17.4. CLOSED SUBSPACES 457

From Corollary 17.3.8,

≤

∫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 ||pLp +

12||g||pLp

)q/p

Now with these Clarkson inequalities, it is not hard to show that all the Lp spaces areuniformly convex.

Theorem 17.3.10 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

∣∣∣∣∣∣∣∣pLp+

∣∣∣∣∣∣∣∣ fn−gn

2

∣∣∣∣∣∣∣∣pLp≤ 1

2(|| fn||pLp + ||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

∣∣∣∣∣∣∣∣qLp+

∣∣∣∣∣∣∣∣ fn−gn

2

∣∣∣∣∣∣∣∣qLp≤(

12|| fn||pLp +

12||gn||pLp

)q/p

≤ 1

which shows that || fn−gn||Lp → 0.

17.4 Closed SubspacesTheorem 17.4.1 Let X be a Banach space and let V = span(x1, · · · ,xn) . Then V is a closedsubspace 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 . (17.4.9)

First suppose cp ≡(cp

1 , · · · ,cpn)

is not bounded in Fn. Then dp ≡ cp/ |cp|Fn is a unit vectorin Fn and so there exists a subsequence, still denoted by dp which converges to d where|d|= 1. Then

0 = limp→∞

x||cp|| = lim

p→∞

n

∑k=1

dpk xk =

n

∑k=1

dkxk