When do we have $\lim_{i\to\infty}\langle\Delta u_i,\phi\rangle=0$ if the Laplacian is distributional and $\phi$ is a test function?

Question

Let $U$ be an open bounded subset of $\mathbb{R}^n$. If $u$ is a function in the Sobolev space $W^{2,p}(U)$, then by introducing weak derivatives, it is possible to define the Laplacian $\Delta u$ of $u$. Now, suppose that $\phi$ is a test function ($\phi\in\mathcal{D}$) and suppose that $\{u_i\}_{i=1}^\infty$ is a sequence in $W^{2,p}(U)$ with $\Delta u_i\to 0$ in $L^p$. Can we conclude that $$\lim_{i\to\infty}\langle\Delta'u_i,\phi\rangle=0?\tag{$*$}$$ Here $\Delta'u_i$ is the distributional Laplacian defined by $$\phi\in\mathcal{D}\mapsto\int u_i\Delta\phi.$$ I know the definition of convergence of sequences of distributions, and it seems that if we can show that $\Delta'u_i\to 0$ in $\mathcal{D}'$, then $(*)$ readily follows. But all we know is that $\Delta u_i\to 0$ in the $L^p$ norm. Is there anything that relates $\Delta$ to $\Delta'$ and hence allows us to attain the desired result? Thank you.

Answer 1

I will do every thing on $\mathbb R$ because I don't like boundary conditions. I think you are confusing yourself by referring to different Laplacians $\Delta'$ and $\Delta$. It is more useful to think of one Laplacian $\Delta$ which we define on the largest space $\mathcal{D}'(\mathbb{R})$ in the usual way by duality. The "other" Laplacians are just restrictions of this one to smaller spaces, i.e. the weak derivative Laplacian is what you obtain when you restrict to $W^{2,p}(\mathbb R)$ functions and the classical Laplacian when you restrict to $C^2(\mathbb R)$ functions. But the operator is one and the same.

To convince your self of this, let's test one of the cases. Pick a $W^{2,p}(\mathbb R)$ function $u$ and assume the Laplacian on this space is different, say $\bar \Delta$. Since it is locally integrable it is also an element of $\mathcal{D}'(\mathbb R)$. So lets take our master Laplacian defined on $\mathcal{D}'(\mathbb R)$ and apply to $u$ and test with an element of $\varphi \in \mathcal{D}(\mathbb R)$. This gives us \begin{align} \langle \Delta u, \varphi \rangle =&\langle u, \Delta \varphi \rangle \\ =& \int_{\mathbb R} \! \bar \Delta u \, \varphi \, \mathrm{d}x \, , \end{align} where in the last step we have just used integration by parts and the definition of the weak derivative. This tells us that $\Delta u =\bar \Delta u$ for all $u$ and so they are the same operator.