📂Lebesgue Spaces

Proving that Lp Spaces are Uniformly Convex and Reflexive

Theorem¹

Let’s say $1 \lt p \lt \infty$. Then, the $L^{p}$ 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 $X$, the norm $\left\| \cdot \right\|$ is said to be uniformly convex if for all $0<\epsilon \le 2$, there exists a positive $\delta (\epsilon)>0$ such that, if $x,y \in X$ and if $| x |=|y|=1$, $| x-y| \ge \epsilon$, then it satisfies $\left\|\dfrac{ x+y}{2} \right\| \le 1-\delta (\epsilon)$.

Auxiliary theorem: Clarkson’s inequality

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

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

If $1<p \le 2$,

$$ \left\| \frac{u+v}{2}\right\|_{p}^{p^{\prime}}+ \left\| \frac{u-v}{2} \right\|_{p}^{p^{\prime}} \le \left( \frac{1}{2}| u|_{p}^{p} + \frac{1}{2}|v|_{p}^{p}\right)^> {p^{\prime}-1} \quad \cdots (2) $$

Proof

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

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

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

• Case 1. $1<p\ \le 2$

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

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

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

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

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

• Case 2. $2 \le p < \infty$

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

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

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

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

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

Auxiliary theorem

A uniformly convex Banach space is reflexive.

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

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