Is the laplacian of these test functions bounded?

Question

Let $B$ an open ball in $\mathbb{R}^{n}$, and $(K_{j})_{j}$ be an increasing sequence of compact subsets of $B$ whose union equals $B$. For each $j$, let $\rho_{j}$ be a cut-off function in $C_{c}^{\infty}(B)$ that equals 1 on a neighborhood of $K_{j}$ and whose support is in $K_{j+1}$. Finally, let $\theta$ be a smooth function whose laplacian $\Delta\theta\equiv1$, and set $$\phi_{j}:=\rho_{j}\theta.$$ My question: can we conclude that $$\sup_{(x,j)\in B\times\mathbb{N}}|\Delta\phi_{j}(x)|$$ is bounded?

Answer 1

Using the divergence theorem,

$$\int_{K_{j+1}\setminus K_j} \Delta \phi_j + \int_{K_j} \Delta \phi_j=\int_B \Delta \phi_j = \int_{\partial B} \nabla \phi_j \cdot dS=0.$$

But $\int_{K_j} \Delta \phi_j = \operatorname{vol}(K_j),$ so

$$\sup_{K_{j+1}\setminus K_j}(-\Delta\phi_j)\geq \frac{1}{\operatorname{vol}(K_{j+1}\setminus K_j)}\int_{K_{j+1}\setminus K_j} -\Delta \phi_j=\frac{\operatorname{vol}(K_j)}{\operatorname{vol}(K_{j+1}\setminus K_j)}\to \infty.$$

Answer 2

I think you cannot conclude that the $\sup$ is finite, because $\nabla\rho_j$ is not bounded.

Consider the following one-dimensional example. Let $B = (-1,1)$, and let $K_j := [-1+1/j, 1-1/j]$. In a first approximation you can think $\rho_j$ a Lipschitz function (piecewise affine) such that $\rho_j = 1$ on $K_j$, $\rho_j = 0$ outside $K_{j+1}$, and $\rho_j$ affine in the two intervals $I_j := (1-1/j, 1-1/(j+1))$, $I_j' := (-1+1/(j+1), -1+1/j)$. (Your "real" $\rho_j$ will be a smooth approximation of this Lipschitz function.)

On these intervals you have $\rho_j' = j(j+1)$. If you compute $$ \phi_j'' = \theta \rho_j'' + \rho_j \theta'' + 2 \rho_j' \theta' $$ you see that you cannot bound the last term. (Probably also the first term cannot be bounded.)