Lecture 5
Econ 2001
2015 August 14
Lecture 5 Outline
1
Boudedness and Extreme Value Theorem
2
Intermediate Value Theorem and Fixed Points
3
Monotonicity
4
Complete Spaces and Cauchy Sequences
5
Contraction Mappings
end of material for exam
We …rst study properties of functions from R to R.
Then, we de…ne a nice class of sequence and the spaces they live in.
Finally, we look at a particularly useful group of functions in general spaces.
Announcements:
- The test will be at 3pm, in WWPH 4716, and there will be recitation at 1pm.
The exam will last an hour.
Properties of Real Functions: Boudedness
De…nition
Let f : X ! R where X
is bounded above.
R. We say f is bounded above if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
De…nition
Let f : X ! R where X
is bounded below.
R. Similarly, we say f is bounded below if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
What does it mean for a function to be bounded above (below)? f (X ) has an
upper (lower) bound.
De…nition
Let f : X ! R where X R. f is bounded if f is both bounded above and
bounded below, that is, if f (X ) is a bounded set.
Properties of Real Functions: Boudedness
De…nition
Let f : X ! R where X
is bounded above.
R. We say f is bounded above if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
De…nition
Let f : X ! R where X
is bounded below.
R. Similarly, we say f is bounded below if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
What does it mean for a function to be bounded above (below)? f (X ) has an
upper (lower) bound.
De…nition
Let f : X ! R where X R. f is bounded if f is both bounded above and
bounded below, that is, if f (X ) is a bounded set.
Properties of Real Functions: Boudedness
De…nition
Let f : X ! R where X
is bounded above.
R. We say f is bounded above if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
De…nition
Let f : X ! R where X
is bounded below.
R. Similarly, we say f is bounded below if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
What does it mean for a function to be bounded above (below)? f (X ) has an
upper (lower) bound.
De…nition
Let f : X ! R where X R. f is bounded if f is both bounded above and
bounded below, that is, if f (X ) is a bounded set.
Properties of Real Functions: Boudedness
De…nition
Let f : X ! R where X
is bounded above.
R. We say f is bounded above if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
De…nition
Let f : X ! R where X
is bounded below.
R. Similarly, we say f is bounded below if
f (X ) = fr 2 R : f (x) = r for some x 2 X g
What does it mean for a function to be bounded above (below)? f (X ) has an
upper (lower) bound.
De…nition
Let f : X ! R where X R. f is bounded if f is both bounded above and
bounded below, that is, if f (X ) is a bounded set.
Extreme Value Theorem
Theorem (Extreme Value Theorem)
Let a; b 2 R with a b and let f : [a; b] ! R be a continuous function.
Then, f assumes its minimum and maximum on [a; b].That is, if
M = sup f (t)
t2[a;b]
and
m = inf f (t)
t2[a;b]
then
9tM ; tm 2 [a; b] such that f (tM ) = M and f (tm ) = m
Extreme Value Theorem
Theorem (Extreme Value Theorem)
Let a; b 2 R with a b and let f : [a; b] ! R be a continuous function.
Then, f assumes its minimum and maximum on [a; b].That is, if
M = sup f (t)
t2[a;b]
and
m = inf f (t)
t2[a;b]
then
9tM ; tm 2 [a; b] such that f (tM ) = M and f (tm ) = m
Extreme Value Theorem
Theorem (Extreme Value Theorem)
Let a; b 2 R with a b and let f : [a; b] ! R be a continuous function.
Then, f assumes its minimum and maximum on [a; b].That is, if
M = sup f (t)
t2[a;b]
and
m = inf f (t)
t2[a;b]
then
9tM ; tm 2 [a; b] such that f (tM ) = M and f (tm ) = m
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Extreme Value Theorem: Proof
Proof.
Let
M = supff (t) : t 2 [a; b]g
If M is …nite, for each n choose tn 2 [a; b] such that M
(if we cannot make such a choice, then M
the supremum).
1
n
f (tn )
M
1
n
is an upper bound and M is not
If M is in…nite, for each n choose tn such that f (tn )
n.
By the Bolzano-Weierstrass Theorem, ftn g contains a convergent subsequence
ftnk g, with
lim tnk = t0 2 [a; b]
Since f is continuous,
k !1
f (t0 ) = lim f (t) = lim f (tnk ) = M
t!t0
Thus M is …nite and
k !1
f (t0 ) = M = supff (t) : t 2 [a; b]g
so f attains its maximum and is bounded above.
The argument for the minimum is similar (do it as exercise).
Continuous Image of Closed Bounded Interval
For a continuous function, the image of a closed and bounded interval
is also a closed and bounded interval
If f : [a; b] ! R is continuous, then there exists two real numbers d
that f ([a; b]) = [c; d].
c such
Assumptions matter
If f is not continuous, then generally there is nothing that can be said about
the image.
If the domain is an open interval, the image could be a closed interval (if f is
constant) or it could be unbounded even if the interval is …nite (for example if
f (x) = 1=x on (0; 1)).
When the domain of f is a closed and bounded interval, the image of f is a
closed and bounded interval; then, f attains both its maximum (the maximum
value is d) and its minimum (the minimum value is c).
Continuous Image of Closed Bounded Interval
For a continuous function, the image of a closed and bounded interval
is also a closed and bounded interval
If f : [a; b] ! R is continuous, then there exists two real numbers d
that f ([a; b]) = [c; d].
c such
Assumptions matter
If f is not continuous, then generally there is nothing that can be said about
the image.
If the domain is an open interval, the image could be a closed interval (if f is
constant) or it could be unbounded even if the interval is …nite (for example if
f (x) = 1=x on (0; 1)).
When the domain of f is a closed and bounded interval, the image of f is a
closed and bounded interval; then, f attains both its maximum (the maximum
value is d) and its minimum (the minimum value is c).
