# Nonseparable Hilbert spaces as quotients of spaces of bounded functions

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 ?).

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@Ady: I made a major addition to my answer. –  Bill Johnson May 10 '10 at 9:11

I don't know who first observed this (maybe Archimedes?) but it is true because $C(\{0,1 \}^\Gamma)$ is a quotient of $\ell_1^\Gamma$ and hence $\ell_1(2^\Gamma)$ embeds into $\ell_\infty(\Gamma)$.

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 $\ell_\infty(\Gamma)$.

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[deleted earlier comment as I had misread Bill's answer] –  Yemon Choi May 2 '10 at 19:52
@Bill Johnson Could you be more specific, please ? Since the Hilbert space is not injective. Some extension result ? –  Ady May 2 '10 at 20:53
I'll give you a hint, Ady. Use a famous theorem of the first Fields Medalist who worked in Banach space theory. –  Bill Johnson May 2 '10 at 21:14
Maybe the second one, historically speaking. And he has many famous theorems. Still, I think that this is a result of Pelczynski or Rosenthal. Perhaps involving the Bohr compactification. –  Ady May 2 '10 at 21:35
@Bill Johnson Grothendieck-Maurey Theorem ? –  Ady May 2 '10 at 21:44