I have read the following lemma :
Let $p>2$. There exists a constant $C>0$ such that if $Y\in W^{1,2}(\mathbb{R}\times ]0,1[,\mathbb{R}^{2n})$ and $DY:=\partial_s Y+J_0\partial_tY+SY\in L^p(\mathbb{R}\times ]0,1[)$ then $Y\in W^{1,p}(\mathbb{R}\times ]0,1[)$ and $||Y||_{W^{1,p}}\leq C||DY||_{L^p}$, where $S$ is a symmetric operator.
Now I was wondering if we could say anything about $W^{1,2}_0$. That is if we add the assumption that $Y\in W^{1,2}_0(\mathbb{R}\times ]0,1[)$ then $Y\in W^{1,p}_0(\mathbb{R}\times ]0,1[)$. I have tried using the definition of the space with the sequences of $\phi_n\in C_{c}^{\infty}(\mathbb{R}\times ]0,1[)$ but I got nowhere. My next idea would be to consider trace operators but I am not very familiar with these.
Does anyone know if this will be true or not ?
Since $W_{0}^{1,p}(\Omega)=\{u\in W^{1,p}(\Omega):Tu=0\}$ where $T$ is the trace operator, I am wandering how does the trace operator behave with respect to different sovolev spaces. That is if we have $u\in W^{1,p}(\Omega)\cap W^{1,q}(\Omega)$ will $T_pu=T_qu$? (where $T_p$ denotes the trace operator defined in the sobolev space $W^{1,p}$).
My Attempt: By the Sobolev embedding theorem we know that for $p>2$ we have $W^{1,p}(\mathbb{R}\times ]0,1[) \hookrightarrow C_b(\mathbb{R}^{2})$. That way we can take the continuous representative in $W^{1,2}_0(\mathbb{R}\times ]0,1[)$ and using the properties of the trace operator we know that $u|_{\partial \Omega}=0$, and so we will have the desired result.
What do you think ? Any enlightment is appreciated, thanks in advance.