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$\DeclareMathOperator{\loc}{\mathrm{loc}}$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??

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  • $\begingroup$ Should it maybe be "$\sup_R \|\phi u_R\|_{\dot{H}^1} < \infty$", so with the homogeneous Sobolev (= gradient) norm? Also, what properties of $\dot{H}^1$ are assumed at the point in the book? For instance, if I see it correctly, if one uses the Sobolev embedding and reflexivity for $\dot{H}^1$, that should do it. (Extract weakly convergent subsequence of $u_R$, use Sobolev embeding, then the weak limit in $\dot{H}^1$ must coincide with $u$ in, say, $L^2_{loc}$.) The UBP argument eludes me, though. $\endgroup$
    – Hannes
    Nov 18, 2020 at 9:30
  • $\begingroup$ The statement $\sup_{R}||\phi u_R||_{{H^1}}<\infty$ holds as well because $u_R\in \dot{H^1}$ and $\phi$ has compact support so their product is in $H^1$. $\endgroup$
    – Fozz
    Nov 18, 2020 at 16:08
  • $\begingroup$ @Fozz what about $u_R=R$ (constant function) which is bounded in $\dot{H}^1$ and for which $\sup_R ||\phi u_R||_{H^1} = \infty$ as soon as $\phi \neq 0$ ? $\endgroup$ Nov 18, 2020 at 16:11
  • $\begingroup$ Strictly speaking, I think the space $\dot{H^1}$ is defined in terms of equivalence classes where functions that differ by polynomials are identified. In particular, all constant functions are equivalent to the zero function. $\endgroup$
    – Fozz
    Nov 18, 2020 at 16:18

1 Answer 1

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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 $u$, $$u_{R_k} \longrightarrow u,\ in \ L^q(R^3),\ \forall 1\leq q<6 $$

  1. 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 $H^1$.

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  • $\begingroup$ The space $\dot{H^1}$ is the space of distributions $f$ such that $\int_{\mathbb{R}^3}|\xi|^2|\mathcal{F}f(\xi)|^2\,d\xi<\infty$ (here $\mathcal{F}$ is the Fourier transform). Basically, it's the space of distributions whose gradient is in $L^2$. $\endgroup$
    – Fozz
    Nov 18, 2020 at 3:43
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    $\begingroup$ Are you viewing $u_R$ as a linear operator from $\dot{H^1}$ to $\mathbb{R}$ ($\phi\mapsto\int\phi u_R\,dx$) or from $\dot{H^1}$ to $H^1$ ($\phi\mapsto \phi u_R$)? $\endgroup$
    – Fozz
    Nov 18, 2020 at 3:44
  • $\begingroup$ from $\dot H^1 \to R$, test function space is dense in $\dot H^1$, so there is a natural extension to make $u_R$ to be a linear operator on $\dot H^1$ $\endgroup$
    – katago
    Nov 18, 2020 at 3:50
  • $\begingroup$ For $\phi\in\mathcal D$, the map $\phi\mapsto\int\phi u_R\,dx$ IS a linear operator, but how do we know it's continuous with respect to $\dot{H^1}$? I guess what I mean is, if we approximate $\phi\in \dot{H^1}$ by $\phi_k\in\mathcal D$ in the space $\dot{H^1}$, how does it follow that $\int \phi_k u_R\,dx\to\int\phi u_R\,dx$ as well? $\endgroup$
    – Fozz
    Nov 18, 2020 at 4:01
  • $\begingroup$ @Fozz, in fact in your case because of priori estimate and application of Leray–Schauder principle we can prove $u_{R} \in \dot{H}^{1}$, so the argument can all live in $\dot{H}^{1} \subset H^ {-1}$ and the limit is also in $\dot{H}^{1}$ $\endgroup$
    – katago
    Nov 18, 2020 at 4:03

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