Proving that Lp Spaces are Uniformly Convex and Reflexive
📂Lebesgue Spaces Proving that Lp Spaces are Uniformly Convex and Reflexive Theorem Let’s say 1 < p < ∞ 1 \lt p \lt \infty 1 < p < ∞ . Then, the L p L^{p} 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 X X , the norm ∥ ⋅ ∥ \left\| \cdot \right\| ∥ ⋅ ∥ is said to be uniformly convex if for all 0 < ϵ ≤ 2 0<\epsilon \le 2 0 < ϵ ≤ 2 , there exists a positive δ ( ϵ ) > 0 \delta (\epsilon)>0 δ ( ϵ ) > 0 such that, if x , y ∈ X x,y \in X x , y ∈ X and if ∣ x ∣ = ∣ y ∣ = 1 | x |=|y|=1 ∣ x ∣ = ∣ y ∣ = 1 , ∣ x − y ∣ ≥ ϵ | x-y| \ge \epsilon ∣ x − y ∣ ≥ ϵ , then it satisfies ∥ x + y 2 ∥ ≤ 1 − δ ( ϵ ) \left\|\dfrac{ x+y}{2} \right\| \le 1-\delta (\epsilon) 2 x + y ≤ 1 − δ ( ϵ ) .
Auxiliary theorem: Clarkson’s inequality
Let’s say u , v ∈ L p ( Ω ) u,v\in {L}^{\ p}(\Omega) u , v ∈ L p ( Ω ) . Also, assume it satisfies 1 p + 1 p ′ = 1 \frac{1}{p}+\frac{1}{p^{\prime}}=1 p 1 + p ′ 1 = 1 . If 2 ≤ p < ∞ 2\le p <\infty 2 ≤ p < ∞ ,
∥ u + v 2 ∥ p p + ∥ u − v 2 ∥ p p ≤ 1 2 ∣ u ∣ p p + 1 2 ∣ v ∣ p p ⋯ ( 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)
2 u + v p p + 2 u − v p p ≤ 2 1 ∣ u ∣ p p + 2 1 ∣ v ∣ p p ⋯ ( 1 )
If 1 < p ≤ 2 1<p \le 2 1 < p ≤ 2 ,
∥ u + v 2 ∥ p p ′ + ∥ u − v 2 ∥ p p ′ ≤ ( 1 2 ∣ u ∣ p p + 1 2 ∣ v ∣ p p ) > p ′ − 1 ⋯ ( 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)
2 u + v p p ′ + 2 u − v p p ′ ≤ ( 2 1 ∣ u ∣ p p + 2 1 ∣ v ∣ p p ) > p ′ − 1 ⋯ ( 2 )
Proof Suppose 0 < ϵ < 2 0 < \epsilon <2 0 < ϵ < 2 is given. And suppose u , v ∈ L p u,\ v \in {L}^{p} u , v ∈ L p is ∥ u ∥ p = ∥ v ∥ q = 1 \| u \|_{p}=\left\| v \right\|_{q}=1 ∥ u ∥ p = ∥ v ∥ q = 1 and satisfy ∣ u − v ∣ p ≥ ϵ |u-v|_{p} \ge \epsilon ∣ u − v ∣ p ≥ ϵ . Then, by the definition of uniform convexity,
∥ x + y 2 ∥ p ≤ 1 − δ
\left\| \frac{x+y}{2} \right\|_{p} \le 1-\delta
2 x + y p ≤ 1 − δ
It suffices to show that there exists δ = δ ( ϵ ) > 0 \delta=\delta (\epsilon)>0 δ = δ ( ϵ ) > 0 satisfying this.
