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This is not a satisfactory answer，should be a comment, but I do not have the reputation to write comment, so I write it here.

1. $u_R$ is uniformly bounded, \begin{equation*} \left\|\vec{u}_{R}\right\|_{\dot{H}^{1}} \leq \frac{1}{\nu}\|\vec{f}\|_{\dot{H}^{-1}} \end{equation*} so a weak limit $u$ of $u_R$ (distribution sense) exist, this can be figure out by take diagonal subsequence of a sequence, further more, in general case the weak limit $u$ is a continuous linear functional defined on $D$ and can not extend to $H^{-1}$ still bounded(or continue), if it can extend as a continuous linear functional on $C^{\infty} \subset H^{1} \subset H^{-1}$, then it is conincide with a $\hat u\in H^1$ and belong to $H^{1}$.

2.by rellich theorem, we can figure out \begin{equation*} H^{1}(R^3)=W^{1, 1}(R^3) \subset \subset L^{q}(R^3) \text { for } 1 \leq q<6 \end{equation*} So by use the method of taking diagonal subsequence of a sequence, we can find a $u$, $$u_{R_k} \longrightarrow u,\ in \ L^q(R^3),\ \forall 1\leq q<6 $$

3. a weak limit of a sequence if exist then it is unique, so by take subsequence of process of 1,2. we can find a subsequence $u_{R_k}$, $u$ is the weak limit of $u_{R_k}$, so $u$ satisfied the equation, \begin{equation*} \nu \Delta \vec{u}+\vec{f}-\operatorname{div}(\vec{u} \otimes \vec{u})=\vec{\nabla} p \end{equation*} in distribution sense, and at the same time $u\in L^{q}(R^3), \forall 1\leq q<6 $

4.Now, the problem is can we go further to get $u\in H^{1}$, in general this is not true for bound sequence $\left\|\vec{u}_{R}\right\|_{\dot{H}^{1}} \leq M$, but when $u_R$ is come from some approximation of a differential equation, this is in general true, if not blow up. I do not know how to firgue out this spetial one, but it seem the general method is use the regularity we already get, i.e. $f\in L^q(R^3), \forall 1\leq q<6$ into the distribution equation, i.e. \begin{equation*} \nu \Delta \vec{u}+\vec{f}-\operatorname{div}(\vec{u} \otimes \vec{u})=\vec{\nabla} p \end{equation*} if change the equation to $\Delta u+u^2=f, f\in L^2(R^n)$, then by the previous argument we can show a suitable limit $u=lim_{k}u_k$ satisfied $u\in L^{q}(R^n), \forall 1\leq q<6$, and then as distribution $\Delta u=f-u^2 \in L^{2}(R^n)$, and then $u\in H^{2,2}(R^3)$, there should be a similar method to gain the information that $u= lim u_{R_k}$ is in $H^1$.

This is not a satisfactory answer，should be a comment, but I do not have the reputation to write comment, so I write it here.

1. $u_R$ is uniformly bounded, \begin{equation*} \left\|\vec{u}_{R}\right\|_{\dot{H}^{1}} \leq \frac{1}{\nu}\|\vec{f}\|_{\dot{H}^{-1}} \end{equation*} so a weak limit $u$ of $u_R$ (distribution sense) exist, this can be figure out by take diagonal subsequence of a sequence, further more, in general case the weak limit $u$ is a continuous linear functional defined on $D$ and can not extend to $H^{-1}$ still bounded(or continue), if it can extend as a continuous linear functional on $C^{\infty} \subset H^{1} \subset H^{-1}$, then it is conincide with a $\hat u\in H^1$ and belong to $H^{1}$.

2.by rellich theorem, we can figure out \begin{equation*} H^{1}(R^3)=W^{1, 1}(R^3) \subset \subset L^{q}(R^3) \text { for } 1 \leq q<6 \end{equation*} So by use the method of taking diagonal subsequence of a sequence, we can find a $u$, $$u_{R_k} \longrightarrow u,\ in \ L^q(R^3),\ \forall 1\leq q<6 $$

3. a weak limit of a sequence if exist then it is unique, so by take subsequence of process of 1,2. we can find a subsequence $u_{R_k}$, $u$ is the weak limit of $u_{R_k}$, so $u$ satisfied the equation, \begin{equation*} \nu \Delta \vec{u}+\vec{f}-\operatorname{div}(\vec{u} \otimes \vec{u})=\vec{\nabla} p \end{equation*} in distribution sense, and at the same time $u\in L^{q}(R^3), \forall 1\leq q<6 $

