Is $\ell^1$ an inner product space?

Question

Considering the parallel result for $L^p$ spaces, I would guess that $\ell^1$ is not an inner product space. The proof would presumably follow by providing counterexample sequences $x = (x_n) $ and $y = (y_n)$ violating the parallelogram law, i.e., satisfying:

$$2\|x\|_1^2+2\|y\|_1^2\neq \|x+y\|_1^2+\|x-y\|_1^2 $$

I can't find a counterexample. Any hints?

Answer 1

$x = (1,1,0,0, \ldots )$

$y = (1,-1,0,0, \ldots) $

By the way, note that $\ell_1$ is complete and separable but has non-separable dual. On the other hand, Hilbert spaces are isomorhic to their dual spaces. This shows that $\ell_1$ is not linearly isomorphic to an inner-product space.

Answer 2

You can just apply the result for $L^1$ spaces, because $\ell^1 = L^1(\mu)$ where $\mu$ is the counting measure.