Case 1. 1 < p ≤ 2 1<p\ \le 2 1 < p ≤ 2
By assumption, ∥ u − v 2 ∥ p ≥ ϵ 2 \displaystyle \left\| \frac{u-v}{2} \right\|_{p} \ge \frac{\epsilon}{2} 2 u − v p ≥ 2 ϵ . Then, by Clarkson’s inequality ( 2 ) (2) ( 2 ) ,
∥ u + v 2 ∥ p p ′ ≤ − ∥ u − v 2 ∥ p p ′ + ( 1 2 ∥ u ∥ p p + 1 2 ∥ v ∥ q p ) p ′ − 1 ≤ − ( ϵ 2 ) p ′ + ( 1 2 + 1 2 ) p ′ − 1 = 1 − ( ϵ 2 ) 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*}
2 u + v p p ′ ≤ ≤ = − 2 u − v p p ′ + ( 2 1 ∥ u ∥ p p + 2 1 ∥ v ∥ q p ) p ′ − 1 − ( 2 ϵ ) p ′ + ( 2 1 + 2 1 ) p ′ − 1 1 − ( 2 ϵ ) p ′
Since 0 < ϵ 2 < 1 0 < \frac{\epsilon}{2} < 1 0 < 2 ϵ < 1 , we have 1 − ( ϵ 2 ) p ′ < 1 1-\left( \frac{\epsilon}{2} \right)^{p^{\prime}}<1 1 − ( 2 ϵ ) p ′ < 1 and
∥ u + v 2 ∥ p ≤ [ 1 − ( ϵ 2 ) p ′ ] 1 p ′ = 1 − δ ( 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)
2 u + v p ≤ [ 1 − ( 2 ϵ ) p ′ ] p ′ 1 = 1 − δ ( ϵ )
A positive δ ( ϵ ) > 0 \delta (\epsilon)>0 δ ( ϵ ) > 0 exists that satisfies this.
Case 2. 2 ≤ p < ∞ 2 \le p < \infty 2 ≤ p < ∞
Similarly, by assumption, ∥ u − v 2 ∥ p ≥ ϵ 2 \displaystyle \left\| \frac{u-v}{2} \right\|_{p} \ge \frac{\epsilon}{2} 2 u − v p ≥ 2 ϵ . Then, by Clarkson’s inequality ( 1 ) (1) ( 1 ) ,
∥ u + v 2 ∥ p p ≤ − ∥ u − v 2 ∥ p p + 1 2 ∥ u ∥ p p + 1 2 ∥ v ∥ q p ≤ − ( ϵ 2 ) p + 1 2 + 1 2 = 1 − ( ϵ 2 ) 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*}
2 u + v p p ≤ ≤ = − 2 u − v p p + 2 1 ∥ u ∥ p p + 2 1 ∥ v ∥ q p − ( 2 ϵ ) p + 2 1 + 2 1 1 − ( 2 ϵ ) p
Similarly, since 0 < ϵ 2 < 1 0 < \frac{\epsilon}{2} < 1 0 < 2 ϵ < 1 , we have 1 − ( ϵ 2 ) p < 1 1-\left( \frac{\epsilon}{2} \right)^{p}<1 1 − ( 2 ϵ ) p < 1 and
∥ u + v 2 ∥ p ≤ [ 1 − ( ϵ 2 ) p ] 1 p = 1 − δ ( ϵ )
\left\| \frac{u+v}{2} \right\|_{p} \le \left[ 1-\left( \frac{\epsilon}{2} \right)^{p}\right]^{\frac{1}{p}} = 1-\delta (\epsilon)
2 u + v p ≤ [ 1 − ( 2 ϵ ) p ] p 1 = 1 − δ ( ϵ )
A positive δ ( ϵ ) > 0 \delta (\epsilon)>0 δ ( ϵ ) > 0 exists that satisfies this. Since there exists 0 < δ = δ ( ϵ ) 0 < \delta =\delta (\epsilon) 0 < δ = δ ( ϵ ) in both cases, the L p L^{p} L p space is uniformly convex.
Auxiliary theorem
A uniformly convex Banach space is reflexive.
The L p {L}^{p} L p space is a Banach space and uniformly convex, so it is reflexive by the auxiliary theorem.
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