LU - Department of Mathematics

Dr H. Gebran

Math 400 Exercises of chapter 1 – Statements and solutions Fall 2022-2023

1. Let A be a subset of a metric space. Show that A is totally bounded if and only if Ā is totally
bounded.

Solution. Suppose first that A is totally bounded Snand let ε > 0 be given. SThen, there are
n ′ (x , ε) ⊂
finitely
Sn many points x 1 , . . . , x n in A such that A ⊂ i=1 B(x i , ε). Then Ā ⊂ i=1 B i
i=1 B(xi , 2ε) and of course x1 , . . . , xn ∈ Ā. Since ε was arbitrary, this means that Ā is totally
bounded.
Suppose conversely that Ā is totally boundedSand let ε > 0 be given. Then there are finitely
many points x1 , . . . , xn ∈ Ā such that Ā ⊂ ni=1 B(xi , ε). Now for each i = 1, . . . , n, there
exists yi ∈ A such that d(xi , yi ) < ε (by a property of the closure). So if x ∈ A, then x ∈ Ā, and
then x ∈ B(xj , ε) for some index j and so d(x, xj ) < ε. By the triangle inequality, d(x, yj ) < 2ε.
This means that A ⊂ ni=1 B(yi , 2ε). Since ε was arbitrary, this means that A is totally bounded
S
(the point here is that the centers of the balls should belong to A).

2. A topological space is said to be separable if it contains a countable dense subset. Show that a
totally bounded metric space is separable.

Solution. Take ε = n1 for n ∈ N∗ in the S definition of total boundedness. Then there is a

finite set of points An in X such that X = a∈An B(a, n1 ). Let A = ∞
S
n=1 A n . Then first, A is
countable as a countable union of finite sets. Second, if x ∈ X, then, d(x, A) ≤ d(x, An ) < n1 .
Since n was arbitrary, we get d(x, A) = 0 and so x ∈ Ā. Thus Ā = X and so A is dense in X.

R1
3. Let F = {f ∈ C 1 [0, 1]; 0 |f ′ (t)|2 dt ≤ 1}. Show that F is equicontinuous.

Solution. Let x, y ∈ [0, 1] and fR∈ F. Assume first that x ≤ y. Then, by the fundamental
y
theorem of calculus, f (y) − f (x) = x f ′ (t) dt. By Cauchy-Schwarz inequality, we get

y 1/2 Z y 1/2 1 1/2

√ √
Z Z
2 ′ 2 ′ 2
|f (y) − f (x)| ≤ 1 dt |f (t)| dt ≤ y−x |f (t)| dt ≤ y − x.
x x 0

Interchanging x and y, we get

|f (x) − f (y)| ≤ |x − y|1/2

for any x, y ∈ [0, 1] and any f ∈ F. Therefore, given ε > 0, let η = ε2 . If |x − y| < η,
then |x − y|1/2 < ε and so |f (x) − f (y)| < ε by the above inequality. Thus, F is uniformly
equicontinuous on [0,1].

4. Let fn (t) = e−nt for n ∈ N∗ and t ∈ [0, 1]. Show that the collection {fn | n ∈ N∗ } is
equicontinuous on ]0,1] but not at zero.

Solution. Let x ∈]0, 1]. We will prove equicontinuity at x. So let y ∈]0, 1]. By the
mean value theorem fn (y) − fn (x) = fn′ (c)(y − x) for some c between x and y (c depends

