The University of Sydney

MATH 3901

Metric Spaces 2004

Tutorial 8

PROBLEM SET 8

1. Let X = (0, 1/4) with d the Euclidean metric and f : X → X be given byf(x) = x2. Prove that f is a contraction mapping and that f has no fixedpoint in X.

Solution.

For any x, y ∈ X, we have

d(f(x), f(y)| = |f(x)− f(y)| = |x2 − y2|

= |x + y| |x− y| ≤ (1/4 + 1/4)d(x, y) =12

d(x, y),

and so f is a contraction mapping.Now if f(x) = x, then x2 = x which implies x = 0 or x = 1. But both 0 and 1are not in X. Hence f has no fixed point in X. Note that X is not completesince it is not closed in R .

2. Let X = {x ∈ Q | x ≥ 1} with d the Euclidean metric and let f : X → X bedefined by

f(x) =x

2+

1x

.

Show that f is a contraction mapping and that f has no fixed point in X.

Solution.

For any x, y ∈ X, we have

d(f(x), f(y)| = |f(x)− f(y)| =∣∣∣∣(x

2+

1x

)−

(y

2+

1y

)∣∣∣∣=

∣∣∣∣12 − 1xy

∣∣∣∣ |x− y| ≤ 12

d(x, y),

since x ≥ 1 and y ≥ 1. Hence f is a contraction mapping.Now, if f(x) = x, then

x

2+

1x

= x

so that x =√

2. But√

2 is not in X. Hence f has no fixed point in X. [Againnote that X is not complete.]

2

3. Let X = [1, ∞) with d the Euclidean metric and let f : X → X be given byf(x) = x + 1/x. Show that, for any distinct x, y ∈ X,

d(f(x), f(y)

)< d(x, y),

and that f has no fixed point in X.

Solution.

For any distinct x, y ∈ X,

d(f(x), f(y)

)= |f(x)− f(y)| = |(x + 1/x)− (y + 1/y)|

=∣∣∣∣1− 1

xy

∣∣∣∣ |x− y| < d(x, y),

since x ≥ 1 and y ≥ 1.

Now if f(x) = x, then x+1/x = x which implies 1/x = 0 which has no solutionin X. Hence f has no fixed point in X.

4. Let X = [1, ∞) with d the Euclidean metric and let f : X → X be given by

f(x) =2526

(x + 1/x).

Show that

d(f(x), f(y)

)≤ 25

26d(x, y),

[and so f is a contraction mapping] and by solving the equation algebraically,show that 5 is the unique fixed point of f .

Solution.

For any x, y ∈ X, we have

d(f(x), f(y)

)= |f(x)− f(y)| =

∣∣∣∣2526

(x + 1/x)− 2526

(y + 1/y)∣∣∣∣

=2526

∣∣∣∣1− 1xy

∣∣∣∣ |x− y| < 2526

d(x, y),

since x ≥ 1 and y ≥ 1. Hence f is a contraction mapping.

If f(x) = x, then2526

(x + 1/x) = x

which implies that x2 = 25 and so x = 5 is the unique fixed point of f in X.

3

5. Let X = [0, 1] with d the Euclidean metric and let f : X → X be given by

f(x) =17(x3 + x2 + 1).

Show thatd(f(x), f(y)

)≤ 5

7d(x, y).

Calculate f(0), f (2)(0), f (3)(0), . . . , and hence find, to three decimal places,the fixed point of f .

Solution.

For any x, y ∈ X, we see that

d(f(x), f(y)

)= |f(x)− f(y)| =

∣∣∣∣17(x3 + x2 + 1)− 17(y3 + y2 + 1)

∣∣∣∣=

17|(x3 − y3) + (x2 − y2)| = 1

7|(x− y)(x2 + xy + y2 + x + y)|

≤ 57|x− y| = 5

7d(x, y),

since 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.

Using the calculator, one can compute that

f(0) = 0.142857, f (2)(0) = f(0.142857) = 0.1457767,

f (3)(0) = f(0.1457767) = 0.145897, · · · .

Hence the fixed point of f is x = 0.146, to three decimal places.

6. Let X = [1, 2] ∩ Q with d the Euclidean metric and let f : X → X be themapping defined by

f(x) = −14(x2 − 2) + x.

Prove that f is a contraction mapping and that f has no fixed point in X.

Solution.

For any x, y ∈ X, we have

d(f(x), f(y)

)= |f(x)− f(y)|

=∣∣∣∣−1

4(x2 − 2) + x +

14(y2 − 2)− y

∣∣∣∣=

∣∣∣∣1− 14(x + y)

∣∣∣∣ |x− y| ≤ 12

d(x, y).

Hence f is a contraction mapping.Now, if f(x) = x, then − 1

4 (x2 − 2) + x = x, which implies that x2 = 2 so thatx =

√2. But

√2 is not in X, it follows that f has no fixed point in X.

4

7. Let f : [a, b] → [a, b] be differentiable over [a, b]. Show that f is a contractionmapping if and only if there exists a number K < 1 such that for all x ∈ (a, b),

|f ′(x)| ≤ K.

Solution.

