Local fractional Sobolev inequality

Question

If $\mathcal{X}$ is a smooth cutoff near 0 in $\mathbb{R}^n$, then $M_0 = \mathcal{X}(-\Delta+Id)\mathcal{X}$ is a self-adjoint operator in $L^2(\mathbb{R}^n)$. Because $M_0$ is semi-positive and the spectral theory of self-adjoint operator, we can define $M_0^{n/4+\epsilon}$. I need an inequality $$\|M_0^{n/4+\epsilon}u\|_{L^2(\mathbb{R}^n)}\geq C\epsilon^{-1/2}|u(0)|\ \,\,\,\,\ \ \forall\epsilon>0$$

The difficulty of this inequality is local and fractional. And the power of $\epsilon$ can be gotten by using Fourier transform and suppose $\mathcal{X}\equiv1$. I met this inequality in a paper of P.Lax, he uses this to get a global inequality in manifold.

Answer 1

For $\epsilon >0$, $u\in H^{\frac{n}{2}+2\epsilon}$, $N_0=-\Delta+1$, $$ \Vert N_0^{\frac{n}{4}+\epsilon} u\Vert_{L^2}=\Vert u\Vert_{H^{\frac{n}{2}+2\epsilon}}\ge c_{n,\epsilon} \Vert u\Vert_{L^\infty}. $$ This implies for $u\in H^{\frac{n}{2}+2\epsilon}_{loc}$ and $\chi_1$ smooth compactly supported, $$ \Vert N_0^{\frac{n}{4}+\epsilon} \chi_1 u\Vert_{L^2}\ge c_{n,\epsilon} \Vert \chi_1 u\Vert_{L^\infty}. $$ Now if $\chi, \chi_1\in C^\infty_c$, $\chi_1=1$ on the support of $\chi$, we get $\chi_1\chi=\chi$ and $$ \Vert \underbrace{(1-\chi_1)N_0^{\frac{n}{4}+\epsilon} \chi}_{\text{Pseudo with order $-\infty$}} u\Vert_{L^2}+ \Vert \chi_1 N_0^{\frac{n}{4}+\epsilon} \chi_1 \chi u\Vert_{L^2}\ge c_{n,\epsilon} \Vert \chi u\Vert_{L^\infty}. $$ This not exactly what you are asking, but it could give you the information you are looking for.