Schauder basis of $(c,\|\cdot\|_{\infty})$

Question

Show that $\{e_{n}\}_{n=0}^{\infty}$ is a Schauder basis of $(c,\|\cdot\|_{\infty})$ under the . Here, $c$ is the collection of all convergent sequences. Also, let $e_{0} = \{1,1,1,\ldots\}$ and $e_{n} = \{0,0,\ldots,1,0,\ldots\}$ where the $1$ is in the $n$th position.

The fact that $e_{0}$ is defined this way is a source of confusion for me. Starting with the typical arguments for a Schauder basis proof, I can't find a way to reconcile this.

Does anyone have any insight into this proof?

Answer 1

Consider a convergent sequence with limit $L$. Begin with $L e_0$ (so that you have a fixed sequence plus a sequence that goes to zero). Then "fix" all the remaining terms: that is, write

$$x=L e_0+\sum_{n=1}^\infty (x_n-L)e_n.$$

This gives existence of the desired sequence of scalars. You now have some two things still to check:

• You need to show this sum converges in the given norm.
• You need to show that this sequence of scalars is unique. To do that, suppose you have some other sequence. For case 1, suppose the coefficient of $e_0$ is $L$. Then it is easy to see (by "essentially" the same argument as in finite dimensions) that you must have the sequence above. Now for case 2, suppose the coefficient of $e_0$ is not $L$. Formally*, you still have just 1 possible sequence given the value of the coefficient of $e_0$, but the resulting sum should now fail to converge.

* Here by "formally" I mean "by just writing out the calculation without thinking about convergence issues". This is contrary to the meaning of the word "formal" to be understood as "rigorous".