5) Contraction mappings and fixed point theorems

Question 1

fixed point

Answer 1

suppose that (X, d) is
a metric space and f : X → X a self-map. Then a point x ∈ X is called fixed point of f
if f(x) = x. For example the function cos defines a self-map on the interval [0, 1], and by
starting with x1 = 0 and inductively computing xn+1 = cos xn one converges to the value
roughly 0.739085 which is a fixed point of cos, i.e. solves the equation cos(x) = x

Question 2

previously discussed Stone Weirstrass theorem

Answer 2

That a collection, space of continuous functions on a compact set will be separable-
proof uses that we can approx any continuous functions by a polynomial

And any polynomial with real coefficients can be nicely approximated by poly with rational coefss as rationals approximate reals

There are a countable number of polynomials with rational coefss,
so we have a countable everywhere dense set which was required for separability of this space

Question 3

Definition 5.1 (Contraction Mapping).

Answer 3

Let (X, d) be a metric space. Then a map f : X → X is called contraction if there exists a constant C < 1 such that d(f(x), f(y)) ≤ Cd(x, y).
we have a strict inequality here

(clearly any contraction map will be uniformly continuous- if not contunous then cant be a contraction!(disprove))

Question 4

(clearly any contraction map will be uniformly continuous)

Answer 4

For all x and for all ε there exists a δ_ε>0 s.t. d(x,y) <δ_ε implies d(f(x),f(y))<ε

A contraction is a special kind of uniformly continuous maps, here we take δ = Cε or ε

Question 5

contraction mapping idea

Answer 5

We look for a fixed point x s.t f(x)=x

E.g for cos x sketch y=x and take say x_1=1, cos(1) then use x_2=cos(2) to find points (spiral formed until fixed point is reached).

x~= lim n tends to infinity of x_n = lim n tends to infinity of x_n+1 = lim as n tends to infinity of cos(x_n) = cos (x~)

sometimes we might have the function dropping much faster, the similar procedure wouldnt work, spirals out and fixed point doesnt exist, there is no convergence. Here what is different is that the derivative |f’| > 1 here

Slope tells us we will have con traction of argument or not

Question 6

contraction mapping
y=ax+b on a:

Answer 6

consider y=x and line y=ax+b. If |a|=1 we will have a closed square
|a|>1 divergence no fixed point
|a|<1 fixed point

y_1=ax_1+b
y_2=ax_2+b
|y_2-y_1| = |a||x_1-x_2|

Question 7

Theorem 5.2 (Banach Fixed Point Theorem)

Answer 7

Suppose that f : X → X is a contraction on a complete metric space (X, d). Then f has a unique fixed point y.

Moreover, for any x ∈ X the sequence (x_n) defined recursively by x_{n+1} = f(x_n), x_1 = x,
converges to y.

(we shrink towards the point using this recursive relation)

Question 8

e.g. unit circle rotated by pi/2

Answer 8

consider unit circle
if we rotate circle by an angle no point will remain in place
But the distance between the mapped points is preserved
(we will say not bigger than)
d(f(x),f(y)) ≤ d(x,y)

Question 9

Theorem 5.2 (Banach Fixed Point Theorem) PROOF

Answer 9

steps: **
Existence of fixed point
Uniqueness of fixed point
this fixed point can be obtained by result of approximation by iterative procedure
**
1)
Assume we have two fixed points x and y. But d(f(X),f(y) = d(x,y) ≤ Cd(x,y)
but since C ≤ 1 we must have d(X,Y) = 0 otherwise inequality not possible, so points coincide.
2)
if f is a contraction we have uniform continuity, in particular its continuous. If we have a convergent sequence x_{n+1} = f(x_n) is convergent to x~

Then x = (limn→ ∞)(x_n)
= (limn→ ∞)(x_{n+1})
by continuity
= = (limn→ ∞)(f(x_n))
=f( = (limn→ ∞)(x_n))
=f(x~)

