$${ \| x-y \|_{p}^{2} =^{?} \| x \|_{p}^{2} -2x^{T}y + \| y\|_{p}^{2}, \; \; \; \; \; \; \; p \neq 2, \; x \in \mathbb{R}^{d} }$$
${x}$ is a column vector, ${x^{T}}$ is the transpose of ${x}$ so is a row vector, ${x^{T}y}$ is the inner product between ${x}$ and ${y}$. ${\| x \|_{p}}$ is the standard Lp-norm as ${ \|x\|_{p} := ({\sum_{i=1}^{d}{ |x_{i}|^{p} }})^{\frac{1}{p}} }$.
It seems not many people have a question on this equality. It is simple that the equality holds when ${p=2}$ because ${|x_{i}|^{2} = x_{i}^{2}}$. But it is already not obvious to me when ${p=1}$ while I'd like to see what happens when ${p \neq 2}$. Could anyone help me to see if this holds or not?
Thanks. And simply giving me a reference is also appreciated.
You already have a counterexample, but this might help you to better understand why this particular equation will most likely only hold for the 2-norm.
If your norm $\Vert x \Vert$ is induced by an inner product over the real numbers, ie $\Vert x \Vert = \sqrt{\langle x, x \rangle}$, then we have:
$$ \Vert x - y \Vert^2 = \langle x - y, x - y\rangle = \langle x, x\rangle - 2 \langle x, y\rangle + \langle y, y\rangle = \Vert x \Vert^2 - 2 \langle x, y\rangle + \Vert y \Vert^2.$$
As the Euclidean norm is induced by the standard dot-product on $\mathbb{R}^d$, this already implies that the occurrence of $-2x^{T}y$ is linked to this norm and probably does not generalize to other p-norms.