|
9
|
|
edited Mar 19 2011 at 6:38 |
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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\lt i\lt j$ , it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ .
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 "well-known". 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.
|
|
|
|
|
8
|
|
edited Mar 18 2011 at 11:54 |
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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\lt i\lt j$ , it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ .
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 on first of all how one defines on the specific definition of the spaces $C^k(\bar\Omega)$ in detailused, 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.
|
|
|
|
|
7
|
|
edited Mar 18 2011 at 11:44 |
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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0{ 0\lt i\lt j$ , it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ .
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 on first of all how one defines the spaces $C^k(\bar\Omega)$ in detail, 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.
|
|
|
|
6
|
|
edited Mar 18 2011 at 11:34 |
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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\le i\le j$ it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ . 0{
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 on first of all how one defines the spaces $C^k(\bar\Omega)$ in detail, 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.
|
|
|
|
|
5
|
|
edited Mar 18 2011 at 11:17 |
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 suitable 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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\le i\le j$ it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ .
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 on first of all how one defines the spaces $C^k(\bar\Omega)$ in detail, 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.
|
|
|
|
|
4
|
|
edited Mar 18 2011 at 11:09 |
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 suitable 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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\le i\le j$ it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ .
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 on first of all how one defines the spaces $C^k(\bar\Omega)$ in detail, 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.
|
|
|
|
|
3
|
|
edited Mar 18 2011 at 11:03 |
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 suitable 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 $|\alpha|=\sum_{i=0}^{N-1}\alpha_i\le k$ . For all $i,j\in\mathbb N_0$ with $0\le i\le j$ it then holds that $\|y\|_i\le(\frac{17}4)^{2^{j-2}}\|y\|_0^{1-\frac ij}\|y\|_j^{\frac ij}$ . 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 $\|y\|_0$ 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 on first of all how one defines the spaces $C^k(\bar\Omega)$ in detail, 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.
|
|
|
|
|
2
|
|
edited Mar 16 2011 at 10:49 |
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 $ 1Now 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 $k\in\lbrace 1,2\rbrace$ .
|
|
|
|
|
1
|
|
answered Mar 16 2011 at 10:36 |
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 $1Now ask, whether there are $M,q$ with $p
|
|
|
