Nonseparable Hilbert spaces as quotients of spaces of bounded functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:47:34Z http://mathoverflow.net/feeds/question/23221 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23221/nonseparable-hilbert-spaces-as-quotients-of-spaces-of-bounded-functions Nonseparable Hilbert spaces as quotients of spaces of bounded functions Ady 2010-05-02T02:37:22Z 2010-05-09T11:43:25Z <p>Is the following result true: the Hilbert space $\ell^{2}\left(2^{\Gamma}\right)$ is a quotient of $\ell^{\infty}\left(\Gamma\right)$ for any uncountable $\Gamma$ ? [I think it is, but cannot remember where I saw it, long time ago.] I would be very grateful for any (freely available, if possible) reference (Pelczynski ? Rosenthal ?).</p> http://mathoverflow.net/questions/23221/nonseparable-hilbert-spaces-as-quotients-of-spaces-of-bounded-functions/23273#23273 Answer by Bill Johnson for Nonseparable Hilbert spaces as quotients of spaces of bounded functions Bill Johnson 2010-05-02T18:57:47Z 2010-05-04T14:45:17Z <p>I don't know who first observed this (maybe Archimedes?) but it is true because <code>$C(\{0,1 \}^\Gamma)$</code> is a quotient of $\ell_1^\Gamma$ and hence $\ell_1(2^\Gamma)$ embeds into $\ell_\infty(\Gamma)$. </p> <p>@Ady </p> <p>Here is a more serious answer to your question. Take a quotient map $Q$ from $\ell_1(2^\Gamma)$ onto $C([0,1]^{2^\Gamma})$ and extend to a norm one mapping $T$ from $\ell_\infty(\Gamma)$ into some injective space $Z$ that contains $C([0,1]^{2^\Gamma})$ (you cannot extend $Q$ to an operator from $\ell_\infty(\Gamma)$ into $C([0,1]^{2^\Gamma})$ because, e.g., $C([0,1]$ is not a quotient of $\ell_\infty$). Use partitions of unity to get a net $(P_a)$ of norm one finite rank projections on $Z$ taking values in $C([0,1]^{2^\Gamma})$ and whose restrictions to $C([0,1]^{2^\Gamma})$ converge strongly to the identity. A weak$^*$ cluster point of $(P_a^* T^*)$ gives an isometric embedding of the dual of $C([0,1]^{2^\Gamma})$ (which contains $L_1([0,1]^{2^\Gamma})$) into the dual of $\ell_\infty(\Gamma)$. Thus if $Y^*$ is any reflexive subspace of $L_1([0,1]^{2^\Gamma})$, such as $\ell_2(2^\Gamma)$, then $Y$ is isometric to a quotient of $\ell_\infty(\Gamma)$.</p>