Integration of laplace operator

Question

Given two functions $f(x,y)$ and $g(x,y)$ defined on $D =]0,a[ \times ]0,b[ \subset \mathbb{R}^2$. $$$$ And given that $$\int _D f(x,y).g(x,y) dxdy =0$$
I have to show the following : $$\int _D \Delta f(x,y).g(x,y) dxdy =0$$ I tried to use green's formula : $$\int _D \Delta f(x,y).g(x,y) dxdy = \int_{\partial D}\frac{\partial f }{\partial n}g d \sigma-\int _D \nabla f(x,y).\nabla g(x,y) dxdy$$
In my problem the derivative of $f$ and $g$ on the boundary is nul so I assumed that $\int_{\partial D}\frac{\partial f }{\partial n}g d \sigma = 0$ which leaves $$\int _D \Delta f(x,y).g(x,y) dxdy = -\int _D \nabla f(x,y).\nabla g(x,y) dxdy$$
I'm not sure about how to continue and wether my solution is correct
I tried using the integration by parts : $$-\int _D \nabla f(x,y).\nabla g(x,y) dxdy = -\biggl[ \nabla f(x,y).g(x,y)\biggr]_D+\int _D \nabla f(x,y).\nabla g(x,y) dxdy$$

Answer 1

This assertion is not correct. Here is a counterexample:

Take $a = b = 1$ and $g(x,y) = 1, \, f(x,y) = x^2 - 1/3$. Then $\Delta f(x,y) = 2$ and thus $$ \int_D f(x,y) g(x,y) dx\, dy = 0 \ne 2 = \int_D \Delta f(x,y) g(x,y) dx\, dy \, . $$