The Metric Space 𝐶(𝐼)
Def. Let 𝑪(𝑰) = {bounded continuous functions 𝒇: 𝑰 ⊆ ℝ → ℝ}
Note: If 𝐼 is closed and bounded then any continuous function is bounded on 𝐼.
𝐶(𝐼) is a metric space with the distance defined as:
𝑑(𝑓(𝑥 ), 𝑔(𝑥 )) = sup |𝑓(𝑥 ) − 𝑔(𝑥 )|
𝑥∈𝐼
1. 𝑑(𝑓(𝑥 ), 𝑔(𝑥 )) = sup |𝑓(𝑥 ) − 𝑔(𝑥 )| ≥ 0 ; and
𝑥∈𝐼
𝑑(𝑓(𝑥 ), 𝑔(𝑥 )) = 0 implies 𝑓 (𝑥 ) = 𝑔(𝑥).
2. 𝑑(𝑓(𝑥 ), 𝑔(𝑥 )) = 𝑑(𝑔(𝑥 ), 𝑓(𝑥 )).
3. 𝑑(𝑓(𝑥 ), 𝑔(𝑥 )) ≤ 𝑑(𝑓(𝑥 ), ℎ(𝑥 )) + 𝑑(ℎ(𝑥 ), 𝑔(𝑥 )).
This is true because if 𝐴(𝑥) = 𝐵(𝑥) + 𝐸(𝑥) then by the triangle inequality:
|𝐴(𝑥)| ≤ |𝐵(𝑥)| + |𝐸 (𝑥)| for any 𝑥 ∈ 𝐼.
Thus we have: sup|𝐴(𝑥)| ≤ sup|𝐵(𝑥)| + sup|𝐸 (𝑥)|.
𝑥∈𝐼 𝑥∈𝐼 𝑥∈𝐼
Now let 𝐴(𝑥) = 𝑓(𝑥) − 𝑔(𝑥), 𝐵(𝑥) = 𝑓 (𝑥) − ℎ(𝑥), 𝐸 (𝑥) = ℎ(𝑥) − 𝑔(𝑥).
This gives us: 𝑑(𝑓(𝑥 ), 𝑔(𝑥 )) ≤ 𝑑(𝑓(𝑥 ), ℎ(𝑥 )) + 𝑑(ℎ(𝑥 ), 𝑔(𝑥 )).
Notice that a sequence of functions 𝑓𝑛 (𝑥) ∈ 𝐶(𝐼) converges to 𝑓(𝑥) with this
metric if given any 𝜖 > 0 there exists a 𝑁 ∈ ℤ+ such that if 𝑛 ≥ 𝑁 then
𝑑(𝑓𝑛 (𝑥 ), 𝑓(𝑥 )) = sup|𝑓𝑛 (𝑥 ) − 𝑓(𝑥 )| < 𝜖.
𝑥∈𝐼
This 𝜖 statement is equivalent to saying that |𝑓𝑛 (𝑥 ) − 𝑓 (𝑥 )| < 𝜖 for all 𝑥 ∈ 𝐼.
Thus convergence in 𝐶(𝐼) is the same as uniform convergence.
, 2
We already know that if 𝑓𝑛 (𝑥) converges uniformly to 𝑓(𝑥) and all of the 𝑓𝑛 (𝑥)
are continuous then so is 𝑓(𝑥). Our goal is to show that 𝐶(𝐼) is complete with the
metric described above. Thus we must show that every Cauchy sequence in 𝐶(𝐼)
converges in 𝐶(𝐼). Later in this section we’ll see that if {𝑓𝑛 (𝑥)} is a Cauchy
sequence in 𝐶(𝐼) then {𝑓𝑛 (𝑥)} converges uniformly to a function 𝑓(𝑥). Since
each 𝑓𝑛 is continuous, 𝑓 must also be continuous. We will then show that since
each 𝑓𝑛 is bounded on 𝐼 and {𝑓𝑛 (𝑥)} and converges uniformly to a function 𝑓(𝑥),
then 𝑓(𝑥) is also bounded on 𝐼. Thus any Cauchy sequence in 𝐶(𝐼) converges to
a function in 𝐶(𝐼). Hence 𝐶(𝐼) is a complete metric space.
Recall that:
Def. Let 𝑉 be a vector (linear) space. A real valued function ‖∙‖ on 𝑉 is called a
norm if for each 𝑣, 𝑤 ∈ 𝑋 and 𝛼 ∈ ℝ:
1. ‖𝑣 + 𝑤‖ ≤ ‖𝑣 ‖ + ‖𝑤‖ (Triangle inequality)
2. ‖𝛼𝑣 ‖ = |𝛼|‖𝑣 ‖ (positive homogenity)
3. ‖𝑣 ‖ ≥ 0 and ‖𝑣 ‖ = 0 if and only if 𝑣 = 0.
Ex. ℝ𝑛 is a normed linear space with 𝑣 =< 𝑎1 , … , 𝑎𝑛 >∈ ℝ𝑛 ; and
‖𝑣 ‖ = √𝑎12 + 𝑎22 + ⋯ + 𝑎𝑛2 .
Given any normed vector space 𝑉 we can always define a metric on 𝑉 by
𝑑 (𝑣, 𝑤) = ‖𝑣 − 𝑤‖.
Ex. 𝐶(𝐼) is a vector space. We can define a norm on 𝐶(𝐼) by
‖𝑓‖∞ = sup |𝑓(𝑥 )| , 𝑓 ∈ 𝐶(𝐼).
𝑥∈𝐼