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