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Let $\gamma : \mathbb{R} \rightarrow \mathbb{R}^2$ be any continuous function, with image given by $C_\gamma$.

  • We can say that $\gamma$ has an image tangent at $t \in \mathbb{R}$ if there exists $\delta \in \mathbb{R}^{>0}$ such that the image of $(t - \delta, t + \delta)$ via $\gamma$ satisfies the following:

there exists a projective unit vector $u \in \mathbb{PS}^1$ such that $\lim_{x \rightarrow \gamma(t), x \in X} \pi(\frac{x - \gamma(t)}{\|x-\gamma(t)\|}) = u$ where $\pi : \mathbb{S}^1 \rightarrow \mathbb{PS}^1$ is the standard map $\pi(x,y) := [x,y]$, and $X$ denotes the image of $(t-\delta,t+\delta)$ via $\gamma$.

This is related to (but different) from saying $\gamma$ is differentiable a $t$. For example if $\gamma(t) = (t^3,|t|^3)$ then $\gamma$ would be everywhere differentiable, but $C_\gamma = \{(x,y) \in \mathbb{R}^2 : y = |x|\}$ which means $\gamma$ wouldn't have an image tangent at $0$.


My questions are:

  1. If $\gamma$ is differentiable a $t \in \mathbb{R}$ and $\gamma'(t) \not = 0$, does $\gamma$ necessarily have an image tangent at $t$?
  2. If $\gamma$ has an image tangent at $t$, does there necessarily exist a reparameterization of $\gamma$ (i.e. a continuous increasing bijection $\phi : \mathbb{R} \rightarrow \mathbb{R}$) such that $\gamma \circ \phi$ is differentiable with non-zero derivative at $\phi^{-1}(t)$?

Edit: In light of Leo Moos's answer, I want to further ask:

  1. If $\gamma$ has an image tangent at $t$ and is injective on some non-empty open interval containing $t$, does there necessarily exist a reparameterization of $\gamma$ (i.e. a continuous increasing bijection $\phi : \mathbb{R} \rightarrow \mathbb{R}$) such that $\gamma \circ \phi$ is differentiable with non-zero derivative at $\phi^{-1}(t)$?
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  • $\begingroup$ Quick comment: let $g: \gamma[(t-\delta,t+\delta)] \setminus \{\gamma(t)\}$ send $g(x) = (x - \gamma(t)) / \|x - \gamma(t)\|$. Given arbitrary $(x_n)$ you have that $(g(x_n))$ lives in a compact set, and hence has a subsequential limit. And hence if you just reformulate to thinking about $g$ taking values in the projective circle, your hypothesis is that every sequence $x_n$ tending to $\gamma(t)$ has $g(x_n)$ converging in $\mathbb{P}\mathbb{S}^1$. This is probably slightly easier to think about. $\endgroup$ Oct 3, 2022 at 17:39
  • $\begingroup$ @WillieWong You are right, that's less clunky than how I worded it. So despite being an equivalent definition, I've edited my question to use that notation. $\endgroup$ Oct 3, 2022 at 17:56
  • $\begingroup$ @PiotrHajlasz I don't follow what you mean? $\gamma(t) + u$ isn't necessarily in the set $X$, so that might not be a valid sequence? $\endgroup$ Oct 3, 2022 at 18:04
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    $\begingroup$ After you ask a question, if an answer is already posted, you should not modify it. It is the lack of respect to those who spend their time answering your questions. Instead, you should spend more time before posting your question so you would not have to modify it. $\endgroup$ Oct 3, 2022 at 23:33
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    $\begingroup$ @PiotrHajlasz Sorry but I don't see how it is disrespectful to ask a follow up question in a way which isn't spamming Overflow with multiple similar questions. I'm very happy and grateful for the time and answers I've gotten for questions one and two. Those questions are still there, only modified due to unclear definitions as you pointed out. $\endgroup$ Oct 3, 2022 at 23:41

2 Answers 2

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The answer to question $1$ is yes: we can suppose $t=0,\gamma(0)=(0,0)$ and $\gamma'(0)=(1,0)$ for the purposes of this question. Then as $\frac{||\gamma(x)||}{|x|}\to 1$ when $x\to0$, there is some $\delta>0$ such that $\forall x\in(-\delta,\delta)\setminus\{0\}$ we have $\frac{||\gamma(x)||}{|x|}>\frac{1}{2}$: this value of $\delta$ will satisfy your definition of image tangent.

