# Injectivity of analytic functions

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.

• 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. Jul 1 '20 at 16:34
• Thanks @Wojowu. Updated the question. Jul 1 '20 at 17:13
• If $n>1$, injectivity would be generic. For n=1, monotonicity seems to be what you are looking for. Jul 1 '20 at 17:37
• @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 $n=2$? Jul 1 '20 at 17:47
• 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 $g(x) = [\sin(x),\sin(2x)]$. Jul 1 '20 at 19:38

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 $$\infty-$$.