Let $U \subseteq \mathbb{R}^n$ be an open set, and $f, g : U \rightarrow \mathbb{R}$ be twice continuously partially differentiable functions. Then we have:
$\Delta(f \cdot g) = f(\Delta g) + 2\langle \nabla f, \nabla g \rangle + (\Delta f)g $
My idea:
Using the product rule for partial derivatives, we expand the expression:
$ \frac{{\partial^2(f \cdot g)}}{{\partial x_i^2}} = \frac{{\partial}}{{\partial x_i}}\left(\frac{{\partial(f \cdot g)}}{{\partial x_i}}\right) = \frac{{\partial}}{{\partial x_i}}\left(f \frac{{\partial g}}{{\partial x_i}} + g \frac{{\partial f}}{{\partial x_i}}\right) $ Applying the product rule again, we get:
$ = \frac{{\partial f}}{{\partial x_i}} \frac{{\partial g}}{{\partial x_i}} + f \frac{{\partial^2 g}}{{\partial x_i^2}} + \frac{{\partial g}}{{\partial x_i}} \frac{{\partial f}}{{\partial x_i}} + g \frac{{\partial^2 f}}{{\partial x_i^2}} $
Summing over all components $i = 1$ to $n$:
$ \sum_{i=1}^{n} \frac{{\partial^2(f \cdot g)}}{{\partial x_i^2}} = \Delta(f \cdot g) $
We can rewrite the expression as follows:
$ \Delta(f \cdot g) = \sum_{i=1}^{n} \left(\frac{{\partial f}}{{\partial x_i}} \frac{{\partial g}}{{\partial x_i}}\right) + \sum_{i=1}^{n} \left(f \frac{{\partial^2 g}}{{\partial x_i^2}} + \frac{{\partial g}}{{\partial x_i}} \frac{{\partial f}}{{\partial x_i}} + g \frac{{\partial^2 f}}{{\partial x_i^2}}\right) $
Using the properties of the dot product and the gradient:
$\Delta(f \cdot g) = \langle \nabla f, \nabla g \rangle + \sum_{i=1}^{n} \left(f \frac{{\partial^2 g}}{{\partial x_i^2}} + \frac{{\partial g}}{{\partial x_i}} \frac{{\partial f}}{{\partial x_i}} + g \frac{{\partial^2 f}}{{\partial x_i^2}}\right) $
Furthermore, we can express the Laplacian of a function using the Laplacian operator:
$ \Delta f = \sum_{i=1}^{n} \frac{{\partial^2 f}}{{\partial x_i^2}} $ Substituting this into the equation, we get:
$ \Delta(f \cdot g) = \langle \nabla f, \nabla g \rangle + \sum_{i=1}^{n} \left(f \frac{{\partial^2 g}}{{\partial x_i^2}} + \frac{{\partial g}}{{\partial x_i}} \frac{{\partial f}}{{\partial x_i}} + g \frac{{\partial^2 f}}{{\partial x_i^2}}\right) = f(\Delta g) + 2\langle \nabla f, \nabla g \rangle + (\Delta f)g $
Is my proof correct?
Yes, that is right.
It is just the definition for the Laplacian, and the product rule for partial differentiation, applied twice.
$$\def\<{\langle}\def\>{\rangle}\begin{align}&~~~~\Delta (f\,g)\\&=\sum_{i=1}^n\tfrac{\partial^2(f\,g)}{{\partial x_i}^2}&&=\small\nabla^2 (f\,g)\\&=\sum_{i=1}^n\tfrac{\partial~~}{\partial x_i}\left(f\tfrac{\partial g}{\partial x_i}+\tfrac{\partial f}{\partial x_i}g\right)&&=\small\<\nabla, f\nabla g+ (\nabla f)g\>\\&=\sum_{i=1}^n\left((f\tfrac{\partial^2 g}{{\partial x_i}^2}+\tfrac{\partial f}{\partial x_i}\tfrac{\partial g}{\partial x_i}){+}(\tfrac{\partial f}{\partial x_i}\tfrac{\partial g}{\partial x_i}+\tfrac{\partial^2 f}{{\partial x_i}^2}g)\right)&&=\small (f\nabla^2 g+\<\nabla f,\nabla g\>)+(\<\nabla f,\nabla g\>+(\nabla^2 f)g)\\&=f~\Delta g + 2\<\nabla f,\nabla g\>+ (\Delta f)\,g\end{align}$$