Can we approximate a vector field on the plane with non-vanishing vector fields in $L^2$? Let $V$ be a compactly-supported smooth 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 $V_n \in C^\infty \cap L^{2}$ on $\mathbb{R}^2$, such that $V_n \to V$ in $L^2$ and $V_n$ do not vanish on  $\mathbb{D}^2$?

 A: I'll assume the zeroes of $V$ are isolated on an open neighborhood of the unit disk.
Here is a procedure for approximating $V$ in $L^2$ by a vector field in the same class - $C^\infty_c$ with isolated zeros in a neighborhood of the unit disk - but with one fewer zero in $\mathbb D^2.$ The idea is to push the zero out. (This is basically what I meant in the linked answer by "composing with a suitable diffeo".)
Pick points $(x_0,y_0)$ and $(x_1,y_1)$ such that:


*

*$(x_0,y_0)\in\mathbb D^2.$

*$(x_1,y_1)\not\in\mathbb D^2.$

*$V(x_0,y_0)=(0,0).$

*The straight line segment from $(x_0,y_0)$ to $(x_1,y_1)$ contains no zeroes of $V$ except $(x_0,y_0).$
Let $C$ be the straight line segment from $(x_0,y_0)$ to $(x_1,y_1).$
Let $C_n$ be a sequence of open neighborhoods of $C$ such that
$$\mu(C_n)\to 0$$
where $\mu$ is Lebesgue measure, and such that $C_n$ contains no zeros of $V$ except $(x_0,y_0).$ Using Whitney's extension theorem pick a function $s:\mathbb R^2\to\mathbb R$ such that:


*

*$s(x,y)=1$ for $(x,y)\in C$

*$s(x,y)=0$ for $(x,y)\not\in C_n$
Since $s$ is compactly supported, $(x,y)\mapsto s(x,y)(x_1-x_0,y_1-y_0)$ is a complete vector field and defines a flow globally: there are diffeomorphisms $\psi_t$ for $t\in\mathbb R$ where $\psi_0$ is the identity and $\frac{d}{dt}\psi_t(x,y)=s(x,y)(x_1-x_0,y_1-y_0).$ Define $$V_n=V \circ \psi_{-1}.$$
Since 
$$\|V-V_n\|_2^2\leq \mu(C_n)(2\max|V|)^2$$
we have $V_n\to V$ in $L^2.$
For $(x,y)\in C_n$ we have $V_n(x,y)=(0,0)$ if and only if $\psi_{-1}(x,y)=(x_0,y_0),$ which is equivalent to $(x,y)=(x_1,y_1).$ And outside $C_n,$ the vector fields $V_n$ and $V$ are the same. So the number of zeros inside the unit disk has decreased by one.
A: 
EDIT. Step 2 is wrong. I asked a separate question on this matter.

I submit that the answer is affirmative, with the $L^2$ convergence. The idea comes from a property of the heat equation, according to which, if a function has a local minimum, then letting it evolve according to the heat equation will fill in such minimum. We are going to apply this idea to the modulus $\lvert \vec{V}\rvert$ of the given vector field, whose minima are precisely the zeros of $\vec V$.

Step 1. This solves the approximation problem, but produces a non-smooth vector field at the zeros of $\vec V$. I attempted to solve the smoothness issue in the forthcoming Step 2, which however contains an error.
Assume that $\vec{V}$ is continuous and $\vec{V}\in L^2(\mathbb R^d; \mathbb R^d)$, and that 
$$Z:=\{x\in \mathbb R^d\ :\ \vec V(x)=0\}$$ 
is a set of measure zero; this is actually a weaker assumption than requested. Write 
$$
\vec{V}(x)=R(x)\omega(x), \quad \text{where }R(x):=\lvert \vec V(x)\rvert,\text{and } \omega(x)\in \mathbb S^{d-1}.$$
We remark that the definition of $\omega(x)$ is ambiguous on $Z$. 
Now let $R(t, x)$ be the unique solution to 
$$
\begin{cases} 
\partial_t R =\Delta R, & t>0, x\in\mathbb R^d,\\ 
R(0, x)=R(x).
\end{cases}$$
Since $R(x)\ge 0$, by the minimum principle $R(t, x)>0$ for all $t>0$.$^{[1]}$ Moreover, $R(t, x)\to R(x)$ both pointwise and in $L^2$ sense. By all these considerations, the time-dependent vector field 
$$
\vec V(t, x):=R(t,x)\omega(x)$$ 
vanishes nowhere and converges to $\vec V$ as $t\downarrow 0$, pointwise and in $L^2$ sense.

Step 2. (wrong) The function $\vec{V}(t, x)$ of the previous step needs not be continuous for $t>0$ and $x\in Z$, where it will point in one of the "ambiguous" directions of $\omega(x)$. 
To circumvent this difficulty, we introduce $\omega(t, x)$, defined as follows. Consider $\omega(x)\in\mathbb S^{d-1}$ as a vector in $\mathbb R^d$, and let 
$$\eta\colon [0, \infty)\times \mathbb R^d\to \mathbb R^d$$ 
be the unique solution to the vector-valued heat equation 
$$
\begin{cases} 
\partial_t \eta = \Delta \eta, &t>0, \\ 
\eta(0, x)=\omega(x).
\end{cases}$$
This makes sense, because $\omega\in L^\infty(\mathbb R^d; \mathbb R^d)$. 

Now, since $\omega(x)\ne 0$ at all $x\in\mathbb R^d$, and since $\eta$ is continuous in $t$, 
  there is a $\delta >0$ such that $\eta(t,x)\ne 0$ for all $t\in [0, \delta]$ and $x\in \mathbb R^d$. 
  (Warning: this needs not be true; for example, consider $\omega(x)=x/\lvert x \rvert$. See the follow-up question.) 

It makes thus sense to define 
$$
\omega(t, x):=\frac{\eta(t, x)}{\lvert \eta(t, x)\rvert}, \qquad t>0.$$
Now, by a standard property of the heat equation known as instantaneous smoothing effect, the function $\eta(t, \cdot)$ is smooth for all $t>0$. Therefore, $\omega(t, \cdot)$ is also smooth. 
Conclusion. By the same argument as in Step 1, the function 
$$\vec V(t, x):= R(t, x) \omega(t, x),\qquad t\in [0, \delta], $$
is such that $\vec V(t, x)\ne 0$ for all $t>0$ and $x\in \mathbb R^d$, and it converges to $\vec V(x)$, both pointwise and in the $L^2$ sense, as $t\downarrow 0$. (It is possible that this convergence can be upgraded with further regularity assumptions on $\vec V$). Also, $\vec V(t, \cdot)$ is smooth for all $t\in (0, \delta)$. 

[1]  Actually, this can also be seen as an immediate consequence of the explicit formula $$R(t, x)=(4\pi t)^{-\frac d 2}\int_{\mathbb R^d} R(y)e^{-\frac{|x-y|^2}{4t}}\, dy.$$ 
