Following the conventions from Heil: "[A Basis Theory Primer](http://people.math.gatech.edu/~heil/papers/bases.pdf)" 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](https://books.google.com/books?id=prfuUT0Sw-AC&pg=PA136) in Heil and [theorem 1.1.3](https://books.google.com/books?id=UG6zDAAAQBAJ&pg=PA3) 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?



Basis vs Schauder basis in normed spaces
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 $(\ell\_1,\|\cdot\|\_2$). 


