Subset of a basis for a normed vector space cannot be Cauchy?

Question

In class, we showed that $e_n \in l^p$ converges weakly but not strongly for $p \in (1,\infty)$ and that $\sin(2\pi nx) \in L^2([0,1])$ converges weakly but not strongly. I was wondering if basis elements cannot converge strongly because if they do they are somehow not linearly independent?

I know that these are Schauder bases vs. Hamel bases for the respective spaces, but don't have a good intuition for whether that should make a difference.

Answer 1

Here is a handy fact (c.f. Diestel, Sequences and Series in Banach Spaces, pg. 36): a sequence of non-zero vectors $(x_n)$ in a Banach space $X$ is a basic sequence (a basis of its closed linear span) if and only if there is a $K>0$ so that for any choice of scalars $(a_n)$ and integers $m<n$ $$ \biggl\Vert \sum_{i=1}^m a_i x_i\biggl\Vert\le K\biggl\Vert \sum_{i=1}^n a_ix_i\Biggl\Vert. $$

This easily implies that a normalized basic sequence cannot be Cauchy.

Answer 2

A Schauder basis can converge in norm to $0$. In fact, for any Schauder basis $u_n$, $u_n/(n \|u_n\|)$ will do.

Answer 3

If $(e_n)$ is a Schauder basis so is $(c_ne_n)$ for any sequence of positive numbers $(c_n)$. This new Schauder basis can converge strongly to $0$. An orthonormal basis cannto converge in norm. Indeed if $(e_i)$ is an orthonormal basis then $\|e_i-e_j\|=\sqrt 2$ for $i \neq j$, so $(e_i)$ is not Cauchy.

Answer 4

It is possible that a sequence consisting of linearly independent vectors is convergent and the norms are bounded away from $ 0.$ For example let $e_n$ form an orthonormal basis in the Hilbert space $\mathcal{H}.$ Then the sequence $$ f_n=e_1+{1\over n}e_n$$ is convergent to $e_1$ and is linearly independent.