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Suppose $f : \mathbb{R} \rightarrow \mathbb{R}^n$ is a real analytic function on $(a, \infty)$. I have two questions:

  1. Suppose $||f(x)|| \rightarrow \infty$ as $x \rightarrow \infty$. I know without further conditions $f$ can be not injective (for e.g. take $f(x) = x + 2 \sin(x)$ for $n=1$). I want to know if any generic conditions are known such that $f$ is eventually injective (meaning that there exists $b > a$ s.t $f$ is injective on $(b, \infty)$.

  2. Suppose $f(x)$ converges as $x \rightarrow \infty$, and $(\partial^m f / \partial x^m)(x) \rightarrow 0$ as $x \rightarrow \infty$, for all $m \geq 1$. Again without further conditions $f$ is not eventually injective (for e.g. $f(x) = e^{-x} \sin (x)$ shows it is not true generically). So again my question is whether there are some simple conditions known that makes $f$ eventually injective.

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    $\begingroup$ In 1, with "cannot be injective" you presumably mean "can be not injective". Your counterexample also doesn't work since $|x\sin(x)|$ doesn't tend to infinity. However, $f(x)=x+2\sin x$ will work for that. $\endgroup$
    – Wojowu
    Jul 1 '20 at 16:34
  • $\begingroup$ Thanks @Wojowu. Updated the question. $\endgroup$ Jul 1 '20 at 17:13
  • $\begingroup$ If $n>1$, injectivity would be generic. For n=1, monotonicity seems to be what you are looking for. $\endgroup$ Jul 1 '20 at 17:37
  • $\begingroup$ @MichaelRenardy I myself would expect injectivity to be generic for $n>2$, but not for $n=2$, by analogy with random paths. I don't know of a way to formalize this statement though, especially if we condition that the functions tend to infinity. Do you have any reason (be it informal) to expect injectivity for $n=2$? $\endgroup$
    – Wojowu
    Jul 1 '20 at 17:47
  • $\begingroup$ It can't be generically injective for $n=2$, e.g. we can force $f$ to be non-injective by requiring $|f(x) - g(x)| < 1/3$ for $-\pi \le x \le \pi$, where $g(x) = [\sin(x),\sin(2x)]$. $\endgroup$ Jul 1 '20 at 19:38
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For $n=1$, such simple (non-tautological) conditions do not exist, because real-analytic functions can mimic any $C^1$ functions in terms of their monotonicity patterns and the limit value at $\infty-$, simultaneously. So, the real-analyticity condition does not help at all; it is far from any kind of a rigidity condition.

Indeed, for any $a\in\mathbb R$, take any $C^1$ function $g\colon[a,\infty)\to\mathbb R$ such that $\exists\ g(\infty-)\in[-\infty,\infty]$ and for some increasing to $\infty$ sequence $(x_j)$ in $[a,\infty)$ we have $(-1)^j g'(x_j)>0$ for all $j$. Extend $g$ to a $C^1$ function on $\mathbb R$ and denote the extension still by $g$.

Take now any continuous function $h\colon\mathbb R\to(0,\infty)$ such that $h(x_j)<|g'(x_j)|$ for all $j$ and $h(\infty-)=0$. Then, according to a theorem by Whitney (see e.g. the first paragraph of this paper), there is a real-analytic function $f\colon\mathbb R\to\mathbb R$ such that $|f-g|+|f'-g'|\le h$ and hence $f(\infty-)=g(\infty-)$ and $(-1)^j f'(x_j)>0$ for all $j$ -- so that the real-analytic function $f$ indeed mimics the monotonicity pattern of the $C^1$ function $g$ and has the same limit value at $\infty-$.

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