Why $H^1(\Omega)\cap H^2(\Omega_-)\cap H^2(\Omega_+) \subset W^1_p(\Omega),\forall p>2$?

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I'm reading a book and it says that:

Let $\Omega=\Omega_-\cup\Omega_+$. Define $X(\Omega)=H^1(\Omega)\cap H^2(\Omega_-)\cap H^2(\Omega_+)$. Then by the Sobolev embedding theorem, $X(\Omega)\subset W^1_p(\Omega)$, for any $p>2$.

I understand that $H^2(\Omega_\pm)\subset W^1_p(\Omega_\pm)$, for any $p>2$ follows from the Sobolev embedding theorem. But how come the statement hold for the whole domain $\Omega$? (since $W^1_p(\Omega)\subset H^1(\Omega),\forall p>2$).

Any help is appreciated. Thank you.

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For $f\in H^1(\Omega)$, you already know that the weak derivatives $f_\alpha := \partial^\alpha f$ exist on all of $\Omega$. Thus, you only need to show $f_\alpha \in L^p(\Omega)$. For this, it suffices to show (why?) $f\alpha \in L^p(\Omega_1)$ and $f_\alpha \in L^p(\Omega_2)$, which follows (apparently) from Sobolev embedding.