24 0. A first look at Banach and Hilbert spaces
The corresponding inner product space is denoted by Lcont(I).
2
Note that
we have
(0.50) f |b a|f

and hence the maximum norm is stronger than the Lcont
2
norm.
Suppose we have two norms .
1
and .
2
on a vector space X. Then
.
2
is said to be stronger than .
1
if there is a constant m 0 such that
(0.51) f
1
m f 2.
It is straightforward to check the following.
Lemma 0.24. If .
2
is stronger than . 1, then every .
2
Cauchy sequence
is also a .
1
Cauchy sequence.
Hence if a function F : X Y is continuous in (X, . 1), it is also
continuous in (X, . 2), and if a set is dense in (X, . 2), it is also dense in
(X, . 1).
In particular, Lcont
2
is separable. But is it also complete? Unfortunately
the answer is no:
Example. Take I = [0, 2] and define
(0.52) fn(x) =

⎪0,



0 x 1
1
n
,
1 + n(x 1), 1
1
n
x 1,
1, 1 x 2.
Then fn(x) is a Cauchy sequence in Lcont,
2
but there is no limit in
Lcont!2
Clearly the limit should be the step function which is 0 for 0 x 1 and 1
for 1 x 2, but this step function is discontinuous (Problem 0.25)!
This shows that in infinite dimensional vector spaces, different norms
will give rise to different convergent sequences! In fact, the key to solving
problems in infinite dimensional spaces is often finding the right norm! This
is something which cannot happen in the finite dimensional case.
Theorem 0.25. If X is a finite dimensional vector space, then all norms
are equivalent. That is, for any two given norms .
1
and . 2, there are
positive constants m1 and m2 such that
(0.53)
1
m2
f
1
f
2
m1 f 1.
Proof. Since equivalence of norms is an equivalence relation (check this!) we
can assume that .
2
is the usual Euclidean norm. Moreover, we can choose
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