$L^p$-space is a Hilbert space if and only if $p=2$

Inspired by $\ell_p$ is Hilbert if and only if $p=2$, I try to prove that a $L^p$-space (provided with the standard norm) is a Hilbert space if and only if $p=2$. I already know that every $L^p$-space is a Banach space with respect to the standard norm.

This is what I got so far.

To prove

Let $(S, \Sigma, \mu)$ be a measure space and assume that $\mu$ is a positive, $\sigma$-finite measure that is not the trivial measure. Let $p\in [1, +\infty]$. The normed space $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} )$ is a Hilbert space if and only if $p=2$.

Proof

Case 1: $p=2$

The standard inner product $\langle \cdot , \cdot \rangle _{L^2}$ induces the standard norm $\| \cdot \| _{L^2}$, so $(L^2 (S, \Sigma , \mu), \| \cdot \| _{L^2})$ is a Hilbert space.

Case 2: $p\in [1, \infty) \backslash \{ 2 \}$

Assume $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} ) $ is a Hilbert space. Then it is also an inner product space, so it must satisfy the parallelogram rule: \begin{align} \forall f,g \in L^p(S,\Sigma , \mu): \ \ \| f + g \| ^2 + \| f -g \| ^2 = 2 ( \| f \| ^2 + \| g \| ^2 ). \end{align} Let $A, B$ be disjoint measurable sets such that $0 < \mu (A), \ \mu (B) < \infty$. Define $f_p, g_p \in \mathcal{L} ^p (S, \Sigma, \mu)$ by \begin{align*} f_p := \frac{1}{(\mu (A) ) ^{1/p}} 1 _A \geq 0 \ \ \ \text{ and } \ \ \ g_p := \frac{1}{(\mu (B) ) ^{1/p}} 1 _B \geq 0.\end{align*} Doing some calculations gives us \begin{align*} 2 \left ( \| f_p \| _{L^p} ^2 + \| g_p \| _{L^p} ^2 \right ) = 4 \ \ \text{ and } \ \ \| f_p + g_p \| _{L^p} ^2 + \| f_p - g_p \| _{L^p} ^2 = 2 \cdot 2 ^{2/p}. \end{align*} The functions $f_p$ and $g_p$ do not satisfy the parallelogram rule, since $4\neq 2 \cdot 2 ^{2/p}$ (remember that $p\neq 2$). But this contradicts our earlier conclusion that all $f,g\in L^p (S, \Sigma , \mu)$ must satisfy the parallelogram rule. Therefore, our assumption that $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} ) $ is a Hilbert space is wrong. Hence, $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} ) $ is not a Hilbert space.

Case 3: $p=\infty$

Assume $(L^{\infty} (S,\Sigma , \mu), \| \cdot \| _{L^{\infty}} ) $ is a Hilbert space. Then it is also an inner product space, so it must satisfy the parallelogram rule.

Let $A$ and $B$ be disjoint measurable sets which are not null sets and consider $F=1 _A$ and $G=1_B$. Then, we have \begin{align*} \| F + G \| _{L^{\infty}} ^2 + \| F - G \| _{L^{\infty}} ^2 = 1 + 1 = 2 \neq 4 = 2(1+1) = 2( \| F \| _{L^{\infty}} ^2 + \| G \| _{L^{\infty}} ^2 ). \end{align*} But this contradicts our earlier conclusion that all $f,g\in L^{\infty} (S, \Sigma , \mu)$ must satisfy the parallelogram rule. Therefore, our assumption that $(L^{\infty} (S,\Sigma , \mu), \| \cdot \| _{L^{\infty}} ) $ is a Hilbert space is wrong. Hence, $(L^{\infty} (S,\Sigma , \mu), \| \cdot \| _{L^{\infty}} ) $ is not a Hilbert space.

Question

I am not completely sure about the following statements "Let $A, B$ be disjoint measurable sets such that $0 < \mu (A), \ \mu (B) < \infty$" and "Let $A$ and $B$ be disjoint measurable sets which are not null sets". How do I know for sure that such measurable sets $A,B$ actually exist? Should I make more assumptions about the measure space in order to make these arguments work?

Solution 1:

Here is an interesting characterization, from which one can see that for the Lebesgue measure, all $L^p$ spaces with $p\ne 2$ are not Hilbert. https://www.docdroid.net/Ib35oGd/lp-pdf