Page 85 - Matrices theory and applications

4. Norms
68
is convex. In other words, if 1/r = θ/p +(1 − θ)/q with θ ∈ (0, 1),then
θ
1−θ
.
A r ≤ A A
q
p
Remark:
1. The proof uses the fact that K = CC. However, the norms induced
by the · p ’s on M n(IR)and M n (CC) take the same values on real
matrices, even though their definitions are different (see Exercise 6).
The statement is thus still true in M n (IR).
2. The case (p, q, r)=(1, ∞, 2) admits a direct proof. See the exercises.
3. The result still holds true in infinite dimension, at the expense of
p
some functional analysis. One even can take different L norms at
the source and target spaces. Here is an example:
D
Theorem 4.3.2 (Riesz–Thorin) Let Ω be an open set in IR and
d
ω an open set in IR .Let p 0 ,p 1 ,q 0 ,q 1 be four numbers in [1, +∞].
Let θ ∈ [0, 1] and p, q be defined by
1 1 − θ θ 1 1 − θ θ
= + , = + .
p p 0 p 1 q q 0 q 1
Consider a linear operator T defined on L ∩L (Ω), taking values in
p 0
p 1
L ∩L (ω). Assume that T can be extended as a continuous operator
q 1
q 0
q j
from L (Ω) to L (ω),withnorm M j , j =1, 2 :
p j
Tf q j
M j := sup .
f =0 f p j
p
Then T can be extended as a continuous operator from L (Ω) to
q
L (ω), and its norm is bounded above by
θ
M 1−θ M .
0 1
4. A fundamental application is the continuity of the Fourier transform

d
d
p
p
from L (IR ) into its dual L (IR )when1 ≤ p ≤ 2. We have only
to observe that (p 0 ,p 1 ,q 0 ,q 1 )= (1, 2, +∞, 2) is suitable. It can be
proved by inspection that every pair (p, q) such that the Fourier trans-
d
d
p
q

form is continuous from L (IR )into L (IR ) has the form (p, p )with
1 ≤ p ≤ 2.
5. One has analogous results for Fourier series. There lies the origin of
Riesz–Thorin theorem.
Proof (due to F. Riesz)
n
Let us fix x and y in K . We have to bound

y
|(Ax, y)| = a jk x j ¯ k .

j,k