Continuous Image of Closed Bounded Interval
For a continuous function, the image of a closed and bounded interval
is also a closed and bounded interval
If f : [a; b] ! R is continuous, then there exists two real numbers d
that f ([a; b]) = [c; d].
c such
Assumptions matter
If f is not continuous, then generally there is nothing that can be said about
the image.
If the domain is an open interval, the image could be a closed interval (if f is
constant) or it could be unbounded even if the interval is …nite (for example if
f (x) = 1=x on (0; 1)).
When the domain of f is a closed and bounded interval, the image of f is a
closed and bounded interval; then, f attains both its maximum (the maximum
value is d) and its minimum (the minimum value is c).
Continuous Image of Closed Bounded Interval
For a continuous function, the image of a closed and bounded interval
is also a closed and bounded interval
If f : [a; b] ! R is continuous, then there exists two real numbers d
that f ([a; b]) = [c; d].
c such
Assumptions matter
If f is not continuous, then generally there is nothing that can be said about
the image.
If the domain is an open interval, the image could be a closed interval (if f is
constant) or it could be unbounded even if the interval is …nite (for example if
f (x) = 1=x on (0; 1)).
When the domain of f is a closed and bounded interval, the image of f is a
closed and bounded interval; then, f attains both its maximum (the maximum
value is d) and its minimum (the minimum value is c).
Continuous Image of Closed Bounded Interval
For a continuous function, the image of a closed and bounded interval
is also a closed and bounded interval
If f : [a; b] ! R is continuous, then there exists two real numbers d
that f ([a; b]) = [c; d].
c such
Assumptions matter
If f is not continuous, then generally there is nothing that can be said about
the image.
If the domain is an open interval, the image could be a closed interval (if f is
constant) or it could be unbounded even if the interval is …nite (for example if
f (x) = 1=x on (0; 1)).
When the domain of f is a closed and bounded interval, the image of f is a
closed and bounded interval; then, f attains both its maximum (the maximum
value is d) and its minimum (the minimum value is c).
Intermediate Value Theorem Remix
Theorem (Intermediate Value Theorem)
Suppose f : [a; b] ! R is continuous, and f (a) < d < f (b). Then there exists
c 2 [a; b] such that f (c) = d.
Proof.
Since f is continuous, the image f ([a; b]) is an interval. Therefore, points between
the extremes are in the image.
NOTE
This theorem is used to show that equations must have solutions.
If f (a) < 0 < f (b) then there must be a c for which the continuous function
f is zero (f (c) = 0).
Intermediate Value Theorem Remix
Theorem (Intermediate Value Theorem)
Suppose f : [a; b] ! R is continuous, and f (a) < d < f (b). Then there exists
c 2 [a; b] such that f (c) = d.
Proof.
Since f is continuous, the image f ([a; b]) is an interval. Therefore, points between
the extremes are in the image.
NOTE
This theorem is used to show that equations must have solutions.
If f (a) < 0 < f (b) then there must be a c for which the continuous function
f is zero (f (c) = 0).
Intermediate Value Theorem Remix
Theorem (Intermediate Value Theorem)
Suppose f : [a; b] ! R is continuous, and f (a) < d < f (b). Then there exists
c 2 [a; b] such that f (c) = d.
Proof.
Since f is continuous, the image f ([a; b]) is an interval. Therefore, points between
the extremes are in the image.
NOTE
This theorem is used to show that equations must have solutions.
If f (a) < 0 < f (b) then there must be a c for which the continuous function
f is zero (f (c) = 0).
Fixed-Points
De…nition
Let f : X ! X . The point x 2 X is called a …xed point if f (x ) = x .
Fixed points have a special place in economics since they are used to establish
existence of a Nash equilibrium in a game, existence of a competitive
equilbrium of an economy, and so on.
This is because there are results showing that …xed points exist in interesting
(to economists) situations.
The easiest is of these is on the next slide.
Fixed-Points
De…nition
Let f : X ! X . The point x 2 X is called a …xed point if f (x ) = x .
Fixed points have a special place in economics since they are used to establish
existence of a Nash equilibrium in a game, existence of a competitive
equilbrium of an economy, and so on.
This is because there are results showing that …xed points exist in interesting
(to economists) situations.
The easiest is of these is on the next slide.
Fixed-Points
De…nition
Let f : X ! X . The point x 2 X is called a …xed point if f (x ) = x .
Fixed points have a special place in economics since they are used to establish
existence of a Nash equilibrium in a game, existence of a competitive
equilbrium of an economy, and so on.
This is because there are results showing that …xed points exist in interesting
(to economists) situations.
The easiest is of these is on the next slide.
Fixed-Points
De…nition
Let f : X ! X . The point x 2 X is called a …xed point if f (x ) = x .
Fixed points have a special place in economics since they are used to establish
existence of a Nash equilibrium in a game, existence of a competitive
equilbrium of an economy, and so on.
This is because there are results showing that …xed points exist in interesting
(to economists) situations.
The easiest is of these is on the next slide.
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
Similarly, since f (b)
b, we have h(b)
0.
0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
0.
Similarly, since f (b) b, we have h(b) 0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
Similarly, since f (b)
b, we have h(b)
0.
0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
Similarly, since f (b)
b, we have h(b)
0.
0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
Similarly, since f (b)
b, we have h(b)
0.
0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
Similarly, since f (b)
b, we have h(b)
0.
0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
(Easy) Fixed-Point Theorem
Theorem
Let [a; b] R be a closed bounded interval and f : [a; b] ! [a; b] a continuous
function. Then, there exists x 2 [a; b] such that f (x ) = x .
Proof.
Consider the function h(x) = f (x)
Since f (a)
x. A …xed point of f is a zero of h.
a (a is the lowest possible value of f ), we have h(a)
Similarly, since f (b)
b, we have h(b)
0.
0.
Since h is a continuous function on a closed bounded interval, there must be
an x such that h(x ) = 0.
This implies f (x ) = x and therefore f has a …xed point.
More general theorems hold when the domain (and range) is a higher
dimensional object, but one needs more than continuity. In fact, there are also
existence of …xed points results for point to set mappings (these are called
correspondences).
