Lecture
38.3. Theorem. If (Y, ρ) is a complete metric space, then the metric space (Y
X, d0
) is complete.
Proof. The metric space (Y, ρ0
) is complete (Example 34.2.3 of Lecture 11).
Consider a fundamental (Cauchy) sequence (fn), fn : X → Y, n ∈ N, in (Y
X, d0
). For
any x ∈ X
ρ
0
(fn(x), fm(x)) ≤ d
0
(fn, fm).
Hence for any point x ∈ X the sequence fn(x), n ∈ N, is fundamental in Y,
and converges to a limit, which we denote f(x). Let us show that the sequence (fn) converges
to f ∈ Y
X, f = (f(x)), x ∈ X.
For an arbitrary ε > 0 there exists such k ∈ N that
ρ
0
(fn(x), fm(x)) <
ε
2
,
for any n, m ≥ k, and any x ∈ X. Fixing n and x, and letting m tend to infinity
we obtain
ρ
0
(fn(x), f(x)) ≤
ε
2
.
This inequality holds for any x ∈ X and n ≥ k. Therefore
d
0
(fn, f) ≤
ε
2
< ε
for any n ≥ k.
If X is a topological space, then in the product (Y
X, d0
) one can consider the subset of continuous mappings, denoted C(X, Y ), and the subset of bounded mappings (i.e. mappings for which diam f(X) = sup{ρ(f(x), f(y)) : x, y ∈ X} < +∞),
denoted B(X, Y ).
38.4. Examples. If X is a discrete space, then C(X, Y ) = Y
X (any mapping
of a discrete space is continuous).
If X is a single point, then C(X, Y ) = Y.
38.5. Theorem. The subsets C(X, Y ) and B(X, Y ) are closed in the space (Y
X, d0
).
If (Y, ρ) is a complete metric space, then
(C(X, Y ), d0
|C(X,Y )) and (B(X, Y ), d0
|B(X,Y )) are complete metric spaces.
Proof. Closedness of the subset C(X, Y ). Let f ∈ Cl(C(X, Y )). There exists
a sequence of continuous mappings (fn), converging to f (see Remark 12.8 of Lecture
2). Let us prove the continuity of f at the point x0.
For any 0 < ε < 1 there exists such k ∈ N that
ρ
0
(f(x), fn(x)) = ρ(f(x), fn(x)) < ε/3 for all n ≥ k and x ∈ X. (38.1)
Since the mapping fk is continuous, there exists a neighborhood Ox0 such that
x ∈ Ox0 ⇒ ρ
0
(fk(x0), fk(x)) = ρ(fk(x0), fk(x)) < ε/3. (38.2)
Then for x ∈ Ox0 we have
ρ(f(x0), f(x)) ≤ ρ(f(x0), fk(x0)) + ρ(fk(x0), fk(x)) + ρ(fk(x), f(x)) <
(according to (38.1) and (38.2)) <
ε
3
+
ε
3
+
ε
3
= ε.
Thus, f is continuous at the point x0, f ∈ C(X, Y ) and C(X, Y ) = Cl(C(X, Y )).
Closedness of the subset B(X, Y ). If f ∈ Cl(B(X, Y )), then there exists a sequence
of bounded mappings (fn), converging to f. There exists such k ∈ N that
ρ
0
(f(x), fk(x)) = ρ(f(x), fk(x)) < 1 for all x ∈ X.
Then, if diam(fk(X)) = sup{ρ(fk(x), fk(y)) : x, y ∈ X} = M, then
diam(f(X)) = sup{ρ(f(x), f(y)) : x, y ∈ X} ≤ sup{ρ(f(x), fk(x)) : x ∈ X}+
+ sup{ρ(fk(x), fk(y)) : x, y ∈ X} + sup{ρ(fk(y), f(y)) : y ∈ X} < M + 2.
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Thus, the mapping f is bounded, and B(X, Y ) = Cl(B(X, Y )).
The second statement of the theorem is a consequence of Example 34.2.2 of Lecture 11.
