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Following the conventions from Heil: "A Basis Theory Primer" and Albiac, Kalton: "Topics in Banach Space Theory", we might define a basis of an (infinite-dimensional) normed space $V$ as a sequence $(e_n)$ in $V$, such that for any $x \in V$ there is a unique sequence of scalars $(\lambda_n)$, such that $x = \sum_n \lambda_n e_n$ (converging in norm), whereas for a Schauder basis we demand that these coefficients are produced by linear continuous functionals $(e^*_n)$, such that $e^*_m(e_n) = \delta_{mn}$ and $\lambda_n = e^*_n(x)$.

Now, if we work in a separable Banach space, these two notions coincide (theorem 4.13 in Heil and theorem 1.1.3 in Albiac, Kalton), but what if $V$ is a separable normed space which is not complete? Is there a simple, instructive example in which these linear functionals $(e^*_n)$ exist but fail to be continuous?

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You ask for an instructive example, so I'll be long winded.

Suppose $(x_n)$ is basis for a normed space $(X,\|\cdot\|)$. The partial sum projections $S_n$ are well defined but might be discontinuous. Define a new, larger, norm on $X$ by $|x| := \sup_n \|S_n x\|$. Then $(x_n)$ is a Schauder basis for $(X,|\cdot|)$. Note that this is the first step in the proof that every basis for a Banach space is a Schauder basis. The harder part of the proof is to show that $(X,|\cdot|)$ is complete when $(X,\|\cdot\|$ is complete. When $(X,\|\cdot\|)$ is not complete, $(X,|\cdot|)$ may not be complete, but if you are building a basis for $(X,\|\cdot\|)$ that is not a Schauder basis, it would be nice to have one such that the basis is a Schauder basis for some natural complete norm on $X$. Also, you would like your non Schauder basis to be in a some natural normed space; let's say an inner product space that is a non closed subspace of $\ell_2$ under the usual $\ell_2$ norm, $\|\cdot\|_2$. Well, $\ell_2$ has lots of non closed subspaces that are complete under a natural norm, the most used being $\ell_1$ under its norm $\|\cdot\|_1$. It should not be too hard to find a Schauder basis for $\ell_1$ such that at least one of the biorthogonal functionals is not in $\ell_2$. Perhaps the simplest such example is $x_1 = e_1$ and $x_n = e_1+e_n$ when $n>1$. (Here $(e_n)$ is the standard unit vector basis.) Notice that the biorthogonal functionals in $\ell_\infty = \ell_1^*$ are given by $x_1^* =(1,-1,-1,\dots)$ and, for $n>1$, $x_n^* = (0,0,\dots,1,0,\dots)$, where the $1$ is in the nth coordinate. Only $x_1^*$ is not continuous in the $\ell_2$ norm--this makes it easy to verify that $(x_n)$ is a basis for $(\ell_1,\|\cdot\|_2$).

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  • $\begingroup$ Great! Thank you for this very nice clarification and neat example! Just one small detail: should it be $x^*_n = (0,...,0,1,0,...)$ for $n>1$? (maybe I'm not reading the notation properly) $\endgroup$ – Ivica Smolić May 10 '20 at 22:27
  • $\begingroup$ Yes; sorry for the typo. I'll fix it. $\endgroup$ – Bill Johnson May 11 '20 at 5:37

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