Why x_n is cauchy?

using series 1+q+q^2+…. = (1-q^(n+1)/(1-q)
the limit of partial sums up to n will exist if |q|<1
limit as n tends to infinity of this will be 1/(1-q) if |q|<1

wlog m>n then m=n+k
d(x_n,x_m) ≤d(x_n,x_(n+1)) + ….d(x_m-1,x_m)
by induction
As d(x_{n+1},x_{n+2}
= d(f(x_n),f(x_{n+1}) ≤Cd(x_n, x_{n+1)}

….
d(x_{n+k}, x_{n+k+1}) ≤ C^k d(x_n, x_{n+1})

By triangle inequality
d(x_{n+k},x_{n+k+1})
≤ d(x_n, x_{n+1)(1 +C + C^2 +…C^k)
≤ C^{n-1}d(x_1, x_2)(1 +C + C^2 +…C^k)
≤ (C(n-1) / (1-c) d(x_1,x_2)< ε

Ensure this happens by:
Taking ε such that
C^{n-1} <
[ ε(1-C)]/[d(x_1,x_2)]
(inequality swaps as logC<0 since C< 1)
n-1 > log([ ε(1-C)]/[d(x_1,x_2)]) /log(C)

N> [log([ ε(1-C)]/[d(x_1,x_2)]) /log(C) ] + 1

Question 10

Corollary 5.3. composition of map several times that wasn’t originally a contraction

Answer 10

Suppose that (X, d) is a complete metric space and f : X → X a map
such that f^ n is a contraction for some n ∈ N. Then f has a unique fixed point.

proof: Since f^n is a contraction it has a unique fixed point x ∈ X, i.e.
f ◦ f . . . ◦ f(x) (n−times) = x.
Now note that
f^n(f(x)) = f^n ◦ f(x) = f^{n+1}(x) = f ◦ f^n(x) = f(f^n(x)) = f(x)
and therefore f(x) is also a fixed point of f^n . By uniqueness we must have f(x) = x

Question 11

Jacobian

Answer 11

We will sue this to verify if a map is a contraction for several variables
Partial derivatives matrix
f: R^n to R^m
( ∂f_1/ ∂x_1…….. ∂f_1/ ∂x_m)
(………………………)
( ∂f_m/ ∂x_1…….. ∂f_m/ ∂x_m)

square matrix and compute the determinant

∥∂f(x)∥≤∥∂f∥*∥x∥
j
(here norm as a linear operator)

Question 12

Theorem 5.4 (Mean Value Inequality).

Used for contractions

Answer 12

Suppose that U ⊂ R^m is an open set with
convex closure overline(U) and let f : overline(U) → R^m be a C^1
-function. Let df be the total derivative
(or Jacobian) understood as a function on U with values in m × m-matrices. Suppose that ∥df(x)∥ ≤ M for all x ∈ U. Then f : U → R
m satisfies
∥f(x) − f(y)∥ ≤ M∥x − y∥
for all x, y ∈ overline(U)

Question 13

The Mean Value Theorem for Integrals

Answer 13

he Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if f(x) is continuous, a point
c exists in [a,b] such that the value of the function at c is equal to the average value off(x) over [a,b]
ie there exists a c in [a,b] s.t
f(c) = (1/(b-a)) . ∫_[a,b] f(x).dx

*since continuous on[a,b]closed interval is bounded
m<f(x)<M
m(b-a)<∫_[a,b] f(x).dx< M(b-a)
*by IVT there is a c s.t f(c) =z for f(z)< z< f(b)

Question 14

norm as linear operator

Answer 14

∥∂ f(x)∥ ≤ ∥∂ f∥ ∥x∥

Question 15

convex domain assumed

convex

Answer 15

for two points in our interval joined by straight line this interval is in our domain

γ(t) = tx + (1 − t)y

d(f(x),f(y)) can be estimated along this interval which reduce our number of variables to 1

