Basic Analysis – Gently Done Topological Vector Spaces

Suppose that we now equip C([0,1]) with the norm
� 1
�f�1 = |f(x)|dx.
0
8: Banach Spaces 85
One can check that this is indeed a norm but C([0,1]) is no longer complete (so
is not a Banach space). In fact, if h n is the function given by
⎧
⎪⎨
hn (x) =
⎪⎩
1,
0, 0 ≤ x ≤ 1
2
n(x− 1
2 ), 1
2
1
2
< x ≤ 1
2
+ 1
n
+ 1
n
< x ≤ 1
then one sees that (h n ) is a Cauchy sequence with respect to the norm � · � 1 .
Suppose that h n → h in (C([0,1]),�·� 1 ) as n → ∞. Then
� 1/2 � 1/2
|h(x)|dx =
0
0
|h(x)−h n (x)|dx ≤ �h−h n � 1 → 0
and so we see that h vanishes on the interval [0, 1
1
]. Similarly, for any 0 < ε < 2 2 ,
we have � 1
1
2 +ε
|h(x)−1|dx ≤ �h−h n�1 → 0
+ ε,1], for
0 < ε < 1
2 . This means that h is equal to 1 on the interval [1
2 ,1]. But such a
function h is not continuous, so we conclude that C([0,1]) is not complete with
respect to the norm �·� 1 .
as n → ∞. Therefore h is equal to 1 on any interval of the form [ 1
2
3. Let S be any (non-empty) set and let X denote the set of bounded complexvalued
functions on S. Then X is a Banach space when equipped with the supremum
norm �f� = sup{|f(s)| : s ∈ S} (and the usual pointwise linear structure).
In particular, if we take S = N, then X is the linear space of bounded complex
sequences. This Banach space is denoted ℓ ∞ (or sometimes ℓ ∞ (N)). With S = Z,
the resulting Banach space is denoted ℓ ∞ (Z).
4. The set of complex sequences, x = (x n ), satisfying
�x� 1 =
∞�
|xn | < ∞
n=1
is a linear space under componentwise operations (and �·� 1 is a norm). Moreover,
one can check that the resulting normed space is complete. This Banach space is
denoted ℓ 1 .