Suppose $p>0$. Prove that there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $\|(x,y)\|=(x^p+y^p)^\frac{1}{p}$ for all $(x,y)\in\mathbb{R}^2$ if and only if $p=2$.
I have already shown that there is no inner product if $p\ne2$, but I am a little stuck on proving that one does exist for $p=2$. Any help is appreciated.
Best Answer
A norm $\|\cdot\|$ in a real vector space comes from a inner product $\langle\cdot,\cdot\rangle$ if and only if the parallelogram law $$\|{\bf x}+{\bf y}\|^2+\|{\bf x}-{\bf y}\|^2 = 2(\|{\bf x}\|^2+\|{\bf y}\|^2)$$holds, in which case the polarization formula holds: $$\langle {\bf x},{\bf y}\rangle = \frac{1}{4}(\|{\bf x}+{\bf y}\|^2-\|{\bf x}-{\bf y}\|^2).$$