Question 16

PROOF OF MVI

Answer 16

Using convex domain, for x,y in U we have γ(t) = tx + (1 − t)y
γ((1)=x and γ(0)=y
(d/dt)γ(t) = x − y.

f(x)-f(y) = ∫[0,1] d/dt(f(γ(t)).dt
= ∫0,1 . (dγ/dt)(t).dt
ftoc
by MVT

using the triangle inequality (useful for Riemann integrals as they are limits of finite sums)
df is the jacobian
so f(x)-f(y) is the integral of product of jacobian and (x-y) (x-y is the direction of the line prev)

∥f(x) − f(y)∥ ≤
∫[0,1] ∥(df).(dγ/dt)(t) ∥.dt
≤
M*
∫[0,1] ∥x-y ∥.dt = M∥x-y ∥

Question 17

triangle inequality on a sum using norms
∥sum of x_k ∥

applied to integrals

Answer 17

∥sum of x_k ∥
≤ sum of ∥x_k∥

∥ ∫f(t).dt∥ ≤ ∫ ∥ f(t).dt∥

Question 18

example 5.5 Consider the map
f:R^2 ⊃ overline(B_1(0)) →overline(B_1(0))

(x,y)→(x²/4 + (y/3) + (1/3) , y²/4 - (x/2))

jacobian df

find the norm of df

Answer 18

df

(x/2 1/3)
(-1/2 y/2)

Finding the norm of linear operator is challenging

∥T∥ =
sup_(∥x∥=1) [ ∥T(x)∥]
but we use HS norm:

df * df^T
and take trace = x²/4 + 1/9 + y²/4 + 1/4
take sqrt < 1
Therefore f is a contraction. We can find the fixed point by starting, for example, with the point (0, 0) and iterating.

Question 19

Hilbert -schmidt norm

Answer 19

in an exercise we found that
∥T∥ ≤∥T∥_HS

∥A∥_HS =
(tr(A∗A^T))^0.5

Question 20

initial value problem (IVP)

Answer 20

dy/dt = f(t, y),

y(t_0) = y0,

family of solutions, initial gives condition for a single or multiple solutions?

For f:K → R
K = [T₁, T₂] × [L₁, L₂] in R^2

y: [T₁,T₂]→ R, t → y(t)
and y_0 in [L₁,L₂]
t_0 in [T₁,T₂]

Question 21

diagrams
family of functions and vector fields and integral curves for next examples

Answer 21

boundaries from the initial conditions

for e.g. 5.8 intitial values give different behaviours

5.9 solution not unique for some different initial values, all cross xaxis

Hence, there are two fundamental questions here: existence and uniqueness of solutions. The
following theorem is one of the basic results in the theorem of ordinary differential equation
and establishes existence and uniqueness under rather general assumptions

Question 22

Example 5.6. Let
f(t, x) = x and
y₀ = 1,
t₀ = 0.
Then the initial value problem is

Answer 22

dy/dt = y,

y(0) = 1
we know there will be a unique sol y(t)=exp(t)

Question 23

Example 5.7. Let
f(t, x) = x² and
y₀ = 1,
t₀ = 0.
Then the initial value problem is

Answer 23

dy/dt = y²,

y(0) = 1
we know there will be a unique sol y(t)= 1/(1-t)

Question 24

Example 5.8. Let
f(t, x) = x²-t and
y₀ = 1,
t₀ = 0.
Then the initial value problem is

Answer 24

dy/dt = y²-t

y(0) = 1
One can show that there exists a solution for small |t|, however this solution cannot be expressed in terms of elementary functions

Question 25

Example 5.9. Let
f(t, x) = x^(2/3) and
y₀ = 0,
t₀ = 0.
Then the initial value problem is

Answer 25

dy/dt = y^(2/3)

y(0) = 0
It has at least two solutions, namely y = 0 and y =t^3/27

Question 26

Theorem 5.10 (Picard–Lindel¨of theorem).

