21.3. UNIFORM CONVEXITY OF Lp 551
However, this is nonnegative because it equals
s2k
p(p−1) · · ·(p−2k+1)(2k)!
−
>0︷ ︸︸ ︷(p−1)(p−2) · · ·(p−2k+1)
(2k−1)!
≥ s2k
(p(p−1) · · ·(p−2k+1)
(2k)!− (p−1)(p−2) · · ·(p−2k+1)
(2k)!
)= s2k (p−1)(p−2) · · ·(p−2k+1)
(2k)!(p−1)> 0. ■
Corollary 21.3.8 Let z,w ∈ C. Then for p ∈ (1,2) ,∣∣∣∣ z+w2
∣∣∣∣q + ∣∣∣∣ z−w2
∣∣∣∣q ≤ (12|z|p + 1
2|w|p
)q/p
Proof: One of |w| , |z| is larger. Say |w| ≥ |z| . Then dividing by |w|q , for t = z/w,showing the above inequality is equivalent to showing that for all t ∈ C, |t| ≤ 1,∣∣∣∣ t +1
2
∣∣∣∣q + ∣∣∣∣1− t2
∣∣∣∣q ≤ (12|t|p + 1
2
)q/p
Now q > 2 and so by the same argument given in proving Corollary 21.3.5, for t = reiθ , theleft side of the above inequality is maximized when θ = 0. Hence, from Lemma 21.3.7,∣∣∣∣ t +1
2
∣∣∣∣q + ∣∣∣∣1− t2
∣∣∣∣q ≤ ∣∣∣∣ |t|+12
∣∣∣∣q + ∣∣∣∣1−|t|2
∣∣∣∣q
≤(
12|t|p + 1
2
)q/p
. ■
From this the hard Clarkson inequality follows. The two Clarkson inequalities aresummarized in the following theorem.
Theorem 21.3.9 Let 2≤ p. Then∥∥∥∥ f +g2
∥∥∥∥p
Lp+
∥∥∥∥ f −g2
∥∥∥∥p
Lp≤ 1
2(∥ f∥p
Lp +∥g∥pLp)
Let 1 < p < 2. Then for 1/p+1/q = 1,∥∥∥∥ f +g2
∥∥∥∥q
Lp+
∥∥∥∥ f −g2
∥∥∥∥q
Lp≤(
12∥ f∥p
Lp +12∥g∥p
Lp
)q/p
Proof: The first was established above. Consider the second.∥∥∥∥ f +g2
∥∥∥∥q
Lp+
∥∥∥∥ f −g2
∥∥∥∥q
Lp=
(∫Ω
∣∣∣∣ f +g2
∣∣∣∣p dµ
)q/p
+
(∫Ω
∣∣∣∣ f −g2
∣∣∣∣p dµ
)q/p