Page 86 - Matrices theory and applications

4.3. An Interpolation Inequality
Let B be the strip in the complex plane defined by z ∈ [0, 1]. Given z ∈ B,

define (conjugate) exponents r(z)and r (z)by
z
1 − z
1
z
1
+
=
=
,
r(z)
p
r (z)
q
q
p

Set

r + 1 − z . 69
−1+r/r(z)
X j (z):= |x j | x j = x j exp − 1 log |x j | ,
r(z)

z
Y j (z):= |y j | −1+r /r (¯) y j .
We then have

r/r(
z) r /r (
z)
X(z) r(
z) = x , Y (z) r (
z) = y .
r r
Next, define a holomorphic map in the strip B by f(z):= (AX(z),Y (z)).
It is bounded, because the numbers X j (z)and Y k (z) are. For example,
r/r(
z)
|X j (z)| = |x j |
r/p r/q
lies between |x j | and |x j | .
Let us set M(θ)= sup{|f(z)|; z = θ}. Hadamard’s three lines lemma
(see [29], Chapter 12, exercise 8) expresses that
θ → log M(θ)
is convex on (0, 1). However, r(0) = q, r(1) = p, r (0) = q , r (1) = p ,

r(θ)= r, r (θ)= r , X(θ)= x,and Y (θ)= y. Hence

θ
|(Ax, y)| = |f(θ)|≤ M(θ) ≤ M(1) M(0) 1−θ .
Now we have
M(1) = sup{|f(z)|; z =1}
≤ sup{ AX(z) r(1) Y (z) r(1) ; z =1}
=sup{ AX(z) p Y (z) p ; z =1}
≤ A p sup{ X(z) p Y (z) p ; z =1}

r/p r /p
= A p x y .
r r

r/q r /q
Likewise, M(0) ≤ A q x r y . Hence
r

θ 1−θ r(θ/p+(1−θ)/q) r (θ/p +(1−θ)/q )
|(Ax, y)| ≤ A A x y
p q r r
θ 1−θ
= A A x r y r .
p q
Finally,
|(Ax, y)| θ 1−θ
Ax r =sup ≤ A A q x r ,
p
y =0 y r
which proves the theorem.