$𝐿 𝑝 (𝑋, 𝜇)$ forms an inner product space iff $𝑝 = 2$

Let $(𝑋, Σ, 𝜇)$ be a measure space with at least two distinct subsets $𝐴, 𝐵 \in Σ$ of finite nonzero measure and let $𝑝 \in [1, \infty]$ . Then the Lebesgue space $𝐿 𝑝 (𝑋, 𝜇)$ satisfies the parallelogram law and therefore has a unique inner product iff $𝑝 = 2$ .

Proof

Without loss of generality we may assume that $𝐴 \cap 𝐵 = \emptyset$ . Let $𝑎 = 𝜇 (𝐴)$ and $𝑏 = 𝜇 (𝐵)$ , and

𝑟 : (0, \infty) \to ℝ 𝑡 \mapsto 𝑡 2 / 𝑝

Then using the indicator functions $1 𝐴$ and $1 𝐵$ we have

2 ‖ 1 𝐴 ‖ 2 𝑝 + 2 ‖ 1 𝐵 ‖ 2 𝑝 = 2 (\int 𝑋 | 1 𝐴 | 𝑝 𝑑 𝜇) 1 / 𝑝 + 2 (\int 𝑋 | 1 𝐴 | 𝑝 𝑑 𝜇) 1 / 𝑝 = 2 𝑟 (𝑎) + 2 𝑟 (𝑏)

and

‖ 1 𝐴 + 1 𝐵 ‖ 2 + ‖ 1 𝐴 - 1 𝐵 ‖ 2 = (\int 𝑋 | 1 𝐴 + 1 𝐵 | 𝑝 𝑑 𝜇) 2 / 𝑝 + (\int 𝑋 | 1 𝐴 - 1 𝐵 | 𝑝 𝑑 𝜇) 2 / 𝑝 = 2 𝑟 (𝑎 + 𝑏)

so for equality we require $𝑟 (𝑎) + 𝑟 (𝑏) = 𝑟 (𝑎 + 𝑏)$ . If $𝑝 > 2$ , then $𝑟$ is a strictly convex function, since its second derivative is positive on its domain, so since strict convexity on the positive reals and $𝑓 (0) \leq 0$ implies superadditivity we have

𝑟 (𝑎) + 𝑟 (𝑏) < 𝑟 (𝑎 + 𝑏)

so equality cannot hold. For $𝑝 \in (1, 2)$ an analogous argument can be made using $- 𝑟$ .

For the converse, see $𝐿 2$ space.

Specific counterexamples

To show that the parallelogram law

2 ‖ 𝑥 ‖ 2 𝑝 + 2 ‖ 𝑦 ‖ 2 𝑝 = ‖ 𝑥 + 𝑦 ‖ 2 𝑝 + ‖ 𝑥 - 𝑦 ‖ 2 𝑝

holds iff $𝑝 = 2$ , the following counterexamples may be used

For $𝑋 = [𝑎, 𝑏]$ take $𝑥 = 1 [𝑎, 𝑚]$ and $𝑦 = 1 [𝑚, 𝑏]$ where $𝑚 = 𝑎 + 𝑏 2$ .
For $𝑋 = ℕ$ , i.e. $ℓ 𝑝$ space, take $𝑥 𝑖 = 𝛿 1 𝑖$ and $𝑦 𝑖 = 𝛿 2 𝑖$

#state/develop | #lang/en | #SemBr

𝐿𝑝(𝑋,𝜇) forms an inner product space iff 𝑝 =2

Specific counterexamples

$𝐿 𝑝 (𝑋, 𝜇)$ forms an inner product space iff $𝑝 = 2$