I'm going to postpone the motivation for this question because the question itself involves no complicated maths and may well have a very simple solution so I don't want to put anyone off with high falutin' symbols.


Here's the question. I have a smooth curve $c \colon (0,1) \to \mathbb{R}^2$ which does not intersect the $x$-axis. As $t \to 0$, this curve approaches the $x$-axis. I want to find out if it has a point of impact. At my disposal, I have any smooth function $f \colon \mathbb{R}^2 \to \mathbb{R}$ which is constant along the $x$-axis. I know that for any such $f$, $f \circ c \colon (0,1) \to \mathbb{R}$ extends to a smooth function $[0,1) \to \mathbb{R}$ (that is, all one-sided derivatives exist at the origin and are the limits of the corresponding derivatives as we approach the origin) with value $f(0,0)$ at $0$. So:



> 
> Is there some $f$ (satisfying the condition) such that the composition $f \circ c$ tells me that $c$ approaches a particular point on the $x$-axis?
> 
> 
> 


If the answer to that is "yes", then my follow-up question is about the derivatives of $c$ at the point of approach.


Here are some comments and partial results:


1. I'm allowed to use **any** information about the compositions $f \circ c$: their values, their derivatives, and so forth.
2. The $y$-value of $c$ easily extends to a smooth function $[0,1) \to \mathbb{R}$ since the second projection $\mathbb{R}^2 \to \mathbb{R}$ is one of our detectors. Moreover, it extends taking the value $0$ at $t = 0$.
3. I can show that the $x$-value of $c$ is bounded as $t$ approaches $0$. If it weren't, I could stick bump functions along the image of $c$ with disjoint support and that were $0$ on the $x$-axis. As they have disjoint support, their sum, call it $f$, is a smooth function and is at the disposal of my detection agency. So there is some sequence $(t\_n) \to 0$ such that $f \circ c(t\_n) = 1$, but $f(0,0) = 0$ so this violates my condition.
4. I can show that if $c$ approaches the $x$-axis with any sort of speed then I can detect its point of impact (and all derivatives). To do this, I use the function $g \colon (x,y) \mapsto x y$. So long as *some* derivative of the (extended) $y$-value of $c$ is non-zero at $0$, I can use this to find out the $x$-value by differentiating $g \circ c$ that many times.
5. If $c$ approaches the $x$-axis infinitely slowly, has a point of impact, and the $x$-value extends to a smooth function $[0,1) \to \mathbb{R}$ (so then $c$ extends to a smooth function $[0,1) \to \mathbb{R}^2$) then I cannot detect the actual point of impact. This is because I can use the chain rule to find the value of any derivative of $f \circ c$ and each term vanishes because either it involves a derivative of the $y$-value (assumed zero) or it involves a pure $x$-derivative of $f$ (zero by assumption on $f$ as we're then on the $x$-axis).


So it seems to me that it's a reasonable conjecture that I can't show that $c$ has a point of impact, if it approaches infinitely slowly. However, the above is not a proof of that fact.




---


**Motivation** I'm trying to finish off the details of an example of a Froelicher space (<http://ncatlab.org/nlab/show/Froelicher+space>). The space in question, let's call it $X$, is the quotient of the plane by the $x$-axis. By the rules for taking quotients, the smooth functions on this space are simply the smooth functions on $\mathbb{R}^2$ which are constant along the $x$-axis. I want to work out the smooth curves. Let $c \colon \mathbb{R} \to X$ be a smooth curve. It's straightforward to show that $c$ lifts to a smooth curve $U\_c \to \mathbb{R}^2$, where $U\_c$ is the complement of the preimage of the collapsed point. As $U\_c$ is (easily shown to be) open, it decomposes as a union of intervals. On each interval, $c$ is a smooth curve in $\mathbb{R}^2$ which approaches the $x$-axis at the end-points. So the question is as to what can be said as $c$ gets near one of these end-points. That's the source of this question.