Indeed, for any sequence $x_n$ in $(-\delta,\delta)$ such that $\gamma(x_n)\to 0$ we have $x_n\to 0$, because $|x_n|<2||\gamma(x_n)||\forall n$. Thus by the definition of derivative, $\frac{\gamma(x_n)}{x_n}\to\gamma'(0)=(1,0)$. This implies that $\lim_n\pi\left(\frac{\gamma(x_n)}{||\gamma(x_n)||}\right)=\lim_n\pi\left(\frac{\frac{\gamma(x_n)}{x_n}}{||\frac{\gamma(x_n)}{x_n}||}\right)=\pi((1,0))$, as we wanted.

The answer to question $3$ is no. An easy counterexample would be the curve $\gamma(t)=(t^3,|t|)$ but I don't think that's in the spirit of the question so in the answer I explain another counterexample that doesn't rely on "changing directions".

Consider the sequences of points $x_n=(-\frac{1}{2^n},\frac{1}{4^n})$ and $y_n=(-\frac{2}{2^n},\frac{1}{4^n})$.

Now let $\gamma$ be a curve with $\gamma(\frac{-1}{2n})=x_n$ and $\gamma(\frac{-1}{2n+1})=y_n$ (you can interpolate linearly) and then $\gamma(0)=0$ and $\gamma(x)=-\gamma(-x)$ for positive $x$. The following picture represents $\gamma(t)$ as $t$ approaches $0$ from below.

enter image description here

Then $\gamma$ satisfies the conditions from question $3$. Now let $\phi:\mathbb{R}\to\mathbb{R}$ be any increasing homeomorphism with $\phi(0)=0$. I claim that $\gamma\circ\phi$ is not differentiable at $0$. Indeed, consider the increasing sequence $a_n:=\phi^{-1}\left(\frac{-1}{n}\right)$, which converges to $0$. Letting $||\cdot||$ be the euclidean vector norm, for each $n\geq2$ we have $\frac{||\gamma\circ\phi(a_{2n})||}{|a_{2n}|} =\frac{||x_n||}{|a_{2n}|} <\frac{\frac{2}{3}||y_n||}{|a_{2n}|} <\frac{\frac{2}{3}||y_n||}{|a_{2n+1}|} =\frac{2}{3}\frac{||\gamma\circ\phi(a_{2n+1})||}{|a_{2n+1}|}$, where $||\cdot||$ is vector norm. So the sequence $\frac{||\gamma\circ\phi\left(a_n\right)||}{|a_n|}$ cannot have a non zero limit, implying that $\gamma\circ\phi$ cannot have a non zero derivative at $0$.

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    $\begingroup$ Thank you for this answer. It seems like it does the trick for the final question, but I will check it more carefully in the morning. $\endgroup$ Oct 3, 2022 at 23:50
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The answer to the first question is yes, for the second it's a rather emphatic no— the curve could be very 'badly behaved'.

  1. To simplify notation, we assume that $\gamma: (-\delta,\delta) \to \mathbf{R}^2$ is differentiable at $t = 0$, with $\gamma(0) = 0$. In addition let us also reparametrise the curve so as to have $\gamma'(0)$ be a unit vector. As $\gamma'(0) \neq 0$, we may assume that $\gamma(t) \neq 0$ if $t \neq 0$, up to shortening the defining interval. For all $t > 0$, \begin{equation} \frac{\gamma(t)}{\lvert \gamma(t) \rvert} = \frac{\gamma(t)}{t} \frac{\lvert t \rvert}{\lvert \gamma(t) \rvert}, \end{equation} with limit as $t \to 0$ equal to $\gamma'(0)$. Similarly, for negative times the limit as $t \to 0$ is $-\gamma'(0)$: the same up to orientation. To really confirm that $\gamma'(0)$ is the image tangent of $\gamma$ at $0$, we argue by contradiction. If this were not so, then by your definition there would be a sequence $\delta_i \to 0$, and for each of these, a corresponding sequence of 'bad' points $(\gamma(t_{n,i}) \mid n \geq 1)$ with \begin{equation} \gamma(t_{n,i}) \to 0 \text{ and } t_{n,i} \in (-\delta_i,\delta_i) \text{ for all $n \geq 1$} \end{equation} but \begin{equation} \frac{\gamma(t_{n,i})}{\lvert \gamma(t_{n,i}) \rvert} \not \to \gamma'(0). \end{equation} Now, as $\gamma$ is differentiable at the origin, \begin{equation} \gamma(t) = (0,0) + t\gamma'(0) + o(\lvert t \rvert) = t \gamma'(0) + o(\lvert t \rvert). \end{equation} In explicit terms, given $\epsilon \in (0,1)$ there is $\delta > 0$ so that for all $t \in (-\delta,\delta)$ \begin{equation} \lvert \gamma(t) - t \gamma'(0) \rvert \leq \epsilon \lvert t \rvert. \end{equation} In particular, there is $I \geq 1$ so that for all $i \geq I$ and all $n \geq 1$, every point in every bad sequence has \begin{equation} \lvert \gamma(t_{n,i}) - t_{n,i} \gamma'(0) \rvert \leq \epsilon \lvert t_{n,i} \rvert. \end{equation} Next note that $\lvert \gamma(t_{n,i}) \rvert = \lvert \gamma(t_{n,i}) - t_{n,i} \gamma'(0) \rvert + \lvert t_{n,i} \rvert$. Therefore \begin{equation} \lvert \lvert \gamma(t_{n,i}) \rvert - t_{n,i} \rvert \leq \epsilon t_{n,i} \text{ for all $i \geq I$ and $n \geq 1$.} \end{equation} So, for a fixed $i \geq I$, the sequence $\gamma(t_{n,i})$ can only go to zero if $t_{n,i}$ does so too. To be clear, suppose that $\gamma(t_{n,i}) \to 0$ as $n \to \infty$, but $\limsup_{n \to \infty} \lvert t_{n,i} \rvert = \tau_i > 0$. Then \begin{equation} \limsup_{n \to \infty} \lvert \lvert \gamma(t_{n,i}) \rvert - t_{n,i} \rvert \leq \limsup_{n \to \infty} \epsilon \lvert t_{n,i} \rvert, \end{equation} that is we would reach the absurd conclusion $\tau_i \leq \epsilon \tau_i$. Therefore, from now on we may assume given a fixed, small $\delta := \delta_i > 0$ and the sequence of 'bad points' $(\gamma(t_{n}) := \gamma(t_{n,i}) \mid n \geq 1)$. We have just proved that $t_{n} := t_{n,i} \to 0$ as $n \to \infty$, which means that necessarily \begin{equation} \frac{\gamma(t_n)}{\lvert \gamma(t_n) \rvert} \to \gamma'(0) \text{ as $n \to \infty$} \end{equation} by the calculations at the incipit: this contradicts the assumed 'bad' nature of the sequence.

