The standard Schauder basis for $l^p$ spaces with $1\leq p < \infty$ is $e_n = \{\delta_{n,j}\}_{j=1}^\infty$ so $$ e_1=(1,0,0,...)\\ e_2=(0,1,0,...)\\ ... $$ $l^\infty$ doesn't have a Schauder basis because it's not separable.

Are there any sequence spaces besides $l^p (1\leq p < \infty)$ that also have the basis $e_n =\{\delta_{n,j}\}_{j=1}^\infty$?

Is there any separable sequence space that has a Schauder basis but it's not $e_n =\{\delta_{n,j}\}_{j=1}^\infty$?

Why are the spaces with the basis $e_n =\{\delta_{n,j}\}_{j=1}^\infty$ easier to work with?


The space $c_0\subset \ell^\infty$ of sequences tending to zero has $e_1,e_2,\dots$ as a Schauder basis.

The space $c\subset \ell^\infty$ of convergent sequences does not have $e_1,e_2,\dots$ as a Schauder basis. In fact the closed linear span of $e_1,e_2,\dots$ is $c_0$. Appending the constant sequence $1$ gives a Schauder basis $1,e_1,e_2,\dots$

A slightly silly example is the space $bv$ of sequences of bounded variation, with norm $|x_1-y_1|+\sum_{i\geq 2}|x_i-x_{i-1}-y_i+y_{i-1}|$. It's isomorphic to $\ell^1$ via the map the sends a sequence $(x_1,x_2,\dots)$ to the differences $(x_1,x_2-x_1,x_3-x_2,\dots)$. This again has $1,e_1,e_2,\dots$ as a Schauder basis, and the closed linear span is the subspace $bv_0$ of sequences of bounded variation tending to zero.

Saying that $e_1,e_2,\dots$ is a Schauder basis implies the property that the set of finitely-supported sequences is dense. This is a very common proof technique: reduce to the finitely-supported case. (This is actually a weaker consequence, for example polynomials are dense in $C([0,1])$ with the sup-norm, but non-analytic functions cannot be uniformly approximated by a series $\sum_{k\geq 0} c_kx^k$.)

  • $\begingroup$ So you mean that the vectors $1, e_1, e_2, ....$ is a Schauder basis for $c$. $\endgroup$ – Neil hawking Jul 17 at 16:52

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