Weak convergence improved by Morrey embedding
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Let $u_n: [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ be a sequence with
begin{equation}
u_n rightharpoonup u text{weakly star in } L^2(0,T;W^{1,infty}(mathbb{R}^3))
end{equation}
and $eta : [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ continuous in time. Assume further $x in mathbb{R}^3$, $psi in L^2(0,T)$. Now it says
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt rightarrow 0
end{equation}
and I'm not sure if I understand why this is true. I think this follows from the Morrey embedding, which states that $W^{1,infty }(mathbb{R}^3) subset C^{0,alpha}(mathbb{R}^3)$ is a compact embedding for $alpha < 1$. What confuses me is that $eta$ depends also on $t$ but I think this does not matter as the convergence in $C^0$ is uniform, we should have something like
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt leq int _0^T sup_{y}|u_n(t, y) - u(t, y) | psi dt rightarrow 0
end{equation}
although I'm not sure if this can be written like that. Is this argument correct?
functional-analysis compactness sobolev-spaces weak-convergence
asked Nov 15 at 18:23
jason paper
12319
add a comment |
up vote
1
down vote
favorite
Let $u_n: [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ be a sequence with
begin{equation}
u_n rightharpoonup u text{weakly star in } L^2(0,T;W^{1,infty}(mathbb{R}^3))
end{equation}
and $eta : [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ continuous in time. Assume further $x in mathbb{R}^3$, $psi in L^2(0,T)$. Now it says
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt rightarrow 0
end{equation}
and I'm not sure if I understand why this is true. I think this follows from the Morrey embedding, which states that $W^{1,infty }(mathbb{R}^3) subset C^{0,alpha}(mathbb{R}^3)$ is a compact embedding for $alpha < 1$. What confuses me is that $eta$ depends also on $t$ but I think this does not matter as the convergence in $C^0$ is uniform, we should have something like
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt leq int _0^T sup_{y}|u_n(t, y) - u(t, y) | psi dt rightarrow 0
end{equation}
although I'm not sure if this can be written like that. Is this argument correct?
functional-analysis compactness sobolev-spaces weak-convergence
asked Nov 15 at 18:23
jason paper
12319
$W^{1,infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding.
– Michał Miśkiewicz
Nov 16 at 19:24
But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough
– jason paper
Nov 17 at 2:32
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $u_n: [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ be a sequence with
begin{equation}
u_n rightharpoonup u text{weakly star in } L^2(0,T;W^{1,infty}(mathbb{R}^3))
end{equation}
and $eta : [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ continuous in time. Assume further $x in mathbb{R}^3$, $psi in L^2(0,T)$. Now it says
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt rightarrow 0
end{equation}
and I'm not sure if I understand why this is true. I think this follows from the Morrey embedding, which states that $W^{1,infty }(mathbb{R}^3) subset C^{0,alpha}(mathbb{R}^3)$ is a compact embedding for $alpha < 1$. What confuses me is that $eta$ depends also on $t$ but I think this does not matter as the convergence in $C^0$ is uniform, we should have something like
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt leq int _0^T sup_{y}|u_n(t, y) - u(t, y) | psi dt rightarrow 0
end{equation}
although I'm not sure if this can be written like that. Is this argument correct?
functional-analysis compactness sobolev-spaces weak-convergence
asked Nov 15 at 18:23
jason paper
12319
Let $u_n: [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ be a sequence with
begin{equation}
u_n rightharpoonup u text{weakly star in } L^2(0,T;W^{1,infty}(mathbb{R}^3))
end{equation}
and $eta : [0,T]times mathbb{R}^3 rightarrow mathbb{R}^3$ continuous in time. Assume further $x in mathbb{R}^3$, $psi in L^2(0,T)$. Now it says
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt rightarrow 0
end{equation}
and I'm not sure if I understand why this is true. I think this follows from the Morrey embedding, which states that $W^{1,infty }(mathbb{R}^3) subset C^{0,alpha}(mathbb{R}^3)$ is a compact embedding for $alpha < 1$. What confuses me is that $eta$ depends also on $t$ but I think this does not matter as the convergence in $C^0$ is uniform, we should have something like
begin{equation}
int _0^T (u_n(t, eta(t,x)) - u(t, eta(t,x)) ) psi dt leq int _0^T sup_{y}|u_n(t, y) - u(t, y) | psi dt rightarrow 0
end{equation}
although I'm not sure if this can be written like that. Is this argument correct?
functional-analysis compactness sobolev-spaces weak-convergence
functional-analysis compactness sobolev-spaces weak-convergence
asked Nov 15 at 18:23
jason paper
12319
asked Nov 15 at 18:23
jason paper
12319
asked Nov 15 at 18:23
jason paper
12319
asked Nov 15 at 18:23
jason paper
12319
asked Nov 15 at 18:23
jason paper
12319
12319
$W^{1,infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding.
– Michał Miśkiewicz
Nov 16 at 19:24
But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough
– jason paper
Nov 17 at 2:32
add a comment |
$W^{1,infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding.
– Michał Miśkiewicz
Nov 16 at 19:24
But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough
– jason paper
Nov 17 at 2:32
$W^{1,infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding.
– Michał Miśkiewicz
Nov 16 at 19:24
$W^{1,infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding.
– Michał Miśkiewicz
Nov 16 at 19:24
But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough
– jason paper
Nov 17 at 2:32
But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough
– jason paper
Nov 17 at 2:32
add a comment |
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$W^{1,infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding.
– Michał Miśkiewicz
Nov 16 at 19:24
But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough
– jason paper
Nov 17 at 2:32