Consider $(X, A, \mu)$ as a measure space where $A$ contains at least two disjoint sets $A_1$ and $A_2$ with positive, finite measures. Let $p$ be in the interval $[1, \infty]\setminus\{2\}$. The goal is to show that there does not exist an inner product $\langle \cdot, \cdot \rangle : L^p(X) \times L^p(X) \rightarrow \mathbb{R}$ that induces the $L^p$-norm, i.e., $\|f\|_{L^p} = \sqrt{\langle f, f \rangle}$ for all $f \in L^p(X)$.
I want to show this by using a proof of contradiction and without using the parallelogramm law. Since we know that we have at least two disjoint sets with positive finite measures, I want to define a function $f$ which takes the value $1$ on $A_1$ and $A_2$ and $0$ elsewhere. Therefore we get
$$\|f\|_{L^p} = \biggl(\int_{X}f^pd\mu\biggr)^\frac{1}{p} = \biggl(\int_{A_1 \cup A_2}1d\mu\biggr)^\frac{1}{p} = (m(A_1) + m(A_2))^{\frac{1}{p}}$$
where $m(A_1)$ and $m(A_2)$ denote the (finite, positive) measure of $A_1$ and $A_2$ respectively. This ultimately gives us
$$\langle f, f \rangle = \|f\|_{L^p}^{2} = (m(A_1) + m(A_2))^{\frac{2}{p}}$$
But unfortunately, I am a little stuck here. I would somehow like to show that we have $\langle f, f \rangle = (m(A_1) + m(A_2))$ so that we can nicely see that $p$ has to be $2$ in order for that to be true.
Or is there any other approach I should rather consider for this specific example? I think I necessarily need to include the two disjoint sets $A_1, A_2$ with finite, positive measure in my proof which is why I came up with the approach above. But I am open for any suggestions.
Hint: $(a,b) \to a\mu(A_1)^{-1/p}1_{A_1}+b\mu(A_2)^{-1/p}1_{A_2}$ is an isomtric isomorphism from $\mathbb R^{2}$ with the $p-$ norm into $L^{p}(X)$. Since $\mathbb R^{2}$ with the $p-$ norm has no inner product, $L^{p}(X)$ cannot have one either.
[Parallelogram Law holds in $\mathbb R^{2}$ with the $p-$ norm only for $p=2$].