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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).
7
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 .
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