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Let $ \mathbb{K}^{\mathbb{N}} $ be the space of all complex sequence. How does one show that there exists no norm on $ \mathbb{K}^{\mathbb{N}} $ such that $ \mathbb{K}^{\mathbb{N}} $ is a Hilbert space. Unfortunately, I am quite lost in this question and don't even know where to start, so any clues would be much appreciated!

Thanks in advance!

Meagain
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    $\mathbb K^{\mathbb N}$ and $\ell^2$ are isomorphic as vector spaces because the have Hamel bases of the same cardinality $c$. If $\Phi:\mathbb K^{\mathbb N} \to \ell^2$ is an isomorphism you can just tranport the Hilbert structure from $\ell^2$ to $\mathbb K^{\mathbb N}$, i.e. $\langle x,y\rangle= \langle \Phi(x),\Phi(y)\rangle_{\ell^2}$. – Jochen Mar 23 '18 at 11:35
  • But as $ \ell^{2} $ is Hilbert that would mean my claim is false? ∑(ﾟДﾟ) – Meagain Mar 23 '18 at 11:40
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    Yes, on the other hand, any such norm will make most of the projections $\mathbb{K}^{\mathbb{N}}\to\mathbb{K}$ discontinuous. So, the space geometrically doesn't look anymore like the original $\mathbb{K}^{\mathbb{N}}$ but as $\ell^2$ for those sequences that happen to be mapped to $e_n\in\ell^{2}$. – blueInk Mar 23 '18 at 11:45
  • Thanks! Let me give it some more thoughts first. – Meagain Mar 23 '18 at 11:46
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    Although I know that one can show the existence of the Hamel basis using the axiom of choice. I'm not sure if my knowledge of Hamel basis is sufficient to talk about its cardinality yet. Is there any reference, with perhaps particular focus on showing that $ \mathbb{K}^{\mathbb{N}} $ and $ \ell^{2} $ have the same Hamel basis cardinality that I could get? – Meagain Mar 23 '18 at 11:59
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    @Meagain Both spaces have cardinality $2^{\aleph_0}$, so a Hamel basis of either has at most $2^{\aleph_0}$ elements. And the family ${ x_{\lambda} = (1, \lambda, \lambda^2, \dotsc, \lambda^n,\dotsc) : \lvert \lambda\rvert < 1}$ is a linearly independent family of cardinality $2^{\aleph_0}$ [in both spaces], so a Hamel basis of either space has at least $2^{\aleph_0}$ elements. This is a related question. – Daniel Fischer Mar 23 '18 at 20:24

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