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Does there exist a $C^\infty$ function $p:\mathbb{R} \to \mathbb{R}$ such that p does not vanish and $p(t)=1$ for small $t$ and $p(t)= \frac{1}{t}$ for large $t$?

Such function $p$ is used in Donaldson's book Riemann Surfaces p.100 Proposition 18. but I can't find such $p$.

Proposition 18. Let $\alpha$ be a non-trivial meromorphic 1-form on a compact Riemann surface $X$ with genus $g$. Then the number of zeros minus number of poles of $\alpha $, counting according to multiplicity, is $2g-2$

Proof. Fix an area form $\omega$ on $X$.

Let $X'$ be the set on which $\alpha$ is holomorphic.

Define a Hermitian metric on $T^*X'$ as $ \vert \alpha \vert ^2 \omega = \alpha \wedge \bar{\alpha}$.

Choose a real-valued functoin $p$ on $\mathbb{R}$ with $p(t)=1$ for small to and $p(t)= \frac{1}{t}$ for large t.

Define $\widetilde{\alpha} := \begin{cases} p(\vert \alpha \vert ^2)\cdot \alpha &\text { on $X'$} \\[4pt] 0 &\text{ on $X-X'$} \end{cases} $

Let $q$ be a pole of $\alpha$, $z$ be a coordinate fucntion centered at $q$, $\alpha = f(z) dz$ near $q$.

Near $q$, $\omega = R \cdot dz\wedge d\bar{z}$ for some strictly positive function $R$.

$\widetilde{\alpha} = \frac{1}{\vert \alpha \vert ^2} f \ dz=\frac{R}{\vert f \vert ^2} f \ dz = \frac{R}{\bar{f}} \ dz$.

Thus $\widetilde{\alpha} $ is smooth 1-form on $X$ and its zero set is the union on zeros and poles of $\alpha$.

Question. For $\alpha$ to be smooth, $p$ has to be a smooth function and for the zero set of $\widetilde{\alpha}$ to be the union on zeros and poles of $\alpha$, $p$ should not vanish.

Does there exist a $C^\infty$ function $p:\mathbb{R} \to \mathbb{R}$ such that p does not vanish and $p(t)=1$ for small $t$ and $p(t)= \frac{1}{t}$ for large $t$?

I know that $(1-g(t)) \cdot k(t)$ is a smooth function with $k(t)=1$ for small $t$ and $k(t)= \frac{1}{t}$ for large $t$ , but $k$ vanishes on $(2/3 , \ 4/3)$

$ g(t) := \frac{\int_{|t-1|}^{2/3} h(u) du }{\int_{1/3}^{2/3} h(u) du}$ is a bump function equal to 1 on $[2/3 , \ 4/3]$ supported in $[1/3, \ 5/3]$

where $h(u) := \begin{cases} exp\{-(u-1/3)^{-1} (u-2/3)^{-1}\} &\text { on $[1/3,\ 2/3]$} \\[4pt] 0 &\text{ otherwise} \end{cases} $,

$k(t):= \begin{cases} 1 &\text{ on $(-\infty,\ 1]$} \\[4pt] \frac{1}{t} &\text{ otherwise} \end{cases}$

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Take the function $f(t):=1$ $(t\leq 1)$ and $:={1\over t}$ $(t\geq1)$, and convolve it with a $C^\infty$ bump function $\phi$ having support in $\bigl[-{1\over2},{1\over2}\bigr]$: $$g(x):=\int_{-\infty}^\infty f(x-t)\>\phi(t)\>dt\ .$$ Here it is assumed that $\phi(t)\geq0$ and $\int_{-\infty}^\infty \phi(t)\>dt=1$.

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