1. Space of Bounded Functions and Space of Continuous functions Let X be a nonempty set. A real-valued function f : X → R is bounded if there exists M > 0 so that |f (x)| ≤ M, for all x ∈ X. The set of bounded real-valued functions on X is denoted by B(X). Given f, g ∈ B(X), and a ∈ R, we define
(f + g)(x) = f (x) + g(x), (af )(x) = af (x) for x ∈ X.
Proposition 1.1. The set B(X) forms a real vector space.
Proof. We leave it to the reader as an exercise.
Let V be a real vector space. A norm on V is a function k · k : V → [0, ∞) such that (1) kavk = |a|kvk for all a ∈ R and v ∈ V.
(2) kvk = 0 if and only if v = 0.
(3) kv + wk ≤ kvk + kwk for v, w ∈ V.
A normed vector space over R is a real vector space together with a norm. It is easy to see that a norm on V induces a metric on V by
d(v, w) = kv − wk.
The metric defined above is called the metric induced from the norm.
Definition 1.1. A normed vector space over R is called a real Banach space if the space V together with the metric induced from the norm is complete.
On B(X), set
kf k∞= sup
x∈X
|f (x)|.
Definition 1.2. If a sequence {fn} converges to f in B(X), we say that {fn} converges uniformly to f on X.
If {fn} converges to f in B(X), by definition, given > 0, there exists N > 0 so that for all n ≥ N, kfn− f k∞< . Hence for all x ∈ X, and n ≥ N, |fn(x) − f (x)| ≤ kfn− f k∞< .
In other words, {fn} converges uniformly to f if given > 0, there exists N > 0 so that for all n ≥ N, and all x ∈ X, |fn(x) − f (x)| < .
Proposition 1.2. The normed space (B(X), k · k∞) is a real Banach space.
Proof. Denote V = B(X). To show that V is complete, we need to show that every Cauchy sequence is convergent.
Let {fn} be a Cauchy sequence in V. Given > 0, there exists N > 0 such that kfn− fmk∞< . Hence for all x ∈ X,
|fn(x) − fm(x)| ≤ kfn− fmk∞< .
This implies that for each x ∈ X, the sequence of real numbers {fn(x)} is a Cauchy sequence.
Since R is complete, {fn(x)} is convergent. Let the limit of {fn(x)} be f (x), i.e. f (x) = limn→∞fn(x). Hence we obtain a function f (x) on X. On the other hand, a Cauchy sequence in a normed space must be bounded. There is M > 0 so that kfnk∞≤ M. In other words, for each x ∈ X, |fn(x)| ≤ kfnk∞≤ M. Taking n → ∞, we find |f (x)| = limn→∞|fn(x)| ≤ M, for each x ∈ X. This shows that f (x) is a bounded function on X and thus f ∈ V. For each x ∈ X, and n ≥ N,
|fn(x) − f (x)| = lim
m→∞|fn(x) − fm(x)| ≤ .
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2
This shows that for n ≥ N, kfn− f k∞ ≤ . We find f is the limit of {fn} in V. We prove
that {fn} is convergent in V.
If (X, d) is a compact metric space, we can talk more about B(X). Let C(X) be the space of all real-valued continuous functions on X. Since X is compact, every continuous function on X is bounded. Therefore C(X) is a subset of B(X). Moreover, since the sum of continuous functions on X is continuous function on X and the scalar multiplication of a continuous function by a real number is again continuous, it is easy to check that C(X) is a vector subspace of B(X).
Definition 1.3. Let (M, d) be a metric space and A be a subset of M. We say that a ∈ M is a limit point of A if there exists a sequence {an} of elements of A whose limit is a. A is said to be closed if A contains all of its limit points.
Proposition 1.3. Let (X, d) be a compact metric space. The space C(X) of real-valued continuous functions is a closed subset of the space B(X) of bounded real-valued functions on X.
Proof. To show that C(X) is closed in B(X), we only need to show that C(X) contains all of its limit points.
Let f be a bounded real-valued function so that f is a limit point of C(X). There exists {fn} in C(X) so that {fn} converges to f in B(X). To show f ∈ C(X), we need to show that f is a continuous function.
Given > 0, we can choose N > 0 so that kfN − f k∞ < /3. Since fN is uniformly continuous1, there exists δ > 0, so that if d(x, y) < δ, |fN(x) − fN(y)| < /3. If d(x, y) < δ, we see
|f (x) − f (y)| ≤ |f (x) − fN(x)| + |fN(x) − fN(y)| + |fN(y) − f (y)| ≤ 2kf − fNk∞+ 3 < .
This shows that f is uniformly continuous on X and hence continuous. Remark. It is equivalent to say that the uniform limit of a sequence of continuous functions is again continuous.
Let (M, d) be a metric space and N be a subset of M. On N, we set dN(x, y) = d(x, y), x, y ∈ N.
Then (N, dN) is again a metric space. We call (N, dN) the metric subspace of (M, d) and dN the metric induced from d.
Proposition 1.4. Let N be a closed subset of a complete metric space (M, d). Then (N, dN) is also a complete metric space.
Proof. To show that N is complete, we show that every Cauchy sequence in N has a limit in N.
Let {an} be a Cauchy sequence in N. Then {an} is a Cauchy sequence in M. Since M is complete, {an} is convergent to a point a ∈ M. This implies that a is a limit point of N.
Since N is closed, a ∈ N. Hence {an} has a limit a in N.
Corollary 1.1. The space C(X) is a real Banach space.
1A continuous function on a compact space is uniformly continuous
3
Proof. By Proposition 1.2, B(X) is complete. By Proposition 1.3, C(X) is a closed subset of B(X). By Proposition 1.4, C(X) is complete.