Answer 26

Suppose that
f: [T₁, T₂] x[y₀-C,y₀+C]→
R is a continuous function such that for some M > 0 we have

**|f(t, y₁) − f(t, y₂)| ≤ M|y₁ − y₂| (Lipschitz condition)
**

for all t ∈[T₁, T₂],
y₁, y₂ ∈[y₀-C,y₀+C]. Then for any t₀ ∈[T₁, T₂], the initial value problem

(dy/dt)(t) = f(t,y(t))
y(t₀)=y₀

Has a unique solution y in C¹[a,b] where [a,b] is the interval [t₀-R, t₀+R] ∩[T₁, T₂] where

R= ∥f∥_∞ ⁻¹C.

(The solution exists for all times t such that |t − t₀| ≤ R).

Question 27

Lipschitz condition)

Answer 27

|f(t, y₁) − f(t, y₂)| ≤ M|y₁ − y₂| (Lipschitz condition)
stronger than uniform continuity

condition that, for any two points, the line

absolute value([f(t,y_1) - f(t,y_2)]/(y_1 - y_2))

<M

Fraction gives slope of the line joining the two points, for any two points this will not be more than this given M

Question 28

Theorem 5.10 (Picard–Lindel¨of theorem). explaining the R term

Answer 28

R is the value of C divided by supremum norm of f

diagram : t against y
region [t₀-R, t₀+R] and [y₀-C,y₀+C].
for the initial value t₀,y₀

we have the direction to follow, we integrate the curve

if F(y)=y then
y(t) = y₀ + integral_[t₀,t] of f(s,y(s)).ds

dy(t)/dt = 0 +f(t,y(t) by ftoc

y(t₀) = y₀ + integral_[t₀,t₀] f(s,y(s)).ds = y₀ + 0 = y₀

(we have a unique solution which exists on an interval size R around t_0 our initial value, C tells us how much up or down we can go)

R= ∥f∥_∞ ⁻¹C.
found by dividing C by the sup norm of f

Question 29

Lipshitz condition implies

Answer 29

uniform continuity and is significantly stronger requirement.
M>0
|f(t, y_1) − f(t, y_2)| ≤ M|y_1 − y_2| (Lipschitz condition)

|f(t, y_1) − f(t, y_2)|/ |y_1 − y_2|
≤ M

absolute value of the slope of the line between any two points is always less than M

Question 30

the ftoc

Answer 30

Useful for proving Useful for proving the Picard-Lindelof thm

Define the map by
F(y)(t) = y₀ + ∫_[t₀,t] f(s,y(s)).ds

F’(y)=0+ f(t,y(t))

Each time if F(y)=y this means that we can set y(t)=y₀ + ∫_[t₀,t] f(s,y(s)).ds
derivative of this is :

dy(t)/dt = 0+ f(t,y(t))

y(t_0)= y₀ + ∫_[t₀,t] f(s,y(s)).ds
=y₀+ 0

The map sends a continuous function in C[T_1,T_2] to another in C[T_1,T_2]

MS: X = C([a, b], [y₀ − C, y₀ + C]

Question 31

Picard iteration

Answer 31

Define the map by
F(y)(t) = y₀ + ∫_[t₀,t] f(s,y(s)).ds

compute the solution by iterating the map F starting for example with the constant function
y(t) = y0. The iteration

y_{n+1}(t) = y₀ + ∫_[t₀,t] f(s,y(s)).ds

Question 32

Picard-lindelof proof gives a bound on the solution

Answer 32

if the assumptions are
satisfied one gets |y(t) − y_0| ≤ C for t ∈ [a, b]

In the proof we also check if its a contraction

Question 33

Basic estimation for integrals using the supremum norm

|∫_[a,b] f(x).dx|

Answer 33

|∫[a,b] f(x).dx|
≤(b-a) sup[a,b] |fx)|

f is continous on a bounded interval attains

We choose an R s.t
R= ∥f∥_∞ ⁻¹C.

