Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

From Wikipedia:

In mathematics, in particular in mathematical analysis, the Whitney extension theorem is a partial converse to Taylor's theorem. Roughly speaking, the theorem asserts that if $A$ is a closed subset of a Euclidean space, then it is possible to extend a given function off $A$ in such a way as to have prescribed derivatives at the points of $A$. It is a result of Hassler Whitney. A related result is due to McShane, hence it is sometimes called the McShane–Whitney extension theorem.

I need a stronger theorem which allows for both the set and the function to vary continuously, and for the resulting extended function to vary continuously.

Consider Euclidean space $\mathbb R^d$ (for any $d \ge 1$), and let $\mathcal K$ denote the metric space of compact subsets of $\mathbb R^d$, equipped with the Hausdorff metric. Select $\alpha \in \mathbb N$, and let $X = C^\alpha(\mathbb R^d, \mathbb R)$ be the Fréchet space of $C^\alpha$-smooth functions.

For any compact set $K \in \mathcal K$, let $X_K = C^\alpha(K, \mathbb R)$ be the Banach space of $C^\alpha$-smooth functions defined on $K$, and let $\rho_K : X \to X_K$ be the restriction map.

Define $Y_K$ to be the Fréchet space of functions which are real analytic on $\mathbb R^d - K$, and $C^\alpha$-smooth on $K$. Let $\iota_K : Y_K \to X$ be the inclusion map, which forgets that a function is analytic and just preserves its $C^\alpha$-smooth character.

The classical Whitney theorem implies that for each $K \in \mathcal K$, there is a function $w_K : X_K \to Y_K$ for which the following diagram commutes: $$\begin{array}[ccc] ~Y_K & \rightarrow^{\iota_K} & X \\ \uparrow {w_K} & & \downarrow {\rho_K} \\ X_K & \rightarrow_{1_{X_K}} & X_K \end{array}$$ where $1_{X_K}$ denotes the identity function on $X_K$. Equivalently, $1_{X_K} = \rho_K \circ \iota_K \circ w_K$. This means that we may extend a function from $K$ analytically, embed it into $X$, then restrict back to $K$, and we're left with the original function.


I would like to ensure not only that the Whitney map is continuous in the extending function, but also that the compact set may be allowed to vary.

Define the coproduct $\mathcal X = \coprod_{K \in \mathcal K} \{K\} \times X_K$, and refine the topology so that the projection map $\mathcal X \to \mathcal K$ is continuous. This way, $\mathcal X$ is kind of like a fibration over $\mathcal K$, except the fibers aren't necessarily homotopically equivalent. A point in $\mathcal X$ encodes both a compact set $K$ and a $C^\alpha$-smooth function defined on that set. (I think this is an admissible construction topologically. Tell me if this is not well-defined.)

Similarly, let $\mathcal Y = \coprod_{K \in \mathcal K} \{K\} \times Y_K$ encode the analytically extended functions, with continuous projection map $\mathcal Y \to \mathcal K$.

Consider the product space $\mathcal K \times X$, along with the inclusion map $\iota : \mathcal Y \to \mathcal K \times X$ and the restriction map $\rho : \mathcal K \times X \to \mathcal X$.

Now for the Whitney-type question. Does there exist a continuous map $w : \mathcal X \to \mathcal Y$ so that the following diagram commutes? $$\begin{array}[ccc] ~\mathcal Y & \rightarrow^{\iota} & \mathcal K \times X \\ \uparrow {w} & & \downarrow {\rho} \\ \mathcal X & \rightarrow_{1_{\mathcal X}} & \mathcal X \end{array}$$

