Derive distributional inequalities from pointwise estimates

Question

My question is how to prove the following claim:

Suppose that $E$ is an algebraic set in $\mathbb{R}^n (n\ge3)$ with dimension $\le n-2$, and $u$ is locally Lipschitz continuous on $\mathbb{R}^n$. If $u\in C^\infty(E^c)$ and there is a positive constant $C$ such that \begin{equation*} \Delta u(x) \le C, \quad \forall\, x\in E^c, %x\in E^c \cap \Omega, \end{equation*} then the above inequality also holds in the sense of distribution, that is, \begin{equation*} \int_{\mathbb{R}^n} u \,\Delta\varphi \,dx \le C \int_{\mathbb{R}^n} \varphi \,dx, \quad\forall\, \varphi\in C^\infty_0(\mathbb{R}^n),\,\varphi\ge0. % \int_\Omega u \,\Delta\varphi \,dx \le C \int_\Omega \varphi \,dx, \quad\forall\, \varphi\in C^\infty_0(\Omega),\,\varphi\ge0. \end{equation*}

In my limited experience with geometry and analysis, though I found some similar estimates in geometric analysis (such as the Laplacian comparison theorem), I couldn't adapt those ideas to the current situation. So I am not sure whether this claim is even correct and do not know how to prove it. Are there any related references?

I already posted this on Mathematics Stack Exchange 2 weeks ago, but didn't get any satisfactory response, so I've cross-posted it here. Any comment would be highly appreciated.

Answer 1

First note that the result is true if the support of $\varphi$ is disjoint from $E$; this follows from integration by parts.

Next note that every algebraic set is a finite union of smooth submanifolds (of possibly different dimensions). So let's first prove this for smooth submanifolds.

Throughout let $K_\varphi$ be the support of $\varphi$, which is compact.

For a closed set $K$, denote by $d_K$ the distance function to $K$.

$\dim(E) \leq n - 3$

Let $E$ be a smooth submanifold of dimension $\leq n -3$. Then in a tubular neighborhood $U$ of $E$, we have that the squared distance function to $E$ is smooth. Let $\chi$ be a fixed bump function on the real line that equals $1$ on $[-1,1]$ and $0$ outside $[-2,2]$.

For $\epsilon > 0$, split $$\varphi(x) = (1 - \chi(\epsilon^{-2}d_E^2(x)) \varphi(x) + \chi(\epsilon^{-2}d_E^2(x)) \varphi(x) $$ The first factor is supported away from $E$ and so we will ignore. For the second factor we can compute its Laplacian to be of size $$ O(\epsilon^{-2}) $$ But as $\dim(E) \leq n - 3$, the set $\{x\in K: d_E(x) < \epsilon\}$ has volume $O(\epsilon^3)$. This means that $$ \int u \triangle (\chi(\epsilon^{-2}d_E^2)\varphi) = O(\epsilon) $$ is negligible as we take $\epsilon\searrow 0$.

A consequence for our original problem, where $E$ is an algebraic set, is:

For every $\epsilon > 0$, there exists open sets $U, V$ with $\bar{U}\subseteq V$ and a smooth function $\eta$ that equals $1$ in $U$ and $0$ outside $V$, such that

• $\left|\int u \triangle (\eta \varphi)\right| < \epsilon$ and $\left| \int \eta\varphi \right| < \epsilon$
• $E \cap (\bar{U})^c$ is a finite union of $n-2$ dimensional smooth manifolds.

Note that for these arguments we just need that $u\in L^\infty_{loc}$, and not any assumption of continuity.

$\dim(E) = n-2$

The previous argument doesn't work, as the volume of the $\epsilon$-tubular neighborhood of $E$ is now size $\epsilon^2$. So here we use Lipschitz continuity.

The details are a bit annoying to write out, but here's the sketch:

1. Since $\bar{E}\cap K_\varphi$ is compact, we can cover it by finitely many small balls on each of which $u$ is approximately constant. Take a partition of unity. This allows us to split $\varphi$ into a part that is far from $E$ (where the result is already know to be true) plus finitely many parts where $u$ is almost constant. Denote by $\tilde{\varphi}$ the smoothly truncated function to one of these small balls.
2. On each small part we have $$ \int u \triangle\tilde{\varphi} = \int (u-u_0) \triangle\tilde{\varphi} + \int u_0 \triangle \tilde{\varphi} $$ the second factor vanishes since it is the integral of the Laplacian of a compactly supported function against a constant.
3. For the first factor we estimate as in the previous section. One part is far from $E$ and hence can be treated trivially. For the other part we use that $\triangle(\chi(\epsilon^{-2} d_E^2)\tilde{\varphi})$ is of size $\epsilon^{-2}$, and that the tube has volume $\epsilon^2$, to conclude that the corresponding integral is bounded by $O(|u-u_0|)$.
4. Continuity of $u$ then finishes the task.