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Jean Duchon
Jean Duchon
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Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}(-||u||_\infty \ ,+||u||_\infty)$$f\in W^{m+1,\infty}_{loc}$ with $f(0)=0$ is enough in this case).

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}(-||u||_\infty \ ,+||u||_\infty)$ with $f(0)=0$ is enough in this case).

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}_{loc}$ with $f(0)=0$ is enough in this case).

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Jean Duchon
Jean Duchon
  • 3.1k
  • 11
  • 17

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}([-||u||_\infty \ ,+||u||_\infty])$$f\in W^{m+1,\infty}(-||u||_\infty \ ,+||u||_\infty)$ with $f(0)=0$ is enough in this case).

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}([-||u||_\infty \ ,+||u||_\infty])$ with $f(0)=0$ is enough in this case).

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}(-||u||_\infty \ ,+||u||_\infty)$ with $f(0)=0$ is enough in this case).

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Jean Duchon
Jean Duchon
  • 3.1k
  • 11
  • 17

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}([-||u||_\infty \ ,+||u||_\infty])$ with $f(0)=0$ is enough in this case).