Properties of Real Functions: Monotonicicty
De…nition
A function f : R ! R is monotonically increasing if
y > x ) f (y )
Can de…ne strictly as well as decreasing.
What about the constant function?
Monotonic functions have nice properties.
f (x)
Properties of Real Functions: Monotonicicty
De…nition
A function f : R ! R is monotonically increasing if
y > x ) f (y )
Can de…ne strictly as well as decreasing.
What about the constant function?
Monotonic functions have nice properties.
f (x)
Properties of Real Functions: Monotonicicty
De…nition
A function f : R ! R is monotonically increasing if
y > x ) f (y )
Can de…ne strictly as well as decreasing.
What about the constant function?
Monotonic functions have nice properties.
f (x)
Properties of Real Functions: Monotonicicty
De…nition
A function f : R ! R is monotonically increasing if
y > x ) f (y )
Can de…ne strictly as well as decreasing.
What about the constant function?
Monotonic functions have nice properties.
f (x)
Limits from Above and Below
De…nition
Given a function f : (a; b) ! R, we say that y is the limit of f at t from above,
written y = limx !t + f (x), if for every " > 0, there is a such that
0<x
t<
)
jf (x)
yj < "
This is sometimes called limit from the right.
De…nition
Given a function f : (a; b) ! R, we say y is the limit of f at t from below,
written y = limx !t f (x), if for every " > 0, there is a such that
0<t
x<
)
jf (x)
This is sometimes called limit from the left.
yj < "
Limits from Above and Below
De…nition
Given a function f : (a; b) ! R, we say that y is the limit of f at t from above,
written y = limx !t + f (x), if for every " > 0, there is a such that
0<x
t<
)
jf (x)
yj < "
This is sometimes called limit from the right.
De…nition
Given a function f : (a; b) ! R, we say y is the limit of f at t from below,
written y = limx !t f (x), if for every " > 0, there is a such that
0<t
x<
)
jf (x)
This is sometimes called limit from the left.
yj < "
Properties of Monotonic Functions
Theorem
Let a; b 2 R with a < b, and let f : (a; b) ! R be monotonically increasing. Then
the one-sided limits
f (t + ) = lim+ f (u)
u!t
and
f (t ) = lim f (u)
u!t
exist and are real numbers for all t 2 (a; b).
The proof is analogous to the proof that a bounded monotone sequence
converges: do it.
Properties of Monotonic Functions
Remember, a function is continuous at a point when the two limits are the same
and are equal to the value of the function at that point.
De…nition
We say that f has a simple jump discontinuity at t if the one-sided limits f (t )
and f (t + ) both exist but f is not continuous at t.
NOTE
There are two ways f can have a simple jump discontinuity at t: either
f (t + ) 6= f (t ), or f (t + ) = f (t ) =
6 f (t).
The previous theorem implies that monotonic functions can have only simple
jump discontinuities.
Monotonicity implies that f (t )
f (t)
f (t + ).
So a monotonic function has a discontinuity at t if and only if f (t + ) 6= f (t ).
Picture
Properties of Monotonic Functions
Remember, a function is continuous at a point when the two limits are the same
and are equal to the value of the function at that point.
De…nition
We say that f has a simple jump discontinuity at t if the one-sided limits f (t )
and f (t + ) both exist but f is not continuous at t.
NOTE
There are two ways f can have a simple jump discontinuity at t: either
f (t + ) 6= f (t ), or f (t + ) = f (t ) =
6 f (t).
The previous theorem implies that monotonic functions can have only simple
jump discontinuities.
Monotonicity implies that f (t )
f (t)
f (t + ).
So a monotonic function has a discontinuity at t if and only if f (t + ) 6= f (t ).
Picture
Properties of Monotonic Functions
Remember, a function is continuous at a point when the two limits are the same
and are equal to the value of the function at that point.
De…nition
We say that f has a simple jump discontinuity at t if the one-sided limits f (t )
and f (t + ) both exist but f is not continuous at t.
NOTE
There are two ways f can have a simple jump discontinuity at t: either
f (t + ) 6= f (t ), or f (t + ) = f (t ) =
6 f (t).
The previous theorem implies that monotonic functions can have only simple
jump discontinuities.
Monotonicity implies that f (t )
f (t)
f (t + ).
So a monotonic function has a discontinuity at t if and only if f (t + ) 6= f (t ).
Picture
Properties of Monotonic Functions
Remember, a function is continuous at a point when the two limits are the same
and are equal to the value of the function at that point.
De…nition
We say that f has a simple jump discontinuity at t if the one-sided limits f (t )
and f (t + ) both exist but f is not continuous at t.
NOTE
There are two ways f can have a simple jump discontinuity at t: either
f (t + ) 6= f (t ), or f (t + ) = f (t ) =
6 f (t).
The previous theorem implies that monotonic functions can have only simple
jump discontinuities.
Monotonicity implies that f (t )
f (t)
f (t + ).
So a monotonic function has a discontinuity at t if and only if f (t + ) 6= f (t ).
Picture
Properties of Monotonic Functions
Remember, a function is continuous at a point when the two limits are the same
and are equal to the value of the function at that point.
De…nition
We say that f has a simple jump discontinuity at t if the one-sided limits f (t )
and f (t + ) both exist but f is not continuous at t.
NOTE
There are two ways f can have a simple jump discontinuity at t: either
f (t + ) 6= f (t ), or f (t + ) = f (t ) =
6 f (t).
The previous theorem implies that monotonic functions can have only simple
jump discontinuities.
Monotonicity implies that f (t )
f (t)
f (t + ).
So a monotonic function has a discontinuity at t if and only if f (t + ) 6= f (t ).
Picture
Properties of Monotonic Functions
Remember, a function is continuous at a point when the two limits are the same
and are equal to the value of the function at that point.
De…nition
We say that f has a simple jump discontinuity at t if the one-sided limits f (t )
and f (t + ) both exist but f is not continuous at t.
