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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 $H$.

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The math doesn't seem to display properly and I cannot fix the error. Could anyone please help? – Orbicular Feb 13 '11 at 12:08
Shouldn't the curve $u$ be injective, at least on $[a,b]$ for the map $f \mapsto \tilde{f}$ to exist? I would even think you need to require $u$ to be something like an embeddeing, for otherwise you could have a sequence $s_i \to \infty$ such that $u(s_i) \to u(c)$ with $c \in [a,b]$. That would imply that $\tilde{f}$ cannot be continuous in general. – Jaap Eldering Feb 13 '11 at 13:39
@eldering: Yes, you are absolutely right. I will correct this. – Orbicular Feb 13 '11 at 13:58
Some more details on your question: do use assume $u(a) = x, u(b) = y$ or does $u$ connect $x,y$ at some other points? Your description of the domain suggests that you're looking for a map (let's give it a name) $F\colon u,f \mapsto \tilde{f}$, not just dependent on $f$. Do you want continuity in both arguments? Finally, what topology do you consider: compact-open, uniform-norm induced, other? I guess this might be irrelevant if you can show that everything relevant is defined on compact domains. – Jaap Eldering Feb 15 '11 at 14:01
The topology is given as follows: Consider $C^k_0(x,y),$ the space of $C^k-$maps $u$ with $u(s)$ converging to $x$ for $s\rightarrow -\infty$ and to $y$ for $s\rightarrow +\infty,$ s.t. $\frac{d^{l}u}{ds^{l}}(s)\rightarrow 0$ as $s\rightarrow \pm\infty$ for $1\le l\le k,$ endowed with the C^k-topology (it is an affine space modelled on $C^k_0(\mathbb{R},H)$). And yes, it should be continuous in both arguments. – Orbicular Feb 15 '11 at 14:26

2 Answers 2

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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''.

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I already thought in those directions. What I do not understand is how to make these choices of neighborhoods "continuous" (and in tow the choices of the cutoff functions). – Orbicular Feb 16 '11 at 10:14
The basic idea of the Omega lemma, which I think you should be able to use here as well: one can lift pointwise continuous dependence on each $u(s)$ to dependence on $u$ as a curve when $u$ and the dependence on $u(s)$ in the tubular neighborhood construction is uniform. As you are considering compact domains, this holds. – Jaap Eldering Feb 16 '11 at 12:30
I am familiar with the Omega-lemma (or Palais' lemma if you want).The problem is the dependence ON the tubular neighborhood. How can you make sure that if two curves are close their tubular neighborhoods are (which, in turn, is still insufficient, as furthermore the cutoff functions constructed should be close as well) – Orbicular Feb 16 '11 at 15:06

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 $\tilde{f}(v_n) = f(n) = n$.

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Sorry, I missed the part about $f$ having compact support. It looks like my counterexample is void for the time being. – ChaosInEquilibrium Feb 14 '11 at 3:05

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