Norm and inner product

Question

Is always square of the norm of a vector is same as the inner product of that vector with itself? In Probability theory we frequently use $L^P$ norm: $\|X\|=E^{1/p}\left(X^p\right)$. But don't we still use $E(XY)$ as the inner product? In that case, $\langle x,x\rangle=EX^2$ which is not equal to the square of the norm when $p\neq2$.

Answer 1

It does not always make sense to refer to "the" norm and "the" inner product.

You should when what you mean is the usual inner product and its corresponding norm - the $L^2$ norm when there is one.

But whenever you have "an" inner product it defines a norm. Spaces can have more than one inner product, so more than one norm that comes from an inner product. For example, on $\mathbb{R}^2$ the inner product $$ \langle (a,b),(c,d)\rangle = ac + 2bd $$ defines the norm in which $$ \lVert (a,b)\rVert^2 = a^2 + 2b^2 $$

but the norm defined by $$ \lVert (a,b) \rVert = |a| + |b| $$ does not come from any inner product.