1
 on x, y and n). But fn′ (t) = −ne−nt . Therefore fn (y) − fn (x) = −ne−nc (y − x) and so
|fn (y) − fn (x)| = ne−nc |y − x|. Let ε > 0 be given.
Suppose first that x < y, then x < c < y so −c < −x and so e−nc < e−nx . Now ne−nx → 0
as n → ∞. In particular, ne−nx is bounded by a certain constant Mx (that depends on x). It
follows that |fn (y) − fn (x)| ≤ Mx |y − x| for every n and every y > x. Suppose now that y < x.
If y sufficiently close to x, then x2 < y < x (this is equivalent to the requirement |y − x| < x2 ).
In this case, x2 < y < c < x and so e−nc < e−nx/2 . Again, as we showed above, there exists a
positive constant Kx = Mx/2 such that |fn (y) − fn (x)| ≤ Kx |y − x| provided |y − x| < x2 .
In conclusion if η = min( Mεx , Kεx , x2 ) then |x − y| < η ⇒ |fn (y) − fn (x)| < ε for every n ∈ N∗ .
This proves the equicontinuity at x of the collection {fn | n ∈ N∗ }.
Remark. Actually ne−nx ≤ ex1
for every x > 0 and n ∈ N∗ . Indeed we claim first that
te ≤ e for every t > 0. Indeed, let g(t) = te−t , then g ′ (t) = (1 − t)e−t . It follows that g has
−t −1

a maximum at t = 1. Thus the claim is proved. In particular nxe−nc ≤ e−1 and so ne−nx ≤
ex 1
.
1 2 εex x
Therefore, we can take Mx = ex and Kx = ex . It follows that we can take η = min 2 , 2 .

Now we show that {fn } is not equicontinuous at 0. Indeed, |fn (x) − fn (0)| = 1 − e−nx . Given
1 > α > 0, let x = α/2. Take n sufficiently large so that e−nα/2 < 1/2. Then 1 − e−nx > 12 .
Thus we proved that
1
∃ε = ∀α > 0 ∃x ∈]0, 1] ∃n ∈ N∗ |x − 0| < α and |fn (x) − fn (0)| > ε.
2
Therefore {fn } is not equicontinuous at 0.

n R1 R1 o
5. Let F = f ∈ C 1 [0, 1]; 0 |f (t)|2 dt ≤ 1 and 0 |f ′ (t)|2 dt ≤ 1 . Show that F is relatively
compact in C[0, 1].

Solution. F is equicontinuous because it is a subset of an equicontinuous family (exercise

3). We show that F is pointwise bounded. Let f ∈ F and let x ∈ [0, 1]. Since |f | is continuous,
it achieves its minimum at a point x0 . ThereforeR|f (x0 )| ≤ |f (t)| and so |f (x0 )|2 ≤ |f (t)|2 .
1
Integrating this inequality and using the condition 0 |f (t)|2 dt ≤ 1, we get |f (x0 )| ≤ 1.
Rx
Now f (x) − f (x0 ) = x0 f ′ (t) dt. Therefore by Cauchy-Schwarz inequality,
Z x 1/2 Z x 1/2
2 ′ 2
|f (x) − f (x0 )| ≤ 1 dt |f (t)| dt ≤ 1.
x0 x0

It follows from the triangle inequality that |f (x)| ≤ |f (x) − f (x0 )| + |f (x0 )| ≤ 2. Since f and x
where arbitrary, F is pointwise bounded. By the Arzela-Ascoli theorem, it is relatively compact.

6. Let X and Y be two metric spaces and let (fn ) be a sequence of functions from X to Y . Suppose
that the collection {fn | n ∈ N∗ } is equicontinuous at a point a ∈ X and that (fn (a)) converges
to some b ∈ Y . Let (xn ) be a sequence converging to a. Show that the sequence (fn (xn ))
converges to b. Show by an example that this result does not hold when the equicontinuity
condition is removed.

Solution. By the triangle inequality, d(fn (xn ), b) ≤ d(fn (xn ), fn (a)) + d(fn (a), b). The first
term can be made arbitrarily small by equicontinuity (because xn → a) and the second term can
be made arbitrarily small by assumption.
1
Take X = [0, 1], a = 0, Y = R, fn (x) = (1 + x)n , xn = n. Then fn (a) = 1 → 1 but
fn (xn ) = (1 + n1 )n → e.