NOTE
There are two ways f can have a simple jump discontinuity at t: either
f (t + ) 6= f (t ), or f (t + ) = f (t ) =
6 f (t).
The previous theorem implies that monotonic functions can have only simple
jump discontinuities.
Monotonicity implies that f (t )
f (t)
f (t + ).
So a monotonic function has a discontinuity at t if and only if f (t + ) 6= f (t ).
Picture
Properties of Monotonic Functions
Theorem
Let a; b 2 R with a < b, and let f : (a; b) ! R be monotonically increasing. Then
D = ft 2 (a; b) : f is discontinuous at tg
is either …nite (possibly empty) or countable.
A monotonic function is continuous “almost everywhere” (that is, “except for
at most countably many points”).
Properties of Monotonic Functions
Theorem
Let a; b 2 R with a < b, and let f : (a; b) ! R be monotonically increasing. Then
D = ft 2 (a; b) : f is discontinuous at tg
is either …nite (possibly empty) or countable.
A monotonic function is continuous “almost everywhere” (that is, “except for
at most countably many points”).
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
Proof.
If t 2 D, then f (t ) < f (t + ) (if they were equal, then by monotonicity they would
have to also equal f (t), so f would be continuous at t).
The set of rational numbers Q is dense in R: that is, if x; y 2 R and x < y
then 9r 2 Q such that x < r < y .
So for every t 2 D we may choose r (t) 2 Q such that
f (t ) < r (t) < f (t + )
This de…nes a function r : D ! Q. Notice that
s > t ) f (s )
f (t + )
so
s > t; s and t 2 D ) r (s) > f (s )
f (t + ) > r (t)
and hence r (s) 6= r (t).
Therefore, r is one-to-one, so it is a bijection from D to a subset of Q.
Since Q is countable, D is either …nite or countable.
end of stu¤ relevant for the test
Cauchy Sequences
We want a metric space well behaved so that every sequence that tries to converge
has a limit to converge to.
Note
Recall that xn ! x means
8" > 0 9N("=2) such that n > N("=2) ) d(xn ; x) <
"
2
Observe that if n; m > N("=2), then
" "
+ ="
2 2
If a sequence converges, its elements are getting closer to each other.
d(xn ; xm )
d(xn ; x) + d(x; xm ) <
This motivates the idea of a Cauchy sequence.
Cauchy Sequences
We want a metric space well behaved so that every sequence that tries to converge
has a limit to converge to.
Note
Recall that xn ! x means
8" > 0 9N("=2) such that n > N("=2) ) d(xn ; x) <
"
2
Observe that if n; m > N("=2), then
" "
+ ="
2 2
If a sequence converges, its elements are getting closer to each other.
d(xn ; xm )
d(xn ; x) + d(x; xm ) <
This motivates the idea of a Cauchy sequence.
Cauchy Sequences
We want a metric space well behaved so that every sequence that tries to converge
has a limit to converge to.
Note
Recall that xn ! x means
8" > 0 9N("=2) such that n > N("=2) ) d(xn ; x) <
"
2
Observe that if n; m > N("=2), then
" "
+ ="
2 2
If a sequence converges, its elements are getting closer to each other.
d(xn ; xm )
d(xn ; x) + d(x; xm ) <
This motivates the idea of a Cauchy sequence.
Cauchy Sequences
We want a metric space well behaved so that every sequence that tries to converge
has a limit to converge to.
Note
Recall that xn ! x means
8" > 0 9N("=2) such that n > N("=2) ) d(xn ; x) <
"
2
Observe that if n; m > N("=2), then
" "
+ ="
2 2
If a sequence converges, its elements are getting closer to each other.
d(xn ; xm )
d(xn ; x) + d(x; xm ) <
This motivates the idea of a Cauchy sequence.
Cauchy Sequences
We want a metric space well behaved so that every sequence that tries to converge
has a limit to converge to.
Note
Recall that xn ! x means
8" > 0 9N("=2) such that n > N("=2) ) d(xn ; x) <
"
2
Observe that if n; m > N("=2), then
" "
+ ="
2 2
If a sequence converges, its elements are getting closer to each other.
d(xn ; xm )
d(xn ; x) + d(x; xm ) <
This motivates the idea of a Cauchy sequence.
Cauchy Sequences
De…nition
A sequence fxn g in a metric space (X ; d) is Cauchy if
8" > 0 9N(") such that n; m > N(") ) d(xn ; xm ) < "
The points along the sequence are close to each other.
Notice that d(xn ; xm ) is the distance between two points far enough along the
sequence.
The Cauchy property depends only on the sequence and the metric, not on
the ambient metric space.
The points along the sequence are, by de…nition, elements of X .
Does every Cauchy sequence converge?
Cauchy Sequences
De…nition
A sequence fxn g in a metric space (X ; d) is Cauchy if
8" > 0 9N(") such that n; m > N(") ) d(xn ; xm ) < "
The points along the sequence are close to each other.
Notice that d(xn ; xm ) is the distance between two points far enough along the
sequence.
The Cauchy property depends only on the sequence and the metric, not on
the ambient metric space.
The points along the sequence are, by de…nition, elements of X .
Does every Cauchy sequence converge?
Cauchy Sequences
De…nition
A sequence fxn g in a metric space (X ; d) is Cauchy if
8" > 0 9N(") such that n; m > N(") ) d(xn ; xm ) < "
The points along the sequence are close to each other.
Notice that d(xn ; xm ) is the distance between two points far enough along the
sequence.
The Cauchy property depends only on the sequence and the metric, not on
the ambient metric space.
The points along the sequence are, by de…nition, elements of X .
Does every Cauchy sequence converge?
Cauchy Sequences
De…nition
A sequence fxn g in a metric space (X ; d) is Cauchy if
8" > 0 9N(") such that n; m > N(") ) d(xn ; xm ) < "
The points along the sequence are close to each other.
Notice that d(xn ; xm ) is the distance between two points far enough along the
sequence.
