456 CHAPTER 17. BANACH SPACES
Corollary 17.3.8 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 17.3.5, for t = reiθ , theleft side of the above inequality is maximized when θ = 0. Hence, from Lemma 17.3.7,∣∣∣∣ 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 17.3.9 Let 2≤ p. Then∣∣∣∣∣∣∣∣ f +g2
∣∣∣∣∣∣∣∣pLp+
∣∣∣∣∣∣∣∣ f −g2
∣∣∣∣∣∣∣∣pLp≤ 1
2(|| f ||pLp + ||g||pLp
)Let 1 < p < 2. Then for 1/p+1/q = 1,∣∣∣∣∣∣∣∣ f +g
2
∣∣∣∣∣∣∣∣qLp+
∣∣∣∣∣∣∣∣ f −g2
∣∣∣∣∣∣∣∣qLp≤(
12|| f ||pLp +
12||g||pLp
)q/p
Proof: The first was established above.∣∣∣∣∣∣∣∣ f +g2
∣∣∣∣∣∣∣∣qLp+
∣∣∣∣∣∣∣∣ f −g2
∣∣∣∣∣∣∣∣qLp≤
(∫Ω
∣∣∣∣ f +g2
∣∣∣∣p dµ
)q/p
+
(∫Ω
∣∣∣∣ f −g2
∣∣∣∣p dµ
)q/p
=
(∫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