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Let us assume that $p=2$, and let us consider $$ \cup_{s>0} H^s(\mathbb R^d)\subset H^0(\mathbb R^d)=L^2(\mathbb R^d). $$ The above inclusion is strict. Let us consider $u\in L^2(\mathbb R^d)$ define …

Too long for comment. take for instance $f=0, g=0$. Then the mapping $h\mapsto u$ is a pseudo-differential operator which will have some Sobolev continuity properties for spaces $W^{s,p}$ with $p\in …

Following Christian Remling suggestion, it seems that $$ \Vert u\Vert_{L^\infty(\mathbb R^d)}\le \gamma(d)\bigl\{ \Vert u\Vert_{L^2(\mathbb R^d)}+ \Vert \vert D\vert^{d/2} L(\vert D\vert) u\Vert_{L^2( …

Let us start with pointing out that $f\in H^\sigma$ is equivalent to $$ (\langle n\rangle^\sigma\hat f(n))_{n\in \mathbb Z}\in \ell^2(\mathbb Z), \quad \text{with $\langle n\rangle=\sqrt{1+n^2}$.} $$ …

Let $f$ be a fonction in $L^1(0,+\infty)$ with norm 1 and let us define for $x\ge 0$ $$ \phi(x)=\int_x^1f(t) dt. $$ Then $\phi$ is continuous since $ -\phi(h)+\phi(0)=\int_0^h f(t) dt, $ which goes t …

You have done half of the job. We have the absolutely convergent integral which is such that $$ f(\xi)=\int_{\mathbb R^d} \frac{\vert e^{2iπ y\cdot \xi}-1\vert^2}{\vert y\vert^{d+2s}} dy =c_{s,d}\vert …

I will more comfortable with the notation $v_\epsilon=\hat{u_\epsilon}$; you have then $$ v_\epsilon(t,x)=\alpha(t,x)+\int_0^t\int e^{2πix(\xi+\epsilon\phi(s,\xi))} \hat{v_\epsilon}(s,\xi) d\xi ds=\al …

From your assumptions, you have a $C^2$ function $\rho:\mathbb R^d\rightarrow \mathbb R$, such that $$ D=\{x\in \mathbb R^d, \rho(x)<0\}, \quad \partial D=\{x\in \mathbb R^d, \rho(x)=0\}, $$ and $ x\i …

In the first place, you must have $s>1/2$. Next you write $$ u(x)=\int_{\mathbb R} e^{2πix\xi}\underbrace{\hat u(\xi)(1+\xi^2)^{s/2}}_{\in L^2(\mathbb R)}\underbrace{(1+\xi^2)^{-s/2}}_{\in L^2(\mathbb …

Up to some normalization, the Fourier transform of the characteristic function $\mathbf 1_I$ of a compact interval $I$ is $$ \widehat{\mathbf 1_I}(\xi)=\frac{\sin \xi}{\xi}. $$ Obviously the function …

You have $ \bigl\vert\vert f_{k+l}(x)\vert-\vert f_{k}(x)\vert\bigr\vert\le \vert f_{k+l}(x)-f_{k}(x)\vert $ and thus $$ \Vert\vert f_{k+l}\vert-\vert f_{k}\vert\Vert_{W^{1,1}} \le \Vert f_{k+l}- f_ …

Since your manifold is compact, $H^s_{loc}$ regularity will be equivalent to $H^s$ regularity. To check the local regularity, you can use cutoff functions and the coordinate charts.

Let $n\ge 1$ be an integer and $s>n/2$. Then you have $H^s(\mathbb R^n)\subset L^\infty(\mathbb R^n)$ and for $f,g\in H^s(\mathbb R^n)$, $$ \Vert fg \Vert_{H^s(\mathbb R^n)}\le c_n\bigl(\Vert f \Vert_ …

More a comment than an answer, but too long for a comment. First a comment on Michael Renardy's remark: there is no homogeneous function in $L^1(\mathbb R^N)$ so the first assumption is not satisfied. …

I guess that your question is ill-formulated: you have to respect the homogeneity on both sides of the inequality. Let us say that for $f$ in the Schwartz space you always have $$ \Vert f\Vert_{W^{t,q …