The Cauchy property depends only on the sequence and the metric, not on
the ambient metric space.
The points along the sequence are, by de…nition, elements of X .
Does every Cauchy sequence converge?
Cauchy Sequences
De…nition
A sequence fxn g in a metric space (X ; d) is Cauchy if
8" > 0 9N(") such that n; m > N(") ) d(xn ; xm ) < "
The points along the sequence are close to each other.
Notice that d(xn ; xm ) is the distance between two points far enough along the
sequence.
The Cauchy property depends only on the sequence and the metric, not on
the ambient metric space.
The points along the sequence are, by de…nition, elements of X .
Does every Cauchy sequence converge?
Cauchy Sequences
De…nition
A sequence fxn g in a metric space (X ; d) is Cauchy if
8" > 0 9N(") such that n; m > N(") ) d(xn ; xm ) < "
The points along the sequence are close to each other.
Notice that d(xn ; xm ) is the distance between two points far enough along the
sequence.
The Cauchy property depends only on the sequence and the metric, not on
the ambient metric space.
The points along the sequence are, by de…nition, elements of X .
Does every Cauchy sequence converge?
Cauchy Sequences
Clearly, if a sequence converges it must be Cauchy.
Theorem
Every convergent sequence in a metric space is Cauchy.
Proof.
We just did it. Let xn ! x.
For every " > 0 9N such that n > N ) d(xn ; x) < "=2.
Then
m; n > N ) d(xn ; xm )
d(xn ; x) + d(x; xm )
" "
+
<
2 2
= "
Cauchy Sequences
Clearly, if a sequence converges it must be Cauchy.
Theorem
Every convergent sequence in a metric space is Cauchy.
Proof.
We just did it. Let xn ! x.
For every " > 0 9N such that n > N ) d(xn ; x) < "=2.
Then
m; n > N ) d(xn ; xm )
d(xn ; x) + d(x; xm )
" "
+
<
2 2
= "
Cauchy Sequences
Clearly, if a sequence converges it must be Cauchy.
Theorem
Every convergent sequence in a metric space is Cauchy.
Proof.
We just did it. Let xn ! x.
For every " > 0 9N such that n > N ) d(xn ; x) < "=2.
Then
m; n > N ) d(xn ; xm )
d(xn ; x) + d(x; xm )
" "
+
<
2 2
= "
Cauchy Sequences
Clearly, if a sequence converges it must be Cauchy.
Theorem
Every convergent sequence in a metric space is Cauchy.
Proof.
We just did it. Let xn ! x.
For every " > 0 9N such that n > N ) d(xn ; x) < "=2.
Then
m; n > N ) d(xn ; xm )
d(xn ; x) + d(x; xm )
" "
+
<
2 2
= "
Cauchy Sequences: Example
Example
Let X = (0; 1], d be the Euclidean metric, and xn = n1 .
We know that xn ! 0 in (R; d), so fxn g is Cauchy.
But fxn g does not converge in (X ; d).
It does not converge in (X ; d) because the point it is trying to converge to is
not an element of X .
A Cauchy sequence is trying to converge, but in a metric space there may not
be anything for it to converge to.
We can de…ne the special class of spaces in which there is something in the
space for the Cauchy sequence to converge to.
Cauchy Sequences: Example
Example
Let X = (0; 1], d be the Euclidean metric, and xn = n1 .
We know that xn ! 0 in (R; d), so fxn g is Cauchy.
But fxn g does not converge in (X ; d).
It does not converge in (X ; d) because the point it is trying to converge to is
not an element of X .
A Cauchy sequence is trying to converge, but in a metric space there may not
be anything for it to converge to.
We can de…ne the special class of spaces in which there is something in the
space for the Cauchy sequence to converge to.
Cauchy Sequences: Example
Example
Let X = (0; 1], d be the Euclidean metric, and xn = n1 .
We know that xn ! 0 in (R; d), so fxn g is Cauchy.
But fxn g does not converge in (X ; d).
It does not converge in (X ; d) because the point it is trying to converge to is
not an element of X .
A Cauchy sequence is trying to converge, but in a metric space there may not
be anything for it to converge to.
We can de…ne the special class of spaces in which there is something in the
space for the Cauchy sequence to converge to.
Cauchy Sequences: Example
Example
Let X = (0; 1], d be the Euclidean metric, and xn = n1 .
We know that xn ! 0 in (R; d), so fxn g is Cauchy.
But fxn g does not converge in (X ; d).
It does not converge in (X ; d) because the point it is trying to converge to is
not an element of X .
A Cauchy sequence is trying to converge, but in a metric space there may not
be anything for it to converge to.
We can de…ne the special class of spaces in which there is something in the
space for the Cauchy sequence to converge to.
Cauchy Sequences: Example
Example
Let X = (0; 1], d be the Euclidean metric, and xn = n1 .
We know that xn ! 0 in (R; d), so fxn g is Cauchy.
But fxn g does not converge in (X ; d).
It does not converge in (X ; d) because the point it is trying to converge to is
not an element of X .
A Cauchy sequence is trying to converge, but in a metric space there may not
be anything for it to converge to.
We can de…ne the special class of spaces in which there is something in the
space for the Cauchy sequence to converge to.
Cauchy Sequences: Example
Example
Let X = (0; 1], d be the Euclidean metric, and xn = n1 .
We know that xn ! 0 in (R; d), so fxn g is Cauchy.
But fxn g does not converge in (X ; d).
It does not converge in (X ; d) because the point it is trying to converge to is
not an element of X .
A Cauchy sequence is trying to converge, but in a metric space there may not
be anything for it to converge to.
We can de…ne the special class of spaces in which there is something in the
space for the Cauchy sequence to converge to.
Complete Metric Spaces and Banach Spaces
De…nition
A metric space (X ; d) is complete if every Cauchy sequence fxn g
to a limit x 2 X .
X converges
A metric space is complete if every sequence that should converge to a limit
has a limit to converge to.
