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

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, 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?

share|improve this question
    
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? –  Loop Space Mar 27 '10 at 2:33

2 Answers 2

up vote 4 down vote accepted

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. $\square$

share|improve this answer
    
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? –  Loop Space 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! –  Loop Space 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.

share|improve this answer
1  
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
2  
@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 ...) –  Loop Space 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.) –  Loop Space Mar 27 '10 at 2:53

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.