Htykuut

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 $pin [1, +infty]$. 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: $pin [1, infty) backslash { 2 }$

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 $4neq 2 cdot 2 ^{2/p}$ (remember that $pneq 2$). But this contradicts our earlier conclusion that all $f,gin L^p (S, Sigma , mu)$ 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 $f,gin L^{infty} (S, Sigma , mu)$ 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?