Normed spaces are not necessarily finite dimensional. So it is important to understand the notion of a basis for an infinite dimensional normed space. Suppose that there is a basis of a normed space $X$ as an infinite sequence $(e_n)$ in $X$. Then any $x\in X$ can be represented as the infinite superposition of the $e_n$’s

\begin{equation}

\label{eq:superposition}

x=\sum_{j=1}^\infty\alpha_je_j,

\end{equation}

where $\alpha_1,\alpha_2,\cdots$ are scalars.

In order for this to make sense, we need to make sure that the infinite sum in \eqref{eq:superposition} converges. Thus we have the following definition of a basis for an infinite dimensional normed space.

*Definition*. Suppose that a normed space $X$ contains a sequence $(e_n)$ with property that $\forall x\in X$, there exists uniquely a sequence of scalars $(\alpha_n)$ such that

$$||x-\sum_{j=1}^n\alpha_je_j||\rightarrow 0\ \mbox{as}\ n\to\infty.$$

Then $(e_n)$ is called a *Schauder basis* for $X$. The infinite sum $\displaystyle\sum_{j=1}^\infty\alpha_je_j$ is called the expansion of $x$.

*Example*. $\ell^p$ has a Schauder basis $(e_n)$, where $e_n=(\delta_{nj})$.

*Theorem*. If a normed space has a Schauder basis, it is separable i.e. it has a countable dense subset.

*Proof*. Recall that $D$ is a dense subset of $X$ if $\bar D=X$. This equivalent to saying that $\forall \epsilon>0$, $\forall x\in X$, $B(x,\epsilon)\cap D\ne\emptyset$.

Let $D$ be the set of all possible finite linear combinations (superpositions) of the $e_n$’s. Then $D$ is countable. Let $x\in X$. Then $\exists$ a sequence of scalars $(\alpha_n)$ such that $x=\displaystyle\sum_{j=1}^\infty\alpha_je_j$. Given $\epsilon>0$, $\exists$ a positive integer $N$ such that

$$||x-(\alpha_1e_1+\cdots+\alpha_ne_n)||<\epsilon$$

for all $n\geq N$. This implies that $\alpha_1e_1+\cdots+\alpha_ne_n\in B(x,\epsilon)\cap D$ for all $n\geq N$.

One question mindful readers may have is does every separable Banach space have a Schauder basis? The answer is negative and a counterexample can be found in

Enflo, P. (1973), A counterexample to the approximation property. Acta Math. **130**, 309–317.

We finish this lecture with the following theorem.

*Theorem*. [Completion] Let $X$ be a normed space. Then there exists a Banach space $\hat X$ and an isometry from $X$ onto $W\subset\hat X$ which is dense in $\hat X$. The space $\hat X$ is unique up to isometries.