4.Now, the problem is can we go further to get $u\in H^{1}$, in general this is not true for bound sequence $\left\|\vec{u}_{R}\right\|_{\dot{H}^{1}} \leq M$, but when $u_R$ is come from some approximation of a differential equation, this is in general true, if not blow up. I do not know how to firgue out this special one, but it seem the general method is use the regularity we already get, i.e. $u\in L^q(R^3), \forall 1\leq q<6$ into the distribution equation, i.e. \begin{equation*} \nu \Delta \vec{u}+\vec{f}-\operatorname{div}(\vec{u} \otimes \vec{u})=\vec{\nabla} p \end{equation*} if change the equation to $\Delta u+u^2=f, f\in L^2(R^n)$, then by the previous argument we can show a suitable limit $u=lim_{k}u_k$ satisfied $u\in L^{q}(R^n), \forall 1\leq q<6$, and then as distribution $\Delta u=f-u^2 \in L^{2}(R^n)$, and then $u\in H^{2,2}(R^3)$, there should be a similar method to gain the information that $u= lim u_{R_k}$ is in $H^1$.

Because $R>0, u_{R}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$, $\left\|u_{R}\right\|_{\dot{H}^{1}}$ is bounded uniformly in $R$. Thus, for every $\phi \in \mathcal{D}=\left\{\right.$ smooth, compactly supported functions in $\left.\mathbb{R}^{3}\right\},$ we have \begin{equation*} \sup _{R}\left\|\phi u_{R}\right\|_{H^{1}}<\infty \end{equation*}

so $\forall R>0$, look $u_R$ as a linear operater on $\dot{H}^{1}$($u_R$ is linear when paring on test function space $\mathcal{D}$, so $u_R$ can extend to a linear operator in $\dot{H}^{1}$ by approximation), and because the dual of $\dot{H}^{1}$ is $H^{-1}$, so $u_{R}, R>0$ as a group of operater in $H^{-1}$ whcih is uniformly bounded, by Uniform boundedness principle. So to prove there is a suhbsequence of $u_{R}, R>0$ converage in $H^{-1}$, it is suffice to find a subsequence that is well-defined(as a unbounded linear operatoer in $H^{-1}$), so coverage in $L_{\text {loc}}^{p}$ for $p<6$ is suffice(I am not very sure what is your $\dot{H}^{1}$ here, if it is ${H}^{1}$, then $L_{\text {loc}}^{2}$ is suffice, the reason is if two function $f,g$ is the same as a $L_{\text {loc}}^{2}$, then they are the same measurable function, i.e. $f-g=0$ as a measurable function, so if we priori have higher regularity of $f,g$, then $f-g=0$ in the higher regularity space. ), and then it is automatally in $\dot{H}^{1}$, beacuse it is a limit of a sequence of uniformly bounded operator.

Reorganize my answer, something is wrong with the previous one,

First, an obersevation is, if there is a strong converage limit of $u_R$ in $H^1$

This is not a satisfactory answer，should be a comment, but I do not have the reputation to write comment, so I write it here.

1. $u_R$ is uniformly bounded, \begin{equation*} \left\|\vec{u}_{R}\right\|_{\dot{H}^{1}} \leq \frac{1}{\nu}\|\vec{f}\|_{\dot{H}^{-1}} \end{equation*} so a weak limit $u$ of $u_R$ (distribution sense) exist, this can be figure out by take diagonal subsequence of a sequence, further more, in general case the weak limit $u$ is a continuous linear functional defined on $D$ and can not extend to $H^{-1}$ still bounded(or continue), if it can extend as a continuous linear functional on $C^{\infty} \subset H^{1} \subset H^{-1}$, then it is conincide with a $\hat u\in H^1$ and belong to $H^{1}$.

2.by rellich theorem, we can figure out \begin{equation*} H^{1}(R^3)=W^{1, 1}(R^3) \subset \subset L^{q}(R^3) \text { for } 1 \leq q<6 \end{equation*} So by use the method of taking diagonal subsequence of a sequence, we can find a $u$, $$u_{R_k} \longrightarrow u,\ in \ L^q(R^3),\ \forall 1\leq q<6 $$

3. a weak limit of a sequence if exist then it is unique, so by take subsequence of process of 1,2. we can find a subsequence $u_{R_k}$, $u$ is the weak limit of $u_{R_k}$, so $u$ satisfied the equation, \begin{equation*} \nu \Delta \vec{u}+\vec{f}-\operatorname{div}(\vec{u} \otimes \vec{u})=\vec{\nabla} p \end{equation*} in distribution sense, and at the same time $u\in L^{q}(R^3), \forall 1\leq q<6 $

