Is the norm on $\ell^\infty$ induced by an inner product?

Question

Let $\ell^\infty$ be the normed space of all bounded sequences $x \colon= \left(\xi_n \right)_{n \in \mathbb{N} }$ of complex numbers, with the norm defined by $$\lVert x \rVert_{\ell^\infty} \colon= \sup_{n \in \mathbb{N}} \lvert \xi_n \rvert.$$

Is this norm induced by an inner product?

That is, can we show that $$\lVert x + y \rVert_{\ell^\infty}^2 + \lVert x - y \rVert_{\ell^\infty}^2 \ = \ 2\left( \lVert x \rVert_{\ell^\infty}^2 + \lVert y \rVert_{\ell^\infty}^2 \right) \ \ \ \mbox{ for all } \ x, y \in \ell^\infty?$$ Or can we find any bounded sequences $x \colon= \left( \xi_n \right)_{n \in \mathbb{N} }$ and $y \colon= \left( \eta_n \right)_{ n \in \mathbb{N} }$ of complex numbers for which the above equality fails?

Answer 1

So I'm just elaborating on the comment by cobber.hat.

Take $x = (1,0,0, 0, \ldots)$ and $y = (0,1,0, 0, 0, \ldots)$. then: $$\lVert x+y\lVert^2_\infty =1 \qquad \lVert x-y\lVert^2_\infty =1 \qquad \lVert x\lVert^2_\infty =1 \qquad \lVert y\lVert^2_\infty =1$$

Thus we have a counter example to your formula.