Suppose that $\Omega$ is a bounded domain in $\mathbb{R}^3$, $F$ is bounded in $L^\infty (\Omega \times (0,T))\cap (\cap_{k=1}^\infty L^{5/3}(0,T;C^k(\bar{\Omega})))$.
Question: Can we say that $F$ is bounded in $L^q(0,T;C^2(\bar{\Omega}))$ for any $q\in [1,\infty)$?

1$\begingroup$ Something is misstated here. Did you mean $k\ge 2$? Or should it be $L^q(0,T;C(\bar\Omega))$? $\endgroup$– Michael RenardyMar 15, 2011 at 14:24

$\begingroup$ By "for an arbitrary integer $k\ge 0$ " do you mean that it holds for some fixed $k\in\mathbb N_0$ or for all such? By "for any $q\in[1,\infty)$ " do you mean that it should hold for some such $q$ or for all? Of course, if in the assumption we take $k$ at least $2$ , then to the question we can answer positively for $1\le q\le\frac 53$ . $\endgroup$– TaQMar 16, 2011 at 0:42

$\begingroup$ By considering functions like $(t,s)\mapsto\sin(t^{r}s)$ , it also seems obvious that if the assumption holds (only) for some $k\ge 2$ , then the question cannot be answered positively for $q>\frac 56 k$ . $\endgroup$– TaQMar 16, 2011 at 1:01

$\begingroup$ Thanks for you all. $F$ is bounded in $L^\infty (0,T;\Omega)$ and $L^{5/3}(0,T;C^k(\bar{\Omega}))$ for all $k\in\mathbb{N}_0$. I have edit the statement. $\endgroup$– jackMar 16, 2011 at 1:30
1 Answer
This is not an answer, but just a formulation of a simpler analogous problem, which might help better understand the original question. Let $I=[0,1]$ and $Q=I\times I$ . Let $ 1 < p < +\infty$ , and let $\langle M_k\rangle\in[1,+\infty[\ ^{\mathbb N}$ satisfy $M_k +1 < M_{k+1}$ for all $k\in\mathbb N$ . Let $\u\=\sup\lbrace u(s):s\in I\rbrace$ for $u\in C(I)$ . Let $S$ be the set of all measurable $z:Q\to[1,1]$ having $\lbrace z(t,\cdot):t\in I\rbrace\subseteq C^\infty(I)$ with $\int_0^1\{\rm D}^k(z(t,\cdot))\^p\ {\rm d\ }t\le M_k$ for all $k\in\mathbb N$ . Now ask, whether there are $M,q$ with $ p < q < +\infty$ and $M<+\infty$ , and such that $\int_0^1\{\rm D}^k(z(t,\cdot))\^q\ {\rm d\ }t\le M$ holds for all $z\in S$ and $k\in\lbrace 1,2\rbrace$ .
Edit. I answer to the original question: We can say that $F$ is bounded in $L^q(0,T;C^2(\bar\Omega))$ for all $q\in[1,+\infty[$ under the additional assumption that $\Omega$ is the interior of an open qube and the spaces $C^k(\bar\Omega)$ are suitably defined. See this preprint for some aspects related to this matter. The deduction is essentially based on
Proposition 1. Let $K=[0,1]^N$ where $N\in\mathbb N$ . For $k\in\mathbb N_0$ and $y\in C^k(K)$ , let $\y\_k$ denote the supremum of all $\partial^\alpha y(s)$ where $s\in K$ and $\alpha\in\mathbb N_0^N$ with $\alpha=\sum_{i=0}^{N1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\lt i\lt j$ , it then holds that $\y\_i\le(\frac{17}4)^{2^{j2}}\y\_0^{1\frac ij}\y\_j^{\frac ij}$ .
Namely, given any $q$ with $\frac 53= p < q < +\infty $ , by taking $k\ge 2p^{1}q$ , we get $\y\_2^q\le C\y\_k^p$ under the assumption that $\y\_0$ remains bounded. For a more general bounded domain $\Omega$ , whether the same holds, depends on whether we can get for it estimates similar to those in Proposition 1. This depends first of all on the specific definition of the spaces $C^k(\bar\Omega)$ used, cf. the preprint referred to above. Once the definition is fixed, if one can construct an extension operator $C^k(\bar\Omega)\to C^k(Q)$ where $Q$ is some compact qube containing $\overline\Omega$ in its interior, then one gets the same result.
Remark. The essential content of Proposition 1 is "wellknown". For example in Richard S. Hamilton's article on "the" Nash−Moser theorem, 2.2.1. Theorem (pp. 143−144) gives a rough sketch of an idea for a proof, without specifying the constant.

$\begingroup$ Thanks. The only problem is the proof of "Proposition 1". How to prove? Or any reference? $\endgroup$– jackMar 18, 2011 at 16:54

$\begingroup$ In order not to increase the number of edits, I here add as a comment that I do not know whether there exists in the literature a published full proof of $\y\_i\le C\y\_0^{1k^{1}i}\y\_k^{k^{1}i}$ . $\endgroup$– TaQMar 19, 2011 at 10:22