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Given a smooth map from $\phi: B \rightarrow M$ where $B$ is a Banach Space and $M$ is a finite dimensional smooth manifold (for example, the end point map for a control system), what is the strongest analogue of Sard's theorem which holds.

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    $\begingroup$ Look at MathSciNet reviews of papers by Sean Bates to see which ones are relevant for you. $\endgroup$
    – Bill Johnson
    Commented May 15, 2015 at 2:11
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    $\begingroup$ I've seen versions for finite dim vector spaces, maps between infinite dim banach space with the requirement that the frechet derivative be a Fredholm operator, maps between finite dim manifolds and a version for maps from a banach space to a finite dim vector space. But I've not seen exactly what the question asks for! $\endgroup$
    – Benjamin
    Commented May 15, 2015 at 2:26

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The classical result is the celebrated Sard-Smale Theorem, see 1.

Theorem (Smale). Let $\phi \colon B \longrightarrow M$ be a $C^q$-Fredholm map between separable Banach manifolds, with $q > \max \{0, \, \textrm{index of } \phi\}$. Then the set of regular values of $\phi$ is residual in $M$.

Here residual set means that it is a countable intersection of open dense sets. The Baire Category Theorem implies that a residual set in a complete metric space is dense, in particular its complement contains no open set.

You cannot expect much better than this, even in your situation. In particular, the Fredholm assumption is an essential one: in fact, Kupka constructed an example of a non-Fredholm, $C^{\infty}$ real function $\phi \colon \ell^2(\mathbb{R}) \longrightarrow \mathbb{R}$ with critical values containing an open set, see 2.

References.

1 S. Smale: An Infinite Dimensional Version of Sard's Theorem, American Journal of Mathematics 87 (1965), 861–866.

2 I. Kupka: Counterexample to the Morse-Sard theorem in the case of infinite-dimensional manifolds, Proc. Amer. Math. Soc. 16 (1965), 954–957.

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  • $\begingroup$ What exactly is meant by the index of a Fredholm map? I know the definition of the index of a Fredholm operator. $\endgroup$
    – Benjamin
    Commented May 15, 2015 at 16:08
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    $\begingroup$ The definition is the natural one, and you can find it in the first page of Smale's paper.Is the Fredholm index of the derivative $$(D \phi)_x \colon T_x B \to T_{\phi(x) }M$$ at some point $x \in B$ (such a index does not depend on $x$). $\endgroup$
    – Francesco Polizzi
    Commented May 15, 2015 at 16:48
  • $\begingroup$ Ok, I was only confused because I couldn't see why it wouldn't depend on $x$. Thanks. $\endgroup$
    – Benjamin
    Commented May 15, 2015 at 16:48

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