Example
Consider the earlier example of X = (0; 1] with d the usual Euclidean metric. The
sequence fxn g with xn = n1 is Cauchy but does not converge, so ((0; 1]; d) is not
complete. The metric space ([0; 1]; d), on the other hand, is complete.
De…nition
A Banach space is a normed vector space that is complete in the metric generated
by its norm.
These are “nice” sets as far as convergence goes.
Complete Metric Spaces and Banach Spaces
De…nition
A metric space (X ; d) is complete if every Cauchy sequence fxn g
to a limit x 2 X .
X converges
A metric space is complete if every sequence that should converge to a limit
has a limit to converge to.
Example
Consider the earlier example of X = (0; 1] with d the usual Euclidean metric. The
sequence fxn g with xn = n1 is Cauchy but does not converge, so ((0; 1]; d) is not
complete. The metric space ([0; 1]; d), on the other hand, is complete.
De…nition
A Banach space is a normed vector space that is complete in the metric generated
by its norm.
These are “nice” sets as far as convergence goes.
Complete Metric Spaces and Banach Spaces
De…nition
A metric space (X ; d) is complete if every Cauchy sequence fxn g
to a limit x 2 X .
X converges
A metric space is complete if every sequence that should converge to a limit
has a limit to converge to.
Example
Consider the earlier example of X = (0; 1] with d the usual Euclidean metric. The
sequence fxn g with xn = n1 is Cauchy but does not converge, so ((0; 1]; d) is not
complete. The metric space ([0; 1]; d), on the other hand, is complete.
De…nition
A Banach space is a normed vector space that is complete in the metric generated
by its norm.
These are “nice” sets as far as convergence goes.
Complete Metric Spaces and Banach Spaces
De…nition
A metric space (X ; d) is complete if every Cauchy sequence fxn g
to a limit x 2 X .
X converges
A metric space is complete if every sequence that should converge to a limit
has a limit to converge to.
Example
Consider the earlier example of X = (0; 1] with d the usual Euclidean metric. The
sequence fxn g with xn = n1 is Cauchy but does not converge, so ((0; 1]; d) is not
complete. The metric space ([0; 1]; d), on the other hand, is complete.
De…nition
A Banach space is a normed vector space that is complete in the metric generated
by its norm.
These are “nice” sets as far as convergence goes.
Complete Metric Spaces and Banach Spaces
De…nition
A metric space (X ; d) is complete if every Cauchy sequence fxn g
to a limit x 2 X .
X converges
A metric space is complete if every sequence that should converge to a limit
has a limit to converge to.
Example
Consider the earlier example of X = (0; 1] with d the usual Euclidean metric. The
sequence fxn g with xn = n1 is Cauchy but does not converge, so ((0; 1]; d) is not
complete. The metric space ([0; 1]; d), on the other hand, is complete.
De…nition
A Banach space is a normed vector space that is complete in the metric generated
by its norm.
These are “nice” sets as far as convergence goes.
Complete and Banach Spaces: Examples
Theorem
R is complete with the usual Euclidean metric (so E1 is a Banach space).
En is the n-dimensinal Euclidean
space: En = (Rn ;dE ) where
qP
n
yi )2
dE (x; y) = kx ykE =
i =1 (xi
Theorem
Rn is complete with the usual Euclidean metric for every n 2 N (so En is a Banach
space).
Complete and Banach Spaces: Examples
Theorem
R is complete with the usual Euclidean metric (so E1 is a Banach space).
En is the n-dimensinal Euclidean
space: En = (Rn ;dE ) where
qP
n
yi )2
dE (x; y) = kx ykE =
i =1 (xi
Theorem
Rn is complete with the usual Euclidean metric for every n 2 N (so En is a Banach
space).
Complete and Banach Spaces: Results
Theorem
Given X Rn , let C (X ) be the set of all bounded continuous functions from X to
R with norm
kf k1 = supfjf (x)j : x 2 X g
Then C (X ) is a Banach space.
This is useful when we are looking for the solutions to the problem: “…nd the
function such that...”.
Theorem
Suppose (X ; d) is a complete metric space and Y
complete if and only if Y is a closed subset of X .
This connects with the previous examples.
X . Then (Y ; d) = (Y ; djY ) is
Complete and Banach Spaces: Results
Theorem
Given X Rn , let C (X ) be the set of all bounded continuous functions from X to
R with norm
kf k1 = supfjf (x)j : x 2 X g
Then C (X ) is a Banach space.
This is useful when we are looking for the solutions to the problem: “…nd the
function such that...”.
Theorem
Suppose (X ; d) is a complete metric space and Y
complete if and only if Y is a closed subset of X .
This connects with the previous examples.
X . Then (Y ; d) = (Y ; djY ) is
Complete and Banach Spaces: Results
Theorem
Given X Rn , let C (X ) be the set of all bounded continuous functions from X to
R with norm
kf k1 = supfjf (x)j : x 2 X g
Then C (X ) is a Banach space.
This is useful when we are looking for the solutions to the problem: “…nd the
function such that...”.
Theorem
Suppose (X ; d) is a complete metric space and Y
complete if and only if Y is a closed subset of X .
This connects with the previous examples.
X . Then (Y ; d) = (Y ; djY ) is
Complete and Banach Spaces: Results
Theorem
Given X Rn , let C (X ) be the set of all bounded continuous functions from X to
R with norm
kf k1 = supfjf (x)j : x 2 X g
Then C (X ) is a Banach space.
This is useful when we are looking for the solutions to the problem: “…nd the
function such that...”.
Theorem
Suppose (X ; d) is a complete metric space and Y
complete if and only if Y is a closed subset of X .
This connects with the previous examples.
X . Then (Y ; d) = (Y ; djY ) is
Contractions
De…nition
Let (X ; d) be a nonempty complete metric space. An operator is a function
T :X !X
An operator maps from a complete metric space to itself.
De…nition
An operator T is a contraction of modulus
d (T (x) ; T (y))
if
< 1 and
d (x; y) 8x; y 2 X
A contraction shrinks distances by a uniform factor
< 1.