Question 34

Picard lindelof applied to higher order differential equations

Answer 34

For example y′′(t) + y(t) = 0,
y(0) = 1, y′(0) = 0 can
be written as

d/dt((y) (w)
(w) = (-y)

(y) (0) = (1)
(w) (0)
here the function f is f(t,(x1, x2)) = (x2, −x1).

Question 35

Example 5.14 . Consider the IVP
dy/dt = y²t + 1,
y(0) = 1.

Answer 35

f(t, x) = x²t + 1

take f to be defined on the square [−T, T] × [1 − C, 1 + C]
we obtain ∥f∥∞ = (1 + C)²T + 1

in this case the
solution will exist up to time

min{ T, [C]/[(1+C)²T+1]
top right value in box or a boundary defined before this

(as we divide C by maximal value, to get minimal

(the function approximates the solution better and better).

e.g. C=2 T=0.5
we get that a unique solution exists up to time|t| ≤ 4/11 . This solution will then satisfy |y(t) − 1| ≤ 2 for |t| ≤ 4/11 .

Question 36

picard iterations for the differential equatuon y’=y
f_0=1

Answer 36

f_0 constant=initial value=1
f_1 linear :
f_1(t)= 1+ integral_[0,t] f_0(s).ds
=1+ t

f_2 quadratic:
f_2(t)=1+ integral_[0,t] (1+s).ds
= 1+t+ t^2/2

f_3(t) = 1 + s +t^2/2! + t^3/3!

dy/dt=f(y,t)

f(x,t)=x here

approx exponential funcion for solution: all picard iterations of taylor expansion of e^t

Question 37

if a function f ∈ C¹[a, b] has nowhere vanishing derivative, then

Answer 37

if a function f ∈ C¹[a, b] has nowhere vanishing derivative, then f is invertible on its image. To be more precise,
f⁻¹: Im(f) → [a, b] exists
and has derivative (f′(x))⁻¹ at the point y = f(x).

In higher dimensions a statement like this can not be correct as the following counterexample shows.

Question 38

Consider a monotonically increasing function

Invertirle?

Answer 38

f’(x)>0
monotonically increasing
implies
bijection
implies
invertible

inverse is found by reflection in y=x

if a function f ∈ C¹[a, b] has nowhere vanishing derivative, then f is invertible on its image.

Question 39

In higher dimensions a statement like this can not be correct as the following counterexample shows.

nowhere vanishing derivative then

Answer 39

Let 0 < a < b and define
f : [a, b] × R → R²

(r, θ) → (r cos θ, r sin θ).
(shape will be a spiral upwards, will we return to the same projection on the plane?)

This maps has invertible derivative
f′(r, θ) =
[cos θ −rsinθ]
[sin θ rcosθ]

det f′(r, θ) = r² > 0
at any point, the map is however not injective

However, for any point we can restrict domain and co-domain, so that the restriction of the function is invertible. In such a case we
say that f is locally invertible.

Question 40

Definition 5.15 (Local Invertibility)

Answer 40

Suppose U₁, U₂ ⊂ R^m are open subsets of R^m.

Then a map f : U₁ → U₂ is called locally invertible at x ∈ U₁ if there exists an open neighbourhood U of x such that
f|ᵤ : U → f(U) is invertible.

The function f is said to be locally invertible it it is locally invertible at x for any x ∈ U₁

Question 41

Definition 5.16 (Diffeomorphism).

Answer 41

. Suppose U1, U2 ⊂ Rᵐ are open subsets of Rᵐ. Then
a map f : U1 → U2 is called Cᴷ
-diffeomorphism if f ∈ Cᴷ(U₁, U₂ ) and if there exists a g ∈ Cᴷ (U₂,U₁) such that

f ◦ g = 1ᵤ_2,

g ◦ f = 1ᵤ_1
.

Question 42

There is also a local version of the above definition.

