(a, b) i

1≤i,j≤n is invertible, then, locally, the solution set

S = {(x, y) ∈ U | f (x, y) = f (a, b)} is a C¹ “manifold” of dimension m.

(b) Let

f : U ⊂ R^m+n → Rⁿ

be a map of the class C¹ on an open subset U of R^(m+n), and assume that the differential df_p = ∂f_i

∂x_j



1≤i≤n;1≤j≤(m+n)

is of the constant rank k at each p ∈ U.

By relabeling, if necessary, we may assume that

f = (fˆ ₁, . . . , f_k) : U ⊂ R^m+n→ R^k is a map with its differential

d ˆf_p = ∂f_i

∂x_j



1≤i≤k;1≤j≤(m+n)

of rank k.

The Implicit Function Theorem says that locally the solution set

S = {x ∈ U | ˆf (x) = 0} is a C¹ “manifold” of dimension (m + n) − k.

Identify S with an open neighborhood W ⊂ R^(m+n)−k of 0 ∈ R^(m+n)−k and identify U with V × W, such that we write each x ∈ V × W ⊂ R^(m+n)as x = (¯x, ˆx), where ¯x = (¯x1, . . . , ¯xk) ∈ V, and ˆx = (ˆx_k+1, . . . , ˆx_m+n) ∈ W. This implies

f (0, ˆˆ x) = 0 and

"

∂ ˆf_i

∂ ¯x_j

#

1≤i,j≤k

is invertible on V × W.

Using the Inverse Function Theorem, we may show that the range

f (V × W ) = f (U ) locally is a k-dimensional “manifold”.ˆ

(c) Two mappings are said to be locally equivalent if under suitable choices of local coordinate systems in the domain and range spaces (with origin at 0) they can be written by the same formulas. The Implicit Function Theorem says that if

f, g : U ⊂ R^m+n → Rⁿ

are maps of the class C¹(U ) , p is a point in U, and rk[df (p)] = rk[dg(p)] = n, then f and g are equivalent, i.e. there exists diffeomorphisms

h : R^m+n → R^m+n and k : Rⁿ → Rⁿ, such that

f ◦ h = k ◦ g.

Definitions

(a) A map f : U ⊂ R^m → Rⁿ is called a Lipschitz map on U if there exists a constant C ≥ 0 such that

kf (x) − f (y)k ≤ Ckx − yk for all x, y ∈ U.

If one can choose a (Lipschitz) constant C < 1 such that the above Lipschitz condition hold on U, then f is called a contraction map.

Example Let U be a convex subset of R^m, and f : U ⊂ R^m→ Rⁿ be a map with bounded kdf k = sup kdf_p(x)k_Rⁿ

kxk_R^m | p ∈ U, x 6= 0

 .

For each i = 1, . . . , m, by setting x = e_i, the i^th unit vector of R^m, we get k∇f_ik ≤ kdf k for each i = 1, . . . , m

which implies that f is Lipscitz on U.

(b) Let U be a subset of R^m, and f be a map that maps U into U, i.e. f : U → U. A point p ∈ U is said to be a fixed point of f if f (p) = p.

Fixed Point Theorem for Contractions Let f : R^m → R^m be a contraction map. Then f has a fixed point.

Note that R^m is complete which implies that any Cauchy sequence (is bounded and has a limiting point by Bolzano-Weierstrass theorem, and Cauchy condition implies that it) converges.