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*. Let $X$ be a normed space with a Schauder basis $(e_n)$. Let $D$ be the set of all possible finite linear combinations (superpositions) of the $e_n$’s with rational scalar coefficients, i.e. $$D=\{\sum_{j=1}^n q_je_j: n\in\mathbb{N},\ q_j\in\mathbb{Q}\}$$Then $D$ is countable. (Why?)

Now, we show that $D$ is dense in $X$. 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 $x\in X$. Since $(e_n)$ is a Schauder basis, $\exists$ 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 $||x-\sum_{j=1}^n\alpha_je_j||<\frac{\epsilon}{2}$ for all $n\geq N$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each $j=1,\cdots,N$, $\exists q_j\in\mathbb{Q}$ such that $|q_j-\alpha_j|<\frac{\epsilon}{2N(||e_j||+1)}$. \begin{align*}||x-\sum_{j=1}^N q_je_j||&<||x-\sum_{j=1}^N\alpha_je_j||+\sum_{j=1}^N|\alpha_j-q_j|( ||e_j||+1)\\&<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\end{align*} This means that $\sum_{j=1}^Nq_je_j\in B(x,\epsilon)\cap D\ne\emptyset$ and hence the proof is complete.

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.