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.
Best Answer
Let $\varphi$ be a continuous function from $[0,1]$ onto the closed unit ball $B$ of $\ell^2$ in the weak topology. Then we define $J: \ell^2 \to C[0,1]$ by $(Jx)(t) = <\phi(t), x>$. Now, how to construct $\varphi$?
Consider convex compact subsets of $B$ of the form $K(a_1,\ldots,a_n) = \{x \in B: a_i/2^n \le x_i \le (a_i+1)/2^n \text{ for } i = 1 \ldots n\}$ for integers $a_i$, $-2^n \le a_i \le 2^n$. Of course many of these are empty, and we disregard those. We can consider the nonempty ones as forming a tree structure $T$ with vertices $v = (a_1, \ldots, a_n)$, and edges joining $(a_1, \ldots, a_n)$ to its children $(b_1, \ldots, b_{n+1})$ where for $1 \le i \le n$, $b_i$ is either $2 a_i$ or $2 a_i + 1$. Thus $K(v) = \bigcup_w K(w)$ where $w$ runs over the children of $v$. The root $r$ of the tree corresponds to $B$ itself (with $n=0$).
Recursively define closed subintervals $J(v)$ of $J(r) = [0,1]$ for $v \in T$ so that
Let $E_n$ be the union of $J(v)$ for all vertices $v$ at level $n$. Define $\varphi_n: [0,1] \to B$ by selecting $x_v \in K(v)$ for each vertex $v$ at level $n$, and defining $\varphi_n(t) = \varphi_{n-1}(t)$ for $t \notin \text{interior}(E_n)$, $\varphi_n(t) = x_v$ for $t \in J(v)$ where $v$ is a vertex at level $n$, and interpolating linearly on the rest. Note that if $m > n$ and $t \in J(v)$ where $v$ is a vertex at level $n$, $\varphi_m(t) \in K(v)$.
Finally, define $\varphi(t) = \lim_{n \to \infty} \varphi_n(t)$. This, I claim, is a continuous surjective function from $[0,1]$ to $B$ with the weak topology.