38.6. Remark. It is easy to check that on the subset B(X, Y ) of the product Y
X a metric is correctly defined
d(f, g) = sup{ρ(f(x), g(x)) : x ∈ X},
and d
0
(f, g) = min{d(f, g), 1}, where d
0 is the uniform metric on Y
X. (If Y = R, then the metric d
is generated by the norm ||f|| = sup{|f(x)| : x ∈ X} on the linear space B(X).)
38.7. Corollary. The space of continuous bounded functions C
∗
(X) = C(X, R) ∩
B(X, R) is a complete metric space in the uniform metric, and in the equivalent metric d, generated by the norm ||f|| = sup{|f(x)| : x ∈ X}.
The topology of pointwise convergence on C(X, Y ) is the topology T |C(X,Y ) of the subspace of the Tychonoff product Y
X.
§ 39. Completion of a metric space.
39.1. Definition. A mapping of metric spaces f : (X, ρ) → (Y, d) is called
an isometric embedding if for any points x, y ∈ X the relation ρ(x, y) =
d(f(x), f(y)) holds. A bijective isometric embedding is called an isometry.
Any isometric embedding is a homeomorphism between X and f(X).
39.2. Theorem. Any metric space (X, ρ) is isometrically embeddable in a
complete metric space.
Proof. Let us show that (X, ρ) is isometrically embeddable in the complete metric
space C
∗
(X) with the metric d(f, g) = sup{|f(x) − g(x)| : x ∈ X}.
Let x0 be a fixed point of X. For a point a ∈ X define the function φa : X → R,
φa(x) = ρ(x, a) − ρ(x, x0).
The functions φa, a ∈ X, are continuous. Let us prove their boundedness. Using the triangle inequality, for any point x ∈ X we have
|φa(x)| = |ρ(x, a) − ρ(x, x0)| ≤ ρ(a, x0).
Define the mapping Φ : X → C
∗
(X), Φ(a) = φa, and let us show that it is an isometric embedding, i.e. that for any pair of points a, b ∈ X we have
ρ(a, b) = d(φa, φb) = sup{|φa(x) − φb(x)| : x ∈ X}.
Since
sup{|φa(x) − φb(x)| : x ∈ X} = sup{|ρ(x, a) − ρ(x, b)| : x ∈ X},
we have d(φa, φb) ≤ ρ(a, b). At x = a we obtain the equality
|φa(a) − φb(a)| = ρ(a, b).
Hence, d(φa, φb) = ρ(a, b).
39.3. Definition. Let X be a metric space, f its isometric embedding in a complete metric space Y. The closure of f(X) in Y is called the completion
of the space X.
39.4. Problem. Verify the correctness of the definition of the completion of a metric space,
i.e. establish the uniqueness of the completion up to isometries.
§ 40. The Stone–Weierstrass theorem.
40.1. Weierstrass approximation theorem. In the space C
∗
(I) in the topology
of uniform convergence the subset of polynomials is everywhere dense.
Let us call a real-valued function p on the Tychonoff cube I
A an elementary polynomial,
if there exist a finite subset {α1, . . . , αk} ∈ A and polynomials p1(t), . . . , pk(t) such
that
p(x) = p1(prα1
(x)) · · · · · pk(prαk
(x)) for any x ∈ I
A.
Let us call a function f polynomial, if it is a finite sum of elementary polynomials f = p
1 +· · ·+p
m (in other words the function f is a polynomial in n variables xα1
, . . . , xαn
,
where n is the number of variables in the elementary polynomials p
1
, . . . , pm).
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It is easy to check that
(a) the family P separates points, i.e. if for any x, y ∈ X, x 6= y there exists f ∈ P
such that f(x) 6= f(y).
(b) the family P contains all constant functions and is a ring of functions, i.e. for
any f, g ∈ P we have fg ∈ P and f + g ∈ P.
Denote by P the closure of the set of polynomial functions P in the space C
∗
(I
A)
in the topology of uniform convergence.
40.2. Lemma. P is a ring of functions. If f ∈ P and inf{f(x) : x ∈ X} > 0, then 1
f
∈ P.