Definition 5.17 (Local Diffeomorphism)

Answer 42

Suppose U₁, U₂ ⊂ R are open subsets of Rᵐ

Then a map f : U₁→ U₂ is called a local-C ᴷ- diffeomorphism at x ∈ U₁
if there exists an open neighbourhood U of x such that

f|ᵤ : U → f(U) is a C ᴷ- diffeomorphism.

It is called a local-C ᴷ- diffeomorphism if it is a local diffeomorphism at any point x ∈ U₁

(for nhd, differentiable up to order k)

Question 43

Not every invertible C ᴷ-map is a diffeomorphism.

Answer 43

An example is the function f(x) = x³ whose inverse g(x) = x¹/³ fails to be differentiable

At zero the derivative of x¹/³ is undefined therefore the inverse is not differentiable

Question 44

Theorem 5.18 (Inverse Function Theorem).
When is f a local C ᴷ-diffeomorphism

Answer 44

Let U ⊂ Rᵐ be an open subset and suppose that f ∈ C ᴷ(U, Rᵐ) such that f′(x) is invertible at every point x ∈ U. Then f is a local C ᴷ-diffeomorphism

thus it not only exists but is k times differentiable

Question 45

WHAT IS THE REQUIREMENT FOR THE INVERSE FUNCTION THM

Theorem 5.18 (Inverse Function Theorem).

applying to map

f:R^m to R^m
f(x_1,..,x_m)
=(f_1(x_1,…,x_m),…,f_m(x_1,…,x_m)

Answer 45

(we restrict:
derivative f’ invertible at x_0
means not equal to 0
1/f’(x_0)
f:R^m to R^m
f(x_1,..,x_m)
=(f_1(x_1,…,x_m),…,f_m(x_1,…,x_m)

jacobian:
J(x)=
(∂f_1/∂x_1 ….. ∂f_1/∂x_m)
(…………)
(∂f_m/∂x_1 ….. ∂f_m/∂x_m)
J(x) numerical matric
linear transform R^m to R^m

When we ask the derivative f’(x) to be invertible (in the thm) at particular point we require the jacobian matrix to be invertible
(det not equal to 0)

We don’t just ask for invertibility at a particular point but in a nhd in thm 5.18

Question 46

Lemma 5.19

under the
assumptions of the inverse function theorem an inverse function must be in C^1
.

RECIPEOCAL OF DERIVATVE

Answer 46

. Suppose that f ∈ C¹
(U₁, U₂) is bijective with continuous inverse. Assume
that the derivative of f is invertible at any point, then f is a C¹-diffeomorphism, and
g′(f(x)) = (f′(x))⁻¹
.

ie derivative of inverse function is given by the reciprocal of derivative of function f

Question 47

Lemma 5.19
example

derivatives of arcsin’x

Answer 47

(arcsin(X) )’

=(sin-1(x))’

Question 48

Mean value thm

Answer 48

Let f:[a,b]→R be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). Then, there exists at least one c in the open interval (a,b) such that

f ′ (c)= b−a
f(b)−f(a)

In other words, there exists a point
c in the open interval (a,b) where the instantaneous rate of change (derivative) of the function f is equal to the average rate of change of f over the interval
[a,b].

as this may be vector in several vars, x_1,x_2,y_1,y_2 might be in R^m

Question 49

MVT with the inverse function thm

Answer 49

g′(f(x)) = (f′(x))⁻¹

used with
f’(c)= (y_2-y_2)/(x_1-x_2)
gives

(x_1-x_1)= f’(c) (y_2 - y_1)
rewritten as
(x_1-x_1)= (f’(c))⁻¹)(y_2 - y_1)

where this inverse is an mxm matrix (f’(c))⁻¹)

Therefore f’(c) would be the mxm jacobian

thus this gives that y_1=f(x_1) implies
x_1=g(y_1) if g is inverse of f

Question 50

We consider limits with the MVT:

for a continous function
for any two points

there exists a c s.t x_1<c<x_2
f’(c)= (y_2-y_2)/(x_1-x_2)

