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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:

  1. $ \|f_n\| =1 $
  2. 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?

Tomasz Kania
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3 Answers3

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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 $X^*$.

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

  2. 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 $C(B_{X^*})$.

  3. 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.
  4. $C(\Delta)$ embeds isometrically into $C[0,1]$ by the Borsuk–Dugundji extension theorem.
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Tomasz Kania
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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:

  1. 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 $\lim\limits_{t \to 1} \varphi(\textbf{x})(t)=\varphi(\textbf{x})(1)=\sum_{k=1}^{+\infty} \vert x_n \vert$.
  2. It is an isometry as $\sup\limits_{t \in [0,1]} \varphi(\textbf{x}) =\sum_{k=1}^{+\infty} \vert x_n \vert= \Vert \textbf{x} \Vert_1$.
  3. $\varphi[\ell^1]$ is therefore an isometric copy of $\ell_1$ in $\mathcal{C}([0,1])$
mathcounterexamples.net
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  • The symbols $\ell_p$ are commonly used in Banach space theory. Myself, I never write $\ell^p$ (this has advantages when dealing with finite-dimensional spaces). – Tomasz Kania Dec 04 '15 at 10:48
  • @TomekKania Nice to know. I'm French and never saw $\ell_p$ in French math literature. Do you mean that you reserve $\ell^p$ for the finite-dimensional spaces? – mathcounterexamples.net Dec 04 '15 at 11:02
  • No, I reserve the upper index for the dimension and lower for the exponent: $\ell_p^4$ is the four-dimensional $\ell_p$-space. – Tomasz Kania Dec 04 '15 at 11:05
  • I'm sorry I didn't mention it explicitly - the maps have to be linear. –  Dec 04 '15 at 11:16
  • Ok and If I'm not missing anything, one can take $a_n\mapsto (\sum_{k=1}^n|x_n|^p)^{1/p}$ for any $1\le p<\infty$, to embed $\ell^n_p$ isometrically into $C([0,1])$ no? – AB_IM Aug 15 '23 at 14:09
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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 $J:\ell_1\to\mathcal C([0,1])$.

Etienne
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  • I think you meant just $(\alpha_n-1)$, not $(2\alpha_n-1)$. –  Dec 05 '15 at 20:52
  • Also, doesn't this proof only work for real $\ell_1$? If I were to extend this map to complex sequences, I think this fails to be an isometry. Is there a way to fix this? –  Dec 05 '15 at 20:55
  • @Amudhan You're right in both cases. As for the complex case, I don't know how to fix this elegantly. – Etienne Dec 05 '15 at 21:36