Proof. The proof of the first statement is left to the reader.
Let us give the proof of the second statement. Let M = ||f|| = sup{|f(x)| : x ∈ X},
g(x) = f(x)
2M , δ = inf{g(x) : x ∈ X} > 0. Then for any point x ∈ X
δ ≤ g(x) ≤
1
2
,
1
g(x)
=
1
1 − (1 − g(x)) =
Xn
j=0
(1 − g(x))j +
(1 − g(x))n+1
g(x)
and
|
(1 − g(x))n+1
g(x)
| ≤ (1 − δ)
n+1
δ
.
Since f ∈ P, g ∈ P, then Pn
j=0(1 − g(x))j ∈ P and, since
(1−δ)
n+1
δ → 0 as n → ∞, then 1
g
∈ P.
Hence 1
f
∈ P.
40.3. Theorem. P = C
∗
(I
A).
Proof. Fix f ∈ C
∗
(I
A) and ε > 0. For any point x ∈ I
A there exist its
neighborhoods Ox =
T
{pr−1
αs
(Oxαs
) : s = 1, . . . , n(x)}, Oxαs
open in Iαs
, and V x =
T
{pr−1
αs
(V xαs
) :
s = 1, . . . , n(x)} such that Cl(Ox) ⊂ V x and diam(f(V x)) < ε.
For any s = 1, . . . , n(x) there exists a function φ
s
x
: Iαs → I such that φ
s
x
(Cl(Oxαs
)) =
1, φ
s
x
(Iαs \ V xαs
) = 0. By Weierstrass's Theorem 40.1 the function φ
s
x belongs to the closure
of the family of polynomials on the segment, hence the function
φx = (φ
1
x ◦ prα1
) · · · · · (φ
s
x ◦ prαs
)
belongs to P, and φx(Cl(Ox)) = 1, φx(X \ V x) = 0.
Let {Oxk : k = 1, . . . , n} be a finite subcover of the cover {Ox : x ∈ I
A}. Set
φ =
Pn
k=1 φxk
. Since φ(x) > 0 for any x ∈ I
A, then taking Lemma 40.2 into account, the functions gk =
φxk
φ
, k = 1, . . . , n, belong to P, and in fact form a partition of unity subordinate
to the cover {V xk : k = 1, . . . , n} (instead of the condition Cl (supp gk) ⊂ Vxk
we have supp gk ⊂ Vxk
).
Consider the function
g(x) = Xn
k=1
f(xk)gk(x).
It is continuous, belongs to P and for any point x ∈ I
A
|f(x) − g(x)| = |f(x)
Xn
k=1
gk(x) −
Xn
k=1
f(xk)gk(x)| ≤ Xn
k=1
|f(x) − f(xk)|gk(x) < ε,
since |f(x) − f(xk)| < ε for x ∈ Oxk and gk(x) = 0 for x 6∈ Oxk, k = 1, . . . , n. Thus
||f − g|| < ε and g ∈ P. Hence f ∈ P.
Assignment N 12
1. Let (Y, ρ) be a metric space. Prove that on the subset B(X, Y ) of the product Y
X a metric is correctly defined
d(f, g) = sup{ρ(f(x), g(x)) : x ∈ X},
and d
0
(f, g) = min{d(f, g), 1}, where d
0 is the uniform metric on Y
X. (If Y = R, then the metric d
is generated by the norm ||f|| = sup{|f(x)| : x ∈ X} on the linear space B(X).)
2. Will the set
U(f, ) = {h ∈ R
X : ρ(f(x), h(x)) < , x ∈ X}
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be open in the topology of uniform convergence on R
X.
Prove that for the ε-neighborhood Oε(f) of the point f in the uniform metric we have
Oε(f) = [
δ<
U(f, δ).