Answer 50

x_2 tends to x_1
c tends to x_1
connected to the derivative at the point n x_1

Thus we get
g(y_2)-g(y_1)= A (y_2-y_1)

The A tends to the derivative of g at y_1
ie the reciprocal derivative
thus g’( f(x_0)) = f’(x_0)⁻¹ as required

f’(x_0)⁻¹ is continuous wrt

Question 51

Inverse function thm:
proof through a diagram

Answer 51

We consider NHD of point x_0:

could be a funct in several variables:
looking for the inverse

y₀=f(x₀)
1) we shift the function: so that this point is translated the point (0,0)
new function
f(x-x₀) - y₀

2)If we have a function which passes (0,0) then we apply the linear change of vars to make the derivative=1
linear transfomation of coords: in one var replace f(x) with h(x)=f(x)/f’(0)
gives required h’(0)=1

in several vars: we apply linear transform change of vars in R^m s.t. the jacobian becomes identity matrix 1_m

by applying jacobian of matrix inverted:
(f’(0))⁻¹

we reduce the thm to f s.t
f(0)=0 and f’(0)=identity or 1

Question 52

Inverse function thm:
proof through a diagram
why we consider neighbourhoods

Answer 52

We imagine line y=x
y=f(g(y))
y-g(g(y))=0
if g is inverse to f

what about a function which isnt an inverse
y-f(u(y)) will tell us about correction to u

in the NHD near to point 0 the function approx maps x to itself as passes through or close to y=x line

we replace u(y) by the function that might be closer

u(y) + y - f(u(y))
the map to look for a fixed point ie
u(y) = u(y) + y - f(u(y))
where y - f(u(y)) =0

We considering the map F
F(u)(y) = u(y) + y - f(u(y))
when F is a contraction

Question 53

Theorem 5.20 (Implicit Function Theorem)

Answer 53

Let U₁ ⊂ Rⁿ × Rᵐ and
U₂ ⊂ Rᵐ be open subsets and let
F : U₁ → U₂,
(x₁, . . . , x₁ₙ, y₁, . . . , yₘ) →
F(x₁, . . . , xₙ, y₁, . . . , yₘ)
be a Cᴷ-map.

y₀=f(x₀)

Suppose that F(x₀, y₀) = 0 for some point (x₀, y₀) ∈ U₁, and that the m × m-matrix

∂ᵧF(x₀, y₀) is invertible.

Then there exists an neighborhood U of (x₀, y₀) ∈ Rⁿ×Rᵐ,
an open neighborhood V of x₀ in Rⁿ,

and a Cᴷ-function f : V → Rᵐ such that
{(x, y) ∈ U | F(x, y) = 0} = {(x, f(x)) ∈ U | x ∈ V}.
The function f has derivative

f′(x₀) =
−(∂ᵧF(x₀, y₀))⁻¹ ∂ₓF(x₀, y₀) at x₀

Question 54

implicit function thm sketch proof

Answer 54

we basically have m +n initial vars
and corresponding m equations

we can exclude n as independent and keep only m as dependent
we can compute derivatives of examples using invertible prev thms

Question 55

illustration of thm 5.10 implicit function thm

Answer 55

consider plot x_1 against x_2
circle

F(x_1,x_2,y_1) = x₁C + x₂² + y₁² + y₂²- R²

Graphs cones with vertex at x_1=x_2=0 y_1=R

Add condition F(x_1,x_2,y_1)=0 correspond to horizontal section in plane corresponding to the circle

Question 56

For the
operator norm
∥ T∥
∥ .∥ m
∥ .∥ op
∥ .∥ infinity

Answer 56

∥ T∥ = sup _ (∥v∥ =1) ∥Tv∥
(operator norm, norm over the unit ball)

∥ T∥
∥ .∥ m in R^m
∥ .∥ op operators
∥ .∥ infinity functions

Question 57

Example 5.21. Consider the system of equations
x₁² + x₂² + y₁² + y₂²= 2,
x₁ + x₂³ + y₁ + y₂³ = 2.

