13
$\begingroup$

Let $V$ be a Fréchet topological vector space. Let $K_0$ and $K_1$ be two closed subsets which are disjoint.

I wish to show the existence of a Fréchet-smooth function $f:V\to [0,1]$ whose restriction to $K_0$ is $0$, and whose restriction to $K_1$ is $1$.

The construction of a continuous function with the above properties is an instance of the usual Urysohn's lemma (and can be written down explicitely in terms of the metric on $V$). But I don't know how to mollify continuous functions on Fréchet spaces….

Note: The same question where $V$ is a Banach space also interests me, and would be sufficient for my purposes.

$\endgroup$
8
  • 5
    $\begingroup$ The fact is that there are Banach spaces with no non-zero C¹ functions with bounded support. In this situation the best partition of unity we can do are loc.lip, and we should be happy, since loc.lip is sufficient for building flows via ODE. $\endgroup$ Nov 1 at 15:59
  • 1
    $\begingroup$ @PietroMajer : Could you provide references to such Banach spaces? $\endgroup$ Nov 1 at 16:10
  • $\begingroup$ @WillieWong Yes, of course, I meant subsets. $\endgroup$ Nov 1 at 17:24
  • 1
    $\begingroup$ For these it works as spaces of smooth function are on compact spaces are know to be smoothly paracompact. All of this is recorder in Detail with Conditions in Kriegl/Michor: The convenient setting of global Analysis. $\endgroup$ Nov 1 at 17:48
  • 1
    $\begingroup$ Actually smooth paracompactness is strenger than what you need. The condition would be smooth regularity (which is weaker than smooth paracompactness). Also for this conditions are know and recorder for example in kriegl/michor $\endgroup$ Nov 1 at 17:52

2 Answers 2

9
$\begingroup$

Warning: Since we will be working outside of Banach spaces, one needs to decide the concept of smoothness applied in the following. Since I am citing from Kriegl/Michor's book 1, the default will be the convenient calculus they use (and the results hold for this choice by their proofs).

Sideremark: As the OP only cared for Frechet spaces this choice still coincides with other common choices in the infinite-dimensional setting (for example convenient smooth on Frechet spaces is equivalent to the popular Bastiani setting of smoothness, cf. 2). In particular all convenient smooth maps on Frechet spaces are automatically continuous.

Recall that a Hausdorff locally convex vector space $X$ is called

  1. smoothly normal, if for two disjoint closed subsets $A_0,A_1$ there exists a smooth function $f\colon X \rightarrow \mathbb{R}$ with $f|_{A_i} = i$ for $i = 0, 1$.
  2. smoothly paracomopact if every open cover of $X$ admits a subordinate partition of unity whose members are smooth functions.

Obviously, every smoothly paracompact space is smoothly normal.

Concerning the spaces $C^\infty (M)=C^\infty(M,\mathbb{R})$ for a compact smooth manifold $M$, the smooth normality can be deduced from from a Theorem originally due to Wells (1973). Wells gives a criterion for smooth paracompactness (so we shall actually establish the stronger property). I cite it here from 1, where the result we are after is a specialised version of the Theorem 16.10 which I dumb down a bit (it is actually much more versatile, so check it out!) for our purposes:

If $X$ is Lindelöf and smoothly regular, then $X$ is smoothly paracompact. In particular, all nuclear Frechet spaces and strict inductive limits of sequences of such spaces are smoothly paracompact.

Hence it suffices to prove that the space $C^\infty (M)$ is a nuclear space. This is a well known fact for these spaces and can be found at many places in the literature. Viewing the space $C^\infty (M)$ as the space of sections into the trivial vector bundle $M \times \mathbb{R}$, we can cite Proposition 4.8 from another classical book by Peter Michor (also freely available on his webpage here: 3) showing nuclearity.

Upshot: $C^\infty (M)$ is smoothly paracompact, whence smoothly normal, whence the smooth Urysohn Lemma is true.

$\endgroup$
11
$\begingroup$

A bump function on a Banach space $X$ is a non-zero function with bounded support. Existence of a bump function of a given smooth regularity has, of course, strong immediate consequences --by translation and rescaling, one can make partitions of unity of that regularity, whence regular approximations; and one can make smooth renorming of that regularity.

A theorem of M. Fabian (see a reference below) states: If $X$ possesses a $C^1$ bump function, it is Asplund (i.e. separable subspaces have separable duals). The simplest non Asplund space is $\ell_1$. Therefore, $\ell_1$ admits no $C^1$ bump functions. This I think was known even before, and there is a possibly simpler proof. (I suggest the above keywords in boldface for a search; some references are listed below).

M. Fabian, On projectional resolution of identity on the duals of certain Banach spaces, Bull. Austral. Math. Soc., 35 (1987), 363–371.

R. Deville, G. Godefroy, V. Zizler, Smoothness and Renormings in Banach Spaces. Pitman Monographs and Surveys in Pure and Applied Mathematics, vol. 64 (1993)

And this survey https://core.ac.uk/download/pdf/81972936.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.