552 CHAPTER 21. BANACH SPACES
=
(∫Ω
(∣∣∣∣ f +g2
∣∣∣∣q)p/q
dµ
)q/p
+
(∫Ω
(∣∣∣∣ f −g2
∣∣∣∣q)p/q
dµ
)q/p
Now p/q < 1 and so the backwards Minkowski inequality applies. Thus
≤
(∫Ω
(∣∣∣∣ f +g2
∣∣∣∣q + ∣∣∣∣ f −g2
∣∣∣∣q)p/q
dµ
)q/p
From Corollary 21.3.8,
≤
∫Ω
((12| f |p + 1
2|g|p)q/p
)p/q
dµ
q/p
=
(∫Ω
(12| f |p + 1
2|g|p)
dµ
)q/p
=
(12∥ f∥p
Lp +12∥g∥p
Lp
)q/p
■
Now with these Clarkson inequalities, it is not hard to show that all the Lp spaces areuniformly convex.
Theorem 21.3.10 The Lp spaces are uniformly convex.
Proof: First suppose p ≥ 2. Suppose ∥ fn∥Lp ,∥gn∥Lp ≤ 1 and∥∥∥ fn+gn
2
∥∥∥Lp→ 1. Then
from the first Clarkson inequality,∥∥∥∥ fn +gn
2
∥∥∥∥p
Lp+
∥∥∥∥ fn−gn
2
∥∥∥∥p
Lp≤ 1
2(∥ fn∥p
Lp +∥gn∥pLp)≤ 1
and so ∥ fn−gn∥Lp → 0.
Next suppose 1 < p < 2 and∥∥∥ fn+gn
2
∥∥∥Lp→ 1. Then from the second Clarkson inequality
∥∥∥∥ fn +gn
2
∥∥∥∥q
Lp+
∥∥∥∥ fn−gn
2
∥∥∥∥q
Lp≤(
12∥ fn∥p
Lp +12∥gn∥p
Lp
)q/p
≤ 1
which shows that ∥ fn−gn∥Lp → 0. ■
21.4 Closed SubspacesTheorem 21.4.1 Let X be a Banach space and let V = span(x1, · · · ,xn) . Then V isa closed subspace of X.
Proof: Without loss of generality, it can be assumed {x1, · · · ,xn} is linearly indepen-dent. Otherwise, delete those vectors which are in the span of the others till a linearlyindependent set is obtained. Let
x = limp→∞
n
∑k=1
cpk xk ∈V . (21.10)