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?
Best Answer
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