4.Now, the problem is can we go further to get $u\in H^{1}$, in general this is not true for bound sequence $\left\|\vec{u}_{R}\right\|_{\dot{H}^{1}} \leq M$, but when $u_R$ is come from some approximation of a differential equation, this is in general true, if not blow up. I do not know how to firgue out this spetial one, but it seem the general method is use the regularity we already get, i.e. $f\in L^q(R^3), \forall 1\leq q<6$ into the distribution equation, i.e. \begin{equation*} \nu \Delta \vec{u}+\vec{f}-\operatorname{div}(\vec{u} \otimes \vec{u})=\vec{\nabla} p \end{equation*} if change the equation to $\Delta u+u^2=f, f\in L^2(R^n)$, then by the previous argument we can show a suitable limit $u=lim_{k}u_k$ satisfied $u\in L^{q}(R^n), \forall 1\leq q<6$, and then as distribution $\Delta u=f-u^2 \in L^{2}(R^n)$, and then $u\in H^{2,2}(R^3)$, there should be a similar method to gain the information that $u= lim u_{R_k}$ is in $H^1$.

Because $R>0, u_{R}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$, $\left\|u_{R}\right\|_{\dot{H}^{1}}$ is bounded uniformly in $R$. Thus, for every $\phi \in \mathcal{D}=\left\{\right.$ smooth, compactly supported functions in $\left.\mathbb{R}^{3}\right\},$ we have \begin{equation*} \sup _{R}\left\|\phi u_{R}\right\|_{H^{1}}<\infty \end{equation*}

so $\forall R>0$, look $u_R$ as a linear operater on $\dot{H}^{1}$($u_R$ is linear when paring on test function space $\mathcal{D}$, so $u_R$ can extend to a linear operator in $\dot{H}^{1}$ by approximation), and because the dual of $\dot{H}^{1}$ is $H^{-1}$, so $u_{R}, R>0$ as a group of operater in $H^{-1}$ whcih is uniformly bounded, by Uniform boundedness principle. So to prove there is a suhbsequence of $u_{R}, R>0$ converage in $H^{-1}$, it is suffice to find a subsequence that is well-defined(as a unbounded linear operatoer in $H^{-1}$), so coverage in $L_{\text {loc}}^{p}$ for $p<6$ is suffice(I am not very sure what is your $\dot{H}^{1}$ here, if it is ${H}^{1}$, then $L_{\text {loc}}^{2}$ is suffice, the reason is if two function $f,g$ is the same as a $L_{\text {loc}}^{2}$, then they are the same measurable function, i.e. $f-g=0$ as a measurable function, so if we priori have higher regularity of $f,g$, then $f-g=0$ in the higher regularity space. ), and then it is automatally in $\dot{H}^{1}$, beacuse it is a limit of a sequence of uniformly bounded operator.

Because $R>0, u_{R}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$, $\left\|u_{R}\right\|_{\dot{H}^{1}}$ is bounded uniformly in $R$. Thus, for every $\phi \in \mathcal{D}=\left\{\right.$ smooth, compactly supported functions in $\left.\mathbb{R}^{3}\right\},$ we have \begin{equation*} \sup _{R}\left\|\phi u_{R}\right\|_{H^{1}}<\infty \end{equation*}

so $\forall R>0$, look $u_R$ as a linear operater on $\dot{H}^{1}$($u_R$ is linear when paring on test function space $\mathcal{D}$, so $u_R$ can extend to a linear operator in $\dot{H}^{1}$ by approximation), and because the dual of $\dot{H}^{1}$ is $H^{-1}$, so $u_{R}, R>0$ as a group of operater in $H^{-1}$ whcih is uniformly bounded, by Uniform boundedness principle. So to prove there is a suhbsequence of $u_{R}, R>0$ converage in $H^{-1}$, it is suffice to find a subsequence that is well-defined(as a unbounded linear operatoer in $H^{-1}$), so coverage in $L_{\text {loc}}^{p}$ for $p<6$ is suffice(I am not very sure what is your $\dot{H}^{1}$ here, if it is ${H}^{1}$, then $L_{\text {loc}}^{2}$ is suffice, the reason is if two function $f,g$ is the same as a $L_{\text {loc}}^{2}$, then they are the same measurable function, i.e. $f-g=0$ as a measurable function, so if we priori have higher regularity of $f,g$, then $f-g=0$ in the higher regularity space. ), and then it is automatally in $\dot{H}^{1}$, beacuse it is a limit of a sequence of uniformly bounded operator.

Reorganize my answer, something is wrong with the previous one,

First, an obersevation is, if there is a strong converage limit of $u_R$ in $H^1$