Composition with Lipschitz map is Lipschitz on Sobolev spaces
$begingroup$
Suppose that $F: mathbb{R}^d rightarrow mathbb{R}^d$ is Lipschitz with some constant $L$ and that $F(0)=0$. Then it is clear that $F$ defines a Lipschitz continuous map $L^2(mathbb{R}^d) rightarrow L^2(mathbb{R}^d)$ by $u mapsto F(u)$ with the same Lipschitz constant $L$.
We can extend this to Sobolev spaces $H^k(mathbb{R}^d)$ when $k in mathbb{N}$, by requiring in addition that $F in C^k$ with the first $k$ derivatives of $F$ Lipschitz and that the first $k$ derivatives vanish at $0$. For instance, if $k=1$, we can compute
begin{align}
| partial_i(F(u)-F(tilde{u})) |_{L^2(mathbb{R}^d)} &= | nabla F(u) partial_i u - nabla F(tilde{u}) partial_i tilde{u} |_{L^2} \
&leq | nabla F(u) |_{L^2} |partial_i u - partial_i tilde{u} |_{L^2} + |partial_i tilde{u} |_{L^2} |nabla F(u) - nabla F(tilde{u}) |_{L^2} \
&leq C left( |u|_{L^2} |u-tilde{u} |_{H^1}+| tilde{u} |_{H^1} | u- tilde{u}|_{L^2} right) \
&leq C (|u|_{H^1} + | tilde{u} |_{H^1}) | u - tilde{u} |_{H^1}
end{align}
where we have used that $nabla F$ is again a Lipschitz map (and I denoted all constants by $C$). Thus we obtain that the map $u mapsto F(u)$ is (at least) locally Lipschitz in $H^1$. Similarly of course for $H^k$, when $k in mathbb{N}$.
Now here is my question: (how) is it possible to extend this result to Sobolev spaces $H^s$ with real index $s in mathbb{R}$? Is the map $u mapsto F(u)$ well-defined and locally Lipschitz continuous on the space $H^s(mathbb{R}^d)$? Is there some interpolation argument that easily accomplishes this? Finally, did I miss something, and is the map perhaps even globally Lipschitz?
functional-analysis analysis sobolev-spaces lipschitz-functions fractional-sobolev-spaces
asked Nov 29 '18 at 10:25
MaxMax
1187
$endgroup$
add a comment |
$begingroup$
Suppose that $F: mathbb{R}^d rightarrow mathbb{R}^d$ is Lipschitz with some constant $L$ and that $F(0)=0$. Then it is clear that $F$ defines a Lipschitz continuous map $L^2(mathbb{R}^d) rightarrow L^2(mathbb{R}^d)$ by $u mapsto F(u)$ with the same Lipschitz constant $L$.
We can extend this to Sobolev spaces $H^k(mathbb{R}^d)$ when $k in mathbb{N}$, by requiring in addition that $F in C^k$ with the first $k$ derivatives of $F$ Lipschitz and that the first $k$ derivatives vanish at $0$. For instance, if $k=1$, we can compute
begin{align}
| partial_i(F(u)-F(tilde{u})) |_{L^2(mathbb{R}^d)} &= | nabla F(u) partial_i u - nabla F(tilde{u}) partial_i tilde{u} |_{L^2} \
&leq | nabla F(u) |_{L^2} |partial_i u - partial_i tilde{u} |_{L^2} + |partial_i tilde{u} |_{L^2} |nabla F(u) - nabla F(tilde{u}) |_{L^2} \
&leq C left( |u|_{L^2} |u-tilde{u} |_{H^1}+| tilde{u} |_{H^1} | u- tilde{u}|_{L^2} right) \
&leq C (|u|_{H^1} + | tilde{u} |_{H^1}) | u - tilde{u} |_{H^1}
end{align}
where we have used that $nabla F$ is again a Lipschitz map (and I denoted all constants by $C$). Thus we obtain that the map $u mapsto F(u)$ is (at least) locally Lipschitz in $H^1$. Similarly of course for $H^k$, when $k in mathbb{N}$.
Now here is my question: (how) is it possible to extend this result to Sobolev spaces $H^s$ with real index $s in mathbb{R}$? Is the map $u mapsto F(u)$ well-defined and locally Lipschitz continuous on the space $H^s(mathbb{R}^d)$? Is there some interpolation argument that easily accomplishes this? Finally, did I miss something, and is the map perhaps even globally Lipschitz?
functional-analysis analysis sobolev-spaces lipschitz-functions fractional-sobolev-spaces
asked Nov 29 '18 at 10:25
MaxMax
1187
$endgroup$
add a comment |
$begingroup$
Suppose that $F: mathbb{R}^d rightarrow mathbb{R}^d$ is Lipschitz with some constant $L$ and that $F(0)=0$. Then it is clear that $F$ defines a Lipschitz continuous map $L^2(mathbb{R}^d) rightarrow L^2(mathbb{R}^d)$ by $u mapsto F(u)$ with the same Lipschitz constant $L$.
