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Lp spaces are uniformly convex. The case p ≥ 2. Lemma: Given p ≥ 2, there exists a constant c = c(p) > 0 such that for all real numbers f, g p p f − g p f + g p + c ≤ |f | + |g| 2 2 2 2 Proof: It is valid with c = 1 if f = 0. We can assume with out loss of generality that f > 0. Dividing through by f and denoting fg by x, we need p 1 − x p 1 + x p + c ≤ 1 + |x| 2 2 2 2 or p 1 − x p 1 |x|p 1 + x = G(x) ≥ c F (x) = + − 2 2 2 2 for −∞ < x < ∞. Clearly F (x) > 0 unless x = 1 when F (1) = 0. Check that F ′′ (1) = p(p−1) > 0. G(x) which is also 0 only at x = 1 vanishes faster there than F (x). Near ±∞, 4 1 they are both asymptotic to multiples |x|p and hence G(x) F (x) is bounded above by some c . Now p p p f (x) + g(x) p + c f (x) − g(x) ≤ |f (x)| + |g(x)| 2 2 2 2 Integrating f − g p f + g p kf kpp kgkpp + c ≤ + 2 2 2 2 p p f −g p f +g p In particular if kf kp = kgkp = 1 and 2 p ≥ 1 − δ then 2 p ≤ c−1 δ. This proves uniform convexity. 2 |1−x| The case p < 2. If we define G(x) = (1+|x|) 2−p , then we can show F (x) ≥ c G(x). The extra (1 + |x|)2−p in the denominator adjusts the behavior at ∞. We start with Z |f (x) − g(x)|2 dµ ≤ c−1 δ |f (x)|2−p + |g(x)|2−p We estimate Z Z p |f (x) − g(x)|p 2−p p + |g(x)|2−p] 2 dµ |f (x) − g(x)| dµ = p [|f (x)| 2−p 2−p [|f (x)| + |g(x)| ] 2 Z p2 p2 |f (x) − g(x)|p ≤ dµ p [|f (x)|2−p + |g(x)|2−p] 2 2 1− p2 Z 2−p 2−p 2−p p dµ × [|f (x)| + |g(x)| ] 2 −1 ≤ (c p 2 δ) × C Z Done! 1 1− p2 [|f (x)| + |g(x)| ]dµ p p