1. hw 8
We have learned several examples of metric spaces in classes. For example, the Euclidean space (Rp, de), the space of bounded functions (B(X), d∞), and the space (lp(N), dp) of p- summable sequences. In this exercise, we are going to learn some abstract properties of metric spaces. Moreover, we will study continuous functions on abstract metric spaces.
(1) Let (X, d) be a metric space. Let f : R → R be an increasing function (s < t implies f (s) < f (t)) so that f (0) = 0 and
f (s + t) ≤ f (s) + f (t) for any s, t ≥ 0.
Define a function ρ : X × X → R by
df(x, y) = f (d(x, y)), x, y ∈ X.
(a) Show that df defines a (new) metric on X.
Proof. Let x, y, z ∈ X. Since d is a metric on X, d(x, y) = d(y, x). Hence df(x, y) = f (d(x, y)) = f (d(y, x)) = df(y, x).
Since d is a metric on X, d(x, y) ≥ 0. Since f is increasing, df(x, y) = f (d(x, y)) ≥ f (0) = 0.
If df(x, y) = 0, f (d(x, y)) = 0. Since f is increasing, d(x, y) = 0. Since d is a metric on X and d(x, y) = 0, x = y.
Since d is a metric on X, d(x, y) ≤ d(x, z) + d(z, y). Since f is increasing and by the property of f ,
df(x, y) = f (d(x, y)) ≤ f (d(x, z) + d(z, y))
≤ f (d(x, z)) + f (d(z, y))
= df(x, z) + df(z, y).
Remark. We do not need to assume that the domain of f is all R. We can assume that f is defined on (a, ∞) where a < 0.
(b) For each x, y ∈ X, define
ρ(x, y) = d(x, y) 1 + d(x, y). Show that ρ also defines a metric on X.
Proof. Let f : (−1, ∞) → R be the function t 7→ t/(1 + t). Then f (0) = 0 and f0(t) = 1/(1 + t)2 > 0 for all t ∈ (−1, ∞). Hence f is an increasing function on (−1, ∞) by mean value theorem. Now, let us prove that for any t, s ≥ 0, f (t + s) ≤ f (t) + f (s). For s, t ≥ 0, 1 + s + t ≥ 1 + s and 1 + s + t ≥ 1 + t.
Moreover,
f (t + s) = t + s
1 + t + s = t
1 + t + s+ s 1 + t + s
≤ t
1 + t+ s
1 + s = f (s) + f (t).
We see that ρ(x, y) = f (d(x, y)) = df(x, y). By the previous result, ρ is a metric on X.
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(2) Let S be the subset of l2(N) consisting of all functions a : N → R such that a(n) = 0 for all but finitely many n ∈ N. We define a function f : S → l2(N) by
f (a1, a2, · · · ) = (a1,√ 2a2,√
3a3, · · · ).
Prove or disprove that f is continuous at 0. Here 0 = (0, 0, · · · ).
Proof. Notice that f is a linear map from S into l2(N). We have seen in class that f is not continuous at 0. In fact, we can replace√
n by any λn> 0 so that limn→∞λn=
∞. The linear map fλ : S → l2(N) defined by fλ(a1, a2, · · · ) = (λ1a1, λ2a2, · · · ) is not continuous at 0. The proof is the same.
(3) It is known that C([0, 1]) is a vector subspace of B([0, 1]). Prove that C([0, 1]) is a closed subset of (B([0, 1]), d∞). (This implies that (C([0, 1], R), k · k∞) is also a Banach space.)
Proof. We proved a more general case in class. See class notes.
(4) Let r > 0 and D(0, r) = {x ∈ Rn: kxk ≤ r}. Here k · k denotes the Euclidean norm
on Rn. Suppose that
f : D(0, r) → Rn is a map satisfying
(a) kf (0)k ≤ r/3
(b) kf (x) − f (y)k ≤ 2kx − yk/3 for x, y ∈ D(0, r).
Show that there exists x0 ∈ D(0, r) so that f(x0) = x0. Hint: use contraction mapping principle.
Proof. We need to show that f (D(0, r)) ⊆ D(0, r). For any x ∈ D(0, r), we know by triangle inequality and (1) and (2) that
kf (x)k ≤ kf (x) − f (0)k + kf (0)k ≤ 2r 3 +r
3 = r.
Hence f (x) ∈ D(0, r). This shows that f (D(0, r)) ⊆ D(0, r). Since D(0, r) is a closed subset of a complete metric space Rn, D(0, r) with the induced metric is also com- plete. By (2), f is a contraction mapping. By contraction mapping principle, f has
a unique fixed point in D(0, r).
(5) The following integral equation for f : [−1, 1] → R arises in a model of a motion of gas particle on a line:
f (x) = 1 + 1 π
Z 1
−1
1
1 + (x − y)2f (y)dy, −1 ≤ x ≤ 1.
Prove that f is a unique solution in C[−1, 1]. Hint: use contraction mapping princi- ple.
Proof. For each h ∈ C[−1, 1], define T (h)(x) = 1 + 1
π Z 1
−1
1
1 + (x − y)2h(y)dy
3
for x ∈ [−1, 1]. Prove that T : C[−1, 1] → C[−1, 1] is a contraction mapping. To do this, you need to use
1 π
Z 1
−1
1
1 + (x − y)2dy = tan−1(1 + x) + tan−1(1 − x)
π , x ∈ [−1, 1].
Prove that for any x ∈ [−1, 1],
0 < tan−1(1 + x) + tan−1(1 − x)
π ≤ 2 tan−11
π = 1
2. This would imply that for any h1, h2∈ C[−1, 1],
kT (h1) − T (h2)k∞≤ 1
2kh1− h2k∞.
(6) Prove that the following differential equation
y0(x) = sin (xy(x)) , y(0) = 0
has a unique solution on C[0, 1] and solve for y(x). Hint: use contraction mapping principle. Consider the map
T : C[0, 1] → C[0, 1], T (f )(x) = Z x
0
sin(tf (t))dt.
Prove that T is a contraction. Prove that f (t) is a solution for the differential equation if and only if T (f ) = f.
Proof. Let h ∈ C[0, 1]. Since the composition of continuous functions continuous, t 7→ sin(th(t)) is continuous on [0, 1]. By fundamental theorem of calculus,
x 7→
Z x 0
sin(th(t))dt
is continuous on [0, 1]. We define T : C[0, 1] → C[0, 1] by T (h)(x) =
Z x 0
sin(th(t))dt.
By mean value theorem,
| sin θ1− sin θ2| ≤ |θ1− θ2| for any θ1, θ2∈ R.
For any h1, h2 ∈ C[0, 1],
|T (h1)(x) − T (h2)(x)| ≤ Z x
0
| sin(th1(t)) − sin(th2(t))|dt
≤ Z x
0
t|h1(t) − h2(t)|dt
≤ kh1− h2k∞ Z x
0
tdt
= x2
2 kh1− h2k∞
≤ 1
2kh1− h2k∞. This shows that
kT (h1) − T (h2)k∞≤ 1
2kh1− h2k∞
4
for any h1, h2 ∈ C[0, 1]. By contraction mapping principle, T has a unique fixed point, say y. Hence
y(x) = Z x
0
sin(ty(t))dt.
This implies that y(0) = 0. By fundamental theorem of calculus, y is differentiable and y0(x) = sin(xy). We find that the fixed point y of T is the solution to the differential equation. Let y0(x) = 0 for x ∈ [0, 1]. Then y0(0) = 0 and T (y0) = y0. We find that y0 is also a fixed point of T. By uniqueness of fixed point y = y0, i.e.
y is the zero function.
