454 CHAPTER 17. BANACH SPACES

a continuous periodic function for θ ∈ R which achieves its maximum value when θ = 0.This follows from the first derivative test from calculus. Therefore, for |t| ≤ 1,∣∣∣∣1+ t

2

∣∣∣∣p + ∣∣∣∣1− t2

∣∣∣∣p ≤ ∣∣∣∣1+ |t|2

∣∣∣∣p + ∣∣∣∣1−|t|2

∣∣∣∣p ≤ 12(1+ |t|p)

by the above lemma.With this corollary, here is the easy Clarkson inequality.

Theorem 17.3.6 Let p≥ 2. Then∣∣∣∣∣∣∣∣ f +g2

∣∣∣∣∣∣∣∣pLp+

∣∣∣∣∣∣∣∣ f −g2

∣∣∣∣∣∣∣∣pLp≤ 1

2(|| f ||pLp + ||g||pLp

)Proof: This follows right away from the above corollary.∫

Ω

∣∣∣∣ f +g2

∣∣∣∣p dµ +∫

Ω

∣∣∣∣ f −g2

∣∣∣∣p dµ ≤ 12

∫Ω

(| f |p + |g|p)dµ

Now it remains to consider the hard Clarkson inequalities. These pertain to p < 2. Firstis the following elementary inequality.

Lemma 17.3.7 For 1 < p < 2, the following inequality holds for all t ∈ [0,1] .∣∣∣∣1+ t2

∣∣∣∣q + ∣∣∣∣1− t2

∣∣∣∣q ≤ (12+

12|t|p)q/p

where here 1/p+1/q = 1 so q > 2.

Proof: First note that if t = 0 or 1, the inequality holds. Next observe that the maps→ 1−s

1+s maps (0,1) onto (0,1). Replace t with (1− s)/(1+ s). Then you get

∣∣∣∣ 1s+1

∣∣∣∣q + ∣∣∣∣ ss+1

∣∣∣∣q ≤ (12+

12

∣∣∣∣1− ss+1

∣∣∣∣p)q/p

Multiplying both sides by (1+ s)q , this is equivalent to showing that for all s ∈ (0,1) ,

1+ sq ≤ ((1+ s)p)q/p(

12+

12

∣∣∣∣1− ss+1

∣∣∣∣p)q/p

=

(12

)q/p

((1+ s)p +(1− s)p)q/p

This is the same as establishing

12((1+ s)p +(1− s)p)− (1+ sq)p−1 ≥ 0 (17.3.8)