Determining origin of norm Reading a script i've found task in which i had to determine whether each norm $$\|x\|_{p}=\left(\sum |x_{i}|^{p}\right)^{1/p}$$ origins from scalar product. Assuming $$p=2$$ i got, it comes from standard scalar product. $$\|x\|_{2}=\langle x,x\rangle^{1/2}$$ no luck for others.
Tip i got was to use Parallelogram law and describe it in norm language.
I get.
$$\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^2+2\|y\|^{2}$$
I just don't see how should i use it. 
 A: You can check that 
$$x = (1, 0, 0, \cdots), \ \ \ y = (0,1,0, 0, \cdots)$$
in $\ell _p$ satisfies the parallelogram law only when $p = 2$. Indeed, the right hand side is $2\cdot 2^{\frac 2p}$ and the left hand side is $4$. Thus $\ell_p$-norm does not come from an inner product. 
A: Let:
$$
x=(1,1,0,0 \cdots) \qquad y=(1,-1,0,0 \cdots)
$$
we have:
$$
\|x\|_p=2^{1/p}=\|y\|_p
$$
and
$$
x+y=(2,0,0\cdots) \Rightarrow \|x+y\|_p=2
$$
$$
x-y=(0,2,0\cdots) \Rightarrow \|x-y\|_p=2
$$
so the parallelogram low becomes:
$$
\|x+y\|_p^2+\|x-y\|_p^2=2\left(\|x\|_p^2+\|y\|_p^2 \right)
$$
i.e:
$$
8=2\left(2^{2/p}+2^{2/p} \right) \iff 2=2^{2/p}
$$
that gives $p=2$.
This means that the parallelogram low is verified only if $p=2$ and only in this case the norm can be derived from an inner product.
A: Consider this an extended comment......
Are you sure that your norm is $\|x\|_p=\left(\sum x_i^{p}\right)^{1/p}$  and not in fact $\|x\|_p=\left(\sum |x_i|^{p}\right)^{1/p}$ 
See for example the wiki entry on p-norm https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm.
If the two answers given assume the p-norm then they are correct.
If your original statement is correct, then it cannot be a norm for any odd value of $p$ since $\|(-1, 0, 0, ...)\|$ is not positive for $p=1$ and is not real for any odd $p > 1$.