---


**Edit** Added in response to Bjorn's answer. The underlying question is:



> 
> What are the smooth functions $c \colon \mathbb{R} \to \mathbb{R}^2/\mathbb{R}$?
> 
> 
> 


(Blah, blah, [Froelicher space structure](http://ncatlab.org/nlab/show/Froelicher+space), blah, blah)


Thus the point of the question is not "I have a curve, what is it?" but "Which curves can I get?". However, I figured that the question "What are the smooth curves in $\mathbb{R}^2/\mathbb{R}$?" wouldn't get much interest, but something about extending smooth curves in $\mathbb{R}^2$ might!


**Also** If, as I suspect, I can get curves with no definite "point of impact", can I limit how bad these curves must be in some way? Can I put some bound on their ($x$-)derivatives, or at least limit how fast they go to infinity?



Can I detect the point of impact without looking at it?
Andrew's comments showed me that in my first answer I was misunderstanding several aspects of his question. Since I am still not entirely sure that I am capturing the spirit of the problem, let be begin this answer by stating in my own words (in very dry mathematical terms) what I interpret the question(s) to be, so that he or someone else can correct me if necessary.




---


Let $\mathcal{F}$ be the set of smooth functions $f \colon \mathbf{R}^2 \to \mathbf{R}$ whose restriction to the $x$-axis is constant. Let $\mathcal{C}$ be the set of smooth functions $c \colon (0,1) \to \mathbf{R}$ such that for every $f \in \mathcal{F}$,


* the function $f \circ c \colon (0,1) \to \mathbf{R}$ extends to a smooth function $[0,1) \to \mathbf{R}$ (in the sense that all one-sided derivatives exist at the origin and equal the one-sided limits of the corresponding derivatives), and
* $\lim\_{t \to 0^+} f(c(t)) = f(0,0)$.


**Question 1:** If $c \in \mathcal{C}$, must $\lim\_{t \to 0^+} c(t)$ exist? (Actually, Andrew is also asking more generally what one can say about $c$ if the limit does not exist.)


**Question 2:** Is there a rule (function) $\mathcal{C} \to \mathcal{F}^r$ for some positive integer $r$, say taking $c$ to $(f\_{1,c},\ldots,f\_{r,c})$, such that $f\_{1,c}$ is independent of $c$, and $f\_{2,c}$ depends only on $f\_{1,c} \circ c$, and $f\_{3,c}$ depends only on $f\_{1,c} \circ c$ and $f\_{2,c} \circ c$, and so on, together with a rule (function) $R$ that takes as input a sequence of smooth functions $(g\_1,\ldots,g\_r)$ from $(0,1)$ to $\mathbf{R}$ and outputs a point in $\mathbf{R}^2$ such that for every $c \in \mathcal{C}$, we have $R(f\_{1,c}\circ c,\ldots,f\_{r,c} \circ c) = \lim\_{t \to 0^+} c(t)$ whenever the latter exists?


**Question 3:** Same as Question 2, but with $\lim\_{t \to 0^+} c'(t)$ in place of $\lim\_{t \to 0^+} c(t)$.




---


**Answer to Question 1:** No. A negative answer was essentially given already by Andrew himself. Namely, let $c(t) := (\sin 1/t,e^{-1/t})$. This $c(t)$ has the following properties: the $x$-coordinate is bounded, the derivatives of the $x$-coordinate grow at most polynomially in $1/t$ as $t \to 0^+$, and the $y$-coordinate and derivatives of the $y$-coordinate decay to $0$ exponentially as $t \to 0^+$. As explained by Andrew, for any $f \in \mathcal{F}$, the chain rule shows that $f(c(t))$ extends to a smooth function $[0,1) \to \mathbf{R}$ whose value at $0$ is $f(0,0)$ and whose higher derivatives at $0$ are all $0$. $\square$


**Answer to Questions 2 and 3:** Yes. In fact, we can do it with $r=2$, and with both $f\_{1,c}$ and $f\_{2,c}$ independent of $c$. Namely, use $f\_1(x,y)=y$ and $f\_2(x,y)=e^x y$. From $f\_1 \circ c$ and $f\_2 \circ c$, we may recover not only the $y$-coordinate of $c$ as $f\_1 \circ c$, but also the $x$-coordinate of $c$ as $\log (f\_2(c(t))/f\_1(c(t)))$. So the whole function $c(t)$, and hence any property of $c(t)$, can be detected. $\square$


