A test function with positive laplacian at a given point

Question

Hi everyone: Let $ B(x,r) $ be a ball in $ \mathbb{R}^{m} $ with $ m\geq2. $ Is there a NON-NEGATIVE infinitly differentiable function $ \phi $ with compact support in $ B(x,r) $ such that $ \phi(x)>0 $? Thanks for your answers. In fact, it suffices to give in $ \mathbb{R} $.

Answer 1

Yes. Use the standard bump function on $\mathbb{R}$ equal to $1$ on $(-r/2,r/2)$ and supported on $(-r,r)$, then extend in polar coordinates to a radially symmetric function on $B(0,r)$, and finally translate to $B(x,r)$.

Edit: If the question is really, "... such that $\Delta\phi > 0$?" then the answer is still "yes." Take the first Dirichlet eigenfunction of the Laplacian, for example, which is a Bessel function.

The standard bump function, if you ensure that it is equal to $1$ only at $0$ instead of on a neighborhood of zero, also has positive Laplacian. You can work this out analytically, and it also follows from the observation that the standard bump function is everywhere at least equal to its arithmetic mean on small balls.

Answer 2

Let $\psi$ be a standard bump function in $B(x,r)$. Then $\psi(x) > 0$ and since $\psi$ has a maximum at $x$ each of the first partial derivatives of $\psi$ is zero at $x$. Let $\eta$ be any smooth function. Then $\eta \psi$ is a compactly supported in $B(x,r)$ and $$\Delta \eta \psi(x) = \Delta \eta(x) \psi(x) + \eta(x) \Delta \psi(x).$$

You can make this positive by making $\eta(x)$ small enough and $\Delta \eta(x)$ big enough. You should be able to work out such an $\eta$.