If $v_n : \mathbb{T}^3 \to \mathbb{R}^3$ is a sequence of uniformly convergent divergence-free vector fields, does its "antiderivative" converge?

Question

Let $\mathbb{T}^3:=[\mathbb{R}/\mathbb{Z}]^3$ be the $3$-dimensional torus and $v_n : \mathbb{T}^3 \to \mathbb{R}^3$ be a sequence of smooth vector fields with the following property:

1. $\sup_{x \in \mathbb{T}^3} \lvert v_n(x) - v_m(x) \rvert \to 0^+$ as $m,n \to \infty$.

2. $\nabla \cdot v_n=0$ for all $n \in \mathbb{N}$.

3. $\int_{\mathbb{T}^3} v_n=0$ for all $n \in \mathbb{N}$.

Then, by Hodge decomposition, it is well-known that there exists another sequence of smooth vector fields $F_n :\mathbb{T}^3 \to \mathbb{R}^3$ such that $\nabla \times F_n = v_n$.

Moreover, I am aware that we can choose the "gauge" such that $\nabla \cdot F_n=0$ for each $n$ as well.

Now, under these conditions, does $F_n$ converge uniformly on $\mathbb{T}^3$ as well? Or are there any additional conditions which make these "antiderivatives" converge uniformly?

I am not sure about this kind of setting at all, so need to ask.