Consider a compact manifold $M$ and a point $q \in M$. Let us say that that the following inequality holds: $$ \Vert \varphi u\Vert_{L^p} \leq C\Vert \varphi u\Vert_{H^1},$$ where $\varphi \in C^\infty_c(M \setminus \{q\})$, that is, $\varphi$ is smooth and compactly supported away from $q$, and $C$ is independent of $\varphi$ and $u$. If we know that $u \in H^1 \cap L^p$, can we conclude from the above that $$\Vert u\Vert_{L^p} \leq C\Vert u\Vert_{H^1}?$$

Edit: It seems that it suffices to claim that there exists a sequence $\varphi_k$ such that $\varphi_k u \to u$ in $L^p$-norm and $\varphi_k u \to u$ in $H^1$-norm. But I cannot justify the existence of such a sequence.

  • $\begingroup$ From which space to which space does $u$ map? Do you mean by $\varphi u$ the multiplication or the composition of $\varphi$ and $u$? $\endgroup$ – user35593 Jun 12 '15 at 14:17
  • $\begingroup$ @user35593 Multiplication. $u$ is usual complex-valued, mapping out of $M$. $\endgroup$ – ginjee Jun 12 '15 at 14:22

I believe that the answer is yes in dimension $d\ge 2$, from the following result.

Theorem. Let $\Omega$ be an open subset of $\mathbb R^d$, let $X$ be a Lipschitz vector field on $\Omega$ and let $u\in L^\infty(\Omega)$. Let $\Omega_0$ be an open subset of $\Omega$ such that $$ \mathcal H^{d-1}(\Omega\backslash \Omega_0)=0,\quad\text{$\mathcal H^{d-1}$ is the Hausdorff measure.} $$ Then if the equation $Xu=f$ holds (weakly) on $\Omega_0$ with $f$ locally $L^1$ then it holds as well on $\Omega$.

In other words, if a first-order PDE holds on $\Omega_0$, it holds as well on $\Omega$. The general proof is not so easy, but for your particular problem, just take a "hole" function $\omega_\epsilon$ equal to 1 outside a shrinking neighborhood of $q$ and compute the $H^1$ norm of $\omega_\epsilon u$, which converges to the $H^1$ norm of $u$.


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