Page 80 - Matrices theory and applications

which gives (4.1).
For p = 2, the norm · 2 is given by a Hermitian form and thus satisfies
the Cauchy–Schwarz inequality:
|(x, y)|≤ x 2 y 2.
o
This is a particular case of H¨lder’s inequality. 4.1. A Brief Review 63

Proposition 4.1.2 For conjugate exponents p, p , one has
(x, y)
x p =sup .
y =0 y p
Proof
The inequality ≥ is a consequence of H¨older’s. The reverse inequality is
p−2
obtained by taking y j =¯x j |x j | if p< ∞.If p = ∞,choose y j =¯x j for
an index j such that |x j | = x ∞ .For k = j,take y k =0.
Definition 4.1.1 Two norms N and N on a (real or complex) vector

space are said to be equivalent if there exist two numbers c, c ∈ IR such
that
N ≤ cN , N ≤ c N.

The equivalence between norms is obviously an equivalence relation, as
its name implies. As announced above, we have the following result.
n
Proposition 4.1.3 All norms on E = K are equivalent. For example,
x ∞ ≤ x p ≤ n 1/p x ∞ .
Proof
It is sufficient to show that every norm is equivalent to · 1 .
Let N be a norm on E.If x ∈ E, the triangle inequality gives
i
N(x) ≤ |x i |N(e ),
i
n
1
where (e ,... , e ) is the canonical basis. One thus has N ≤ c · 1 for
i
c := max i N(e ). Observe that this first inequality expresses the fact that
N is Lipschitz (hence continuous) on the metric space X =(E, · 1 ).
For the reverse inequality, we reduce ad absurdum: Let us assume that
the supremum of x 1 /N(x) is infinite for x = 0. By homogeneity, there
m
m
would then exist a sequence of vectors (x ) m∈IN such that x 1 =1 and
m
N(x ) → 0when m → +∞. Since the unit sphere of X is compact, one
may assume (up to the extraction of a subsequence) that x m converges to
a vector x such that x 1 =1. In particular, x =0. Since N is continuous
m
on X, one has also N(x) = lim m→+∞ N(x ) = 0. Since N is a norm, we
deduce x = 0, a contradiction.