Let $X$ be any separable Banach space. The Banach-Mazur theorem states (astonishingly) that $X$ is isometric to a closed subspace of $C(\Delta)$, the space of continuous functions on the cantor set $\Delta$, or alternatively, $X$ is isometric to a closed subspace of $C([0,1])$.

Having checked a few online resources, I have come to understand that not every such isometric embedding of a banach space $X$ will be complemented (i.e. possess a closed complement) in $C(\Delta)$ (or $C([0,1])$). In fact, Banach and Mazur themselves showed that any isometric copy of $l^1 \subset C([0,1])$ is noncomplemented.

I understand that this question might have a long answer in the literature, but here it goes: is there any kind of simple sufficient condition for a closed subspace of $C([0,1])$ to be complemented?

Failing at that, I'm also interested in isomorphic copies of $X$ in $C([0,1])$ where the norm of the isomorphism and its inverse are controlled in some way. For instance, what properties of $X$ guarantee the following:

There exists a constant $C > 0$ such that for any $X$ satisfying some property, there always exists an isomorphism $$L : X \rightarrow Y \subset C([0,1])$$ where $L^{-1} : Y \to X$ exists, $Y$ is complemented in $C([0,1])$, and $\| L \| \|L^{-1}\| \leq C$.

  • $\begingroup$ "...there exists a 'universal' constant ...".$$$$ Universal constant with respect to what? $\endgroup$ – Norbert Sep 17 '13 at 16:08
  • $\begingroup$ What I mean is this: 'there is a constant $C > 0$ such that for all Banach spaces $X$ with some property...' $\endgroup$ – A Blumenthal Sep 17 '13 at 17:53

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