We would like to know if this system implicitly determines functions y₁(x₁, x₂) and y₂(x₁, x₂)
near the point (0, 0, 1, 1), which solves the equation.

Answer 57

one simply applies the implicit
function theorem to

F(x₁,x₂,y₁,y₂)=
(x₁² + x₂² + y₁² + y₂²- 2, x₁ + x₂³ + y₁ + y₂³ -2)

F: R^3= R^1 x R^2 to R^2

F(x,y,x)=(0,0,0)
1 independent R
2 dependent vars R^2
The derivatives are
∂ₓF =
[2x₁ 2x₂]
[1 3x₂²]

∂ᵧF=
[2y₁ 2y₂]
[ 1 3y₂²]
The values of these derivatives at the point (0, 0, 1, 1) are
∂ₓF(0, 0, 1, 1) =
[0 0]
[1 0]

∂ᵧF(0, 0, 1, 1) =
[2 2]
[1 3]

The latter matrix is invertible and one computes
−(∂yF(x0, y0))−1
∂xF(x0, y0)(0, 0, 1, 1)
=
−(∂yF(x0, y0))−1
∂xF(x0, y0)(0, 0, 1, 1)
We conclude that there is an implicitly defined function (y1, y2) = f(x1, x2) whose derivative
at (0, 0) is given by

[1/2 0]
[−1/2 0]
.
The geometric meaning is that near the point (0, 0, 1, 1) the system defines a two dimensional manifold that is locally given by the graph of a function. Its tangent plane
is spanned by the vectors (1/2, 0, 1, 0) and (−1/2, 0, 0, 1).

Question 58

Example 5.22. Consider the system of equations
x² + y²+ z² = 1,
x + yz + z³ = 1.

Answer 58

This is the intersection of a sphere (drawn in light green on Figure 4) with some cubic
surface defined by the second equation (drawn in light blue). The point (0, 0, 1) solves
the equation and is pictured as a little orange dot. By the implicit function theorem the
intersection is a smooth curve (drawn in red) near this point which can be parametrised by
x coordinate. Indeed, we can express y and z along the curve as functions of x because the
resulting matrix
∂(y,z)F(0, 1) =
[2y 2z]
[z y + 3z²] y=0,z=1

=
[0 2]
[1 3]
is invertible

Question 59

Exercise 5.23. Fig. 4 suggests that the intersection curve can be alternatively parametrised
by the coordinates y and cannot by z (why?). Check these claims by verifying conditions
of Thm. 5.20.

Answer 59

…

geometrically means intersection

point 0,0,1 in the nhd we see if intesection dep on one var only as it moves along the curve

Question 60

T is a contraction why?

If we want to prove it…

proof for 5.1 continued

Answer 60

T is a contraction if norm
∥T(x) -T’(x)∥ <C ∥x-x’∥

(To prove we use the MVT:
f’(c)= (y_2-y_2)/(x_1-x_2)
f’(c)* (x_1-x_2)= (y_2-y_2)

e.g. our map in the proof

∥x+y-f(x) - (x’+y-f(x’))∥ as y=y’

=∥x-f(x) - (x’ -f(x’))∥
=∥(f’(c)-I)(x-x’)∥
(identity matrix, f′(x_0) = 1 properties of map give)
≤0.5 ∥x-x’∥

Question 61

contraction of functions:
prev consider functions of vectors

Answer 61

∥F(u)(y) − F(v)(y)∥ =
∥u(y) − f(u(y)) − (v(y) − f(v(y))) ∥
≤1/2∥u(y) − v(y)∥ [by (9)]
≤1/2∥u − v∥∞.

bounded by supremum norm as true for all values of y
Hence, there exists a unique fixed point g.