0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\sum_{k\geq 0} c_kx^k$.)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.