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$L^{p}$ is uniformly convex for $1 < p < \infty$

Statement

Assume countable choice ${A C}_{ω}$ . Let $(S, A, μ)$ be a measure space and let $1 < p < \infty$ . Then both $L^{p} (μ; R)$ and $L^{p} (μ; C)$ , with their usual norms, are uniformly convex.

More precisely, for $0 < ε \leq 2$ one may use

$δ_{p} (ε) = {\begin{cases} 1 - (1 - (ε / 2)^{q})^{1 / q}, & 1 < p \leq 2, q = p / (p - 1), \\ 1 - (1 - (ε / 2)^{p})^{1 / p}, & 2 \leq p < \infty . \end{cases}$

At $p = 2$ the two formulas agree.

Facts & Assumptions

Given: Countable choice, a measure space $(S, A, μ)$ , a real number $1 < p < \infty$ , a scalar field $K \in {R, C}$ , and $0 < ε \leq 2$ .

[F1]

A real or complex Banach space is uniformly convex if every $ε \in (0, 2]$ admits a $δ > 0$ such that unit-ball vectors $x, y$ with $∥ x - y ∥ \geq ε$ satisfy $∥ (x + y) / 2 ∥ \leq 1 - δ$ (Uniformly convex Banach space).

[F2]

The usual quotient formulas give norms on real and complex $L^{p}$ . Under countable choice these spaces are complete for $1 \leq p \leq \infty$ (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Complex Holder, Minkowski, and the quotient norm, Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ , Complex Lp completeness and almost-everywhere subsequences).

[F3]

If $f, g \in L^{p} (μ; K)$ , the Clarkson inequalities (Clarkson inequalities in both exponent ranges) say

${∥ \frac{f + g}{2} ∥}_{p}^{p} + {∥ \frac{f - g}{2} ∥}_{p}^{p} \leq \frac{∥ f ∥_{p}^{p} + ∥ g ∥_{p}^{p}}{2} (p \geq 2),$

and, when $1 < p \leq 2$ and $q = p / (p - 1)$ ,

${∥ \frac{f + g}{2} ∥}_{p}^{q} + {∥ \frac{f - g}{2} ∥}_{p}^{q} \leq {(\frac{∥ f ∥_{p}^{p} + ∥ g ∥_{p}^{p}}{2})}^{q / p} .$

These assertions hold for both real and complex scalars.

[F4]

Norms are nonnegative and absolutely homogeneous (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[F5]

For $a > 0$ , $x^{a}$ is positive when $x > 0$ , has derivative $a x^{a - 1} > 0$ on $(0, \infty)$ , and is therefore strictly increasing there; the convention $0^{a} = 0$ extends this strict increase to $[0, \infty)$ (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, Continuity and derivatives of positive-base real powers, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed).

Proof

technique · Insert the lower bound on the half-difference into the appropriate Clarkson inequality and retain the resulting quantitative modulus

1.1

By [F2], $L^{p} (μ; K)$ with its usual norm is complete for either choice of $K$ . Its normed-space structure and completeness make it a real or complex Banach space, as required in [F1]. This includes the empty and zero measure spaces, whose $L^{p}$ spaces are the zero Banach space.

F1F2

1.2

Let $f, g$ lie in its closed unit ball and suppose $∥ f - g ∥_{p} \geq ε$ . Put $m = \frac{f + g}{2}, d = \frac{f - g}{2} .$ Absolute homogeneity gives $∥ d ∥_{p} = ∥ f - g ∥_{p} / 2 \geq ε / 2$ .

F4given

2.1

Suppose $p \geq 2$ . The first inequality in [F3] and $∥ f ∥_{p}, ∥ g ∥_{p} \leq 1$ give $∥ m ∥_{p}^{p} + ∥ d ∥_{p}^{p} \leq 1.$ By step 1.2 and strict increase of the positive $p$ th power, $∥ m ∥_{p}^{p} \leq 1 - ∥ d ∥_{p}^{p} \leq 1 - (ε / 2)^{p} .$ Both sides are nonnegative. If $1 - (ε / 2)^{p} = 0$ , then $∥ m ∥_{p}^{p} = 0$ and hence $∥ m ∥_{p} = 0$ . Otherwise the iterated-power law gives $((1 - (ε / 2)^{p})^{1 / p})^{p} = 1 - (ε / 2)^{p}$ , and strict increase of the positive $p$ th power yields $∥ m ∥_{p} \leq (1 - (ε / 2)^{p})^{1 / p} = 1 - δ_{p} (ε) .$ Here $0 < ε / 2 \leq 1$ . Thus $0 < (ε / 2)^{p} \leq 1$ , so $0 \leq 1 - (ε / 2)^{p} < 1$ ; applying strict increase of the positive $1 / p$ power, including its zero-base convention, shows $(1 - (ε / 2)^{p})^{1 / p} < 1$ . Hence $δ_{p} (ε) > 0$ . At $ε = 2$ the root is $0$ and $δ_{p} (2) = 1$ .

step 1.2F3F4F5

3.1

Suppose $1 < p \leq 2$ and put $q = p / (p - 1)$ . Then $q \geq 2$ . The second inequality in [F3] has right side at most $1$ , because $(∥ f ∥_{p}^{p} + ∥ g ∥_{p}^{p}) / 2 \leq 1$ and the positive $q / p$ power is increasing. Consequently $∥ m ∥_{p}^{q} + ∥ d ∥_{p}^{q} \leq 1.$ Repeating step 2.1 with $q$ in place of $p$ gives $∥ m ∥_{p} \leq (1 - (ε / 2)^{q})^{1 / q} = 1 - δ_{p} (ε),$ and the same strict-power argument proves $δ_{p} (ε) > 0$ , with $δ_{p} (2) = 1$ . When $p = 2$ , one has $q = 2$ , so this is the same modulus as in step 2.1.

step 1.2step 2.1F3F4F5

4.1

For the fixed $ε$ , choose the displayed number $δ_{p} (ε)$ by the applicable exponent range. It is an explicitly defined positive real, so this is no use of a choice principle. Steps 2.1 and 3.1 prove the implication required by [F1] for arbitrary unit-ball $f, g$ . Therefore $L^{p} (μ; K)$ is uniformly convex. The only use of ${A C}_{ω}$ is through the real and complex completeness results in [F2]; Clarkson's inequalities and the modulus calculation are choice-free.

step 1.1step 2.1step 3.1F1F2∎

Source notes

Kuriyama--Miyagi--Okada--Miyoshi prove the real and complex Clarkson inequalities and state the resulting uniform convexity on printed p. 124. The explicit modulus and endpoint check above are derived from their inequalities. The countable-choice hypothesis is added because this library's definition of uniform convexity is a property of Banach spaces and its published real and complex $L^{p}$ completeness interfaces both carry that hypothesis.

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Sources

Ken Kuriyama, Mitsuhiro Miyagi, Mari Okada and Tetsuhiko Miyoshi, Elementary proof of Clarkson's inequalities and their generalization (standard reference, not scraped)

LpLp is uniformly convex for 1<p<∞1<p<∞