Banach-Mazur applied to a Hilbert space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:46:49Z http://mathoverflow.net/feeds/question/82720 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82720/banach-mazur-applied-to-a-hilbert-space Banach-Mazur applied to a Hilbert space Laurent Berger 2011-12-05T18:40:39Z 2011-12-06T01:13:46Z <p>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.</p> <p>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. </p> <p>My question is can one write down explicitly such a subspace of $C^0([0;1],R)$?</p> <p>I'm just curious, that's all.</p> http://mathoverflow.net/questions/82720/banach-mazur-applied-to-a-hilbert-space/82725#82725 Answer by Igor Rivin for Banach-Mazur applied to a Hilbert space Igor Rivin 2011-12-05T19:18:40Z 2011-12-05T19:18:40Z <p><a href="http://dl.dropbox.com/u/5188175/BanachMazur.pdf" rel="nofollow">This proof</a> of Banach-Mazur seems quite constructive (well, as constructive as you can get in this business) -- this is from <em>Banach Space Theory: the basis for linear and Nonlinear analysis.</em></p> http://mathoverflow.net/questions/82720/banach-mazur-applied-to-a-hilbert-space/82737#82737 Answer by Robert Israel for Banach-Mazur applied to a Hilbert space Robert Israel 2011-12-05T20:40:24Z 2011-12-05T20:40:24Z <p>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 $(Jx)(t) = &lt;\phi(t), x>$. Now, how to construct $\varphi$? </p> <p>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 $n=0$).</p> <p>Recursively define closed subintervals $J(v)$ of $J(r) = [0,1]$ for $v \in T$ so that</p> <ol> <li>If $w$ is a child of $v$, $J(w) \subset \text{interior}(J(v))$ </li> <li>If $w_1$ and $w_2$ are disjoint children of $v$, $J(w_1)$ and $J(w_2)$ are disjoint.</li> <li>If $v$ is at level $n$ in the tree, $J(v)$ has length at most $2^{-n}$.</li> </ol> <p>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 $n$, $\varphi_m(t) \in K(v)$. </p> <p>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.</p> http://mathoverflow.net/questions/82720/banach-mazur-applied-to-a-hilbert-space/82746#82746 Answer by Bill Johnson for Banach-Mazur applied to a Hilbert space Bill Johnson 2011-12-05T22:52:51Z 2011-12-05T22:52:51Z <p>If you want a reasonably explicit embedding of $\ell_2^n$ into $C[0,1]$ with distortion independent of $n$ (but not isometric) you can do the following. Take the first $2^n$ terms of your favorite lacunary sequence of characters and form the Rademacher functions over them, divided by $2^n$--call them $x_1,\dots,x_n$. A lacunary sequence of characters is equivalent to the unit vector basis of $\ell_!$ and hence $x_1,\dots,x_n$ is, by Khintchine's inequality, well equivalent to an orthonormal basis for $\ell_2^n$.</p> <p>The better thing to do is to embed isometrically $\ell_2$ into $C(X)$, where $X$ is the unit ball of $\ell_2$ in its weak topology, in the way suggested by Robert Israel. By Milutin's theorem, $C(X)$ is isomorphic (but, of course, not isometric) to $C[0,1]$.</p> http://mathoverflow.net/questions/82720/banach-mazur-applied-to-a-hilbert-space/82759#82759 Answer by fedja for Banach-Mazur applied to a Hilbert space fedja 2011-12-06T01:13:46Z 2011-12-06T01:13:46Z <blockquote> <p>By "explicit", I was really asking for a sequence of continuous functions...</p> </blockquote> <p>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. </p> <p>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. </p>