3. Which of the necessary and sufficient conditions hold in the following statements:
(a) the mappings fα of the family of mappings {fα : X → Y : α ∈ A}, where Y is a metric space, are continuous if and only if the diagonal product of the mappings fα
in Y
X in the topology of uniform convergence is continuous;
(b) the mappings fα of the family of mappings {fα : Xα → Y : α ∈ A}, where Y is a
metric space, are continuous if and only if the product of the mappings Q
α∈A fα of the Tychonoff product Q
α∈A Xα in Y
X in the topology of uniform convergence is continuous.
4. Will the product R
X in the topology of uniform convergence be a linear topological space (i.e. the naturally defined operations of addition and multiplication by scalars
are continuous)?
5. Prove that the space R
N in the topology of uniform convergence does not satisfy
the second axiom of countability and is not separable.
6. Prove that the set C([0, 1], R) is not closed in the Tychonoff product R
[0,1]
.
7. In which of the topologies on the set C(R, R) is the sequence of functions fn =
x
n
, n ∈ N,
convergent?
8. A family of continuous functions F ⊂ C(X) on a metric space X is called
equicontinuous at the point x, if for any > 0 there exists δ > 0 such that
|f(x) − f(t)| < for any f ∈ F, t ∈ Oδ(x). A family F ⊂ C(X), equicontinuous
at all points of X, is called equicontinuous.
Prove that any finite subfamily F ⊂ C(X) is equicontinuous.
Prove that any uniformly convergent sequence of functions fn ∈ C(X) is an equicontinuous family.
Will the sequence of functions fn : [0, 1] →
R, fn = x
n, n ∈ N, be an equicontinuous family?
9. Which of the families of mappings C(R, R)
(a) {fn = x + sin(nx) : n ∈ N},
(b) {fn = n + sin(x) : n ∈ N},
(c) {fn = x
1
n : n ∈ N},
(d) {fn = n sin( x
n
) : n ∈ N}
are equicontinuous?
10. (Arzela–Ascoli theorem for mappings into R) Let K be a metrizable compact space,
the space C(K) with the topology of uniform convergence. Prove that a subset F ⊂
C(K) is compact if and only if it is closed, equicontinuous and
bounded.
11. (Arzela's theorem for mappings into R) Let K be a metrizable compact space, the space C(K) with the topology of uniform convergence, fn ∈ C(K), n ∈ N. If the sequence (fn) is an equicontinuous family and for any point x ∈ K
the set {fn(x) : n ∈ N} is bounded, then the sequence (fn) has a convergent subsequence.
12. Let a sequence of functions fn ∈ C(X, R) on a compact space X converge in the topology of pointwise convergence to a function f. Prove that if the family fn, n ∈ N,
is equicontinuous, then f is continuous, and fn converges to f in the topology of uniform convergence.
13. Prove the uniqueness of the completion of a metric space (up to isometries).
14. Find the completion of R with the metrics ρ? : R × R → R+
a) ρ1(x, y) = |e
x − e
y
|;
b) ρ2(x, y) = |arctg(x) − arctg(y)|.
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15. Find the completions of the rational and irrational numbers with the usual metric.
16. Prove that R
n is not isometric to R
m (with Euclidean metrics) for n, m ∈ N, n 6= m.
17. Prove that the family P of polynomial functions on the Tychonoff cube I
A
(a) separates points, i.e. if for any x, y ∈ X, x 6= y there exists f ∈ P such that
f(x) 6= f(y),
(b) contains all constant functions and is a ring of functions, i.e. for any f, g ∈ P
we have fg ∈ P and f + g ∈ P.
(c) the closure P of the set of polynomial functions P in the space C
∗
(I
A) in the topology
of uniform convergence is a ring of functions.
Additional problems for Assignment N 12
18. Will the product R
X in the topology of uniform convergence be a normed (Euclidean) topological space (i.e. is the topology given by a norm, an inner product)?
19. Does there exist a metric on the space of rational (irrational) numbers, whose completion is the Euclidean space R
n, n ∈ N, the Tychonoff product R
N?
20. Prove that for any n ∈ N the space R
n (with the Euclidean metric) can be
isometrically embedded in the Hilbert space `
2
.
21. For every isometry f of the Hilbert space `
2
does there exist a fixed point
of the mapping f?
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