Application of uniform boundedness (Banach-Steinhaus) principle

Question

$\DeclareMathOperator{\loc}{\mathrm{loc}}$This is from Lemarié-Rieusset's book "The Navier-Stokes problem in the 21st century", from the proof of a result about stationary solutions to Navier-Stokes (Theorem 16.2). There's a step where he uses the uniform boundedness theorem in a way that I can't comprehend. I will outline the relevant part of the proof and also add some steps that he doesn't explicitly give.

We're considering functions parameterized by $R>0$, $u_R:\mathbb{R}^3\to\mathbb{R}^3$ such that $\|u_R\|_{\dot{H^1}}$ is bounded uniformly in $R$. Thus, for every $$\phi\in \mathcal{D}=\{\text{smooth, compactly supported functions in }\mathbb{R}^3\},$$ we have $$\sup_{R}\|\phi u_R\|_{H^1}<\infty.$$ For any choice of test function $\phi$, by Rellich Kondrachov, we can extract a subsequence $R_k\to \infty$ such that $\phi u_{R_k}$ converges in $L^p$ for $p\in[1,6)$. Taking as test functions a sequence of functions $\phi_N\in \mathcal D$ that approximate the constant function $1$ in the limit $N\to\infty$, we obtain, via a diagonlization argument, a subsequence of $R_k$, which we continue to call $R_k$, such that $u_{R_k}$ converges to some function $u$ in $L^p_{\loc}$. (That is, we have a $3\epsilon$ argument: we have that $\phi_Nu_{R_k}$ converges as $R_k\to \infty$ to some function $u_N$ in $L^p_{\loc}$, then we show that $u_N$ form a Cauchy sequence in $L^p_{\loc}$. On the other hand, $u_{R_k}$ can be approximated by $\phi_N u_{R_k}$.).

Then, it is claimed that "by Banach-Steinhaus" (i.e. uniformed bounded principle), we have $u\in \dot {H^1}$.

I simply don't see in what way Banach-Steinhaus was used to reach such a conclusion. The way I see it, we started with a sequence $u_R$ which we know a priori to be bounded in $\dot H^1$, uniformly in $R$. From this, a subsequence $u_{R_k}$ has been extracted such that $u_{R_k}\to u$. The latter convergence, as far as I can see is only in $L^p_{\loc}$ for $p<6$. Of course, if this convergence were to hold in $\dot{H^1}$, then by definition, $u\in\dot{H^1}$, and no appeal to Banach-Steinhaus is needed. So I'm assuming that indeed we only have convergence in the aforementioned local $L^p$ spaces. But by some separate reasoning involving Banach Steinhaus, we are somehow able to conclude that $u$ nonetheless is in $\dot{H^1}$... how is this done??

Answer 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 $H^1$.