Isometric copy of $\ell_1$ in $C[0,1]$? How do I show that there is an isometric copy of $\ell_1$ in $C[0,1]$?
If I obtain a sequence of functions $(f_n) \in C[0,1]$ with the following properties, then I am done:


*

*$ \|f_n\| =1 $

*For all sequences $ (x_n) $, with $|x_n|=1$, there is a $t \in [0,1]$ such that 
$f_n(t)=x_n$


Does this help? Or is the proof non-constructive?
 A: Every seprable Banach space embeds into $C[0,1]$ isometrically. This is the Banach-Mazur theorem.
Sketch of the proof. Let $X$ be a separable Banach space and let $B_{X^*}$ the unit ball of $X^*$.


*

*$B_{X^*}$ is weak*-compact (by the Banach–Alaoglu theorem) and
metrisable in the weak*-topology.

*The map $T\colon X \to C(B_{X^*})$ given by $Tx = \hat{x}$ ($x\in X$), where $\hat{x}(f)=\langle f,x\rangle (f\in B_{X^*})$ is isometric. Hence $X$ embeds isometrically into $C(B_{X^*})$.

*Every compact metric space is a continuous image of the Cantor set
$\Delta$. Let $\varphi\colon \Delta \to B_{X^*}$ be a continuous
surjection. Then the map $S\colon C(B_{X^*})\to C(\Delta)$ given by $Sf=f\circ \varphi$ ($f\in C(B_{X^*})$ is isometric.

*$C(\Delta)$ embeds isometrically into $C[0,1]$ by the Borsuk–Dugundji extension theorem.

A: I imagine that by $\ell_1$, you're referring to the $\ell^1$ space of sequences whose series is absolutely convergent. My answer below is based on this assumption.
Consider a strictly increasing sequence $(\alpha_n) \in [0,1]^{\mathbb N}$ such that $\lim\limits_{n \to +\infty} \alpha_n=1$. The sequence $(1-\frac{1}{n})_{n \in \mathbb N}$ would do the job.
Now consider $\varphi : \ell^1 \to \mathcal{C}([0,1])$ defined as follows for $\textbf{x}=(x_n) \in \ell^1$:
$$\begin{array}{l|rcl}
\varphi(\textbf{x}) : & [0,1] & \longrightarrow & \mathbb R \\
    & 0 & \longmapsto & 0\\
    & 1 & \longmapsto & \sum_{k=1}^{+\infty} \vert x_n \vert\\
    & \alpha_n & \longmapsto & \sum_{k=1}^n \vert x_n \vert \end{array}$$ and $\varphi(\textbf{x})$ is piecewise linear for the other values of $[0,1]$.
$\varphi$ has following properties:


*

*It is well defined as for $\textbf{x}=(x_n) \in \ell^1$, $\varphi(\textbf{x})$ is continuous on $[0,1)$. $\varphi(\textbf{x})$ is also continuous at $1$ as $\lim\limits_{t \to 1} \varphi(\textbf{x})(t)=\varphi(\textbf{x})(1)=\sum_{k=1}^{+\infty} \vert x_n \vert$.

*It is an isometry as $\sup\limits_{t \in [0,1]} \varphi(\textbf{x}) =\sum_{k=1}^{+\infty} \vert x_n \vert= \Vert \textbf{x} \Vert_1$.

*$\varphi[\ell^1]$ is therefore an isometric copy of $\ell_1$ in $\mathcal{C}([0,1])$

A: Tomek' answer is of course the "correct" one, because the Banach-Mazur theorem is both beautiful and completely general. If one particularize its proof to the case of $\ell_1$, one gets 
an explicit isometric embedding of $\ell_1$ into $\mathcal C([0,1])$, as follows.
Let $K\subseteq [0,1]$ be the classical triadic Cantor set. Every $t\in K$ can be written in one and only one way as
$$t=\sum_{n=1}^\infty \frac{\alpha_n(t)}{3^n}\qquad\hbox{where $\alpha_n(t)\in \{ 0,2\}$};$$
and conversely, every sequence $(\alpha_n)\in\{ 0,2\}^{\mathbb N}$ defines a point $t\in K$.
Now, for any $u=(u_n)_{n\in\mathbb N}\in \ell_1$, let us denote by $J_0u :K\to \mathbb R$ the function defined by 
$$J_0u(t)=\sum_{n=1}^\infty (\alpha_n(t)-1) \, u_n\, .$$
It is not difficult to show that $J_0u$ is a continuous function, so we have a map $J_0: \ell_1\to\mathcal C(K)$, which is obviously linear. Moreover, $J_0$ is also an isometry because, as $(\alpha_n)$ varies over $\{ 0, 2\}^{\mathbb N}$, the sequence $(\alpha_n-1)$ varies over $\{ -1,1\}^{\mathbb N}$ and because we have $$\Vert u\Vert =\sup\,\Bigl\{ \sum_{n=1}^\infty \varepsilon_n u_n;\; (\varepsilon_n)\in\{ -1,1\}^{\mathbb N}\Bigr\}\qquad\hbox{ for every $u\in\ell_1$}.$$
Now, for any $u\in\ell_1$, extend $J_0u$ to a continuous function $Ju$ on $[0,1]$ in the natural way, by declaring $Ju$ to be affine on each interval "contiguous" to the Cantor set $K$. This gives the required isometric embedding $J:\ell_1\to\mathcal C([0,1])$. 
