Below is what i have proved:
If $V$ is a normed vector space over $\mathbb{R}$ satisfies parallelogram equality, then there exists an inner product $\langle \bullet,\bullet\rangle$ such that $\langle \bullet,\bullet \rangle = \lVert \bullet \rVert^2$
If $V$ is an inner product space over $\mathbb{R}$ and $\lVert \bullet \rVert \triangleq \sqrt{\langle \bullet \rangle}$, then $\lVert \bullet \rVert$ is a norm and $\langle x,y\rangle = \frac{1}{4}(\lVert x+y\rVert^2 - \lVert x-y\rVert^2)$.
I thought defining a norm as square root of inner product $\langle x,x\rangle$ is just one possinle way to define norm on an inner product space.
However, the below is the article in wikipedia about impossibility of defining $p-norm$ on an inner product space:
"p ≠ 2 is a normed space but not an inner product space, because this norm does not satisfy the parallelogram equality required of a norm to have an inner product associated with it"
I don't think what i proved above give this conclusion in wikipeia.. What should i prove to conclude this?
Plus, it's on wikipedia that $2$-notm is completely natural to define as so, but it's not that natural to take square root to inner product to define norm on an inner product space to me. Why do we specifically use 2-norm?
Moreover, what's the point of relating two concepts inner product and norm, literally?