2
7. Let X be a metric space and let F be an equicontinuous subset of C(X, R). Let

Fx = {f (x); f ∈ F} and A = {x ∈ X; Fx is bounded} .

(a) Show that A is open and closed in X.

(b) Suppose that X is compact and connected and Fx0 is bounded for some x0 ∈ X. Show
that F is relatively compact.

Solution. a) A is open. Let x ∈ A. Then there exists M > 0 such that |f (x)| ≤ M for all
f ∈ F. By the equicontinuity condition there exists α > 0 such that |f (x) − f (y)| < 1 (∀f ∈ F)
whenever d(x, y) < α. By the triangle inequality |f (y)| ≤ |f (y) − f (x)| + |f (x)| < 1 + M for
all y ∈ B(x, α). So Fy is bounded for all y in a neighborhood of x.

A is closed. Let x ∈ Ā. Then there exists a sequence (xn ) in A converging to x. Again
by equicontinuity, there exists α > 0 such that |f (x) − f (y)| < 1 whenever d(x, y) < α.
Now choose and fix an integer n large enough so that d(xn , x) < α. Since xn ∈ A there
exists a constant Mn depending on n such that |f (xn )| ≤ Mn . By the triangle inequality
|f (x)| ≤ |f (x) − f (xn )| + f (xn )| ≤ 1 + Mn . This holds for every f ∈ F. Thus Fx is bounded
and so x ∈ A.

b) Connectedness implies that A = X. So the assumptions of the Arzela-Ascoli theorem are

satisfied. Hence the conclusion.

8. Let E denote the space of continuous and bounded functions f : [0, ∞[→ R. We
√ equip E with
∗
the norm of uniform convergence. For x ≥ 0 and n ∈ N , we set fn (x) = sin x + n2 π 2 and
we consider the collection F = {f1 , f2 , . . .} ⊂ E.

(a) Show that (fn ) converges pointwise to 0. Hint. Use the facts that sin(nπ) = 0 and
| sin a − sin b| ≤ |a − b|.
(b) Check that for every x ≥ 0, Fx is bounded.
(c) Show that F is equicontinuous.
(d) Show that ||fn ||∞ = 1.
(e) Deduce that F is not relatively compact in E. Hint. Reason by contradiction.
(f) Does this contradict the Arzela-Ascoli theorem?

Solution. (a) We can write

p
|fn (x) − 0| = sin x + n2 π 2 − sin nπ
p
≤ x + n2 π 2 − nπ
x
=√ −→ 0 as n → ∞.
x + n2 π 2 + nπ

(b) We have p
Fx = {sin x + n2 π 2 ; n ∈ N∗ } ⊂ [−1, 1].
This means that Fx is bounded.

3
 (c) We can write
p p
|fn (x) − fn (y)| = sin x + n2 π 2 − sin y + n2 π 2
p p
≤ x + n2 π 2 − y + n2 π 2
|x − y|
=√ p
x + n2 π 2 + y + n2 π 2
|x − y|
≤
2nπ
|x − y|
≤ .
2π
This proves that F is uniformly equicontinuous on [0, +∞[ (we can take η = 2πε in the definition
of uniform equicontinuity).

(d) For every x ≥ 0, We have

p
|fn (x)| = | sin x + n2 π 2 | ≤ 1.

This implies that ||fn ||∞ ≤ 1. On the other hand, if x = ( π2 + 2nπ)2 − n2 π 2 , we have
√
x + n2 π 2 = π2 + 2nπ and so fn (x) = 1. This implies that ||fn ||∞ ≥ 1. Hence equality.

(e) Suppose that F is relatively compact. Then the sequence (fn ) has a subsequence (fnk ) that
converges uniformly (and so pointwise) to a limit f . Since (fn ) converges pointwise to 0, then
(fnk ) also converges pointwise to 0. By uniqueness of limits, f = 0. However by the previous
question ||fnk ||∞ = 1 and so ||f ||∞ = 1. A contradiction.

