Do non-zero derivatives imply tangent lines (and vice versa)?

Question

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 $0$.

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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 $\phi^{-1}(t)$?

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

3. 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 $\phi^{-1}(t)$?

Answer 1

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.

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 $0$.

Answer 2

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.