Sobolev embedding theorem in the homogeneous case

Question

We know that if $s>\frac{n}{2}$ the following inclusion holds $$H^s(\mathbb{R}^n)\subset L^\infty(\mathbb{R}^n)$$ Is it also true in the case we deal with the homogeneous space ${\dot{H}}^s(\mathbb{R}^n)$? According to me in this case I can use scaling argument, for instance with the function $f^\lambda(x)=f(\lambda x)$, to show that an inequality of the following type $$\Vert f\Vert_{L^q}\leq C\Vert f\Vert_{{\dot{H}}^s}$$ cannot hold if $s>\frac{n}{2}$ and $q>2$. Is it correct?

Answer 1

If $u_{\lambda}=u(\lambda\cdot)$, then $\widehat{u_{\lambda}}=\lambda^{-n}\widehat{u}_{\lambda^{-1}}$ and dilation invariance yields $$\left(\int_{\mathbb{R}^{n}}|\xi|^{2s}|\widehat{u_{\lambda}}(\xi)|^{2}d\xi\right)^{1/2}=\left(\lambda^{-2n+2s}\int_{\mathbb{R}^{n}}|\lambda^{-1}\xi|^{2s}|\widehat{u}_{\lambda^{-1}}(\xi)|^{2}d\xi\right)^{1/2}=\lambda^{-\frac{n}{2}+s}\|u\|_{\dot{H}^{s}}$$ Noting that $\|u_{\lambda}\|_{L^{\infty}}=\|u\|_{L^{\infty}}$ should answer your question.

If $u\in\dot{H}^{s}$ and $\widehat{u}$ is locally integrable, then one can make the qualitative statement that $u$ is bounded and continuous. To show this, it suffices by Fourier inversion to show that $\widehat{u}\in L^{1}(\mathbb{R}^{n})$. Writing $\widehat{u}$ as $\widehat{u}=\widehat{u}\chi_{|\xi|\leq 1}+\widehat{u}\chi_{|\xi|\geq 1}$, we obtain \begin{align*} \int_{\mathbb{R}^{n}}|\widehat{u}(\xi)|&\leq\int_{|\xi|\leq 1}|\widehat{u}(\xi)|d\xi+\int_{|\xi|\geq 1}|\widehat{u}(\xi)|d\xi\\ &\leq\|\widehat{u}\chi_{|\xi|\leq 1}\|_{L^{1}}+\int_{|\xi|\geq 1}|\xi|^{-s}(|\xi|^{s}|\widehat{u}(\xi)|)d\xi\\ &\leq\|\widehat{u}\chi_{|\xi|\leq 1}\|_{L^{1}}+\left(\int_{|\xi|\geq 1}|\xi|^{-2s}d\xi\right)^{1/2}\|u\|_{\dot{H}^{s}} \end{align*} by Cauchy-Schwarz. Note that the integral in the second converges as $2s>n$.

One can make a quantitative statement about the Holder continuity of $u$, though. More precisely, if $0<\rho:=(s-\dfrac{n}{2})<1$, then $$\sup_{x\neq y}\dfrac{|u(x)-u(y)|}{|x-y|^{\rho}}\lesssim_{n,s}\|u\|_{\dot{H}^{s}}$$ To prove this estimate, it suffices by density to consider the case where $u$ is Schwartz. We split $u$ into a $u$ into a low frequency component $P_{\leq N}u$ and a high frequency component $P_{>N}u$ as follows. Let $\phi$ be a $C_{c}^{\infty}$ function such that $\phi\equiv 1$ on $|\xi|\leq 1$ and supported in the ball $|\xi|\leq 2$. Set $P_{\leq N}u:=(\phi(\xi/N)\widehat{u})^{\vee}$ and $P_{>N}u:=u-P_{\leq N}u$.

Since $P_{\leq N}u$ is the convolution of a Schwartz function with a tempered distribution, it is smooth. By the mean value theorem, $$|(P_{\leq N}u)(x)-(P_{\leq N}u)(y)|\leq\|\nabla(P_{\leq N}u)\|_{L^{\infty}}|x-y|,\quad\forall x,y\in\mathbb{R}^{n}$$ To estimate the $L^{\infty}$ norm, Fourier analysis gives \begin{align*} \|\nabla(P_{\leq N}u)\|_{L^{\infty}}\leq\int_{\mathbb{R}^{n}}|2\pi\xi||\phi(\xi/N)\widehat{u}(\xi)|d\xi&\lesssim\int_{\mathbb{R}^{n}}|\xi|^{1-s}|\xi|^{s}|\phi(\xi/N)\widehat{u}(\xi)|d\xi\\ &\leq\left(\int_{|\xi\leq 2N}|\xi|^{2(1-s)}d\xi\right)^{1/2}\|u\|_{\dot{H}^{s}}\\ &\leq_{s,n}N^{1-s}N^{n/2}\|u\|_{\dot{H}^{s}}=N^{1-\rho}\|u\|_{\dot{H}^{s}} \end{align*} by Cauchy-Schwarz and dilation invariance. The integral in the penultimate line above is convergent since $1-s>1-(s-\frac{n}{2})-\frac{n}{2}>-\frac{n}{2}$.

For the high frequency component, we have the estimate \begin{align*} \|(P_{>N}u)\|_{L^{\infty}}\leq\int_{\mathbb{R}^{n}}|(1-\phi(\xi/N)||\widehat{u}(\xi)|d\xi&\leq\left(\int_{|\xi|\geq N}|\xi|^{-2s}d\xi\right)^{1/2}\|u\|_{\dot{H}^{s}}\\ &=N^{-s+\frac{n}{2}}\left(\int_{|\xi|\geq 1}|\xi|^{-2s}d\xi\right)^{1/2}\|u\|_{\dot{H}^{s}}\\ &\lesssim_{n,s}N^{-\rho}\|u\|_{\dot{H}^{s}} \end{align*} by Cauchy-Schwarz and dilation invariance. Choose $N=|x-y|^{-1}$. Then by the triangle inequality and the above estimates, we have that \begin{align*} |u(x)-u(y)|&\leq\|\nabla(P_{\leq N}u)\|_{L^{\infty}}|x-y|+2\|P_{>N}u\|_{L^{\infty}}\\ &\lesssim_{n,s}N^{1-\rho}\|u\|_{\dot{H}^{s}}|x-y|+2N^{-\rho}\|u\|_{\dot{H}^{s}}\\ &\lesssim \|u\|_{\dot{H}^{s}}|x-y|^{\rho} \end{align*}