Basis vs Schauder basis in normed spaces

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?

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 $$(\ell_1,\|\cdot\|_2$$).
• 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) – Ivica Smolić May 10 '20 at 22:27