Let's consider $L_2(\mathbb{R}^n)$. Let $Y$ be a non empty closed subspace of $L_2(\mathbb{R}^n)$.
Let $x\notin Y$. Let $y^*$ be the best approximation of $x$ on $Y$, i.e., $\|x-y^*\|_2=\inf_{y\in Y}\|x-y\|_2$.
We know then that, $x-y^*$ would be orthogonal to $Y$ and hence from parallelogram law, one can deduce the pythagoras theorem: $\|x-y\|_2^2=\|x-y^*\|_2^2+\|y^*-y\|_2^2 \text{ for } y\in Y$
I'm wondering whether the same kind of result would be true for $L_p(\mathbb{R}^n)$, $p\ge 1$, $p\neq 2$ also, i.e., whether $\|x-y\|_p^p=\|x-y^*\|_p^p+\|y^*-y\|_p^p $
I think its not possible to deduce from the parallelogram law as we have only inequality in parallelogram law in $L_p(\mathbb{R}^n)$ and there's no notion of orthogonality in $L_p(\mathbb{R}^n)$ for $p\neq 2$. But I think there may be some other way to get the result. At least mentioning some reference is appreciated.