Page 79 - Matrices theory and applications

4. Norms
62
Furthermore, one has H¨older’s inequality
1
1
|(x, y)|≤ x p y p ,
=1.
(4.2)
+
p
p

The numbers p, p are called conjugate exponents.

Proof
o
Everything except the H¨lder and Minkowski inequalities is obvious.
When p =1 or p = ∞, these inequalities are trivial. We thus assume
that 1 <p< ∞.
Let us begin with (4.2). If x or y is null, it is obvious. Indeed, one can
even assume, by decreasing the value of n,that noneof the x j ,y j ’s is null.

Likewise, since |(x, y)|≤ |x j ||y j |, one can also assume that the x j ,y j are
j
real and positive. Dividing by x p and by y p , one may restrict attention
to the case where x p = y p = 1. Hence, x j ,y j ∈ (0, 1] for every j.Let
us define

a j = p log x j , b j = p log y j .
Since the exponential function is convex,
1 1

a j /p+b j /p
b j
e ≤ e a j + e ,
p p
that is,
1 p 1 p
y .
x j y j ≤ x + j
j
p p
Summing over j,weobtain
1 p 1 p 1 1
(x, y) ≤ x + y = + =1,
p
p
p p p p
which proves (4.2).
We now turn to (4.1). First, we have
p p−1 p−1
p
x + y = |x k + y k | ≤ |x k ||x k + y k | + |y k ||x k + y k | .
p
k k k
Let us apply H¨older’s inequality to each of the two terms of the right-hand
side. For example,
1/p
p−1 (p−1)p

|x k ||x k + y k | ≤ x p |x k + y k | ,
k k
which amounts to
p−1 p−1

|x k ||x k + y k | ≤ x p x + y .
p
k
Finally,
p p−1
x + y ≤ ( x p + y p) x + y ,
p p