Picture.
Contractions are useful in many economic settings; for example in dynamic
programming problems (why?).
Contractions
De…nition
Let (X ; d) be a nonempty complete metric space. An operator is a function
T :X !X
An operator maps from a complete metric space to itself.
De…nition
An operator T is a contraction of modulus
d (T (x) ; T (y))
if
< 1 and
d (x; y) 8x; y 2 X
A contraction shrinks distances by a uniform factor
< 1.
Picture.
Contractions are useful in many economic settings; for example in dynamic
programming problems (why?).
Contractions
De…nition
Let (X ; d) be a nonempty complete metric space. An operator is a function
T :X !X
An operator maps from a complete metric space to itself.
De…nition
An operator T is a contraction of modulus
d (T (x) ; T (y))
if
< 1 and
d (x; y) 8x; y 2 X
A contraction shrinks distances by a uniform factor
< 1.
Picture.
Contractions are useful in many economic settings; for example in dynamic
programming problems (why?).
Contractions
De…nition
Let (X ; d) be a nonempty complete metric space. An operator is a function
T :X !X
An operator maps from a complete metric space to itself.
De…nition
An operator T is a contraction of modulus
d (T (x) ; T (y))
if
< 1 and
d (x; y) 8x; y 2 X
A contraction shrinks distances by a uniform factor
< 1.
Picture.
Contractions are useful in many economic settings; for example in dynamic
programming problems (why?).
Contractions
De…nition
Let (X ; d) be a nonempty complete metric space. An operator is a function
T :X !X
An operator maps from a complete metric space to itself.
De…nition
An operator T is a contraction of modulus
d (T (x) ; T (y))
if
< 1 and
d (x; y) 8x; y 2 X
A contraction shrinks distances by a uniform factor
< 1.
Picture.
Contractions are useful in many economic settings; for example in dynamic
programming problems (why?).
Contractions
De…nition
Let (X ; d) be a nonempty complete metric space. An operator is a function
T :X !X
An operator maps from a complete metric space to itself.
De…nition
An operator T is a contraction of modulus
d (T (x) ; T (y))
if
< 1 and
d (x; y) 8x; y 2 X
A contraction shrinks distances by a uniform factor
< 1.
Picture.
Contractions are useful in many economic settings; for example in dynamic
programming problems (why?).
Contractions
Theorem
Every contraction is uniformly continuous.
Proof.
Fix " > 0. Let = " .
Then 8x; y such that d(x; y) < ,
d(T (x); T (y))
d(x; y)
<
= "
A contraction is, by de…nition, Lipschitz continuous with Lipschitz constant
< 1 (and hence also uniformly continuous).
Contractions
Theorem
Every contraction is uniformly continuous.
Proof.
Fix " > 0. Let = " .
Then 8x; y such that d(x; y) < ,
d(T (x); T (y))
d(x; y)
<
= "
A contraction is, by de…nition, Lipschitz continuous with Lipschitz constant
< 1 (and hence also uniformly continuous).
Contractions
Theorem
Every contraction is uniformly continuous.
Proof.
Fix " > 0. Let = " .
Then 8x; y such that d(x; y) < ,
d(T (x); T (y))
d(x; y)
<
= "
A contraction is, by de…nition, Lipschitz continuous with Lipschitz constant
< 1 (and hence also uniformly continuous).
Contractions
Theorem
Every contraction is uniformly continuous.
Proof.
Fix " > 0. Let = " .
Then 8x; y such that d(x; y) < ,
d(T (x); T (y))
d(x; y)
<
= "
A contraction is, by de…nition, Lipschitz continuous with Lipschitz constant
< 1 (and hence also uniformly continuous).
Contractions and Fixed Points
De…nition
A …xed point of an operator T is point x 2 X such that T (x ) = x .
Theorem
Let (X ; d) be a nonempty complete metric space and T : X ! X a contraction
with modulus < 1. Then
1
T has a unique …xed point x .
2
For every x0 2 X , the sequence fxn g where
x1 = T (x0 ); x2 = T (x1 ) = T (T (x0 )); ::::; xn = T (xn
converges to x .
1)
for each n
(Draw a picture)
The theorem asserts both the existence and uniqueness of the …xed point, as
well as giving an algorithm to …nd the …xed point of a contraction.
The algorithm generates a sequence that converges to the …xed point for any
initial point x0 .
More general …xed point theorems guarantee existence but are not
constructive.
Contractions and Fixed Points
De…nition
A …xed point of an operator T is point x 2 X such that T (x ) = x .
Theorem
Let (X ; d) be a nonempty complete metric space and T : X ! X a contraction
with modulus < 1. Then
1
T has a unique …xed point x .
2
For every x0 2 X , the sequence fxn g where
x1 = T (x0 ); x2 = T (x1 ) = T (T (x0 )); ::::; xn = T (xn
converges to x .
1)
for each n
(Draw a picture)
The theorem asserts both the existence and uniqueness of the …xed point, as
well as giving an algorithm to …nd the …xed point of a contraction.
The algorithm generates a sequence that converges to the …xed point for any
initial point x0 .
More general …xed point theorems guarantee existence but are not
constructive.
Contractions and Fixed Points
De…nition
A …xed point of an operator T is point x 2 X such that T (x ) = x .
Theorem
Let (X ; d) be a nonempty complete metric space and T : X ! X a contraction
with modulus < 1. Then
1
T has a unique …xed point x .
2
For every x0 2 X , the sequence fxn g where
x1 = T (x0 ); x2 = T (x1 ) = T (T (x0 )); ::::; xn = T (xn
converges to x .
1)
for each n
(Draw a picture)
The theorem asserts both the existence and uniqueness of the …xed point, as
well as giving an algorithm to …nd the …xed point of a contraction.
The algorithm generates a sequence that converges to the …xed point for any
initial point x0 .
More general …xed point theorems guarantee existence but are not
constructive.
