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In the space $(\mathscr{l}^1, ||\cdot||)$, it can be shown that no inner product can be induced by the given norm. However, any vector space can admit an inner product, so I'm trying to construct one such operator for $\mathscr{l}^1$. One way to do so is to find a basis for $\mathscr{l}^1$, and this part is troublesome to me.

From this question, I understand that there's no formula for such explicit basis in the general case $\mathscr{l}^p$, but how about $\mathscr{l}^1$?

Tran Khanh
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    There is no effective method, as far as I am aware, to find an algebraic basis for any given infinite-dimensional Banach space, because such a basis always consists of uncountably many elements. – Dean Miller Jan 26 '24 at 04:35

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The obvious $\ell^2$ product $$ x\cdot y=\sum_nx_ny_n $$ works because every element in $\ell^1$ is in $\ell^2$. Of course this gives you the $2$-norm, and $\ell^1$ is not complete with respect to this norm.

As for a Hamel basis, the existence depends on the axiom of choice, it is uncountable, and you cannot expect to show one explicitly.

Martin Argerami
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