[[Lebesgue space]]
# $L^p(X,\mu)$ forms an inner product space iff $p=2$

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

> [!check]- Proof
> Without loss of generality we may assume that $A \cap B = \0$.
> Let $a = \mu(A)$ and $b = \mu(B)$, and
> $$
> \begin{align*}
> r : (0,\infty) &\to \mathbb{R} \\
> t &\mapsto t^{2/p}
> \end{align*}
> $$
> Then using the [[indicator function|indicator functions]] $1_{A}$ and $1_{B}$ we have
> $$
> \begin{align*}
> 2\| 1_{A}\|_{p}^2 + 2\|1_{B}\|_{p}^2 
> &= 2\left( \int _{X} |1_{A}|^p \, d\mu  \right)^{1/p} + 2\left( \int _{X} |1_{\underset{  }{ A }}|^p \, d\mu  \right)^{1/p} \\
> &= 2r(a) + 2r(b)
> \end{align*}
> $$
> and
> $$
> \begin{align*}
> \|1_{A}+1_{B}\|^2 + \|1_{A} - 1_{B}\|^2 
> &= \left( \int _{X}|1_{A} + 1_{B}|^p \, d\mu  \right) ^{2/p} + \left( \int _{X}|1_{A} - 1_{B}|^p \, d\mu  \right) ^{2/p}
> \\
> &= 2 r(a+b)
> \end{align*}
> $$
> so for equality we require $r(a) + r(b) = r(a+b)$.
> If $p > 2$, then $r$ is a [[Convex function|strictly convex function]], since [[A twice-differentiable function is convex iff its second derivative is nonnegative everywhere|its second derivative is positive on its domain]],
> so since [[Convexity on the positive reals and negative f(0) implies superadditivity|strict convexity on the positive reals and $f(0) \leq 0$ implies superadditivity]] we have
> $$
> \begin{align*}
> r(a) + r(b) < r(a+b)
> \end{align*}
> $$
> so equality cannot hold.
> For $p \in (1,2)$ an analogous argument can be made using $-r$. 
> 
> For the converse, see [[L2 space]]. <span class="QED"/>

## Specific counterexamples

To show that the parallelogram law
$$
\begin{align*}
2 \|x\|_{p}^2 + 2 \|y\|_{p}^2 = \|x+y\|_{p}^2 + \|x-y\|_{p}^2
\end{align*}
$$
holds iff $p=2$, the following counterexamples may be used

- For $X=[a,b]$ take $x=1_{\left[ a, m \right]}$ and $y=1_{\left[ m, b \right]}$ where $m = \frac{a+b}{2}$.
- For $X = \mathbb{N}$, i.e. [[Lebesgue sequence space]], take $x_{i} = \delta_{1i}$ and $y_{i}=\delta_{2i}$

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