A noncompact embedded curve with no uniform tubular radius

The first coordinate of $\gamma$ is $t$, so $\gamma$ is injective with a continuous inverse on its image; also ${\gamma }^{\mathrm{\prime }}(t)=(1,f^{\mathrm{\prime }}(t))\mathrm{\ne }0$. Thus it is a smooth embedding, and its image is noncompact because its first-coordinate projection is all of $\mathbb{R}$.

For $k\ge 0$ put $t_k=\sqrt{\pi \mathrm{/}2+2\pi k}$. Direct differentiation gives $f^{\mathrm{\prime }}(t_k)=0$ and $f^{\mathrm{\prime }\mathrm{\prime }}(t_k)=-4t_k^2$, so $\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}(t_k)\mathrm{\mid }\to \mathrm{\infty }$.

In the normal-bundle coordinates $(t,v)$, normal addition is $E(t,v)=\gamma (t)+v\nu (t)$. At $t_k$, $\nu (t_k)=(0,1)$ and ${\nu }^{\mathrm{\prime }}(t_k)=(-f^{\mathrm{\prime }\mathrm{\prime }}(t_k),0)$, so ${\mathrm{\partial }}_{t}E(t_k,v)=(1-v{f}^{\mathrm{\prime }\mathrm{\prime }}(t_k),0)$ and ${\mathrm{\partial }}_{v}E(t_k,v)=(0,1)$. At $v_k=1\mathrm{/}{f}^{\mathrm{\prime }\mathrm{\prime }}(t_k)$ the differential of $E$ is singular.

Given $r>0$, choose $k$ with $\mathrm{\mid }{v}_k\mathrm{\mid }=1\mathrm{/}(4t_k^2)<r$. Then $(t_k,v_k)$ lies inside the constant-radius normal tube, but its normal addition map is not a local diffeomorphism there. Hence that tube cannot be a tubular diffeomorphism for any fixed $r>0$.