# Show that $l^2$ is the only $l^p$ space which norm is induced by the inner product

I want to use the theorem of Jordan and von Neumann which states that norm is induced by inner product if and only if the parallelogram law is true

Let $\Vert x \Vert_p$ be the norm in $l^p, 1\le p < \infty$. In parallelogram law we have,

$\Vert x+y \Vert_p^2 + \Vert x-y \Vert_p^2 = 2\Vert x \Vert_p^2 + 2\Vert y \Vert_p^2$

which is equivalent

$(\sum|x_i+y_i|^p)^{(2/p)} + (\sum|x_i-y_i|^p)^{(2/p)} = 2(\sum|x_i|^p)^{(2/p)} +2(\sum|y_i|^p)^{(2/p)}$.

But now how can we get that $p=2$ is the only correct one? Is it possible to construct a sequence for which parallelogram fails for each $p$ different than 2?

• yes, take $x=(1,1)$, $y=(1,0)$
– Surb
Apr 11 '17 at 11:54

Choose $x=(1,0,0,...), y=(0,1,0,...)$. We have $\|x+y\|_p^{2}+\|x-y\|_p^{2}=2 \cdot2^{2/p}$ and $2\|x\|_p^{2/p}+2\|x\|_p^{2/p}=2 + 2 =4$. From parallelogram equality we get $2^{2/p}=2$ and then $p=2$.