$\displaystyle \left( \int_{0}^{1} 2^{p} dx \right)^{\frac{1}{p}}$ =

$\displaystyle \left( [2^{p}x]_{0}^{1} \right)^{\frac{1}{p}}$ =

2

What do you think ?

So far I have that:

$\displaystyle \Vert f \Vert_{p}$ = 1

$\displaystyle \Vert g \Vert_{p}$ = 1

$\displaystyle \Vert f+g \Vert_{p}$ = 2

$\displaystyle \Vert f-g \Vert_{p}$ = 0

My next question is: How do I show that $\displaystyle L^{p}(\mathbb{R})$ does not form a Hilbert-space for $\displaystyle p\neq 2$.

$\displaystyle \text{Thanks}^{\infty}$