TITLE: Banach-Mazur applied to a Hilbert space
QUESTION [16 upvotes]: The Banach-Mazur theorem says that every separable Banach space is isometric to a subspace of $C^0([0;1],R)$, the space of continuous real valued functions on the interval $[0;1]$, with the sup norm.
If we apply this to $\ell^2(R)$, then we see that $C^0([0;1],R)$ has a subspace which is a Hilbert space for the sup norm. 
My question is can one write down explicitly such a subspace of $C^0([0;1],R)$?
I'm just curious, that's all.

REPLY [8 votes]: By "explicit", I was really asking for a sequence of continuous functions...

Ah, OK. Take a Peano curve $(f,g)$ from $[-1,1]$ to $[-1,1]^2$. Now define inductively $F_1=f$, $F_{k+1}=F_k\circ g$. Note that for any finite sequence of points $y_j\in[-1,1]$ ($j=1,\dots, n$), we can find $x\in[-1,1]$ such that $F_j(x)=y_j$ (induction: the last $n-1$ functions determine $g(x)$ after which $f(x)$ is free to choose). Now just define $H_1=F_1$, $H_{k+1}=F_{k+1}\sqrt{1-H_1^2-\dots-H_k^2}$. This is your sequence. 
It just remains to construct the Peano curve "explicitly". It is not going to be nice, so an elementary formula is out of question. However, you can take the Fourier series $\sum_{k} v_k\cos(2\pi A_k x)$, where $v_k$ is a sequence of vectors in $\mathbb R^2$ and $A_k$ is a sequence of integers, with properly chosen (explicit) $v_k$ and $A_k$ to get the image to contain the unit square. Now just truncate the values to $[-1,1]$ in the usual way.