Suppose first that f is a contraction mapping, i.e. that there exists K < 1 suchthat

|f(x)− f(y)| ≤ K|x− y|,

for all x, y ∈ [a, b]. Then, in particular, for and x and x + δx in [a, b], we have

|f(x + δx)− f(x)| ≤ K|(x + δx)− x| = K|δx|.

Hence for δx 6= 0, ∣∣∣∣f(x + δx)− f(x)δx

∣∣∣∣ ≤ K,

and the limit of the left-hand expression as δx → 0 is also less than or equal toK. But that limit is precisely |f ′(x)|. Hence there exists K < 1 such that forall x ∈ (a, b),

|f ′(x)| ≤ K.

Conversely, asssume that for all x ∈ (a, b),

|f ′(x)| ≤ K < 1.

Now for any x 6= y in [a, b], there is c between x and y such that

f(x)− f(y)x− y

= f ′(c).

But |f ′(c)| ≤ K so that ∣∣∣∣f(x)− f(y)x− y

∣∣∣∣ = |f ′(c)| ≤ K.

Hence, for all x, y ∈ [a, b],

|f(x)− f(y)| ≤ K|x− y|,

and so f is a contraction mapping.

5

8. Let X = C[0, 1] be the metric space with d given by

d(f, g) = supx∈[0, 1]

|f(x)− g(x)|.

Define F : X → X by

{F (f)}(x) =∫ x

0

f(t) dt (f ∈ X).

Show that

(i) {F (f)}(x)− {F (g)}(x) ≤ x d(f, g),

(ii) {F (2)(f)}(x)− {F (2)(g)}(x) ≤ x2

2d(f, g),

for all f, g ∈ X, and deduce that F (2) is a contraction mapping. Show, however,that F is not a contraction mapping.

Solution.

(i) We have

{F (f)}(x)− {F (g)}(x) =∫ x

0

(f(t)− g(t)

)dt

≤∫ x

0

d(f, g) dt ≤ x d(f, g).

(ii) We have

{F (2)(f)}(x)− {F (2)(g)}(x) =∫ x

0

[{F (f)}(t)− {F (g)}(t)

]dt

≤∫ x

0

t d(f, g) dt =x2

2d(f, g).

It follows that

d(F (f), F (g)

)= sup

x∈[0, 1]

|{F (f)}(x)− {F (g)}(x)|

≤ supx∈[0, 1]

x d(f, g) ≤ d(f, g),

andd(F (2)(f), F (2)(g)

)= sup

x∈[0, 1]

|{F (2)(f)}(x)− {F (2)(g)}(x)|

≤ supx∈[0, 1]

x2

2d(f, g) ≤ 1

2d(f, g).

Hence F (2) is a contraction mapping.But F is not, since by taking f = 1 and g = 0, we have d

(F (f), F (g)

)=

supx∈[0,1] x = 1 = d(f, g).

6

9. (Column sum criterion) Let X = Rn with metric d1 given by

d1(x, y) = |x1 − y1|+ |x2 − y2|+ · · ·+ |xn − yn|.

Show that instead of the condition (2.2) in Theorem 2.1, we obtain the condition

n∑j=1

|cjk| < 1 (k = 1, 2, . . . , n).

Solution.

Recall that, if z = f(x), then

zj =n∑

k=1

cjkxk + bj .

Set w = (wj) = f(y), then we have

d1

(f(x), f(y)

)= d(z, w) =

n∑j=1

|zj − wj |

=n∑

j=1

∣∣∣∣∣n∑

k=1

cjk(xk − yk)

∣∣∣∣∣≤

(max

k

n∑j=1

|cjk|) n∑

k=1

|xk − yk|

=(

maxk

n∑j=1

|cjk|)

d1(x, y)

= K d1(x, y) ,

where

K = maxk

n∑j=1

|cjk|.

We see that f is a contraction if K < 1. Hence we obtain the condition

n∑j=1

|cjk| < 1 (k = 1, 2, . . . , n).

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10. (Square sum criterion) Let X = Rn with metric d2 given by

d2(x, y) =√

(x1 − y1)2 + (x2 − y2)2 + · · ·+ (xn − yn)2.

Show that instead of the condition (2.2) in Theorem 2.1, we obtain the condition

n∑j=1

n∑k=1

c2jk < 1 .

Solution.

Recall that, if z = f(x), then

zj =n∑

k=1

cjkxk + bj .

Set w = (wj) = f(y), then we have

d2

(f(x), f(y)

)= d(z, w) =

{ n∑j=1

(zj − wj)2}1/2

={ n∑

j=1

[ n∑k=1

cjk(xk − yk)]2}1/2

≤{ n∑

j=1

[ n∑k=1

c2jk

n∑k=1

(xk − yk)2]}1/2

(by Cauchy’s inequality)

=( n∑

j=1

n∑k=1

c2jk

)1/2{ n∑k=1

(xk − yk)2}1/2

=( n∑

j=1

n∑k=1

c2jk

)1/2

d2(x, y)

= K1/2 d2(x, y) ,

where

K =n∑

j=1

n∑k=1

c2jk .

We see that f is a contraction if K < 1. Hence we obtain the condition

n∑j=1

n∑k=1

c2jk < 1 .