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Let $V$ be a compactly-supported vector field on $\mathbb{R}^2$, whose zeros inside some open neighbourhood of the closed unit disk $\mathbb{D}^2$ are isolated.

Does there exist a sequence of vector fields $V_n \in C^\infty \cap W^{1,2}$ on $\mathbb{R}^2$, such that $V_n \to V$ in $W^{1,2}$ and the $V_n$ do not vanish on $\mathbb{D}^2$?

If we replace the $W^{1,2}$ convergence with $L^2$ convergence, than the answer is positive. The idea is to push the zeroes out of the disk by composing $V$ with a diffeomorphism which affects a region of very small measure.

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    $\begingroup$ The strongest convergence you want must be strictly weaker of the uniform convergence, because if $V_n\to V$ uniformly on $\overline{\mathbb{D}^2}$, and $V$ has non-zero topological degree in $\mathbb{D}^2$ wrto $0$, the same holds eventually for the $V_n$, and they will vanish as well. But $W^{2,2}(\mathbb{D}^2,\mathbb{R}^2)$ is continuously embedded in $C^0(\mathbb{D}^2,\mathbb{R}^2)$: this convergence is too strong. $\endgroup$
    – Pietro Majer
    Commented Nov 17, 2019 at 15:07
  • $\begingroup$ Minor side comment: compact support and isolated zeros are not compatible. Perhaps you meant that it only has isolated zeros on $\mathbb{D}^2$? $\endgroup$
    – Willie Wong
    Commented Nov 18, 2019 at 14:26
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    $\begingroup$ Deleted my answer below because Pietro's comment made me doubt an essential part of my reasoning. For convenience, however, a survey of the VMO degree theory (which would apply in your case because $W^{1,2}$ embeds in $VMO$ in two dimensions) can be found here (Brezis, "New questions related to the topological degree".) $\endgroup$
    – Willie Wong
    Commented Nov 18, 2019 at 14:57
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    $\begingroup$ @PietroMajer: I completely forgot about the $C^\infty$ part. Ok, I will undelete. $\endgroup$
    – Willie Wong
    Commented Nov 18, 2019 at 16:24
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    $\begingroup$ I think this was already answered by Pietro Majer's answer here: mathoverflow.net/a/307832/112284. Set $p_n$ there to $V_n$ here. (And apply a cutoff function i.e. multiply the counterexample $p_n$ by a $C^\infty$ function that is $1$ on a neighborhood of $\mathbb D^2,$ and restrict $r_0$ to $(0,1).$) $\endgroup$
    – Dap
    Commented Nov 20, 2019 at 5:23

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Going from the topological index idea: let $C$ be a circle around one of the isolated zeros, and let $D$ be a disk containing $C$. By the trace theorem, your vector fields in $W^{1,2}(D)$ restricts to vector fields $W^{1/2,2}(C)$ on the circle. Since they do not vanish, you can regard them as vector fields $W^{1/2,2}(C, \mathbb{S}^1)$. (If there's any problem with my argument, it would be this step, passing from an $\mathbb{R}^2$ valued function to $\mathbb{S}^1$ valued one.)

The $W^{1/2,2}(C,\mathbb{S}^1)$ functions are continuously embedded in $VMO(C,\mathbb{S}^1)$ since $C$ is one dimensional, and one can use the $VMO$-degree theory (originally due to Boutet de Monvel and Gabber, and extended by Brezis and Nirenberg, see this survey by Brezis). The upshot is that for continuous functions the VMO degree coincides with the the usual topological degree, and the $VMO$-degree is continuous under $VMO$-convergence. So I think Pietro's argument against the $W^{2,2}$ case using degree theory should carry over also, telling you that you shouldn't be able to do your approximation.

To be more precise: the chain of arguments should go something like: since your $V_n$ are in $C^\infty$ and has no zeros, their corresponding topological degree would be zero, measured either in $VMO$ or classically; but $V_n$ converges to $V$ in $VMO$ (by argument above), and this means that $V$ has to also have degree zero, a contradiction.

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