Schauder basis

Remarks

Example: in $\ell^p$ with $1 \le p < \infty$, see spaces ℓ^p, the unit vectors $e_n$ (with $1$ in place $n$ and $0$ elsewhere) form a Schauder basis. For $x \in \ell^p$ and $\alpha_n = x_n$, one has $\norm{x - \sum_{n \le N} x_n e_n}_p = \left( \sum_{n > N} \abs{x_n}^p \right)^{1/p} \to 0$. The coefficients are unique, because a norm-convergent series converges coordinatewise, so the $n$-th coordinate of $\sum_k \alpha_k e_k$ is $\alpha_n$.

Non-example: in $\ell^\infty$ the unit vectors are not a Schauder basis. For $x = (1, 1, 1, \dots)$ the only possible coefficients are $\alpha_n = 1$, and $\norm{x - \sum_{n \le N} e_n}_\infty = 1$ for every $N$. In fact $\ell^\infty$ has no Schauder basis at all, because it is not separable.

The order of the basis matters, since the series need not converge after a rearrangement. A Schauder basis of an infinite-dimensional Banach space is never a Hamel basis. The coefficient functionals $x \mapsto \alpha_n$ are always continuous.