share|improve this question
1  
Out of curiousity: Whitney's construction is very explicit in its estimates and builds out of dyadic cube decomposition of the exterior domain. Have you checked that Whitney's original construction in fact does not do what you want? –  Willie Wong Jun 21 '13 at 7:34
1  
If you just use the standard (coproduct of products) topology on $\mathcal{X}$ then the projection $\pi\:\mathcal{X}\to\mathcal{K}$ is already continuous. You want to discard some open sets to get a smaller topology with some kind of minimality property such that $\pi$ is still continuous. I don't think that there is an obvious way to do that. –  Neil Strickland Jun 21 '13 at 11:38
1  
I think you should be able to turn "$\coprod_{K\in\mathcal K}(K\times X_K)$" into a bundle requiring suitable local trivializations. This requires that you should be able to construct a homeomorphism between $C^\alpha(K_1)$ and $C^\alpha(K_2)$ when $K_1$ and $K_2$ are nearby. How could you manage this? Think about e.g. a cube $K_1$ and a nearby $K_2$ with wildly saw-toothed boundary. Already the definition of $C^\alpha(K)$ is not obvious when $K$ has irregular boundary, cf. arxiv.org/abs/math/0505380 . –  TaQ Jun 23 '13 at 10:32
1  
Note that for every $K\in\mathcal{K}$ there are arbitrarily close points $F\in\mathcal{K}$ such that $F$ is a finite, discrete subset of $K$. Given this, I don't think you can hope to make $\mathcal{X}$ into a bundle over $\mathcal{K}$. –  Neil Strickland Jun 24 '13 at 11:19
add comment

1 Answer 1

This is not an answer but a suggestion of how to formulate the question.

I will simplify things as follows. Let $B$ be a closed ball in $\mathbb{R}^d$ (or $B$ could be be any other compact metric space). Let $\mathcal{K}$ be the space of closed subsets of $B$ with the Hausdorff metric. Let $\mathcal{X}$ be the set of pairs $(K,f)$, where $K\in\mathcal{K}$ and $f:K\to [-1,1]$ is continuous. Let $\mathcal{Y}$ be the set of pairs $(K,f)$ where $K\in\mathcal{K}$ and $f:B\to [-1,1]$ is continuous. There is a map $\rho:\mathcal{Y}\to\mathcal{X}$ given by $\rho(K,f)=(K,f|_K)$, and the Tietze extension theorem says that this is surjective.

For any point $(K,f)\in\mathcal{X}$ we can put $$ G(f)=\{(x,f(x)):x\in K\}\subset K\times [-1,1], $$ and define $d((K_0,f_0),(K_1,f_1))$ to be the Hausdorff distance between $G(f_0)$ and $G(f_1)$. This defines a metric on $\mathcal{X}$. We can introduce a similar metric on $\mathcal{Y}$, and I think it is equivalent to the obvious product metric. The map $\rho$ is continuous, and one can ask whether there is a continuous section. I think that that is true. Indeed, one just needs to show that the ingredients in the standard proof of the Tietze extension theorem depend continuously on the input data with respect to suitable metrics.

One of the main ingredients is Urysohn's lemma, which says that if $P$ and $Q$ are disjoint closed subsets of $B$ then there is a continuous map $f:B\to [0,1]$ with $f=0$ on $P$ and $f=1$ on $Q$. As we have a metric on $B$ we can just take $f(x)=d(x,P)/(d(x,P)+d(x,Q))$, and this depends continuously on the pair $(P,Q)$.

The other main ingredient is the map $\phi_a:\text{Map}(B,[-1,1])\to\mathcal{K}$ given by $\phi_a(f)=f^{-1}([a,1])$. Given $f$ and $\epsilon>0$ I think one can find $\delta>0$ such that the sets $d(f^{-1}([a-\delta,1]),f^{-1}([a+\delta,1]))<\epsilon$, and then $d(\phi_a(f),\phi_a(g))<\epsilon$ whenever $d(f,g)<\delta$. Thus, $\phi_a$ is also continuous.

For the Whitney theorem you probably want to define $J_\alpha(f)(x)\in\mathbb{R}^N$ (for suitable $N$) to be the vector of all derivatives of $f$ at $x$ of orders less than or equal to $\alpha$, then let $G_\alpha(f)\subset K\times\mathbb{R}^N$ be the graph of $J_\alpha(f)$. You can then use the Hausdorff distance between these sets as a metric on $\mathcal{X}$. If you are willing to work with $C^\alpha$ extensions rather than real analytic ones then there is an obvious parallel metric on $\mathcal{Y}$. This could probably be adapted to cover the analytic case as well.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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