  2. Define the continuous curve $\gamma: t \in (-1,1) \mapsto (t \operatorname{sin}\frac{1}{t},0)$, where we set $\gamma(0) = (0,0)$. Note that for all $\delta > 0$, $\gamma(t)$ crosses the origin infinitely often as $t$ varies through $(-\delta,\delta)$. (And a homeomorphism $\phi: \mathbf{R} \to \mathbf{R}$ maps intervals to intervals.) If $\gamma \circ \phi = (x(\gamma \circ \phi),y(\gamma \circ \phi))$ were differentiable at $\phi^{-1}(0)$ with $\gamma'(0) \neq 0$, then the function $t \mapsto x(\gamma \circ \phi)(t)$ would be strictly monotone in a neighbourhood of $\phi^{-1}(0)$. As it changes sign infinitely often in every neighbourhood, this is absurd.

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  • $\begingroup$ Thank you for the answer, however I feel there are some gaps in your proof unless I am mistaken. For example in your first answer just because $x_n \rightarrow 0$ how do you know that $t_n \rightarrow 0$? See this crude drawing I made i.imgur.com/MNlLaWa.jpg where the red circles are the $x_n$. $\endgroup$ Oct 3, 2022 at 20:07
  • $\begingroup$ I have also added a third question in response to your second answer, as your counterexample makes me realize I made an oversight of what I was trying to capture. Thank you. $\endgroup$ Oct 3, 2022 at 20:14
  • $\begingroup$ @SamForster I've clarified the argument. My impression is that the new question you've added is trickier. $\endgroup$
    – Leo Moos
    Oct 3, 2022 at 21:46
  • $\begingroup$ Again I think there is a flaw in your first argument, I don't think your $t_n$ sequence necessarily exists. Take a look at this second (beautiful) crude drawing i.imgur.com/nWY7ywv.jpg If these coloured regions are what the image of the curve looks like (due to it being some pathological space-filling curve), then $\lim_{x \rightarrow \gamma(t), x \in X} \pi(\frac{x-\gamma)}{\|x-\gamma(t)\|}) \not = \hat{i}$ no matter how small $\delta$ is, but it still seems possible that $\gamma'(0) = \hat{i}$. $\endgroup$ Oct 3, 2022 at 22:36
  • $\begingroup$ In your definition, you say that '$\gamma$ has an image tangent at $t \in \mathbf{R}$ if there exists $\delta > 0$ [...]'. By definition, if $\gamma$ were not to have an image tangent, then there would exist a sequence $\delta_n \to 0$ and a corresponding sequence of 'bad' points that are images of $t_n \in (t - \delta_n,t+\delta_n)$. You can ask the question without this restriction, but it's not the question you asked. $\endgroup$
    – Leo Moos
    Oct 3, 2022 at 22:50

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