# Can I detect the point of impact without looking at it?

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.

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

(Blah, blah, Froelicher space structure, 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?

• Two clarification questions: (1) I assume that $t$ is the variable parameterizing $(0,1)$? (2) What is a "point of impact"? – Theo Johnson-Freyd Mar 27 '10 at 1:48
• @Theo: (1) Yes. (2) An attempt at making it sound more interesting than it possibly is! By "point of impact" I mean the point (if it exists) where the curve hits the x-axis. Does that make sense? – Andrew Stacey Mar 27 '10 at 2:33

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. $\square$

• Excellent! I hadn't thought of using two (or r) functions simultaneously. That's an extremely useful tool to have learnt. Question (4) is still outstanding: is there some intrinsic characterisation of those curves c satisfying the condition (ie without reference to f). However, although that is my implicit question, it wasn't really the question in the question so I know I'm being a bit cheeky tacking it on the end! I'll have a think about what you've written and check that I understand it all. Incidentally, why use e^x y and not just x y? – Andrew Stacey Mar 27 '10 at 12:48
• You're right; I could have just used xy. (I was worrying about a problem that doesn't exist.) – Bjorn Poonen Mar 27 '10 at 15:43
• Thinking about this a little, I realised that the counterexample to Question 1 is what I was really looking for. Questions 2 and 3 seem still in the mindset of your first question (that c is unknown, rather than which c satisfy the condition). Working through the details of the counterexample then showed me how to adapt my proof that I can't detect the point of impact into a characterisation of those curves that satisfy the condition-the basic idea is that the y-value goes to 0 faster than any of the derivatives of the x-value. I'll put the details at the nlab. Thanks for the help! – Andrew Stacey Apr 6 '10 at 10:28

The answer is that everything can be recovered, if $f$ is chosen suitably!

There are $2^{\aleph_0}$ possibilities for $c$, so we can fix an injection $c \mapsto u_c$ from the set of possible $c$ into the set of real numbers. Let $y_c$ be the $y$-coordinate of $c(1/2)$, so $y_c \ne 0$.

In the examples you give, you allow $f$ to depend on $c$, so I will take $f(x,y) := \frac{u_c}{y_c} y$.

Claim: I can recover $c$ from $f \circ c$. (So in particular, any information you want about its limiting behavior as $t \to 0$ can be recovered too.)

Proof: Evaluating $f \circ c$ at $1/2$ gives $u_c$, which tells me what $c$ is.

• If you're willing to commit such abuses, why not just let $f$ be the constant function $u_c$? – Tom Church Mar 27 '10 at 1:18
• @Tom: I had misread the sentence before the blockquote as requiring $f(0,0)=0$, but after re-reading, I think that he is not requiring this, so you are right. Your solution is even better! Anyway, I did not mean for this to be an obnoxious answer, but rather one that would encourage Andrew to make precise what he is really looking for (since I am sure it is not this!) – Bjorn Poonen Mar 27 '10 at 1:46
• I've voted for your answer because of your comment! I think that answers that expose vagueness are worthwhile. I do have a mild problem with the answer: to know what function to use, I have to know c already, which means that I know what c is! But more fundamentally, I'm trying to define the set of possible curves but to do what you suggest I need to already know it. In addition, whilst I can pick f how I like, I can only really pick it based on prior information from other f's. (ctd ...) – Andrew Stacey Mar 27 '10 at 2:52
• ... (ctd) Thus, for example, I first probe with (x,y) -> y to get the y-value, and then based on that information I pick some other f (e.g. if y has some non-zero derivative, I can pick (x,y) -> xy). Is that any clearer? (I've also added a comment to the question to hopefully make it a little clearer.) – Andrew Stacey Mar 27 '10 at 2:53