(f) This does not contradict the Arzela-Ascoli theorem because the underlying space [0, +∞[ is
not compact.

9. Let X be a compact metric space and let {f1 , . . . fn } ⊂ C(X, R) be a finite collection that
separates the points of X. Show that X is homeomorphic to a subset of Rn .

Solution. Let Φ : X → Rn be defined by Φ(x) = (f1 (x), . . . , fn (x)). The Φ is continuous

because each component is continuous. Next, Φ is injective because if x ̸= y, then fk (x) ̸= fk (y)
for some k ∈ {1, . . . , n} and so Φ(x) ̸= Φ(y). So Φ is a continuous bijection from X to Φ(X).
Since X is compact, a theorem from topology implies that Φ is a homeomorphism (Φ is a closed
map).

10. Let a and b be two real numbers such that a < b and let f : [a, b] → R be a continuous function.

(a) Show that there exists a sequence (Pn ) of polynomials such that
Z b Z b
lim f (x)Pn (x) dx = f 2 (x) dx.
n→∞ a a

Z b
(b) Suppose that f (x)xk dx = 0 for all k ∈ N. Show that f = 0.
a

4
 Solution. (a) By the Weierstrass approximation theorem, there exists a sequence (Pn ) of
polynomials such that ||Pn − f ||∞ := maxa≤x≤b |Pn (x) − f (x)| → 0. Now,
Z b Z b Z b
2
f (x)Pn (x) dx − f (x) dx = f (x) (Pn (x) − f (x)) dx
a a a
Z b
≤ |f (x)||Pn (x) − f (x)| dx
a
Z b
≤ |f (x)|||Pn − f ||∞ dx
a
= ||Pn − f ||∞ ||f ||1 → 0.

Hence the conclusion.

Z b
(b) The assumption and the linearity of the integral imply that f (x)P (x) dx = 0 for any
a
polynomial P . In particular if (Pn ) is a sequence of polynomials converging uniformly to f , then
Z b
Rb
f (x)Pn (x) dx = 0. According to the previous question a f 2 (x) dx = 0. Continuity of f
a
implies that f = 0.

11. Let (X, d) be a compact metric space and let

A = {f : X → R; ∃kf , |f (x) − f (y)| ≤ kf d(x, y) ∀x, y ∈ X} .

(a) Show that the A is a sub-algebra of C(X, R).

(b) Show that A separates the points of X.
(c) Show that A is dense in C(X, R).

Solution. (a) Observe first A is just the set of all Lipschitz continuous functions on X. So
A ⊂ C(X, R). Next, let f, g ∈ A, α ∈ R and x, y ∈ X. Then

|(f + g)(x) − (f + g)(y)| ≤ (kf + kg )d(x, y),

and
|(αf )(x) − (αf )(y)| ≤ |α|kf d(x, y).
This proves that f + g and αf ∈ A. Finally,

|f (x)g(x) − f (y)g(y)| ≤ |f (x)g(x) − f (x)g(y)| + |f (x)g(y) − f (y)g(y)|

≤ ||f ||∞ |g(x) − g(y)| + ||g||∞ |f (x) − f (y)|
≤ ||f ||∞ kg d(x, y) + ||g||∞ kf d(x, y)
= (||f ||∞ kg + ||g||∞ kf ) d(x, y).

This proves that f g ∈ A.

(b) Let x ̸= y. Consider the function f : X → R defined by f (z) = d(x, z). Then f ∈ A
because
|f (z1 ) − f (z2 )| = |d(x, z1 ) − d(x, z2 )| ≤ d(z1 , z2 ).
On the other hand, f (x) = d(x, x) = 0 and f (y) = d(x, y) > 0.

(c) Constant functions are indeed Lipschitz continuous. It follows from the Stone-Weierstrass
theorem that A is dense in C(X, R).