We can extend this to Sobolev spaces $H^k(mathbb{R}^d)$ when $k in mathbb{N}$, by requiring in addition that $F in C^k$ with the first $k$ derivatives of $F$ Lipschitz and that the first $k$ derivatives vanish at $0$. For instance, if $k=1$, we can compute
begin{align}
| partial_i(F(u)-F(tilde{u})) |_{L^2(mathbb{R}^d)} &= | nabla F(u) partial_i u - nabla F(tilde{u}) partial_i tilde{u} |_{L^2} \
&leq | nabla F(u) |_{L^2} |partial_i u - partial_i tilde{u} |_{L^2} + |partial_i tilde{u} |_{L^2} |nabla F(u) - nabla F(tilde{u}) |_{L^2} \
&leq C left( |u|_{L^2} |u-tilde{u} |_{H^1}+| tilde{u} |_{H^1} | u- tilde{u}|_{L^2} right) \
&leq C (|u|_{H^1} + | tilde{u} |_{H^1}) | u - tilde{u} |_{H^1}
end{align}
where we have used that $nabla F$ is again a Lipschitz map (and I denoted all constants by $C$). Thus we obtain that the map $u mapsto F(u)$ is (at least) locally Lipschitz in $H^1$. Similarly of course for $H^k$, when $k in mathbb{N}$.
Now here is my question: (how) is it possible to extend this result to Sobolev spaces $H^s$ with real index $s in mathbb{R}$? Is the map $u mapsto F(u)$ well-defined and locally Lipschitz continuous on the space $H^s(mathbb{R}^d)$? Is there some interpolation argument that easily accomplishes this? Finally, did I miss something, and is the map perhaps even globally Lipschitz?
functional-analysis analysis sobolev-spaces lipschitz-functions fractional-sobolev-spaces
asked Nov 29 '18 at 10:25
MaxMax
1187
$endgroup$
Suppose that $F: mathbb{R}^d rightarrow mathbb{R}^d$ is Lipschitz with some constant $L$ and that $F(0)=0$. Then it is clear that $F$ defines a Lipschitz continuous map $L^2(mathbb{R}^d) rightarrow L^2(mathbb{R}^d)$ by $u mapsto F(u)$ with the same Lipschitz constant $L$.
We can extend this to Sobolev spaces $H^k(mathbb{R}^d)$ when $k in mathbb{N}$, by requiring in addition that $F in C^k$ with the first $k$ derivatives of $F$ Lipschitz and that the first $k$ derivatives vanish at $0$. For instance, if $k=1$, we can compute
begin{align}
| partial_i(F(u)-F(tilde{u})) |_{L^2(mathbb{R}^d)} &= | nabla F(u) partial_i u - nabla F(tilde{u}) partial_i tilde{u} |_{L^2} \
&leq | nabla F(u) |_{L^2} |partial_i u - partial_i tilde{u} |_{L^2} + |partial_i tilde{u} |_{L^2} |nabla F(u) - nabla F(tilde{u}) |_{L^2} \
&leq C left( |u|_{L^2} |u-tilde{u} |_{H^1}+| tilde{u} |_{H^1} | u- tilde{u}|_{L^2} right) \
&leq C (|u|_{H^1} + | tilde{u} |_{H^1}) | u - tilde{u} |_{H^1}
end{align}
where we have used that $nabla F$ is again a Lipschitz map (and I denoted all constants by $C$). Thus we obtain that the map $u mapsto F(u)$ is (at least) locally Lipschitz in $H^1$. Similarly of course for $H^k$, when $k in mathbb{N}$.
Now here is my question: (how) is it possible to extend this result to Sobolev spaces $H^s$ with real index $s in mathbb{R}$? Is the map $u mapsto F(u)$ well-defined and locally Lipschitz continuous on the space $H^s(mathbb{R}^d)$? Is there some interpolation argument that easily accomplishes this? Finally, did I miss something, and is the map perhaps even globally Lipschitz?
functional-analysis analysis sobolev-spaces lipschitz-functions fractional-sobolev-spaces
functional-analysis analysis sobolev-spaces lipschitz-functions fractional-sobolev-spaces
asked Nov 29 '18 at 10:25
MaxMax
1187
asked Nov 29 '18 at 10:25
MaxMax
1187
asked Nov 29 '18 at 10:25
MaxMax
1187
asked Nov 29 '18 at 10:25
MaxMax
1187
asked Nov 29 '18 at 10:25
MaxMax
1187
1187
add a comment |
add a comment |
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