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I am trying to answer the following question:

Let $\Omega$ be open, bounded subset of $\mathbb{R}^n$ with Lipschitz boundary. If $f \in W^{1,2}_0(\Omega)$ and we also have that $\nabla f \in L^p(\Omega)$ for some $p > 2$, do we automatically get that $f \in W^{1,p}_0(\Omega)$?

I can't use the Poincaré inequality directly because that supposes that $f$ is $L^p$ already. Instead I tried approximating $f$ by smooth functions of compact support, but because I can only do that in $W^{1,2}_0(\Omega)$, I don't know whether such a sequence also converges in $W^{1,p}_0(\Omega)$.

Localth
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El_lautinho
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1 Answers1

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Yes. You can first show that $f \in W^{1,p}(\Omega)$ by taking an explicit approximation, then use the existence of a trace operator to argue that $f \in W^{1,p}_0(\Omega)$.

Let $\tilde f$ denote the extension of $f$ by zero to $\mathbb R^n$. Then by assumption we have $\tilde f \in W^{1,2}(\mathbb R^n)$ and $\nabla\tilde f \in L^p(\mathbb R^n)$. Given a standard mollifier $\rho_{\varepsilon}$, we then approximate $f$ by the functions $f_{\varepsilon} = f \ast \rho_{\varepsilon} |_{\Omega}$. Since $\tilde f$ is defined on all of $\mathbb R^n$ there is no problem with defining this mollification globally, however one should note that $f_{\varepsilon}$ need not vanish on $\partial\Omega$.

Standard properties of the mollification shows that $f_{\varepsilon} \in C^{\infty}(\overline\Omega)$ and we have the convergence $\nabla f_{\varepsilon} = \nabla f \ast \rho_{\varepsilon} \to f$ in $L^p(\Omega)$ as $\varepsilon \to 0$. Now the Poincaré-Wirtinger inequality (noting $\Omega$ is a Lipschitz domain) applied to $f_{\varepsilon}$ gives $$ \lVert f_{\varepsilon} - (f_{\varepsilon})_{\Omega} \rVert_{L^p(\Omega)} \leq C \lVert \nabla f_{\varepsilon} \rVert_{L^p(\Omega)}, $$ which is bounded as $\varepsilon \to 0$ since the gradients converge in $L^p$.

Now for $\varepsilon,\delta>0$, applying the above to $f_{\varepsilon} - f_{\delta}$ and using the triangle inequality we have $$ \lVert f_{\varepsilon} - f_{\delta}\rVert_{L^p(\Omega)} \leq \lVert (f_{\varepsilon})_{\Omega} - (f_{\delta})_{\Omega}\rVert_{L^p(\Omega)} + \lVert \nabla f_{\varepsilon} - \nabla f_{\delta} \rVert_{L^p(\Omega)}.$$ Moreover since $f \in L^1(\Omega)$ (since $f \in W^{1,2}(\Omega)$ and $\Omega$ is bounded), we have convergence of the averages $(f_{\varepsilon})_{\Omega} \to (f)_{\Omega}$ as $\varepsilon \to 0$. Hence the above tends to zero as $\varepsilon,\delta \to 0$, so $\{f_{\varepsilon}\}$ is Cauchy in $L^p(\Omega)$ and hence the limit $f$ (which is uniquely determined by $L^2$-convergence) also lies in $L^p(\Omega)$. Thus $f \in W^{1,p}(\Omega)$.

Now to show that $f \in W^{1,p}_0(\Omega)$, we crucially use the fact that $\Omega$ is a Lipschitz domain, so we have the existence of a trace operator $$ \operatorname{Tr}_p : W^{1,p}(\Omega) \to L^p(\partial\Omega).$$ Now for the same sequence $f_{\varepsilon}$, we know that $\operatorname{Tr}_2(f_{\varepsilon}) \to 0$ in $L^2(\partial\Omega)$ as $\varepsilon \to 0$. Since $\operatorname{Tr}_p(f_{\varepsilon})$ converges in $L^p(\partial\Omega)$ as $\varepsilon \to 0$, and that $\operatorname{Tr}_p(f_{\varepsilon}) = \operatorname{Tr}_2(f_{\varepsilon})$ for all $\varepsilon>0$ as the trace operator is classically defined for smooth functions, we deduce that $\operatorname{Tr}_p(f) = \lim_{\varepsilon \to 0} \operatorname{Tr}_p(f) = 0$ in $L^p(\Omega)$ also. Therefore $f \in W^{1,p}_0(\Omega)$.

Side-note: In light of this counterexample, I believe it is necessary to argue using the characterisation of $W^{1,p}_0(\Omega)$ via traces. For general domains one should still be able to show that $W^{1,p}(\Omega)$, though one would need to apply the Poincaré-Wirtinger inequality on a possibly larger domain $\widetilde\Omega \supset \Omega$ which is bounded and Lipschitz (e.g. a sufficiently large ball). Note this does not cause any problems since $\tilde f$ vanishes on $\widetilde\Omega \setminus \Omega$.

ktoi
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