5
12. The purpose of the exercise is to give a counterexample of the complex Stone-Weierstrass theo-
rem. Let S 1 denote the unit circle |z| = 1. Consider the set A of all polynomials in the complex
variable z (restricted to S 1 ).

(a) Show that A is a subalgebra of C(S 1 , C) that contains the constant functions and separates
the points of S 1 .
Z
1
(b) Consider the function f : S → C defined by f (z) = z. Compute the line integral f (z) dz
S1
where S 1 is positively oriented.
Z
(c) Let P ∈ A. Compute P (z) dz.
S1
(d) Deduce that f cannot be approximated uniformly by elements of A and conclude.

Solution. (a) The sum and product of two polynomials in z is again a polynomial in z. A
complex multiple of a polynomial in z is again a polynomial in z. The constant function is a
polynomial in z. Finally let z1 , z2 be two distinct points in S 1 . Let P (z) = z. Then P is a
polynomial in z and P (z1 ) ̸= P (z2 ).

(b) A positive parametrization of S 1 is given by γ(t) = eit for t ∈ [0, 2π]. Therefore
Z Z Z 2π
f (z) dz = z dz = e−it ieit dt = 2πi.
S1 γ 0

Z
(c) Since P has an antiderivative on C, we know from M3304 that P (z) dz = 0.
S1

(d) Suppose that f can be approximated uniformly on S 1 by polynomials and let 0 < ε < 1.
Then there exists P ∈ A such that supz∈S 1 |f (z) − P (z)| ≤ ε. It follows from the basic estimate
of line integrals that
Z Z Z


f (z) dz − P (z) dz = f (z) − P (z) dz ≤ 2πε.
S1 S1 S1

However, by the previous two questions

Z Z
f (z) dz − P (z) dz = 2πi.
S1 S1

It follows that 2π ≤ 2πε and so 1 ≤ ε, contradicting our choice of ε. Therefore, f (which is

continuous) cannot be uniformly approximated by polynomials and so A is not dense in C(S 1 , C)
with the supremum distance.

Therefore an additional assumption is needed in order to save to complex form of the Stone-
Weierstrass theorem and this is the purpose of the next exercise.

13. The complex form of the Stone-Weierstrass theorem. Let X be a compact metric space
and let A be a subalgebra of C(X, C). Suppose that A separates the points of X, contains the
constant functions and moreover is closed under complex conjugation (that is f ∈ A ⇒ f ∈ A).
Let A1 = {Re(f ); f ∈ A} and A2 = {Im(f ); f ∈ A}.

(a) Show that A1 = A2 ⊂ A.

(b) Show that A1 is dense in C(X, R) (with the usual distance).

6
(c) Deduce that A is dense in C(X, C).

Solution. (a) Let f ∈ A. Then Re(f ) = 12 (f + f ) ∈ A. This means that A1 ⊂ A. Similarly

for A2 . Since Im(f ) = Re(−if ) and Re(f ) = Im(i f ) we see that A1 = A2 .

(b) First A1 is a real subalgebra of C(X, R). This is because

• Re(f ) + Re(g) = Re(f + g);

• αRe(f ) = Re(αf ) if α ∈ R;
• Re(f )Re(g) = Re 12 f + f g .



Next, A1 contains the real constant functions, and it separates the points of X. Indeed, let
x ̸= y. Then there is f ∈ A such that f (x) ̸= f (y); so either Re(f )(x) ̸= Re(f )(y) or
Im(f )(x) ̸= Im(f )(y). In both cases, x and y can be separated by a function in A1 = A2 . By
the Stone-Weierstrass theorem A1 is dense in C(X, R).

(c) Let f ∈ A. Then Re(f ) and Im(f ) can be approximated uniformly respectively by functions
g and h in A1 . Therefore f can be approximated by g + ih ∈ A (use the triangle inequality).