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Proving that Lp Spaces are Uniformly Convex and Reflexive 📂Lebesgue Spaces

Proving that Lp Spaces are Uniformly Convex and Reflexive

Theorem1

Let’s say 1<p<1 \lt p \lt \infty. Then, the LpL^{p} space is uniformly convex and reflexive.

Explanation

It can be proven using the definition of uniform convexity and the Clarkson inequality. Thanks to the Clarkson inequality, the proof ends easily and briefly. It feels like a finishing move.

Uniformly convex

On the norm space XX, the norm \left\| \cdot \right\| is said to be uniformly convex if for all 0<ϵ20<\epsilon \le 2, there exists a positive δ(ϵ)>0\delta (\epsilon)>0 such that, if x,yXx,y \in X and if x=y=1| x |=|y|=1, xyϵ| x-y| \ge \epsilon, then it satisfies x+y21δ(ϵ)\left\|\dfrac{ x+y}{2} \right\| \le 1-\delta (\epsilon).

Auxiliary theorem: Clarkson’s inequality

Let’s say u,vL p(Ω)u,v\in {L}^{\ p}(\Omega). Also, assume it satisfies 1p+1p=1\frac{1}{p}+\frac{1}{p^{\prime}}=1. If 2p<2\le p <\infty,

u+v2pp+uv2pp12upp+12vpp(1) \left\| \frac{u+v}{2}\right\|_{p}^{p}+ \left\| \frac{u-v}{2} \right\|_{p}^{p} \le \frac{1}{2}| u|_{p}^{p} + \frac{1}{2}|v|_{p}^{p} \quad \cdots (1)

If 1<p21<p \le 2,

u+v2pp+uv2pp(12upp+12vpp)>p1(2) \left\| \frac{u+v}{2}\right\|_{p}^{p^{\prime}}+ \left\| \frac{u-v}{2} \right\|_{p}^{p^{\prime}} \le \left( \frac{1}{2}| u|_{p}^{p} + \frac{1}{2}|v|_{p}^{p}\right)^> {p^{\prime}-1} \quad \cdots (2)

Proof

Suppose 0<ϵ<20 < \epsilon <2 is given. And suppose u, vLpu,\ v \in {L}^{p} is up=vq=1\| u \|_{p}=\left\| v \right\|_{q}=1 and satisfy uvpϵ|u-v|_{p} \ge \epsilon . Then, by the definition of uniform convexity,

x+y2p1δ \left\| \frac{x+y}{2} \right\|_{p} \le 1-\delta

It suffices to show that there exists δ=δ(ϵ)>0\delta=\delta (\epsilon)>0 satisfying this.

  • Case 1. 1<p 21<p\ \le 2

    By assumption, uv2pϵ2\displaystyle \left\| \frac{u-v}{2} \right\|_{p} \ge \frac{\epsilon}{2}. Then, by Clarkson’s inequality (2)(2),

    u+v2ppuv2pp+(12upp+12vqp)p1(ϵ2)p+(12+12)p1= 1(ϵ2)p \begin{align*} \left\| \frac{u+v}{2} \right\|_{p}^{p^{\prime}} \le & -\left\| \frac{u-v}{2}\right\|_{p}^{p^{\prime}} + \left( \frac{1}{2} \| u \|_{p}^{p} +\frac{1}{2}\left\| v \right\|_{q}^{p} \right)^{p^{\prime}-1} \\ \le& -\left( \frac{\epsilon}{2}\right)^{p^{\prime}} +\left( \frac{1}{2} + \frac{1}{2} \right)^{p^{\prime}-1} \\ =&\ 1-\left( \frac{\epsilon}{2} \right)^{p^{\prime}} \end{align*}

    Since 0<ϵ2<10 < \frac{\epsilon}{2} < 1, we have 1(ϵ2)p<11-\left( \frac{\epsilon}{2} \right)^{p^{\prime}}<1 and

    u+v2p[1(ϵ2)p]1p=1δ(ϵ) \left\| \frac{u+v}{2} \right\|_{p} \le \left[ 1-\left( \frac{\epsilon}{2} \right)^{p^{\prime}}\right]^{\frac{1}{p^{\prime}}} = 1-\delta (\epsilon)

    A positive δ(ϵ)>0\delta (\epsilon)>0 exists that satisfies this.

  • Case 2. 2p<2 \le p < \infty

    Similarly, by assumption, uv2pϵ2\displaystyle \left\| \frac{u-v}{2} \right\|_{p} \ge \frac{\epsilon}{2}. Then, by Clarkson’s inequality (1)(1),

    u+v2ppuv2pp+12upp+12vqp(ϵ2)p+12+12= 1(ϵ2)p \begin{align*} \left\| \frac{u+v}{2} \right\|_{p}^{p} \le& -\left\| \frac{u-v}{2}\right\|_{p}^{p} + \frac{1}{2} \| u \|_{p}^{p} +\frac{1}{2}\left\| v \right\|_{q}^{p} \\ \le & -\left( \frac{\epsilon}{2}\right)^{p} +\frac{1}{2} + \frac{1}{2} \\ =&\ 1-\left( \frac{\epsilon}{2} \right)^{p} \end{align*}

    Similarly, since 0<ϵ2<10 < \frac{\epsilon}{2} < 1, we have 1(ϵ2)p<11-\left( \frac{\epsilon}{2} \right)^{p}<1 and

    u+v2p[1(ϵ2)p]1p=1δ(ϵ) \left\| \frac{u+v}{2} \right\|_{p} \le \left[ 1-\left( \frac{\epsilon}{2} \right)^{p}\right]^{\frac{1}{p}} = 1-\delta (\epsilon)

    A positive δ(ϵ)>0\delta (\epsilon)>0 exists that satisfies this. Since there exists 0<δ=δ(ϵ)0 < \delta =\delta (\epsilon) in both cases, the LpL^{p} space is uniformly convex.

Auxiliary theorem

A uniformly convex Banach space is reflexive.

The Lp{L}^{p} space is a Banach space and uniformly convex, so it is reflexive by the auxiliary theorem.


  1. Robert A. Adams and John J. F. Foutnier, Sobolev Space (2nd Edition, 2003), p45 ↩︎