Contractions and Fixed Points
De…nition
A …xed point of an operator T is point x 2 X such that T (x ) = x .
Theorem
Let (X ; d) be a nonempty complete metric space and T : X ! X a contraction
with modulus < 1. Then
1
T has a unique …xed point x .
2
For every x0 2 X , the sequence fxn g where
x1 = T (x0 ); x2 = T (x1 ) = T (T (x0 )); ::::; xn = T (xn
converges to x .
1)
for each n
(Draw a picture)
The theorem asserts both the existence and uniqueness of the …xed point, as
well as giving an algorithm to …nd the …xed point of a contraction.
The algorithm generates a sequence that converges to the …xed point for any
initial point x0 .
More general …xed point theorems guarantee existence but are not
constructive.
Contractions and Fixed Points
De…nition
A …xed point of an operator T is point x 2 X such that T (x ) = x .
Theorem
Let (X ; d) be a nonempty complete metric space and T : X ! X a contraction
with modulus < 1. Then
1
T has a unique …xed point x .
2
For every x0 2 X , the sequence fxn g where
x1 = T (x0 ); x2 = T (x1 ) = T (T (x0 )); ::::; xn = T (xn
converges to x .
1)
for each n
(Draw a picture)
The theorem asserts both the existence and uniqueness of the …xed point, as
well as giving an algorithm to …nd the …xed point of a contraction.
The algorithm generates a sequence that converges to the …xed point for any
initial point x0 .
More general …xed point theorems guarantee existence but are not
constructive.
Contractions and Fixed Points
De…nition
A …xed point of an operator T is point x 2 X such that T (x ) = x .
Theorem
Let (X ; d) be a nonempty complete metric space and T : X ! X a contraction
with modulus < 1. Then
1
T has a unique …xed point x .
2
For every x0 2 X , the sequence fxn g where
x1 = T (x0 ); x2 = T (x1 ) = T (T (x0 )); ::::; xn = T (xn
converges to x .
1)
for each n
(Draw a picture)
The theorem asserts both the existence and uniqueness of the …xed point, as
well as giving an algorithm to …nd the …xed point of a contraction.
The algorithm generates a sequence that converges to the …xed point for any
initial point x0 .
More general …xed point theorems guarantee existence but are not
constructive.
Blackwell’s Su¢ cient Conditions
These are su¢ cient conditions for an operator to be a contraction.
Let X be a set, and let B(X ) be the set of all bounded functions from X to
R. Then (B(X ); k k1 ) is a normed vector space.
Theorem
(Blackwell’s Su¢ cient Conditions) Consider B(X ) with the sup norm k k1 .
Let T : B(X ) ! B(X ) be an operator satisfying
1
Monotonicity. For any f ; g 2 B(X ),
f (x)
2
g (x) 8x 2 X ) (Tf )(x)
(Tg )(x) 8x 2 X
Discounting. 9 2 (0; 1) such that for every f 2 B(X ); a
(T (f + a)) (x)
(Tf )(x) + a
0 and x 2 X ,
Then T is a contraction with modulus .
Sloppy notation alert: I wrote interchangeably a 2 R and a : X ! R to denote
the function de…ned as a(x) = a 8x 2 X .
Blackwell’s Su¢ cient Conditions
These are su¢ cient conditions for an operator to be a contraction.
Let X be a set, and let B(X ) be the set of all bounded functions from X to
R. Then (B(X ); k k1 ) is a normed vector space.
Theorem
(Blackwell’s Su¢ cient Conditions) Consider B(X ) with the sup norm k k1 .
Let T : B(X ) ! B(X ) be an operator satisfying
1
Monotonicity. For any f ; g 2 B(X ),
f (x)
2
g (x) 8x 2 X ) (Tf )(x)
(Tg )(x) 8x 2 X
Discounting. 9 2 (0; 1) such that for every f 2 B(X ); a
(T (f + a)) (x)
(Tf )(x) + a
0 and x 2 X ,
Then T is a contraction with modulus .
Sloppy notation alert: I wrote interchangeably a 2 R and a : X ! R to denote
the function de…ned as a(x) = a 8x 2 X .
Blackwell’s Su¢ cient Conditions
These are su¢ cient conditions for an operator to be a contraction.
Let X be a set, and let B(X ) be the set of all bounded functions from X to
R. Then (B(X ); k k1 ) is a normed vector space.
Theorem
(Blackwell’s Su¢ cient Conditions) Consider B(X ) with the sup norm k k1 .
Let T : B(X ) ! B(X ) be an operator satisfying
1
Monotonicity. For any f ; g 2 B(X ),
f (x)
2
g (x) 8x 2 X ) (Tf )(x)
(Tg )(x) 8x 2 X
Discounting. 9 2 (0; 1) such that for every f 2 B(X ); a
(T (f + a)) (x)
(Tf )(x) + a
0 and x 2 X ,
Then T is a contraction with modulus .
Sloppy notation alert: I wrote interchangeably a 2 R and a : X ! R to denote
the function de…ned as a(x) = a 8x 2 X .
Blackwell’s Su¢ cient Conditions
These are su¢ cient conditions for an operator to be a contraction.
Let X be a set, and let B(X ) be the set of all bounded functions from X to
R. Then (B(X ); k k1 ) is a normed vector space.
Theorem
(Blackwell’s Su¢ cient Conditions) Consider B(X ) with the sup norm k k1 .
Let T : B(X ) ! B(X ) be an operator satisfying
1
Monotonicity. For any f ; g 2 B(X ),
f (x)
2
g (x) 8x 2 X ) (Tf )(x)
(Tg )(x) 8x 2 X
Discounting. 9 2 (0; 1) such that for every f 2 B(X ); a
(T (f + a)) (x)
(Tf )(x) + a
0 and x 2 X ,
Then T is a contraction with modulus .
Sloppy notation alert: I wrote interchangeably a 2 R and a : X ! R to denote
the function de…ned as a(x) = a 8x 2 X .