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?

  • 2
    $\begingroup$ Just take $\mathbb{R}^2$, with $x=(1,0), y=(0,1)$. $\endgroup$
    – copper.hat
    Apr 29, 2015 at 7:45

1 Answer 1


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

Take $x = (1,0,0,\ldots)$ and $y = (0,1,0,\ldots)$. 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.


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