continuity of extension of maps along curves - MathOverflow

continuity of extension of maps along curves

2011-02-13T12:06:31Z

Let $a\le b$ and $k\ge 0$ be given and fixed. Let furthermore $x$ and $y$ denote two different elements of a Hilbert space $H$. Suppose $u:\mathbb{R}\rightarrow H$ is a $C^k$-embedding connecting $x$ and $y$, s.t. the derivatives up to order $k$ vanish at infinity and $f:\mathbb{R} \rightarrow \mathbb{R}$ a given $C^k$-map with support in $[a,b].$ Then it it possible to construct a $C^k$map $\tilde{f}:H\rightarrow \mathbb{R}$ satisfying $\tilde{f}(u(s))=f(s)$ for any $s\in \mathbb{R}.$ (This can be done using a tubular neighborhood and a suitable cutoff function.)

Is it possible to make the map $f\mapsto\tilde{f}$ continuous? The domain of this map should be the space of $C^k$-embeddings (as above) times $\mathbb{R}$-valued $C^k$- functions supported in $[a,b]$. And the codomain should be the space of $C^k$-maps on $H$.

Answer by Arie Israel for continuity of extension of maps along curves

2011-02-14T02:38:12Z

Let $\gamma = u(\mathbb{R})$ be the curve, and consider your $f$ as a function on the subset $\gamma$ of $H$. A minor note is that if the curve is almost straight, then you can make a nice extension by foliating $H$ through translations of $\gamma$. More precisely, set $v = u'(0)$, and define $Tf \in C^k(H)$ by $Tf(y) = f(y-t)$, where $t \in H$ is the unique element perpendicular to $v$ and with $y-t$ in $\gamma$. On the other hand, for more curvy curves I think it is impossible, at least for any $k \geq 1$. There exists a $C^k$ curve $u:\mathbb{R} \rightarrow H$ contained in the unit ball of $H$ which satisfies $u(n) = v_n$, for $v_n$ an orthonormal basis for $H$, and $n \in \mathbb{Z}$. Let $f(x) = x$. Now note that there cannot exist a $C^k$ function $\tilde{f}:H \rightarrow \mathbb{R}$ which extends $f$ as you want. In fact, by the mean value theorem $|\tilde{f}(v_0)-\tilde{f}(v_n)| \leq C |v_0 - v_n| \leq 2 C$, but by the extension property, $\tilde{f}(v_0) - f(0) = 0$, and $\tilde{f}(v_n) = f(n) = n$.

Answer by Jaap Eldering for continuity of extension of maps along curves

2011-02-16T09:49:51Z

Some rough ideas to construct a continuous map. I'm not sure that there aren't any obstructions this might run into.

First, it's nice to have an (explicit) construction of the map $F:u,f \mapsto \tilde{f}$ to be able to say anything about continuity.

An explicit construction of a tubular neighborhood of $u([a,b])$ can be made via the exponential mapping of the normal bundle, but this loses one degree of smoothness. One could use a smoothing operator on $u([a,b])$ first. Both $u([a,b])$ and $u(\mathbb{R}\setminus(a-\delta,b+\delta))$ are compact, so these are separated by a finite distance $\epsilon$ and therefore there should be a neighborhood of $u$ which still has this property for a finite distance, say $\epsilon/2$. I think that such compactness arguments should allow you to proof continuity of the map $F$. You may want to look into the Omega lemma, which states conditions for the composition mapping $f,g \mapsto f \circ g$ to be continuous. As a reference, you may look into Abraham, Marsden, Ratiu ``Manifolds, tensor analysis, and applications''.